SOLUTIONS MANUAL for Physics for Scientists & Engineers with Modern Physics 5th Edition by Douglas C by welldoneassistant

Physics for Scientists & Engineers with Modern Physics 5th Edition by Douglas C. Giancoli.

TABLE OF CONTENTS CHAPTER 1 Introduction, Measurement, Estimating CHAPTER 2 Describing Motion: Kinematics in One Dimension CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors CHAPTER 4 Dynamics: Newton’s Laws of Motion CHAPTER 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces CHAPTER 6 Gravitation and Newton’s Synthesis CHAPTER 7 Work and Energy CHAPTER 8 Conservation of Energy CHAPTER 9 Linear Momentum CHAPTER 10 Rotational Motion CHAPTER 11 Angular Momentum; General Rotation CHAPTER 12 Static Equilibrium; Elasticity and Fracture CHAPTER 13 Fluids CHAPTER 14 Oscillations CHAPTER 15 Wave Motion CHAPTER 16 Sound CHAPTER 17 Temperature, Thermal Expansion, and the Ideal Gas Law CHAPTER 18 Kinetic Theory of Gases CHAPTER 19 Heat and the First Law of Thermodynamics CHAPTER 20 Second Law of Thermodynamics

CHAPTER 21 Electric Charge and Electric Field CHAPTER 22 Gauss’s Law CHAPTER 23 Electric Potential CHAPTER 24 Capacitance, Dielectrics, Electric Energy Storage CHAPTER 25 Electric Current and Resistance CHAPTER 26 DC Circuits CHAPTER 27 Magnetism CHAPTER 28 Sources of Magnetic Field CHAPTER 29 Electromagnetic Induction and Faraday’s Law CHAPTER 30 Inductance, Electromagnetic Oscillations, and AC Circuits CHAPTER 31 Maxwell’s Equations and Electromagnetic Waves CHAPTER 32 Light: Reflection and Refraction CHAPTER 33 Lenses and Optical Instruments CHAPTER 34 The Wave Nature of Light: Interference and Polarization CHAPTER 35 Diffraction CHAPTER 36 The Special Theory of Relativity CHAPTER 37 Early Quantum Theory and Models of the Atom CHAPTER 38 Quantum Mechanics CHAPTER 39 Quantum Mechanics of Atoms CHAPTER 40 Molecules and Solids CHAPTER 41 Nuclear Physics and Radioactivity CHAPTER 42 Nuclear Energy; Effects and Uses of Radiation CHAPTER 43 Elementary Particles CHAPTER 44 Astrophysics and Cosmology

CHAPTER 1: Introduction, Measurement, Estimating Responses to Questions 1.

(a) A particular person’s foot. Merits: reproducible. Drawbacks: not accessible to the general public; not invariable (could change size with age, time of day, etc.); not indestructible. (b) Any person’s foot. Merits: accessible. Drawbacks: not reproducible (different people have different size feet); not invariable (could change size with age, time of day, etc.); not indestructible. Neither of these options would make a good standard.

The distance in miles is given to one significant figure, and the distance in kilometers is given to five significant figures! The value in kilometers indicates more precision than really exists or than is meaningful. The last digit represents a distance on the same order of magnitude as a car’s length! The sign should perhaps read “7.0 mi (11 km),” where each value has the same number of significant figures, or “7 mi (11 km),” where each value has about the same % uncertainty.

The number of digits you present in your answer should represent the precision with which you know a measurement; it says very little about the accuracy of the measurement. For example, if you measure the length of a table to great precision, but with a measuring instrument that is not calibrated correctly, you will not measure accurately. Accuracy is a measure of how close a measurement is to the true value.

If you measure the length of an object and you report that it is “4,” you haven’t given enough information for your answer to be useful. There is a large difference between an object that is 4 meters long and one that is 4 km long. Units are necessary to give meaning to a numerical answer.

You should report a result of 8.32 cm. Your measurement had three significant figures. When you multiply by 2, you are really multiplying by the integer 2, which is an exact value. The number of significant figures is determined by the measurement.

The correct number of significant figures is three: sin 30.0º = 0.500.

Useful assumptions include the population of the city, the fraction of people who own cars, the average number of visits to a mechanic that each car makes in a year, the average number of weeks a mechanic works in a year, and the average number of cars each mechanic can see in a week. (a) There are about 800,000 people in San Francisco. Assume that half of them have cars. If each of these 400,000 cars needs servicing twice a year, then there are 800,000 visits to mechanics in a year. If mechanics typically work 50 weeks a year, then about 16,000 cars would need to be seen each week. Assume that on average, a mechanic can work on 4 cars per day, or 20 cars a week. The final estimate, then, is 800 car mechanics in San Francisco. (b) Answers will vary. But following the same reasoning, the estimate is 1/1000 of the population.

© 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Responses to MisConceptual Questions 1.

(c) As stated in the text, scientific laws are descriptive – they are meant to describe how nature behaves. Since our understanding of nature evolves, so do the laws of physics, when evidence can convince the community of physicists. The laws of physics are not permanent, and are not subject to political treaties. In fact, there have been major changes in the laws of physics since 1900 – particularly due to relativity and quantum mechanics. The laws of physics apply in chemistry and other scientific fields, since those areas of study are based on physics. Finally, the laws of physics are man-made, not a part of nature. They are our “best description” of nature as we currently understand it. As stated in the text, “Laws are not lying there in nature, waiting to be discovered.”

(e) The first product is 142.08 m, which is only accurate to the 10’s place, since 37 m/s has only two significant figures. The second product is 74.73 m, which is only accurate to the 1’s place, since 5.3 s has only two significant figures. Thus the sum of the two terms can only be accurate to the 10’s place. 142.08 + 74.73 = 216.81, which to the 10’s place is 220 m.

(a) The total number of digits present does not determine the precision, as the leading zeros in (c) and (d) are only place holders. Rewriting all the measurements in units of meters shows that (a) implies a precision of 0.0001m, (b) and (c) both imply a precision of 0.001 m, and (d) implies a precision of 0.01 m. Note that since the period is shown, the trailing zeros are significant. If all the measurements are expressed in meters, (a) has 4 significant figures, (b) and (c) each have 3 significant figures, and (d) has 2 significant figures.

(b) The leading zeros are not significant. Rewriting this number in scientific notation, 7.8  10−3 , shows that it only has two significant digits.

(b) When you add or subtract numbers, the final answer should contain no more decimal places than the number with the fewest decimal places. Since 25.2 has one decimal place, the answer must be rounded to one decimal place, or to 26.6. Thus the answer has 3 significant figures.

(b) The word “accuracy” is often misused. If a student repeats a measurement multiple times and obtains the same answer each time, it is often assumed to be accurate. In fact, students are frequently given an “ideal” number of times to repeat the experiment for “accuracy.” However, systematic errors may cause each measurement to be inaccurate. A poorly working instrument may also limit the accuracy of your measurement.

(a) Quoting the textbook, “precision” refers to the repeatability of the measurement using a given instrument. Precision and accuracy are often confused. “Accuracy” is defined by answer (b).

(d) This addresses misconceptions about squared units and about which factor should be in the numerator of the conversion. This error can be avoided by treating the units as algebraic symbols that must be cancelled out.

(e) When making estimates, the estimator may frequently believe that their answers are more significant than they actually are. This question helps the estimator realize what an order-ofmagnitude estimation is NOT supposed to accomplish.

10. (d) This addresses the fact that the generic unit symbol, like [L], does not indicate a specific system of units. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 1

Introduction, Measurement, Estimating

Solutions to Problems 1.

(a) 777

3 significant figures

(b) 81.60

4 significant figures

(c)

7.03

3 significant figures

(d) 0.03

1 significant figure

(e)

0.0086

2 significant figures

(f)

6465

4 significant figures

(g) 8700

2 significant figures

(a) 5.859 = 5.859  100 (b) 21.8 = 2.18  101 (c)

0.0068 = 6.8  10−3

(d) 328.65 = 3.2865  102

(e)

0.219 = 2.19  10−1

(f)

444 = 4.44  102

(a) 8.69  105 = 869, 000 (b) 9.1103 = 9100 (c)

2.5  10−1 = 0.25

(d) 4.76 102 = 476 (e) 3.62 10−5 = 0.0000362 0.35 m

% uncertainty =

(a) % uncertainty =

3.25 m

(b) % uncertainty =

 100% = 11%

0.2 s 4.5s

 100% = 4.444%  4%

0.2 s

 100% = 0.4444%  0.4% 45s (c) The time of 4.5 minutes is 270 seconds. 0.2 s % uncertainty =  100% = 0.0741%  0.07% 270 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

To add values with significant figures, adjust all values to be added so that their exponents are all the same. ( 9.2  103 s ) + ( 6.3  104 s ) + ( 0.008  106 s ) = ( 9.2  103 s ) + ( 63  103 s ) + (8  103 s ) = ( 9.2 + 63 + 8 )  103 s = 80.2  103 s = 8.0  104 s

When adding, keep the least accurate value, and so keep to the “ones” place in the last set of parentheses. 7.

When you multiply, the result should have as many digits as the number with the least number of significant digits used in the calculation.

( 4.079  10 m )(0.057  10 m ) = 2.325m  2.3 m 2

−1

The “best value” uncertainty is taken to be 0.01 m. 0.01m 2 % uncertainty =  100% = 0.787%  0.8% 1.27 m 2

 (radians) 0 0.10 0.12

sin( ) 0.00 0.10 0.12

tan( ) 0.00 0.10 0.12

0.20 0.24 0.25

0.20 0.24 0.26

Keeping 2 significant figures in the angle, and expressing the angle in radians, the largest angle that has the same sine and tangent is 0.24 radians . In degrees, the largest angle (keeping 2 significant figure) is 12 .

10. (a) To find the number of students, multiply the sample size times the percentage. ( 215 students )( 0.372 ) = 79.98 students But the number of students must be an integer, so the number is 80 students . (b) The original statement gave the percentage with too many significant figures. The calculation is 80 / 215 = 0.372093023… , but that result should only have 2 significant figures, since the value of 80 (as an integer, and not an estimate) has only 2 significant figures. The percentage should have been quoted as 37%. That would still give the correct number of students. ( 215 students )( 0.37 ) = 79.55 students, which rounds to 80 students. 11. We calculate what fraction 9 seconds is of an entire year. 1yr  9s   −5 −5  1yr   3.156  107s   100% = 2.85  10 %  3  10 %    12. In order to find the approximate uncertainty in the area, calculate the area for the specified radius, the minimum radius, and the maximum radius. Subtract the extreme areas. The uncertainty in the area is then half this variation in area. The uncertainty in the radius is assumed to be 0.1  104 cm .

(

2 Aspecified =  rspecified =  5.1  104 cm

) = 8.17  10 cm 2

( ) = 7.85  10 cm =  ( 5.2  10 cm ) = 8.49  10 cm

2 Amin =  rmin =  5.0  104 cm

2 Amax =  rmax

Chapter 1

Introduction, Measurement, Estimating

(

)

A = 12 ( Amax − Amin ) = 12 8.49  109 cm 2 − 7.85  109 cm 2 = 0.32  109 cm 2

Note that the last value above has only 1 significant figure. Thus the area should be quoted as

A = (8.2  0.3)  109 cm2 . 13. In order to find the approximate uncertainty in the volume, calculate the volume for the minimum radius and the volume for the maximum radius. Subtract the extreme volumes. The uncertainty in the volume is then half of this variation in volume. 3 3 Vspecified = 43  rspecified = 43  ( 0.64 m ) = 1.098 m3 3 Vmin = 43  rmin = 43  ( 0.60 m ) = 0.905 m 3 3

3 Vmax = 43  rmax = 43  ( 0.68 m ) = 1.317 m 3 3

(

)

V = 12 (Vmax − Vmin ) = 12 1.317 m3 − 0.905 m3 = 0.206 m3

Note that the last value above has only 1 significant figure. Thus the percent uncertainty is V 0.206 m 3 =  100 = 18.6  20 % . Vspecified 1.098 m 3 14. (a) 286.6 mm

286.6  10−3 m

0.286 6 m

(b) 74  V

74  10−6 V

0.000 074 V

(c)

430 mg

430  10 −3 g

0.43g (if last zero is not significant)

(d) 47.2 ps

47.2  10−12 s

0.000 000 000 047 2 s

(e)

22.5 nm

22.5  10−9 m

0.000 000 022 5 m

(f) 2.50 gigavolts

2.5  109 volts

2, 500, 000, 000 volts

Note that in part (f) in particular, the correct number of significant digits cannot be determined when you write the number in this format. 15. (a) 3  106 volts

3megavolt = 3 Mvolt

(b) 2  10−6 meters

2 micrometers = 2  m

5 kilodays = 5 kdays

(d) 18  102 bucks

18 hectobucks = 18 hbucks or 1.8 kilobucks

(e)

−7

9  10 seconds

900 nanoseconds = 900 ns or 0.9 s

16. Assuming a height of 5 feet 10 inches, then 5'10" = ( 70 in )(1 m 39.37 in ) = 1.8 m . Assuming a weight of 165 lbs, then (165 lbs )( 0.456 kg 1 lb ) = 75.2 kg . Technically, pounds and mass measure two separate properties. To make this conversion, we have to assume that we are at a location where the acceleration due to gravity is 9.80 m/s2.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

17. (a)

(b)

SA Earth SA Moon

2 4 REarth 2 4 RMoon

4 3

( 6.38  10 km ) = 13.4 = (1.74  10 km ) ( 6.38  10 km ) = 49.3 = (1.74 10 km )

2 REarth

2 RMoon

3 3  REarth REarth = 4 = 3 3 RMoon VMoon 3  RMoon

VEarth

Instructor Solutions Manual

18. To answer this question, convert 15 m/s to mi/h, and compare to the speed limit.  1 mi   3600s  15 m s    = 33.56 mi h  34 mi h  1609 m   1h  Since the conversion only gives about 34 mi/h, the driver is not exceeding the speed limit . We could also change 35 mi/h to m/s, and find that 35 mi/h is more than 15 m/s. 19. (a) 14 billion years = 1.4 1010 years (b)

 3.156  107s  = 4.4  1017s )  1yr  

(

1.4  1010 yr 

(

)

20. (a) 93 million miles = 93 106 miles (1610 m 1 mile ) = 1.5  1011 m (b) 1.5  1011 m = (1.5  1011 m )(1km 103 m ) = 1.5  108 km 21. To add values with significant figures, adjust all values to be added so that their units are all the same. 1.90 m + 142.5cm + 6.27  105  m = 1.90 m + 1.425 m + 0.627 m = 3.952 m  3.95 m When you add, the final result is to be no more accurate than the least accurate number used. In this case, that is the first measurement, which is accurate to the hundredths place when expressed in meters.

(

)

22. (a) 1.0 10−10 m = 1.0 10−10 m ( 39.37 in 1 m ) = 3.9 10 −9 in (b)

(1.0 cm ) 

1 m   1 atom  8   = 1.0  10 atoms −10  100 cm   1.0  10 m 

23. (a)

(1km h ) 

(b)

(1m s ) 

(c)

(1km h ) 

0.621 mi 

 = 0.621mi h , so the conversion factor is  1 km  3.28 ft 

 = 3.28 ft s, so the conversion factor is  1m 

0.621mi h 1km h

3.28 ft s 1m s

1000 m   1 h

0.278 m s  .   = 0.278 m s, so the conversion factor is 1km h  1 km   3600 s 

Note that if more significant figures were used in the original factors, such as 0.6214 miles per kilometer, then more significant figures could have been included in the answers.

Chapter 1

Introduction, Measurement, Estimating

(

)

(

24. (a) 1 ft 2 = 1 ft 2 (1 yd 3 ft ) = 0.111 yd 2, and so the conversion factor is (b) 1 m = 1 m 2

) ( 3.28 ft 1 m ) = 10.8 ft , and so the conversion factor is 2

0.111 yd 2 1 ft 2

10.8ft 2 1m 2

25. (a) Find the distance by multiplying the speed by the time.

(

)(

)

1.00 ly = 2.998  108 m s 3.156  107 s = 9.462  1015 m  9.46  1015 m (b) Do a unit conversion from ly to AU.  9.462  1015 m   1 AU  4 (1.00 ly )    = 6.31 10 AU  11  1.00 ly   1.50  10 m  26. One mile is 1609 m, according to the unit conversions in the front of the textbook. Thus it is 109 m longer than a 1500-m race. The percentage difference is calculated here. 109 m  100% = 7.3% 1500 m 27. From Example 1–6, the thickness of a page of this book is about 6  10−5 m. The wavelength of orange krypton-86 light is found from the fact that 1,650,763.73 wavelengths of that light is the definition of the meter.  6  10−5 m   1, 650, 763.73 wavelengths  1page    = 99  100 1m  1page    28. The original definition of the meter was that 1 meter was one ten-millionth of the distance from the Earth’s equator to either pole. The distance from the equator to the pole would be one-fourth of the circumference of a perfectly spherical Earth. Thus the circumference would be 40 million meters:

C = 4  107 m . We use the circumference to find the radius. C = 2 r → r =

4  107 m

= 6.37  106 m 2 2 The value in the front of the textbook is 6.38  106 m. Since the circumference and the radius are proportional to each other, the % error would be the same in both cases. We calculate it for the radius. R 0.01  106 m =  100 = 0.1567  0.2 % R 6.38  106 m

29. The value of 12 liters per km can also be expressed as 1 km per 12 liters. We convert the units of km per liter to miles per gallon.

1km  3.785 L   0.6214 mi 

  12 L  1gal  

1km

 = 0.196 gal  0.20 gal 

30. Since the meter is longer than the yard, the soccer field is longer than the football field. 1.094 yd Lsoccer − Lfootball = 100.0 m  − 100.0 yd = 9.4 yd 1m © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Lsoccer − Lfootball = 100.0 m − 100.0 yd 

= 8.6 m 1.094 yd Since the soccer field is 109.4 yd compared to the 100.0-yd football field, the soccer field is

9.4 % longer than the football field.

 3.156  107s  7  = 3.16  10 s 1 yr   7  3.156  10 s   1  109 ns  (b) # of nanoseconds in 1.00 yr: 1.00 yr = (1.00 yr )  = 3.16  1016 ns    1yr   1 s  1 yr   = 3.17  10−8 yr 1.00 s = (1.00 s )  (c) # of years in 1.00 s: 7   3.156  10 s  1.00 yr = (1.00 yr ) 

31. (a) # of seconds in 1.00 yr:

 10−15 kg  1 proton or neutron  = 1012 protons or neutrons   −27 10 kg   1 bacterium   −17   1 proton or neutron  10 kg (b)  = 1010 protons or neutrons   −27 10 kg   1 DNA molecule   2  10 kg  1 proton or neutron  (c)  = 1029 protons or neutrons   −27 10 kg   1 human   41  10 kg  1 proton or neutron  68 (d)    = 10 protons or neutrons −27 1 galaxy 10 kg   

32. (a) 

33. The surface area of a sphere is found by A = 4 r 2 = 4 ( d 2 ) =  d 2 . 2

(

) = 7.478  10 m 2

(a)

2 =  4.879  106 m AMercury =  DMercury

(b)

2  DEarth   REarth   6.38  106 m   DEarth = =  = D  =R  = 6.84 2 6 AMercury  DMercury  Moon   Mercury   2.4395  10 m 

AEarth

34. The radius of the ball can be found from the circumference (represented by “c” in the equations below), and then the volume can be found from the radius. Finally, the mass is found from the volume of the baseball multiplied by the density ( = mass/volume) of a nucleon.

cball = 2 rball → rball =

c  3 ; Vball = 43  rball = 43   ball  2  2 

cball

   mnucleon   m  mnucleon V = Vball  nucleon =   ball   3 3 4 4 1  Vnucleon   3  rnucleon   3  ( 2 d nucleon ) 

mball = Vball  nucleon = Vball 

3  0.23m    c  mnucleon c   =   ball   = mnucleon  ball  = (10−27 kg )   3   (10−15 m )   2   43  ( 12 d nucleon )    d nucleon    3

4 3

= 3.9  1014 kg  4  1014 kg

Chapter 1

Introduction, Measurement, Estimating

35. (a) 3200 = 3.28  103  1  103 = 103 (b) 86.30  103 = 8.630  104  10  104 = 105 (c)

0.076 = 7.6  10−2  10  10−2 = 10−1

(d) 15.0  108 = 1.5  109  1 109 = 109 36. The textbook is slightly smaller than 25 cm deep and 5 cm wide. With books on both sides of a shelf, the shelf would then need to be about 50 cm deep. If the aisle is 1.5 m wide, then about 1/4 of the floor space is covered by shelving. The number of books on a single shelf level is then 1 4









= 1.3  10 books. With 8 shelves of books, the total number of ( 6500 m )  ( 0.25 1mbook )( 0.05 m )  2

books stored is as follows:  1.3  105 books  8 shelves = 1.04  106 books  1  106 books )  ( shelf level   37. The distance across the U.S. is about 3000 miles.

( 3000 mi )(1 km 0.621 mi )(1 hr 10 km )  500 hr

Of course, it would take more time on the clock for a runner to run across the U.S. The runner obviously could not run for 500 hours non-stop. If he or she could run for 5 hours a day, then it would take about 100 days to cross the country. 38. A commonly accepted measure is that a person should drink eight 8-oz. glasses of water each day. That is about 2 quarts, or 2 liters of water per day. Approximate the lifetime as 70 years.

( 70 y )( 365 d 1 y )( 2 L 1 d )  5  104 L 39. To determine an estimation for the number of cells in the human body we will make several assumptions. First, we assume that the cell is spherical, so we can find its volume using the formula for the volume of a sphere, V = 43  r 3. And then we will assume that the mass of an “average” human is 65 kg, which corresponds to a weight of about 140 lbs. We can use the given density to find the volume of our typical human, and then compare that to the volume of a cell. kg 1m3 1cell cells 65   = 1.24  1014 3 6 − human 1000 kg 43  5  10 m human

(

)

So for an order-of-magnitude estimate, we would say there are about 1 1014 cells in a human . 40. An NCAA-regulation football field is 360 feet long (including the end zones) and 160 feet wide, which is about 110 meters by 50 meters, or 5500 m2. We assume the mower has a cutting width of 0.5 meters and that a person mowing can walk at about 4.5 km/h, which is about 3 mi/h. Thus the distance to be walked is as follows: area 5500 m 2 d= = = 11000 m = 11 km width 0.5 m 1h  2.5 h to mow the field. At a speed of 4.5 km/h, it will take about 11km  4.5 km

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

41. There are about 3  108 people in the U.S. Assume that half of them have cars, that they drive an average of 12,000 miles per year, and that their cars get an average of 20 miles per gallon of gasoline.  1 automobile   12, 000 mi auto   1 gallon  11 3  108 people     20 mi   1  10 gal y 2 people 1 y    

(

)

42. In estimating the number of dentists, the assumptions and estimates needed are: • the population of the city • the number of patients that a dentist sees in a day • the number of days that a dentist works in a year • the number of times that each person visits the dentist each year We estimate that a dentist can see 10 patients a day, that a dentist works 225 days a year, and that each person visits the dentist twice per year. (a) For San Francisco, the population is approximately 800,000 (according to the U.S. Census Bureau). The number of dentists is found by the following calculation:  2 visits        1 yr 1 dentist   year 8  105 people      700 dentists   1 person   225 workdays   10 visits     workday  (b) For Marion, Indiana, the population is about 30,000. The number of dentists is found by a similar calculation to that in part (a), and would be about 30 dentists . There were about 40 dentists listed in a recent “yellow pages” phone book.

(

)

43. Make the estimate that each person has 1.5 loads of laundry per week, and that there are 300 million people in the United States. loads week 52 weeks 0.1kg kg kg   = 2.34  10  2  10 ( 300  10 people )  1.5 1person 1 yr 1load yr yr 6

1/ 3

 3V  . For a 1000-kg rock, 44. The volume of a sphere is given by V =  r , so the radius is r =    4  3

4 3

which weighs about 2200 lb, the volume is calculated from the density, and then the diameter from the volume.  2200 lb   1ft 3  3 V = (1T )    186 lb  = 11.8ft 1T    3 1/3  3V  = 2  3 (11.8ft )  = 2.82 ft  3ft d = 2r = 2     4  4    1/3

45. The covering of 49 km in two and a half days can be used to convert km into days. They have 270 km – 49 km = 221 km remaining. 2.5days 221km  = 11.3days  11days 49 km

Chapter 1

Introduction, Measurement, Estimating

46. The person walks 4 km h , 12 hours each day. The radius of the Earth is about 6380 km, and the distance around the Earth at the equator is the circumference, 2 REarth. We assume that the person can “walk on water,” so we ignore the existence of the oceans.  1 h   1 day  = 835days  800 days 2 ( 6380 km )     4 km   12 h  47. We approximate the jar as a cylinder with a uniform cross-sectional area. In counting the jelly beans in the top layer, we find about 25 jelly beans. Thus we estimate that one layer contains about 25 jelly beans. In counting vertically, we see that there are about 20 rows. Thus we estimate that there are 25  20 = 500 jellybeans in the jar. 48. The maximum number of buses would be needed during rush hour. We assume that a bus can hold 50 passengers. (a) The current population of Washington, D.C. is about 660,000 as of 2014 statistics. We estimate that 10% of them ride the bus during rush hour. 1bus 1driver 66, 000 passengers   = 1320 drivers  1000 drivers 50 passengers 1bus (b) For Marion, Indiana, the population is about 30,000. Because the town is so much smaller geographically, we estimate that only 5% of the current population rides the bus during rush hour. 1bus 1driver   30 drivers 1500 passengers  50 passengers 1bus 49. Consider the diagram shown (not to scale). The balloon is a distance h = 300 m above the surface of the Earth, and the tangent line from the balloon height to the surface of the Earth indicates the location of the horizon, a distance d away from the balloon. Use the Pythagorean theorem. ( r + h )2 = r 2 + d 2 → r 2 + 2rh + h 2 = r 2 + d 2

2 rh + h 2 = d 2 → d = 2 rh + h 2

(

)

d = 2 6.4  106 m ( 300 m ) + ( 300 m ) = 6.20  10 4 m  6  10 4 m (  40 mi ) 2

50. At $1000 per day, you would earn $30,000 in 30 days. With the other pay method, you would get $0.01 2t −1 on the tth day. On the first day, you get $0.01 21−1 = $0.01. On the second day, you get

( ) ( ) $0.01( 2 ) = $0.02. On the third day, you get $0.01( 2 ) = $0.04. On the 30th day, you get $0.01( 2 ) = $5.4  10 , which is over 5 million dollars. Get paid by the second method. 2 −1

30 −1

3 −1

51. Assume that the tires last for 5 years, and so there is a tread wearing of 0.2 cm/year. Assume the average tire has a radius of 40 cm, and a width of 10 cm. Thus the volume of rubber that is becoming pollution each year from one tire is the surface area of the tire times the thickness per year that is wearing. Also assume that there are 1.5  108 automobiles in the country – approximately one automobile for every two people. And there are 4 tires per automobile. The mass wear per year is given by the following calculation. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

 mass   surface area   thickness wear    year  =   ( density of rubber )( # of tires ) tire year       2 ( 0.4 m )( 0.1m )   0.002 m   1200 kg  8 8  1yr   1m3  ( 6.0  10 tires ) = 4  10 kg y  1 tire    

=

52. In the figure in the textbook, the distance d is perpendicular to the radius that is drawn approximately vertically. Thus there is a right triangle, with legs of d and R, and a hypotenuse of R + h. Since h R, h 2 2 Rh. d 2 + R 2 = ( R + h ) = R 2 + 2 Rh + h 2 → d 2 = 2 Rh + h 2 → d 2  2 Rh → 2

d2 2h

( 4400 m ) = 6.5  106 m 2 (1.5 m ) 2

A better measurement gives R = 6.38  106 m. 53. For you to see the Sun “disappear,” your line of sight to the top of the Sun must be tangent to the Earth’s surface. Initially, your eyes are a distance h1 = 20 cm above the sand and you see the first sunset at a (very small) angle of 1 relative to perfectly horizontal. Then you stand up, elevating your eyes to the height h2 = 130 cm. When you stand, your line of sight is tangent to the Earth’s surface at a slightly different angle  2 , so that is the direction to the second sunset. The change in angle  2 − 1 is the angle through which the Sun appears to move relative to the Earth during the time to be measured. The distance d is the distance from your eyes to a point on the horizon where you “see” the sun set. d1

To 2nd sunset

To 1st sunset

1

2

Earth center

Use the Pythagorean theorem for the following relationship: 2 d 2 + R 2 = ( R + h ) = R 2 + 2 Rh + h 2 → d 2 = 2 Rh + h 2 The distance h is much smaller than the distance R, so h 2 2 Rh which leads to d 2  2 Rh. We also have from the same triangle that d R = tan  , so d = R tan  . Combining these two relationships gives d 2  2 Rh = R 2 tan 2  , so R =

2h tan 2 

The change in angle,  =  2 − 1 , can be found from the two heights and the radius of the Earth. The elapsed time between the two sightings can then be found from the change in angle, because we know that a full revolution takes 24 hours. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 1

Introduction, Measurement, Estimating

2h tan  2

 2 = tan −1

2h1

→ 1 = tan −1 2h2

= tan −1

R t sec  = o 3600 s 360 24 h  1h

2 ( 0.20 m )

= tan −1

2 (1.5 m ) 6.38  106 m

6.38  10 m 6

(

= 3.93  10 −2

(

= 1.43  10 −2

)

) ;  =  −  = ( 2.50  10 ) o

−2

→

−2 o 3600 s   ( 2.50  10 )   3600 s        24 h  24 h  t = =  = 6.0 s  o  o  1h   360 1h   360    

54. Density units =

M  =  3  . SI units for density are kg m 3. volume units  L  mass units

55. (a) For the equation v = At 3 − Bt , the units of At 3 must be the same as the units of v. So the units 4 of A must be the same as the units of v t 3 , which would be L T . Also, the units of Bt must

be the same as the units of v. So the units of B must be the same as the units of v t , which would be L T 2 . (b) For A, the SI units would be m s4 , and for B the SI units would be m s2 . 56. (a) The quantity vt 2 has units of ( m s ) ( s 2 ) = m  s, which do not match with the units of meters for x. The quantity 2at has units ( m s 2 ) ( s ) = m s , which also do not match with the units of meters for x. Thus this equation cannot be correct .

(b) The quantity v 0 t has units of ( m s )( s ) = m, and 12 at 2 has units of ( m s 2 )( s 2 ) = m. Thus, since each term has units of meters, this equation can be correct .

(c) The quantity v 0 t has units of ( m s )( s ) = m, and 2at 2 has units of ( m s 2 )( s 2 ) = m. Thus, since each term has units of meters, this equation can be correct . 57. We consider the “Planck time.”

(a) tP =

(b) tP =

Gh c5

 L3   ML2  3 2 5 5  MT 2   T     =  L L T M  =  T  = T 2  = T  5    MT 3 L5  T 3      L  T 

→ units of t P =

( 6.67  10 m kg  s )( 6.63  10 kg  m s ) = 1.35  10 s  10 s = c ( 3.00  10 m s )

Gh 5

−11

−34

−43

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

58. The percentage accuracy is

Instructor Solutions Manual

 100% = 1  10−5% . The distance of 20,000,000 m needs to

2  10 m be distinguishable from 20,000,002 m, which means that 8 significant figures are needed in the distance measurements. 7

59. Divide the number of atoms by the Earth’s surface area.

# atoms m

6.02  1023 atoms 4 R

2 Earth

6.02  1023 atoms

(

4 6.38  10 m 6

)

= 1.18  109

atoms m2

This is more than one billion atoms per square meter. 60. Multiply the number of chips per wafer by the number of wafers that can be made from a cylinder. We assume the number of chips per wafer is more accurate than 1 significant figure.

 750 chips   1 wafer   250 mm  = 6.25  105  6.3  105 chips     wafer   0.300 mm   1 cylinder  cylinder  61. We assume that there are 40 hours of work per week and that you type 50 weeks out of the year. (Everybody deserves a vacation!) 1char 1min 1hour 1week 1year 1.0  1012 bytes      = 5.556  104 years 1byte 150 char 60 min 40 hour 50 weeks

(

)

 56, 000 years 62. The volume of water used by the people can be calculated as follows: 3

L day   365day   1000 cm   1km  3 −3 ( 4  10 people )  1200   1 y   1L   105cm  = 4.38  10 km y 4 people      3

The depth of water is found by dividing the volume by the area. 5 V 4.38  10−3 km3 yr  −5 km   10 cm  7.3 10 d= = =     = 7.3cm yr  7 cm yr 60 km 2 yr   1 km  A  63. We do a “units conversion” from bytes to minutes, using the given CD reading rate. 8 bits 1sec 1min 783.216  106 bytes    = 74.592 min  75 min 6 1byte 1.4  10 bits 60 sec

(

)

 10−10 m   1nm    −9  = 0.10 nm  1Å   10 m 

64. (a) 1.0Å = (1.0Å ) 

 10−10 m   1fm  5   10−15 m  = 1.0  10 fm 1Å   

(b) 1.0Å = (1.0Å ) 

 1Å  = 1.0  1010 Å  −10  10 m   9.46  1015 m   1Å  25 (d) 1.0 ly = (1.0 ly )   −10  = 9.5  10 Å  1 ly    10 m  (c) 1.0 m = (1.0 m ) 

Chapter 1

Introduction, Measurement, Estimating

65. Assume that the alveoli are spherical and that the volume of a typical human lung is about 2 liters, which is 0.002 m3. The diameter can be found from the volume of a sphere, 43  r 3. 4 3

 r =  ( d 2) = 3

4 3

d3 6

( 3 10 )  6 = 2 10 m 8

−3

 6 ( 2  10−3 ) 3  → d= m  8  3  10  

1/ 3

= 2  10 −4 m

66. A pencil has a diameter of about 0.7 cm. If held about 0.75 m from the eye, it can just block out the Moon. The ratio of pencil diameter to arm length is the same as the ratio of Moon diameter to Moon distance. From the diagram, we have the following ratios. Pencil

Moon

Pencil Distance

Moon Distance

Pencil diameter

Pencil distance

Moon diameter =

Moon diameter

→

Moon distance Pencil diameter

Pencil distance The actual value is 3480 km.

( Moon distance ) =

7  10−3 m

( 3.8  10 km )  3500 km 0.75 m 5

67. To calculate the mass of water, we need to find the volume of water and then convert the volume to 2 mass. The volume of water is the area of the city 35 km times the depth of the water (1.0 cm).

(

)

2 5    10−3 kg   1 metric ton  2  10 cm  5 5 35 km 1.0 cm ( ) )  1 km   (  1 cm3   103 kg  = 3.5  10  4  10 metric tons       

To find the number of gallons, convert the volume to gallons. 2 5    1 L   1 gal  = 9.26  107 gal  9  107 gal 2  10 cm  ( 35 km )   (1.0 cm )    3 3   1  10 cm   3.78 L   1 km   

68. We assume that the amount the ocean rises will be very small compared to the radius of the Earth. With this assumption, we calculate the volume of the water melted and set it equal to the volume of a thin “shell” of water, whose volume is 2/3 of the Earth’s surface area, times the water thickness. 2 Volume of Greenland ice = 23 4 REarth  water thickness, 3

2 m = 23 4 ( 6.38  106 m ) t → (1.7 10 km ) ( 2.5 km )  1000   1km  6

m (1.7 10 km ) ( 2.5 km )  1000   1km  = 12.46 m  12 m t= 2 2 4 ( 6.38  106 m ) 3 6

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

69. A cubit is about a half of a meter, by measuring several people’s forearms. Thus the dimensions of Noah’s ark would be 150 m long, 25 m wide, 15 m high . The volume of the ark is found by multiplying the three dimensions.

V = (150 m )( 25 m )(15 m ) = 5.625  104 m 3  6  104 m 3 70. The volume of the oil will be the area times the thickness. The area is  r =  ( d 2 ) . 2

V =  ( d 2) t → d = 2 2

t

 1m    100 cm  = 3  103 m −10

1000 cm 3 

 ( 2  10 m )

This is approximately 2 miles. 71. Utilize the fact that walking totally around the Earth along the meridian would trace out a circle whose full 360° would equal the circumference of the Earth. 3 1o    2 6.38  10 km   0.621mi  (1minute of arc )      = 1.15mi 360o  60 minutes of arc     1km 

(

)

72. (a) Note that sin15.0o = 0.259 and sin15.5o = 0.267, so  sin  = 0.267 − 0.259 = 0.008.

 0.5o     100 =    15.0o 100 = 3%     

 8  10−3    sin   100 =    0.259 100 = 3%  sin    

(b) Note that sin 75.0o = 0.966 and sin 75.5o = 0.968, so  sin  = 0.968 − 0.966 = 0.002.

 2  10−3    sin   100 =    0.966 100 = 0.2%  sin    

 0.5o     100 =    75.0o 100 = 0.7%     

A consequence of this result is that when you use a protractor, and you have a fixed uncertainty in the angle ( 0.5o in this case), you should measure the angles from a reference line that gives a large angle measurement rather than a small one. Note above that the angles around 75° had only a 0.2% error in sin , while the angles around 15° had a 3% error in sin . 73. Consider the diagram as shown in the text. Let l represent 85 strides that he walks upstream. Then from the diagram find the distance d across the river. d → tan 60o = l  0.8 m  o d = l tan 60o = ( 85 strides )   tan 60 = 117.8m  120 m  stride  But since the problem asks for an estimate, we might just keep 1 significant figure, and call the distance 100 meters. 74. Consider the body to be a cylinder, about 170 cm tall (  57 ) , and about 12 cm in cross-sectional radius (which corresponds to a 30-inch waist). The volume of a cylinder is given by the area of the cross section times the height.

V =  r 2 h =  ( 0.12 m ) (1.7 m ) = 7.69  10−2 m 3  8  10 −2 m 3 2

Chapter 1

Introduction, Measurement, Estimating

75. According to census data, there are about 800,000 people in San Francisco. We then approximate each household as having 4 people, that each household needs 1 plumber visit per year, that a plumber can make 4 visits per day, and that a plumber works 250 days each calendar year.  1 visit        1 household 1 plumber 1year    year 8  105 people       =  4 people   1 household   4 visits   250 work days     work day    

(

)

200 plumbers 76. We use values from Table 1–3. mhuman 102 kg = −17 = 1019 mDNA 10 kg molecule

77. The units for each term must be in liters, since the volume is in liters.

 units of 4.1 m =  L →  units of 4.1 =

L m

 units of 0.018 year  =  L →  units of 0.018 =

L year

 units of 2.7 = L 78. The density is the mass divided by volume. There will be only 1 significant figure in the answer. mass 6g density = = = 2.118 g cm 3  2 g cm 3 3 volume 2.8325cm 79. Multiply the volume of a spherical universe times the density of matter, adjusted to ordinary matter. The volume of a sphere is 43  r 3 .

(

m = V = 1  10−26 kg m 3

)  (13.0  10 m ) ( 0.04 ) 25

4 3

= 3.68  1051 kg  4  1051 kg

CHAPTER 2: Describing Motion: Kinematics in One Dimension Responses to Questions 1.

A car speedometer measures only speed. It does not give any information about the direction of motion, and so does not measure velocity.

If the velocity of an object is constant, the speed must also be constant. (A constant velocity means that the speed and direction are both constant.) If the speed of an object is constant, the velocity CAN vary. For example, a car traveling around a curve at constant speed has a varying velocity, since the direction of the velocity vector is changing.

If the velocity of an object is constant, the speed and the direction of travel must also be constant. If that is the case, then the average velocity is the same as the instantaneous velocity, because nothing about its velocity is changing. The ratio of displacement to elapsed time will not be changing, no matter the actual displacement or time interval used for the measurement.

There is no general relationship between the magnitude of speed and the magnitude of acceleration. For example, one object may have a large but constant speed. The acceleration of that object is then 0. Another object may have a small speed but be gaining speed, and therefore have a positive acceleration. So in this case the object with the greater speed has the lesser acceleration. Or consider two objects that are dropped from rest at different times. If we ignore air resistance, then the object dropped first will always have a greater speed than the object dropped second, but both will have the same acceleration of 9.80 m/s2.

The accelerations of the motorcycle and the bicycle are the same, assuming that both objects travel in a straight line. Acceleration is the change in velocity divided by the change in time. The magnitude of the change in velocity in each case is the same, 10 km/h, so over the same time interval the accelerations will be equal.

Yes, for example, a car that is traveling northward and slowing down has a northward velocity and a southward acceleration.

The velocity of an object can be negative when its acceleration is positive. If we define the positive direction to be to the right, then an object traveling to the left that is slowing down will have a negative velocity with a positive acceleration. The velocity of an object can also be positive when its acceleration is negative. If again we define the positive direction to be to the right, then an object traveling to the right that is slowing down will have a positive velocity and a negative acceleration.

If north is defined as the positive direction, then an object traveling to the south and increasing in speed has both a negative velocity and a negative acceleration. Or, if “up” is defined as the positive direction, then an object falling due to gravity has both a negative velocity and a negative acceleration (if air resistance is ignored).

If the two cars emerge side by side, then the one moving faster is passing the other one. Thus car A is passing car B. With the acceleration data given for the problem, the ensuing motion would be that car A would pull away from car B for a time, but eventually car B would catch up to and pass car A.

Chapter 2

Describing Motion: Kinematics in One Dimension

10. Yes. Remember that acceleration is a change in velocity per unit time, or a rate of change in velocity. So, velocity can be increasing while the rate of increase goes down. For example, suppose a car is traveling at 40 km/h and one second later is going 50 km/h. One second after that, the car’s speed is 55 km/h. The car’s speed was increasing the entire time, but its acceleration in the second time interval was lower than in the first time interval. Thus its acceleration was decreasing even as the speed was increasing. Another example would be an object falling WITH air resistance, but with a speed less than its terminal velocity. Let the downward direction be positive. As the object falls, it gains speed, and the air resistance increases. As the air resistance increases, the acceleration of the falling object decreases, and so it gains speed less quickly the longer it falls. 11. If there were no air resistance, the ball’s only acceleration during flight would be the acceleration due to gravity, so the ball would land in the catcher’s mitt with the same speed it had when it left the bat, 120 km/h. Since the acceleration is the same through the entire flight, the time for the ball’s speed to change from 120 km/h to 0 on the way up is the same as the time for its speed to change from 0 to 120 km/h on the way down. In both cases the ball has the same magnitude of displacement. 12. (a) If air resistance is negligible, the acceleration of a freely falling object stays the same as the object falls. The object’s speed increases, but since it increases at a constant rate, the acceleration is constant. (b) In the presence of air resistance, the acceleration decreases. Air resistance increases as the speed increases. If the object falls far enough, the acceleration will become zero and the velocity will become constant. That velocity is often called the terminal velocity. 13. Average speed is the displacement divided by the time. Since the distances from A to B and from B to C are equal, you spend more time traveling at 70 km/h than at 90 km/h, so your average speed should be less than 80 km/h. If the distance from A to B (or B to C) is x km, then the total distance traveled is 2x. The total time required to travel this distance is x/70 h plus x/90 h. Then 2x 2(90)(70) d v= = = = 78.75 km/h  79 km/h. t x 70 + x 90 90 + 70 14. Yes. Consider an object thrown straight up in the air. It has a nonzero acceleration due to gravity for its entire flight (neglecting air resistance). However, at the highest point it momentarily has a zero velocity. A car, at the moment it starts moving from rest, has zero velocity and nonzero acceleration. 15. Yes. Any time the velocity is constant, the acceleration is zero. For example, a car traveling at a constant 90 km/h in a straight line has nonzero velocity and zero acceleration. 16. A rock falling from a cliff has a constant acceleration if air resistance is neglected. An elevator moving from the second floor to the fifth floor making stops along the way does NOT have a constant acceleration. Its acceleration will change in magnitude and direction as the elevator starts and stops. The dish resting on a table has a constant (zero) acceleration. 17. The two conditions are that the motion needs to be near the surface of the Earth, and that there is no air resistance. An example where the second condition is not even a reasonable approximation is that of parachuting. The air resistance caused by the parachute results in the acceleration not being constant, and having values much different than 9.8 m s 2. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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18. The slope of the position versus time curve is the velocity. The object starts at the origin with a constant velocity (and therefore zero acceleration), which it maintains for about 20 s. For the next 10 s, the positive curvature of the graph indicates the object has a positive acceleration; its speed is increasing. From 30 s to 45 s, the graph has a negative curvature; the object uniformly slows to a stop, changes direction, and then moves backwards with increasing speed. During this time interval its acceleration is negative, since the object is slowing down while traveling in the positive direction and then speeding up while traveling in the negative direction. For the final 5 s shown, the object continues moving in the negative direction but slows down, which gives it a positive acceleration. During the 50 s shown, the object travels from the origin to a point 20 m away, and then back 10 m to end up 10 m from the starting position. 19. Initially, the object begins with a speed of 14 m/s, moving in the positive direction with a constant acceleration, until t = 45 s, when it has a velocity of about 37 m/s in the positive direction. The acceleration then decreases, reaching an instantaneous acceleration of 0 at t = 50 s, when the object has its maximum speed of about 38 m/s. The object then begins to slow down, but continues to move in the positive direction. The object stops moving at t = 90 s and stays at rest until t = 108 s. Then the object begins to move in the positive direction again, at first with a larger acceleration, and then a lesser acceleration. At the end of the recorded motion, the object is still moving to the right and gaining speed.

Responses to MisConceptual Questions 1.

(d) It is a common misconception that a positive acceleration always increases the speed, as in (b) and (c). However, when the velocity and acceleration are in opposite directions the speed will decrease.

(d) Since the velocity and acceleration are in opposite directions, the object will slow to a stop. However, since the acceleration remains constant, the object will only momentarily stop before moving toward the left.

(c) Since the distances are the same, a common error is to assume that the average speed will be halfway between the two speeds, or 40 km/h. However, it takes the car longer to travel the 4 km at 30 km/h than to travel the other 4 km at 50 km/h. Since more time is spent at 30 km/h the average speed will be closer to 30 km/h than to 50 km/h.

(d) From Eq. 2–12(c), for an object with constant acceleration starting from rest, the final speed is given by v = 2ad . Thus if aB = 4aA, then v B = 2v A.

(a) A common misconception is that the acceleration of an object in free-fall depends upon the motion of the object. If there is no air resistance, the accelerations is the same throughout the flight, even at the top. If the acceleration were 0 at the top, the object would not move since it also has a velocity of 0 at the top. The velocity is constantly changing throughout the flight.

(c) A common misconception is that the acceleration of an object in free-fall depends upon the motion of the object. If there is no air resistance, the accelerations for the two balls are the same magnitude and direction throughout both of their flights.

(a) Since the distance between the rocks increases with time, a common misconception is that the velocities are increasing at different rates. However, both rocks fall under the influence of gravity, so their velocities increase at the same rate.

Chapter 2

Describing Motion: Kinematics in One Dimension

(e) Both objects experience the same acceleration, so the rock that is dropped first always has a higher speed than the second rock. Since the first rock is always moving faster, it is always getting farther away from the second rock.

(b, c) Each of the given equations are based on Eq. 2–12(a–d). Answer (a) has the acceleration replaced properly with –g, but the initial velocity is downward and as such should be negative. Answer (d) is incorrect because the initial velocity has been inserted for the average velocity. Answers (b) and (c) have the correct signs for each variable and the known values are inserted properly.

10. (a) Increasing speed means that the slope must be getting steeper over time. In graphs (b) and (e) the slope remains constant, so these are cars moving at constant speed. In graph (c), as time increases x decreases. However the rate at which it decreases is also decreasing. This is a car slowing down. In graph (d) the car is moving away from the origin, but again it is slowing down. The only graph in which the slope is increasing with time is graph (a). 11. (c, e) The two lines have different slopes at point A, so they don’t have the same instantaneous velocity or speed. The object represented by the dashed line has traveled more distance, since it has traveled “forward and back” to get to the point represented by A. But they are both at the same position after the same amount of time, and so they have the same average velocity.

Solutions to Problems 1.

The distance of travel (displacement) can be found by rearranging Eq. 2–2 for the average velocity. Also note that the units of the velocity and the time are not the same, so the speed units will be converted.  1h  v = x t → x = vt = ( 85 km h )   ( 2.0s ) = 0.0472 km  47 m  3600 s 

The average speed is given by Eq. 2–1, using d to represent distance traveled.

v = d t = 235 km 2.85 h = 82.5 km h 3.

The average velocity is given by Eq. 2–2. x 8.5 cm − 5.2 cm 3.3cm v= = = = 0.61cm s 3.4 s − ( −2.0 s ) 5.4 s t The average speed cannot be calculated. To calculate the average speed, we would need to know the actual distance traveled, and it is not given. We only have the displacement.

(a) The speed of sound is intimated in the problem as 1 mile per 5 seconds. The speed is calculated as follows. distance  1mi   1610 m  speed = =  = 322 m s  300 m s  time  5s   1 mi  (b) The speed of 322 m s would imply the sound traveling a distance of 966 meters (which is approximately 1 km) in 3 seconds. So the rule could be approximated as 1 km every 3 s .

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

The time for the first part of the trip is calculated from the initial speed and the first distance, using d to represent distance. d d 210 km v1 = 1 → t1 = 1 = = 2.211h t1 v1 95 km h The time for the second part of the trip is now calculated. t2 = t total − t1 = 4.5 h − 2.211h = 2.289 h The distance for the second part of the trip is calculated from the average speed for that part of the trip and the time for that part of the trip. d v 2 = 2 → d 2 = v2 t2 = ( 65 km h )( 2.289 h ) = 148.8 km  150 km t2 (a) The total distance is then d total = d1 + d 2 = 210 km + 148.8 km = 358.8 km  360 km . (b) The average speed is NOT the average of the two speeds, and so is NOT 85 km/h. Use the definition of average speed, Eq. 2–1. d 358.8 km v = total = = 79.73 km h  80 km h t total 4.5 h Note that the result has 2 significant figures, and so could be expressed as 8.0  101 km h.

The distance traveled is 38 m + 12 ( 38 m ) = 57 m, and the displacement is 38 m − 12 ( 38 m ) = 19 m. The total time is 7.4 s + 1.8 s = 9.2 s. distance 57 m (a) Average speed = = = 6.2 m s time elapsed 9.2 s (b) Average velocity = vavg =

displacement time elapsed

19 m 9.2 s

= 2.1m s

The distance traveled is 3200 m ( 8 laps  400 m lap ). That distance probably has either 3 or 4 significant figures, since the track distance is probably known to at least the nearest meter for competition purposes. The displacement is 0 because the ending point is the same as the starting point. 3200 m  1 min  d (a) Average speed = =   = 3.68 m s  t 14.5 min  60 s  (b) Average velocity = v = x t = 0 m s

The average speed is the distance divided by the time. d  1  109 km   1yr  1 d  5 5 v = =    = 1.141 10 km h  1 10 km h  t  1yr   365.25 d  24 h 

Both objects will have the same time of travel. If the truck travels a distance xtruck , then the distance the car travels will be xcar = xtruck + 310 m. Use Eq. 2–2 for average speed, v = x t , solve for time, and equate the two times. x x xtruck x + 310 m t = truck = car = truck 75 km h 95 km h v truck vcar

Chapter 2

Describing Motion: Kinematics in One Dimension

Solving for xtruck gives xtruck = ( 310 m )

( 75 km h )

( 95 km h − 75 km h )

= 1163m.

 1163 m  60 min    = 0.9304 min = 55.82 s  56 s . v truck  75000 m h  1h  x  1163 m + 310 m  60 min  Also note that t = car =    = 0.9303 min = 56 s. vcar  95000 m h  1h  The time of travel is t =

xtruck

=

ALTERNATE SOLUTION: The speed of the car relative to the truck is 95 km h − 75 km h = 20 km h. In the reference frame of the truck, the car must travel 310 m to catch it. 0.31 km  3600 s  t =   = 56s 20 km h  1 h  10. The distance traveled is 560 km (280 km outgoing, 280 km return). The displacement ( x ) is 0 because the ending point is the same as the starting point. (a) To find the average speed, we need the distance traveled (560 km) and the total time x x 280 km elapsed. During the outgoing portion, v1 = 1 and so t1 = 1 = = 2.947 h. 95 km h v1 t1 During the return portion, v 2 =

x 2 t 2

, and so t2 =

x2 v2

280 km 55 km h

= 5.091h. Thus the

total time, including lunch, is ttotal = t1 + tlunch + t2 = 2.947 h + 1h + 5.091h = 9.038 h. v=

xtotal

560 km

= 61.96 km h  62 km h t total 9.038 h (b) To find the average velocity, use the displacement and the elapsed time. The displacement is 0 since the person returns to their starting point.

v = x t = 0

11. Since the locomotives have the same speed, they each travel half the distance, 4.75 km. Find the time of travel from the average speed. x x 4.75 km  60 min  v= → t = = = 0.03065 h   = 1.839 min  1.8 min v 155 km h t  1h  12. (a) The area between the concentric circles is equal to the length times the width of the spiral path.  R22 −  R12 = w l →

l =

 ( R22 − R12 ) w

 ( 0.058 m ) − ( 0.025 m )  2

−6

1.6  10 m

= 5.378  103 m  5400 m

 1s  1min    = 74.69 min  75 min  1.2 m  60 s 

(b) 5.378  103 m 

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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13. (a)

(b) The average velocity is the displacement divided by the elapsed time.

x ( 3.0 ) − x ( 0.0 )

 27 + 10 ( 3.0 ) − 2 ( 3.0 )3  m − ( 27 m )  = = −8.0 m s

3.0s − 0.0s 3.0s (c) The instantaneous velocity is given by the derivative of the position function.

(

)

= 10 − 6t 2 m s

10 − 6t 2 = 0 → t =

s = 1.3s dt 3 This can be seen from the graph as the “highest” point on the graph. 14. The average speed for each segment of the trip is given by v = For the first segment, t1 =

For the second segment, t2 =

1900 km 720 km h

d2 v2

d t

, so t =

d v

for each segment.

= 2.639 h.

2700 km 990 km h

= 2.727 h.

Thus the total time is ttot = t1 + t2 = 2.639 h + 2.727 h = 5.366 h  5.4 h . The average speed of the plane for the entire trip is d 1900 km + 2700 km v = tot = = 857.2 km h  860 km h . t tot 5.366 h Note that Eq. 2–12d does NOT apply in this situation. The numeric average of the two speeds is 855 km/h. 15. In order for the two ships to meet “in the middle,” they must both have the same average velocity (ignoring their different directions). We can find the time for each segment of the trip using d d , so t = for each segment. Both ships travel 600 km + 800 km + 600 km = 2000 km. v= t v d d d 600 km 800 km 600 km + + = 80 h First ship: t = 1 + 2 + 3 = v1 v 2 v3 20 km h 40 km h 20 km h Second ship: v =

d t

2000 km 80 h

= 25 km h

Chapter 2

Describing Motion: Kinematics in One Dimension

16. Slightly different answers may be obtained since the data comes from reading the graph. (a) The instantaneous velocity is given by the slope of the tangent line to the curve. At t = 10.0s, 3m − 0 the slope is approximately v (10 )  = 0.3 m s . 10.0 s − 0 (b) At t = 30.0s, the slope of the tangent line to the curve, and thus the instantaneous velocity, is 20 m − 8 m = 1.2 m s . approximately v ( 30 )  35 s − 25 s (c) The average velocity is given by v = (d) The average velocity is given by v = (e) The average velocity is given by v =

x ( 5.0 ) − x ( 0 ) 5.0 s − 0 s

1.5 m − 0

x ( 30.0 ) − x ( 25.0 ) 30.0 s − 25.0 s x ( 50.0 ) − x ( 40.0 ) 50.0 s − 40.0 s

5.0 s = =

= 0.30 m s .

16 m − 9 m 5.0 s

= 1.4 m s .

10 m − 19.5 m 10.0 s

= −0.95 m s .

17. Slightly different answers may be obtained since the data comes from reading the graph. (a) The indication of a constant velocity on a position-time graph is a constant slope, which occurs from t = 0 s to t  18s . (b) The greatest velocity will occur when the slope is the highest positive value, which occurs at

t  27 s . (c) The indication of a 0 velocity on a position-time graph is a slope of 0, which occurs at

t  38s . (d)

The object moves in both directions . When the slope is positive, from t = 0 s to t  38 s,

the object is moving in the positive direction. When the slope is negative, from t  38 s to t = 50 s, the object is moving in the negative direction. 18. The v vs. t graph is found by taking the slope of the x vs. t graph. Both graphs are shown here.

19. The average speed of sound is given by vsound = x t , and so the time for the sound to travel from the end of the lane back to the bowler is tsound =

x vsound

16.5 m 340 m s

= 4.85  10 −2 s. Thus the time for

the ball to travel from the bowler to the end of the lane is given by tball = ttotal − tsound = © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

2.75s − 4.85  10 −2 s = 2.7015s. And so the speed of the ball is as follows.

v ball =

x tball

16.5 m 2.7015 s

= 6.1077 m s  6.11m s

20. For the car to pass the train, the car must travel the length of the train AND the distance the train travels. The distance the car travels can thus be written as either d car = vcar t = ( 95 km h ) t or

d car = l train + v train t = 1.50 km + ( 75 km h ) t. To solve for the time, equate these two expressions for

the distance the car travels.

( 95 km h ) t = 1.50 km + ( 75 km h ) t → t =

1.50 km 20 km h

 60 min   = 4.5 min  1h 

= 0.075 h 

Note that this is the same as calculating from the reference frame of the train, in which the car is moving at 20 km/h and must travel the length of the train. The distance the car travels during this time is d = ( 95 km h )( 0.075 h ) = 7.125 km  7.1km . If the train is traveling in the opposite direction from the car, then the car must travel the length of the train MINUS the distance the train travels. Thus the distance the car travels can be written as either d car = ( 95 km h ) t or d car = 1.50 km − ( 75 km h ) t. To solve for the time, equate these two expressions for the distance the car travels. ( 95 km h ) t = 1.50 km − ( 75 km h ) t →

1.50 km 170 km h

 3600s   = 31.8s  32 s  1h 

= 8.824  10−3 h 

The distance the car travels during this time is d = ( 95 km h ) ( 8.824  10−3 h ) = 0.84 km . 21. (a) The average acceleration of the sprinter is a =

v t

9.00 m s − 0.00 m s 1.48 s

= 6.08 m s 2 .

(b) We change the units for the acceleration. 2

 1km   3600s  a = ( 6.08 m s )  = 7.88  104 km h 2     1000 m   1h  2

22. The initial velocity is v0 = 15 km h, the final velocity is v = 65 km h, and the displacement is x − x0 = 4.0 km = 4000 m. Find the average acceleration from Eq. 2–12c.

v 2 = v02 + 2a ( x − x0 ) → 2

v 2 − v02

2 ( x − x0 )

 1m s  ( 65 km h ) 2 − (15 km h ) 2     3.6 km h    = 3.9  10−2 m s 2 = 2 ( 4000 m )

23. The initial velocity of the car is the average velocity of the car before it accelerates. x 120 m v= = = 24 m s = v0 5.0 s t The final velocity is v = 0, and the time to stop is 3.7 s. Use Eq. 2–12a to find the acceleration. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

v = v0 + at → a =

v − v0 t

0 − 24 m s 3.7 s

= −6.486 m s 2

(

)  9.801 mg s  = 0.66 g’s .

Thus the magnitude of the acceleration is 6.5m s2 , or 6.486 m s 2 





24. We assume that the speedometer can read to the nearest km/h, and so the value of 120 km/h has three significant digits. The time can be found from the average acceleration, a = v t.

t =

v a

120 km h − 65 km h 1.8 m s 2

 1m s    3.6 km h  = 8.488s  8.5s

( 55 km h ) 

1.8 m s 2

25. (a) The average velocity is the displacement divided by the elapsed time. x 385 m − 25 m v= = = 21.2 m s t 20.0 s − 3.00 s (b) a =

v t

45.0 m s − 11.0 m s 20.0 s − 3.00 s

= 2.00 m s 2

26. The acceleration is the second derivative of the position function. dx d 2x dv 2 x = 4.8t + 7.3t → v = = 4.8 + 14.6t → a = 2 = = 14.6 m s 2 dt dt dt 27. (a) Since the units of A times the units of t must equal meters, the units of A must be m s . Since the units of B times the units of t 2 must equal meters, the units of B must be

m s2 . (b) The acceleration is the second derivative of the position function. dx d 2x dv x = At + Bt 2 → v = = A + 2 Bt → a = 2 = = 2B m s2 dt dt dt (c)

v = A + 2 Bt → v ( 5 ) = ( A + 12 B ) m s

a = 2B m s2

(d) The velocity is the derivative of the position function. dx x = At + Bt −3 → v = = A − 3Bt −4 dt

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

28. To estimate the velocity, find the average velocity over each time interval, and assume that the car had that velocity at the midpoint of the time interval. To estimate the acceleration, find the average acceleration over each time interval, and assume that the car had that acceleration at the midpoint of the time interval. A sample of each calculation is shown. From 2.00 s to 2.50 s, for average velocity: 2.50 s + 2.00 s tmid = = 2.25 s 2 x 13.79 m − 8.55 m 5.24 m vavg = = = = 10.48 m s 2.50 s − 2.00 s 0.50 s t From 2.25 s to 2.75 s, for average acceleration: 2.25 s + 2.75 s = 2.50 s tmid = 2 v 13.14 m s − 10.48 m s 2.66 m s = = aavg = t 2.75 s − 2.25 s 0.50 s

= 5.32 m s 2

v (m/s)

a (m/s )

t (s)

29. We call the location where the cars are initially next to each other x0 = 0. The position of the first

car can be written as x = x0 + v0t + 12 at 2 = 25.0t − 12 ( 2.0 ) t 2 . The position of the second car can be written as x = x0 + v0t + 12 at 2 = 12 ( 2.0 ) t 2 . We equate the two position equations and solve for the time.

25.0t − 12 ( 2.0) t 2 = 12 ( 2.0) t 2 → 25.0t = 2.0t 2 → t = 0 or 12.5 s

The value of 0 is the time when the cars were initially next to each other, so the “next” time they are next to each other is t = 12.5s  13s .

Chapter 2

Describing Motion: Kinematics in One Dimension

30. Slightly different answers may be obtained since the data comes from reading the graph. (a) The greatest velocity is found at the highest point on the graph, which is at t  48 s . (b) The indication of a constant velocity on a velocity–time graph is a slope of 0, which occurs from t = 90 s to t  108 s . (c) The indication of a constant acceleration on a velocity–time graph is a constant slope, which occurs from t = 0 s to t  42 s , again from t  65 s to t  83 s , and again from t = 90 s to t  108 s . (d) The magnitude of the acceleration is greatest when the magnitude of the slope is greatest, which occurs from t  65 s to t  83 s .

31. Slightly different answers may be obtained since the data comes from reading the graph. We assume that the short, nearly horizontal portions of the graph are the times that shifting is occurring, and those times are not counted as being “in” a certain gear. v 2 24 m s − 14 m s = = 2.5 m s 2 . (a) The average acceleration in 2nd gear is given by a2 = 8 s−4 s t2 (b) The average acceleration in 4th gear is given by a4 =

v 4 t4

44 m s − 37 m s 27 s − 16 s

= 0.6 m s 2 .

32. (a) The train’s constant speed is v train = 5.0 m s, and the location of the empty box car as a

function of time is given by xtrain = vtrain t = ( 5.0 m s ) t. The fugitive has v0 = 0 m s and a = 1.4 m s 2 until his final speed is 6.0 m s . The elapsed time during the acceleration is

tacc =

v − v0

6.0 m s

= 4.286 s. Let the origin be the location of the fugitive when he starts a 1.4 m s 2 to run. The first possibility to consider is, “Can the fugitive catch the empty box car before he reaches his maximum speed?” During the fugitive’s acceleration, his location as a function of time is given by Eq. 2–12b, xfugitive = x0 + v0t + 12 at 2 = 0 + 0 + 12 1.4 m s 2 t 2 . For him to catch

(

)

( 5.0 m s ) t = 12 (1.4 m s 2 ) t 2 . The solutions of this

the train, we must have xtrain = xfugitive →

are t = 0s, 7.1s. Thus the fugitive cannot catch the car during his 4.286 s of acceleration. Now the equation of motion of the fugitive changes. After the 4.286 s of acceleration, he runs with a constant speed of 6.0 m s . Thus his location is now given (for times t  4.286s ) by the following. 2 xfugitive = 12 1.4 m s 2 ( 4.286s ) + ( 6.0 m s )( t − 4.286s ) = ( 6.0 m s ) t − 12.86 m

(

)

So now, for the fugitive to catch the train, we again set the locations equal. xtrain = xfugitive → ( 5.0 m s ) t = ( 6.0 m s ) t − 12.86 m → t = 12.86 s  13s (b) The distance traveled to reach the box car is given by the following. xfugitive ( t = 12.86s ) = ( 6.0 m s )(12.86s ) − 12.86 m = 64 m 33. The acceleration can be found from Eq. 2–12c.

v 2 = v02 + 2a ( x − x0 ) → a =

v 2 − v02

2 ( x − x0 )

0 − ( 26 m s ) 2 ( 88 m )

= −3.8 m s 2

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

34. The acceleration can be found from Eq. 2–12a. v − v0 0 − 28 m s v = v0 + at → a = = = −4.4 m s 2 t 6.3s 35. By definition, the acceleration is a =

v − v0

22 m s − 13 m s

= 1.38 m s 2  1.4 m s 2 . t 6.5 s The distance of travel can be found from Eqs. 2–8 and 2–9. v +v 13 m s + 22 m s x − x0 = vt = 0 t = ( 6.5s ) = 114 m  110 m 2 2 It can also be found from Eq. 2–12b.

(

)

x − x0 = v0t + 12 at 2 = (13 m s )( 6.5s ) + 12 1.38 m s 2 ( 6.5s ) = 113.7 m  110 m 2

It can also be found from Eq. 2–12c.

v = v + 2a ( x − x0 ) → x − x0 = 2

2 0

v 2 − v02 2a

( 22 m s ) − (13 m s ) 2

(

2 1.38 m s 2

)

= 114.1m  110 m

There are slight differences in the answers because of rounding the acceleration. 36. The sprinter starts from rest. The average acceleration is found from Eq. 2–12c.

(11.5 m s ) − 0 = = 3.674 m s 2  3.67 m s 2 v = v + 2a ( x − x0 ) → a = 2 ( x − x0 ) 2 (18.0 m ) 2

2 0

v 2 − v02

Her elapsed time is found by solving Eq. 2–12a for time. v − v0 11.5 m s − 0 v = v0 + at → t = = = 3.13s a 3.674 m s 2 37. The words “slowing down uniformly” imply that the car has a constant acceleration. The distance of travel is found from combining Eqs. 2–8 and 2–9. v +v  28.0 m s + 0 m s  2 x − x0 = 0 t =  ( 8.60 sec ) = 1.20  10 m 2 2   38. The final velocity of the truck is zero. The initial velocity is found from Eq. 2–12c with v = 0 and solving for v 0. Note that the acceleration is negative.

v 2 = v02 + 2a ( x − x0 ) → v0 = v 2 − 2a ( x − x0 ) = 0 − 2 ( −6.00 m s 2 ) ( 45 m ) = 23m s 39. For the baseball, v0 = 0, x − x0 = 3.5 m, and the final speed of the baseball (during the throwing motion ) is v = 43 m s. The acceleration is found from Eq. 2–12c.

v 2 = v02 + 2a ( x − x0 ) → a =

( 43 m s ) − 0 = 264 m s 2  260 m s 2 2 ( x − x0 ) 2 ( 3.5 m ) v 2 − v02

Chapter 2

Describing Motion: Kinematics in One Dimension

40. The final velocity of the driver is zero. The acceleration is found from Eq. 2–12c with v = 0 and solving for a. 2

v 2 − v02

2 ( x − x0 )

  1000 m  1h   0 − ( 95 km h )    1km  3600 s     = = −435.2 m s 2 2 ( 0.80 m )

Converting to “g’s”: a =

435.2 m s 2

( 9.80 m s ) g 2

= 44 g’s .

41. (a) The final velocity of the car is 0. The distance is found from Eq. 2–12c with an acceleration of a = −0.50 m s 2 and an initial velocity of 85 km h . 2

  1000 m  1h   0 − ( 85 km h )    2 2 1km  3600 s   v − v0   x − x0 = = = 557 m  560 m 2

(

2 −0.50 m s

)

(b) The time to stop is found from Eq. 2–12a.



v − v0 a

 1000 m  1h      1km  3600 s   = 47.22 s  47 s

0 − ( 85 km h )  =



( −0.50 m s ) 2

(c) Take x0 = x ( t = 0 ) = 0 m. Use Eq. 2–12b, with a = −0.50 m s 2 and an initial velocity of 85 km h . The first second is from t = 0 s to t = 1s, and the fifth second is from t = 4 s to t = 5s.

 1m s  2 (1s ) + 12 ( −0.50 m s 2 ) (1s ) = 23.36 m →   3.6 km h 

x ( 0 ) = 0 ; x (1) = 0 + ( 85 km h )  x (1) − x ( 0 ) = 23 m

 1m s  2 ( 4 s ) + 12 ( −0.50 m s 2 ) ( 4 s ) = 90.44 m   3.6 km h 

x ( 4 ) = 0 + ( 85 km h ) 

 1m s  2 ( 5s ) + 12 ( −0.50 m s 2 ) ( 5s ) = 111.81m   3.6 km h 

x ( 5 ) = 0 + ( 85 km h ) 

x ( 5 ) − x ( 4 ) = 111.81m − 90.44 m = 21.37m  21m

42. The origin is the location of the car at the beginning of the reaction time. The initial speed of the car

 1000 m   1h    = 26.39 m s . The location where the brakes are applied is  1km   3600s 

is ( 95 km h ) 

found from the equation for motion at constant velocity. x0 = v0 t R = ( 26.39 m s )( 0.40 s ) = 10.56 m This is now the starting location for the application of the brakes. In each case, the final speed is 0.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(a) Solve Eq. 2–12c for the final location. v 2 = v02 + 2a ( x − x0 ) → x = x0 +

v 2 − v02 2a

= 10.56 m +

0 − ( 26.39 m s )

(

2 −2.5 m s 2

= 149.84 m  150 m

)

(b) Solve Eq. 2–12c for the final location with the second acceleration.

x = x0 +

v 2 − v02 2a

= 10.56 m +

0 − ( 26.39 m s )

(

2 −5.5 m s 2

)

= 74 m

43. Calculate the distance that the car travels during the reaction time and the deceleration. x1 = v0 t = (18.0 m s )( 0.380 s ) = 6.84 m

v = v + 2ax2 → x2 = 2

2 0

v 2 − v02 2a

0 − (18.0 m s )

(

2 −3.65 m s 2

)

= 44.38 m

x = 6.84 m + 44.4 m = 51.22 m ( only 3 significant figures ) Since she is only 24.0 m from the intersection, she will NOT be able to stop in time. She will be 27.2 m past the intersection. 44. The average velocity is defined by Eq. 2–2, v =

x

x − x0

. Compare this expression to t t Eq. 2–12d, v = 12 ( v + v0 ) . A relation for the velocity is found by integrating the expression for the acceleration, since the acceleration is the derivative of the velocity. Assume the velocity is v0 at time t = 0. v t dv a = A + Bt = → d v = ( A + Bt ) dt →  d v =  ( A + Bt ) dt → v = v0 + At + 12 Bt 2 dt v 0 0

Find an expression for the position by integrating the velocity, assuming that x = x0 at time t = 0.

v = v0 + At + 12 Bt 2 = x

(

)

→ dx = v0 + At + 12 Bt 2 dt →

(

)

x − x0

to 12 ( v + v0 ) .

2 2 3  dx =  v0 + At + 12 Bt dt → x − x0 = v0t + 12 At + 16 Bt

Compare v= 1 2

t x − x0 t

( v + v0 ) =

v0 t + 12 At 2 + 16 Bt 3

t v0 + v0 + At + 12 Bt 2 2

= v0 + 12 At + 16 Bt 2 = v0 + 12 At + 14 Bt 2

They are different, so v  12 ( v + v0 ) . 45. Use the information for the first 180 m to find the acceleration, and the information for the full motion to find the final velocity. For the first segment, the train has v0 = 0 m s, v1 = 18 m s, and a displacement of x1 − x0 = 180 m. Find the acceleration from Eq. 2–12c. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

v12 = v02 + 2a ( x1 − x0 ) → a =

(18 m s ) − 0 = 0.90 m s 2 2 ( x1 − x0 ) 2 (180 m ) 2

v12 − v02

Find the speed of the train after it has traveled the total distance (total displacement of x2 − x0 = 180 m + 85 m = 265 m ) using Eq. 2–12c.

v22 = v02 + 2a ( x2 − x0 ) → v2 = v02 + 2a ( x2 − x0 ) = 2 ( 0.90 m s 2 ) ( 265 m ) = 22 m s 46. Calculate the acceleration from the velocity–time data using Eq. 2–12a, and then use Eq. 2–12b to calculate the displacement at t = 2.0s and t = 6.0s. The initial velocity is v0 = 85 m s. a=

v − v0 t

162 m s − 85 m s 10.0 s

= 7.7 m s 2

x = x0 + v0t + 12 at 2 →

(

x ( 7.0 s ) − x ( 2.0 s ) =  x0 + v0 ( 7.0 s ) + 12 a ( 7.0 s )

) − ( x + v ( 2.0 s ) + a ( 2.0 s ) ) 0

1 2

(

= v0 ( 7.0 s − 2.0 s ) + 12 a ( 7.0 s ) − ( 2.0 s )  = ( 85 m s )( 5.0 s ) + 12 7.7 m s 2 2

)( 45s ) 2

= 598.25 m  6.0  102 m 47. During the final part of the race, the runner must have a displacement of 1200 m in a time of 180 s (3.0 min). Assume that the starting speed for the final part is the same as the average speed thus far. 8800 m x v= = = 5.432 m s = v0 t ( 27  60 ) s The runner will accomplish this by accelerating from speed v0 to speed v for t seconds, covering a

distance d1 , and then running at a constant speed of v for (180 − t ) seconds, covering a distance d 2 . We have these relationships from Eq. 2–12a and Eq. 2–12b. v = v0 + at d1 = v0t + 12 at 2 d 2 = v (180 − t ) = ( v0 + at )(180 − t )

1200 m = d1 + d 2 = v0t + 12 at 2 + ( v0 + at )(180 − t ) → 1200 m = 180v0 + 180at − 12 at 2 →

(

)

(

)

1200 m = (180 s )( 5.432 m s ) + (180 s ) 0.20 m s 2 t − 12 0.20 m s 2 t 2 → 0.10t − 36t + 222.24 = 0 → t = 2

36  362 − 4 ( 0.10 )( 222.24 ) 2 ( 0.10 )

= 353.7 s, 6.28s

Since we must have t  180 s, the solution is t = 6.3s . 48. For the runners to cross the finish line side-by-side means they must both reach the finish line in the same amount of time from their current positions. Take Mary’s current location as the origin. Use Eq. 2–12b. For Sally:

22 = 5.0 + 5.0t + 12 ( −0.40 ) t 2 → t 2 − 25t + 85 = 0 → t=

25  252 − 4 ( 85 ) 2

= 4.059 s, 20.94 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

The first time is the time she first crosses the finish line, and so it is the time to be used for the problem. Now find Mary’s acceleration so that she crosses the finish line in that same amount of time. 22 − 4t 22 − 4 ( 4.059 ) 22 = 0 + 4t + 12 at 2 → a = 1 2 = = 0.70 m s 2 For Mary: 2 1 t 4.059 ) 2 2( 49. Define the origin to be the location where the speeder passes the police car. Start a timer at the instant that the speeder passes the police car, and find another time that both cars have the same displacement from the origin. For the speeder, traveling with a constant speed, the displacement is given by the following.  1m s  xs = vs t = (135 km h )   ( t ) = ( 37.5 t ) m  3.6 km h  For the police car, the displacement is given by two components. The first part is the distance traveled at the initially constant speed during the 1 second of reaction time.  1m s  xp1 = v0p (1.00 s ) = ( 95 km h )   (1.00 s ) = 26.39 m  3.6 km h  The second part of the police car’s displacement is that which occurs during the accelerated motion, which lasts for ( t − 1.00 ) s. So this second part of the police car displacement, using Eq. 2–12b, is given as follows. Note that this part of the displacement is NOT VALID for t < 1.00 s. 2 2 xp2 = v0p ( t − 1.00 ) + 12 ap ( t − 1.00 ) = ( 26.39 m s )( t − 1.00 ) + 12 ( 2.60 m s 2 ) ( t − 1.00 )  m   So the total police car displacement is:

(

xp = xp1 + xp 2 = 26.39 + 26.39 ( t − 1.00 ) + 1.30 ( t − 1.00 )

Now set the two displacements equal, and solve for the time. 26.39 + 26.39 ( t − 1.00 ) + 1.30 ( t − 1.00 ) = 37.5 t 2

10.55 

(10.55 ) − 4.00

→ t 2 − 10.55t + 1.00 = 0

= 9.57  10 −2 s, 10.5s

The first answer from the quadratic formula is a “fake” answer, because the equation for the police car’s displacement while accelerating, which was used to find that solution, was not valid for times t < 1.00 s. Thus we may ignore that answer. The answer of 10.5 s is the time for the police car to overtake the speeder. As a check on the answer, the speeder travels xs = ( 37.5 m s ) (10.5 s ) = 394 m, and the police car travels xp =  26.39 + 26.39 ( 9.5) + 1.30 ( 9.5 )  m = 394 m.





50. Define the origin to be the location where the speeder passes the police car. Start a timer at the instant that the speeder passes the police car. Both cars have the same displacement 8.00 s after the initial passing by the speeder. But the police car only is accelerating for 7.00 s. For the speeder, traveling with a constant speed, the displacement is given by xs = vs t = ( 8.00vs ) m. For the police car, the displacement is given by two components. The first part is the distance traveled at the initially constant speed during the 1.00 s of reaction time.  1m s  xp1 = v0p (1.00 s ) = ( 95 km h )   (1.00 s ) = 26.39 m  3.6 km h  © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

The second part of the police car displacement is that during the accelerated motion, which lasts for 7.00 s. So this second part of the police car displacement, using Eq. 2–12b, is given by the following. 2 2 xp 2 = v0p ( 7.00 s ) + 12 ap ( 7.00 s ) = ( 26.39 m s )( 7.00 s ) + 12 ( 2.60 m s 2 ) ( 7.00 s ) = 248.43 m Thus the total police car displacement is xp = xp1 + xp2 = ( 26.39 + 248.43) m = 274.82 m. Now set the two displacements equal, and solve for the speeder’s velocity.  3.6 km h  ( 8.00vs ) m = 274.82 m → vs = 34.35 m s = ( 34.35 m s )   = 124 km h  1m s  51. From the given information, we can break the race up into three segments: Segment # 1: Time of 0.15 seconds, no displacement. Segment # 2: Starting speed of 0, final speed v, constant acceleration a, displacement 30.0 m, elapsed time t2 . Thus v = at 2 and 30 m = 12 at 22 . Segment # 3: Constant velocity v, displacement of 370 m, elapsed time t3 . Thus 370 m = v t3 . We also know that t2 + t3 = 54.85s. So we have 4 relationships defining the 4 unknowns of

v , a, t2 , and t3. We solve that system of 4 equations. v = at 2 ; 30 = 12 at 22 ; 370 = vt3 ; t2 + t3 = 54.85

Substitute the 1st equation into the 3rd.

30 = 12 at 22 ; 370 = at 2t3 ; t2 + t3 = 54.85

Divide the 2nd equation by the 1st.

at t 2t = 1 2 32 = 3 ; t2 + t3 = 54.85 30 at 2 t2 2

Substitute the 1st equation into the 2nd.

370 370 30

2t3 t2

→ t2 =

60 370

t3 ;

60 370

t3 + t3 = 54.85

54.85

= 47.20 s → t2 = 54.85 − t3 = 7.65s 1.162 v 7.839 m s 370 m 370 m v= = = 7.839 m s ; a = = = 1.025 m s 2 t3 t2 47.20 s 7.65s 370

t3 + t3 = 1.162t3 = 54.85 → t3 =

Solve for t3 .

The requested results are v = 7.8 m s ; a = 1.0 m s 2 . 52. Choose downward to be the positive direction, and take y0 = 0 at the top of the cliff. The initial velocity is v0 = 0, and the acceleration is a = 9.80 m s 2. The displacement is found from Eq. 2–12b, with x replaced by y. 2 y = y0 + v0t + 12 at 2 → y − 0 = 0 + 12 ( 9.80 m s 2 ) ( 3.25s ) → y = 51.8 m 53. Choose downward to be the positive direction, and take y0 = 0 to be at the top of the Empire State Building. The initial velocity is v0 = 0, and the acceleration is a = 9.80 m s . (a) The elapsed time can be found from Eq. 2–12b, with x replaced by y. 2

y − y0 = v0t + 12 at 2

→

2y a

2 ( 380 m ) 9.80 m s 2

= 8.806 s  8.8 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The final velocity can be found from Eq. 2–12a. v = v0 + at = 0 + ( 9.80 m s 2 ) ( 8.806 s ) = 86 m s 54. Choose downward to be the positive direction. The initial velocity is v0 = 0, the final velocity is

 1m s  2 v = ( 55 km h )  = 15.28 m s , and the acceleration is a = 9.80 m s . Solve Eq. 2–12a   3.6 km h  for the time.

v = v0 + at → t =

v − v0 a

15.28 m s − 0 9.80 m s 2

= 1.6 s

55. Choose upward to be the positive direction, and take y0 = 0 to be the height from which the ball was thrown. The acceleration is a = −9.80 m s 2. The displacement upon catching the ball is 0, assuming it was caught at the same height from which it was thrown. The starting speed can be found from Eq. 2–12b, with x replaced by y. y = y0 + v0t + 12 at 2 = 0 →

v0 =

y − y0 − 12 at 2

(

)

= − 12 at = − 12 −9.80 m s 2 ( 2.6 s ) = 12.74 m s  13 m s

t The height can be calculated from Eq. 2–12c, with a final velocity of v = 0 at the top of the path.

v = v + 2 a ( y − y0 ) → y = y 0 + 2

2 0

v 2 − v02 2a

= 0+

0 − (12.74 m s )

(

2 −9.80 m s 2

)

= 8.3 m

56. Choose upward to be the positive direction, and take y0 = 0 to be at the height where the ball was hit. For the upward path, v0 = 22 m s, v = 0 at the top of the path, and a = −9.80 m s 2. (a) The displacement can be found from Eq. 2–12c, with x replaced by y.

v 2 = v02 + 2a ( y − y0 ) → y = y0 +

v 2 − v02 2a

=0+

0 − ( 22 m s )

(

2 −9.80 m s 2

)

= 25 m

(b) The time of flight can be found from Eq. 2–12b, with x replaced by y, using a displacement of 0 for the displacement of the ball returning to the height from which it was hit. y = y0 + v0t + 12 at 2 = 0 → t ( v0 + 12 at ) = 0 →

t = 0, t =

2v0

2 ( 22 m s )

= 4.5s − a −9.80 m s 2 The result of t = 0 s is the time for the original displacement of zero (when the ball was hit), and the result of t = 4.5 s is the time to return to the original displacement. Thus the answer is t = 4.5 seconds. (c) This is an estimate primarily because the effects of the air have been ignored. There is a nontrivial amount of air effect on a baseball as it moves through the air – that’s why pitches like the “curve ball” work, for example. So ignoring the effects of air make this an estimate. Another effect is that the problem says “almost” straight up, but the problem was solved as if the initial velocity was perfectly upwards. Finally, we assume that the ball was caught at the same height as which it was hit. That was not stated in the problem either, so that is an estimate.

Chapter 2

Describing Motion: Kinematics in One Dimension

57. Choose downward to be the positive direction, and take y0 = 0 to be at the maximum height of the kangaroo. Consider just the downward motion of the kangaroo. Then the displacement is y = 1.45 m, the acceleration is a = 9.80 m s 2 , and the initial velocity is v0 = 0 m s . Use Eq. 2–12b to calculate the time for the kangaroo to fall back to the ground. The total time is then twice the falling time. y = y0 + v0t + 12 at 2 = 0 ttotal = 2

2y a

→

y = 12 at 2 → tfall =

2 (1.45 m )

( 9.80 m s ) 2

2y a

→

= 1.09 s

58. Choose upward to be the positive direction, and take y0 = 0 to be at the floor level, where the jump starts. For the upward path, y = 1.2 m, v = 0 at the top of the path, and a = −9.80 m s 2. (a) The initial speed can be found from Eq. 2–12c, with x replaced by y. v 2 = v02 + 2a ( y − y0 ) →

v0 = v 2 − 2a ( y − y0 ) = −2ay = −2 ( −9.80 m s 2 ) (1.2 m ) = 4.8497 m s  4.8 m s (b) The time of flight can be found from Eq. 2–12b, with x replaced by y, using a displacement of 0 for the displacement of the jumper returning to the original height. y = y0 + v0t + 12 at 2 = 0 → t ( v0 + 12 at ) = 0 → t = 0, t =

2v0

2 ( 4.897 m s )

= 0.99 s −a 9.80 m s 2 The result of t = 0 s is the time for the original displacement of zero (when the jumper started to jump), and the result of t = 0.99 s is the time to return to the original displacement. Thus the answer is t = 0.99 seconds.

59. Choose upward to be the positive direction, and y0 = 0 to be the height from which the stone is thrown. We have v0 = 18.0 m s , a = −9.80 m s 2 , and y − y0 = 11.0 m. (a) The velocity can be found from Eq, 2–11c, with x replaced by y. v 2 = v02 + 2a ( y − y0 ) = 0 →

v =  v02 + 2ay = 

(18.0 m s ) + 2 ( −9.80 m s 2 ) (11.0 m ) = 10.4 m s 2

Thus the speed is v = 10.4 m s . (b) The time to reach that height can be found from Eq. 2–12b. 2 (18.0 m s ) 2 ( −11.0 m ) y = y0 + v0 t + 12 at 2 → t 2 + t+ =0 2 −9.80 m s −9.80 m s 2

t − 3.6735t + 2.245 = 0 → t =

3.6735 

→

( 3.6735 ) − 4 ( 2.2449 ) 2

= t = 2.90 s, 0.774 s 2 (c) There are two times at which the object reaches that height – once on the way up ( t = 0.774 s ) , 2

and once on the way down ( t = 2.90 s ).

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

60. Choose downward to be the positive direction, and take y0 = 0 to be the height from which the object is released. The initial velocity is v0 = 0, and the acceleration is a = g . Then we can calculate the position as a function of time from Eq. 2–12b, with x replaced by y, as y ( t ) = 12 gt 2 . At the end of each second, the position would be as follows. 2 2 y ( 0 ) = 0 ; y (1) = 12 g ; y ( 2 ) = 12 g ( 2 ) = 4 y (1) ; y ( 3 ) = 12 g ( 3 ) = 9 y (1) The distance traveled during each second can be found by subtracting two adjacent position values from the above list. d (1) = y (1) − y ( 0 ) = y (1) ; d ( 2 ) = y ( 2 ) − y (1) = 3 y (1) ; d ( 3 ) = y ( 3 ) − y ( 2 ) = 5 y (1) We could do this in general. Let n be a positive integer, starting with 0. y ( n ) = 12 gn 2

y ( n + 1) = 12 g ( n + 1)

(

d ( n + 1) = y ( n + 1) − y ( n ) = 12 g ( n + 1) − 12 gn 2 = 12 g ( n + 1) − n 2 2

(

)

= 12 g n 2 + 2n + 1 − n 2 = 12 g ( 2n + 1)

The value of ( 2n + 1) is always odd, in the sequence 1, 3, 5, 7, …. 61. Choose downward to be the positive direction, and the origin to be at the location of the plane. The 2 parachutist has v0 = 0, a = g = 9.80 m s , and will have y − y0 = 3800 m − 450 m = 3350 m when she pulls the ripcord. Eq. 2–12b, with x replaced by y, is used to find the time when she pulls the ripcord.

2 ( y − y0 ) a =

y = y0 + v0t + 12 at 2 → t =

(

)

2 ( 3350 m ) 9.80 m s 2 = 26.147 s  26.1s

The speed is found from Eq. 2–12a.

v = v0 + at = 0 + ( 9.80 m s 2 ) ( 26.147 s ) = 256.24 m s  256 m s  3.6 km h    922 km h  1m s 

( 256.24 m s ) 

This is well over 550 miles per hour! 62. Choose downward to be the positive direction, and y0 = 0 to be at the start of the pelican’s dive. The pelican has an initial velocity is v0 = 0, an acceleration of a = g , and a final location of y = 16.0 m. Find the total time of the pelican’s dive from Eq. 2–12b, with x replaced by y. y = y0 + v0 t + 12 at 2 → y = 0 + 0 + 12 at 2 → tdive =

2y a

2 (16.0 m ) 9.80 m s 2

= 1.81 s.

The fish can take evasive action if he sees the pelican at a time of 1.81 s – 0.20 s = 1.61 s into the dive. Find the location of the pelican at that time from Eq. 2–12b. 2 y = y0 + v0t + 12 at = 0 + 0 + 12 9.80 m s 2 (1.61 s ) = 12.7 m

(

)

Thus the fish must spot the pelican at a minimum height from the surface of the water of 16.0 m − 12.7 m = 3.3m .

Chapter 2

Describing Motion: Kinematics in One Dimension

63. Choose upward to be the positive direction, and y0 = 0 to be at the throwing location of the stone. The initial velocity is v0 = 15.5 m s , the acceleration is a = −9.80 m s 2 , and the final location is y = −75 m . (a) Using Eq. 2–12b and substituting y for x, we have the following. y = y0 + v0 t + 12 at 2 → 4.9 m s 2 t 2 − (15.5 m s ) t − 75 m = 0 →

(

15.5 

)

(15.5 ) − 4 ( 4.9 )( −75 ) = 5.802 s, − 2.638s 2 ( 4.9 ) 2

The positive answer is the physical answer: t = 5.80 s . (b) Use Eq. 2–12a to find the velocity just before hitting. v = v0 + at = 15.5 m s + ( −9.80 m s 2 ) ( 5.802 s ) = −41.4 m s →

v = 41.4 m s .

(c) The total distance traveled will be the distance up plus the distance down. The distance down will be 75 m more than the distance up. To find the distance up, use the fact that the speed at the top of the path will be 0. Then using Eq. 2–12c we have the following.

v = v + 2 a ( y − y0 ) → y = y 0 + 2

2 0

v 2 − v02 2a

= 0+

0 − (15.5 m s )

(

2 −9.80 m s 2

)

= 12.26 m.

Thus the distance up is 12.26 m, the distance down is 87.26 m, and the total distance traveled is 99.5 m . 64. (a) Choose upward to be the positive direction, and y0 = 0 at the ground. The rocket has v0 = 0, a = 3.2 m s 2 , and y = 725m when it runs out of fuel. Find the velocity of the rocket when it runs out of fuel from Eq. 2–12c, with x replaced by y. 2 v725 = v02 + 2a ( y − y0 ) → m

v725 m =  v02 + 2a ( y − y0 ) =  0 + 2 ( 3.2 m s 2 ) ( 725 m ) = 68.12 m s  68 m s The positive root is chosen since the rocket is moving upwards when it runs out of fuel. (b) The time to reach the 725 m location can be found from Eq. 2–12a. v725 m − v0 68.12 m s − 0 v725 m = v0 + at725 m → t725 m = = = 21.29 s  21s a 3.2 m s 2 (c) For this part of the problem, the rocket will have an initial velocity v0 = v725m = 68.12 m s , an acceleration of a = −9.80 m s 2, and a final velocity of v = 0 at its maximum altitude. The altitude reached from the out-of-fuel point can be found from Eq. 2–12c. 2 v 2 = v725 + 2a ( y − 725 m ) → m

ymax = 725 m +

2 0 − v725 m

= 725 m +

− ( 68.12 m s )

(

2 −9.80 m s 2

)

= 725 m + 237 m = 962 m  960 m

(d) The time for the “coasting” portion of the flight can be found from Eq. 2–12a. v − v725 m 0 − 68.12 m s = = 6.95s v = v725 m + atcoast → tcoast = a −9.80 m s 2 Thus the total time to reach the maximum altitude is t = 21.29 s + 6.95s = 28.24 s  28s . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(e) For this part of the problem, the rocket has v0 = 0 m s, a = −9.80 m s 2, and a displacement of −962 m (it falls from a height of 962 m to the ground). Find the velocity upon reaching the Earth from Eq. 2–12c. v 2 = v02 + 2a ( y − y0 ) →

v =  v02 + 2a ( y − y0 ) =  0 + 2 ( −9.80 m s 2 ) ( −962 m ) = −137.3 m s  −137 m s (f)

The negative root was chosen because the rocket is moving downward, which is the negative direction. The time for the rocket to fall back to the Earth is found from Eq. 2–12a. v − v0 −137.3 m s − 0 = = 14.01s v = v0 + at → tfall = a −9.80 m s 2 Thus the total time for the entire flight is t = 28.24 s + 14.01s = 42.25s  42 s .

65. Choose upward to be the positive direction, and y0 = 0 to be the location of the nozzle. The initial velocity is v0 , the acceleration is a = −9.80 m s 2 , the final location is y = −1.8 m, and the time of flight is t = 2.5s. Using Eq. 2–12b and substituting y for x gives the following.

y = y0 + v0t + 12 at 2 →

v0 =

y − 12 at 2 t

(

)

−1.8 m − 12 −9.80 m s 2 ( 2.5s )

= 11.53 m s  12 m s

2.5s

66. Choose downward to be the positive direction, and take y0 = 0 to be the height where the package was released. The initial velocity is v0 = −6.40 m s , the acceleration is a = 9.80 m s 2 , and the displacement of the package will be y = 105 m. The time to reach the ground can be found from Eq. 2–12b, with x replaced by y. 2 ( −6.40 m s ) 2 (105 m ) 2v 2y y = y0 + v0t + 12 at 2 → t 2 + 0 t − = 0 → t2 + t− =0 → 2 9.80 m s 9.80 m s 2 a a

t − 1.306t − 21.43 = 0 → t = 2

1.306 

(1.306 ) − 4 ( −21.43) 2

= 5.33s, − 4.02 s

The correct time is the positive answer, t = 5.33s . 67. (a) Choose y = 0 to be the ground level, and positive to be upward. Then y = 0 m, y0 = 15 m, a = − g , and t = 0.83s describe the motion of the balloon. Use Eq. 2–12b.

y = y0 + v0 t + 12 at 2 →

v0 =

y − y0 − 12 at 2 t

(

)

0 − 15 m − 12 −9.80 m s 2 ( 0.83s )

( 0.83s )

= −14.01m s  −14 m s

So the speed is 14 m s .

Chapter 2

Describing Motion: Kinematics in One Dimension

(b) Consider the change in velocity from being released to being at Roger’s room, using Eq. 2–12c.

v 2 = v02 + 2ay → y =

v 2 − v02 2a

− ( −14.01m s )

(

2 −9.80 m s 2

)

= 10.01m

Thus the balloons are coming from 25 m above the ground. 68. Choose upward to be the positive direction, and y0 = 0 to be the level from which the ball was thrown. The initial velocity is v 0 , the instantaneous velocity is v = 13 m s, the acceleration is a = −9.80 m s 2, and the location of the window is y = 18 m . (a) Using Eq. 2–12c and substituting y for x, we have v 2 = v02 + 2a ( y − y0 ) →

v 0 =  v 2 − 2 a ( y − y0 ) = 

(13 m s ) − 2 ( −9.80 m s 2 ) (18 m ) = 22.84 m s  23 m s 2

Choose the positive value because the initial direction is upward. (b) At the top of its path, the velocity will be 0, and so we can use the initial velocity as found above, along with Eq. 2–12c. v 2 − v02

v = v + 2 a ( y − y0 ) → y = y 0 + 2

2 0

= 0+

0 − ( 22.84 m s )

(

2 −9.80 m s 2

)

= 26.62 m  27 m

(c) We want the time elapsed from throwing (velocity v0 = 22.84 m s) to reaching the window (velocity v = 13 m s ). Use Eq. 2–12a. v − v0 13 m s − 22.84 m s = = 1.004 s  1.0 s v = v0 + at → t = a −9.80 m s 2 (d) We want the time elapsed from the window ( v0 = 13m s ) to reaching the street

( v = −22.84 m s ). Use Eq. 2–12a. v = v0 + at → t =

v − v0

−22.84 m s − 13 m s

= 3.657 s  3.7 s . a −9.80 m s 2 The total time from throwing to reaching the street again is 1.004s + 3.657 s = 4.661s  4.7 s. 69. Choose downward to be the positive direction, and y0 = 0 to be the height from which the stone is dropped. Call the location of the top of the window ytop . From the distance and time given you can use Eq. 2–12b to find the velocity at the top of the window, vtop . That speed can then be used to find the height from which the stone was dropped. ( ybottom − ytop ) = vtopt + 12 at 2 → v top

(y =

bottom

− ytop ) − 12 at 2 t

(

)

2.2 m − 12 9.80 m s 2 ( 0.28 s )

2 vtop = v02 + 2a ( ytop − y0 ) → ytop =

( 0.28 s )

2 v top − v02

− y0 =

= 6.485 m s

( 6.485 m s )

(

2 9.80 m s 2

)

= 2.146 m  2.1m

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

70. Choose up to be the positive direction, so a = − g . Let the ground be the y = 0 location. As an intermediate result, the velocity at the bottom of the window can be found from the data given. Assume the rocket is at the bottom of the window at t = 0, and use Eq. 2–12b. 2 ytop of = ybottom of + vbottom of tpass + 12 atpass → window

window

(

)

10.0 m = 8.0 m + vbottom of ( 0.15s ) + 12 −9.80 m s 2 ( 0.15s ) window

→ vbottom of = 14.07 m s window

Now use the velocity at the bottom of the window with Eq. 2–12c to find the launch velocity, assuming the launch velocity was achieved at the ground level. 2 2 vbottom = vlaunch + 2 a ( y − y0 ) → of window 2 vlaunch = vbottom − 2 a ( y − y0 ) = of window

(14.07 m s ) − 2 ( −9.80 m s 2 ) ( 8.0 m ) = 18.84 m s 2

 18.8 m s

The maximum height can also be found from Eq. 2–12c, using the launch velocity and a velocity of 0 at the maximum height. 2 2 vmaximum = vlaunch + 2a ( ymax − y0 ) → height 2 2 v maximum − vlaunch

ymax = y0 +

height

− (18.84 m s )

(

2 −9.80 m s 2

)

= 18.1m

71. Choose up to be the positive direction. Let the bottom of the cliff be the y = 0 location. The equation of motion for the dropped ball is yball = y0 + v0 t + 12 at 2 = 55.0 m + 12 ( −9.80 m s 2 ) t 2 . The equation of motion for the thrown stone is ystone = y0 + v0t + 12 at 2 = ( 24.0 m s ) t + 12 ( −9.80 m s 2 ) t 2 . Set the two equations equal and solve for the time of the collision. Then use that time to find the location of either object. yball = ystone → 55.0 m + 12 −9.80 m s 2 t 2 = ( 24.0 m s ) t + 12 −9.80 m s 2 t 2 →

(

55.0 m = ( 24.0 m s ) t → t =

)

55.0 m 24.0 m s

(

)

= 2.292 s

(

)

yball = y0 + v0t + 12 at 2 = 55.0 m + 0 + 12 −9.80 m s 2 ( 2.292 s ) = 29.3 m 2

(

)

ystone = y0 + v0t + 12 at 2 = 0 + ( 24.0 m s ) ( 2.292 s ) + 12 −9.80 m s 2 ( 2.292 s ) = 29.3 m 2

72. The displacement is found from the integral of the velocity, over the given time interval. t2

x =  v dt = t1

t = 3.6 s

 ( 25 + 18t ) dt = ( 25t + 9t ) 2

t =1.3s

t = 3.6 s t =1.3s

=  25 ( 3.6 ) + 9 ( 3.6 )  −  25 (1.3 ) + 9 (1.3 ) 



 



= 158.93 m  1.6  10 2 m

Chapter 2

Describing Motion: Kinematics in One Dimension

73. (a) The velocity is the integral of the acceleration. dv a= → d v = adt → d v = A tdt → dt

 d v = A tdt →

v − v0 = 23 At 3/ 2 → v = v0 + 23 At 3/ 2 → v = 7.5 m s + 23 ( 3.0 m s 5/2 ) t 3/ 2 (b) The displacement is the integral of the velocity. dx v= → dx = v dt → dx = v0 + 23 At 3/ 2 dt → dt

(

)

(

)

3/ 2 5/ 2 5/2 5/ 2  dx =  v0 + 23 At dt → x = v0t + 23 52 At = ( 7.5 m s ) t + 154 3.0 m s t

(c)

(

) 5.0s = 6.7 m s v ( t = 5.0 s ) = 7.5 m s + ( 3.0 m s ) ( 5.0 s ) = 29.86 m s  3.0  10 m s x ( t = 5.0 s ) = ( 7.5 m s )( 5.0 s ) + ( 3.0 m s ) ( 5.0 s ) = 82.22 m  82 m a ( t = 5.0s ) = 3.0 m s5/2

3/ 2

5/2

2 3

5/ 2

5/2

4 15

74. (a) The velocity is found by integrating the acceleration with respect to time. Note that with the substitution given in the hint, the initial value of u is u0 = g − k v0 = g . a=

→ d v = adt → d v = ( g − k v ) dt →

dt Now make the substitution that u  g − k v.

u  g − kv → d v = − u

g − kv

u

= − k  dt → ln u g = − kt → ln

(1 − e ) k

= dt →

u g

dv g − kv

−

du 1 k u

= dt

= dt →

du u

= − kdt

= − kt → u = ge − kt = g − k v →

− kt

(b) As t goes to infinity, the value of the velocity is vterm = lim t →

g 1− e ) = . We also note that ( k k

− kt

if the acceleration is zero (which happens at terminal velocity), then a = g − k v = 0 →

vterm =

g . k

75. Choose the upward direction to be positive, and y0 = 0 to be the level from which the object is thrown. The initial velocity is v0 and the velocity at the top of the path is v = 0 m s. The height at the top of the path can be found from Eq. 2–12c with x replaced by y. −v 2 v 2 = v02 + 2a ( y − y0 ) → y − y0 = 0 . 2a From this we see that the displacement is inversely proportional to the acceleration, and so if the acceleration is reduced by a factor of 6 by going to the Moon, and the initial velocity is unchanged, then the displacement increases by a factor of 6 . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

76. (a) For the free-falling part of the motion, choose downward to be the positive direction, and y0 = 0 to be the height from which the person jumped. The initial velocity is v0 = 0, the acceleration is a = 9.80 m s 2 , and the location of the net is y = 18.0 m. Find the speed upon reaching the net from Eq. 2–12c with x replaced by y. v 2 = v02 + 2a ( y − y0 ) →

v =  0 + 2a ( y − 0 ) =  2 ( 9.80 m s 2 ) (18.0 m ) = 18.78 m s The positive root is selected since the person is moving downward. For the net-stretching part of the motion, choose downward to be the positive direction, and y0 = 18.0 m to be the height at which the person first contacts the net. The initial velocity is

v0 = 18.78 m s, the final velocity is v = 0, and the location at the stretched position is y = 19.0 m. Find the acceleration from Eq. 2–12c with x replaced by y.

v 2 = v02 + 2a ( y − y0 ) → a=

v 2 − v02

02 − (18.78 m s )

= −176 m s 2  −180 m s 2

2 ( y − y0 ) 2 (1.0 m ) This is about 18 g’s. (b) For the acceleration to be smaller, in the above equation we see that the displacement would have to be larger. This means that the net should be “loosened.” 77. We treat the value of 30 g’s as if it had 2 significant figures. The initial velocity of the car is  1m s  v0 = ( 95 km h )   = 26.39 m s. Choose x0 = 0 to be location at which the deceleration  3.6 km h  begins. We have v = 0 m s and a = −30 g = −294 m s 2 . Find the displacement from Eq. 2–12c.

v = v + 2a ( x − x0 ) → x = x0 + 2

2 0

v 2 − v02 2a

= 0+

0 − ( 26.39 m s )

(

2 −294 m s 2

)

= 1.18 m  1.2 m

78. We are given that x ( t ) = 2.0 − 3.6t + 1.7t 2 , where x is in meters and t is in seconds. (a) The value of 2.0 must have units of meters, and is the initial position of the ball. The value of 3.6 must have units of m/s, and is the initial speed of the ball – the speed at t = 0. Note that the ball is initially moving in the negative direction, since – 3.6 is used. The value of 1.7 must have units of m s 2 , and based on Eq. 2–12b is ½ the acceleration of the ball. (b) As given above, the units of 2.0 are meters, the units of 3.6 are m/s, and the units of 1.7 are m s 2. (c)

( ) x ( 2.0s ) = 2.0 m − ( 3.6 m s )( 2.0s ) + (1.7 m s ) ( 2.0s ) = 1.6 m x ( 3.0s ) = 2.0 m − ( 3.6 m s )( 3.0s ) + (1.7 m s ) ( 3.0s ) = 6.5 m

x (1.0s ) = 2.0 m − ( 3.6 m s )(1.0s ) + 1.7 m s 2 (1.0s ) = 0.1m

(d) v =

x t

6.5m − 0.1m 2.0 s

= 3.2 m s

Chapter 2

Describing Motion: Kinematics in One Dimension

79. (a) The sounds will not occur at equal time intervals because the longer any particular bolt falls, the higher its speed. The bolts all start from rest and all have the same acceleration, so at any moment in time, they will all have the same speed. However, they have different distances to travel in reaching the floor and therefore will be falling for different lengths of time. The farther a bolt falls, and thus the later a bolt hits the tin plate, the longer it has been accelerating and therefore the faster it is moving. (b) With equal distances between bolts, each successive bolt, having fallen a longer time when its predecessor reaches the plate, will have a higher average velocity and thus travel the inter-bolt distance in shorter periods of time. Thus the sounds will occur with smaller and smaller intervals between sounds. Thus the time between “clinks” will decrease as the string falls. (c) To hear the sounds at equal intervals, the bolts would have to be tied at distances corresponding to equal time intervals. The first bolt (call it bolt # 0) is touching the plate. Since each bolt has an initial speed of 0, the distance of fall and time of fall for each bolt are related to each other by di = 12 gti2. For bolt # 1, d1 = 12 gt12. For bolt # 2, we want t 2 = 2t1, so d 2 = 12 gt22 = 12 g ( 2t1 )

(

)

= 4 12 gt12 = 4d1 . Likewise t3 = 3t1, which leads to d 3 = 9d1 ; t 4 = 4t1, which leads to d 4 = 16d1 , and so

on. If the distance from the bolt initially on the pan to the next bolt is d1, then the distance from that bolt to the next one is 3d1, the distance to the next bolt is 5d1, and so on. The accompanying table shows these

relationships in a simpler format. 80. Choose downward to be the positive direction, and the origin to be at the top of the building. The 2 barometer has y0 = 0, v0 = 0, and a = g = 9.8 m s . Use Eq. 2–12b to find the height of the building, with x replaced by y. y = y0 + v0 t + 12 at 2 = 0 + 0 + 12 9.8 m s 2 t 2

(

)

yt = 2.0 = 12 9.8 m s 2 ( 2.0 s ) = 20 m 2

)

(

)

yt = 2.3 = 12 9.8 m s 2 ( 2.3 s ) = 26 m 2

The difference in the estimates is 6 m. If we assume the height of the building is the average of the 6m two measurements, then the % difference in the two values is  100 = 26% . 23m

 1m s   = 11.11m s.  3.6 km h 

81. The speed limit is 40 km h 

(a) For your motion, you would need to travel (10 + 15 + 50 + 15 + 70 ) m = 160 m to get the front of the car to the third stoplight. The time to travel the 160 m is found using the distance and the speed limit. 160 m x x = v  t → t = = = 14.40 s 11.11m s v No , you cannot make it to the third light without stopping, since it takes you longer than 13.0 seconds to reach the third light.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The second car needs to travel 165 m before the third light turns red. This car accelerates from v0 = 0 m s to a maximum of v = 11.11m s with a = 2.00 m s 2. Use Eq. 2–12a to determine the duration of that acceleration. v − v0 11.11m s − 0 m s v = v0 + at → tacc = = = 5.556 s a 2.00 m s 2 The distance traveled during that time is found from Eq. 2–12b. 2 ( x − x0 )acc = v0tacc + 12 atacc2 = 0 + 12 ( 2.00 m s 2 ) ( 5.556 s ) = 30.87 m Since 5.556 s have elapsed, there are 13.0 – 5.556 = 7.444 s remaining to clear the intersection. The car travels another 7.444 s at a speed of 11.11 m/s, covering a distance of xconstant = vavg t = speed

(11.11m s )( 7.444 s ) = 82.70 m. Thus the total distance is 30.87 m + 82.70 m = 113.57 m. No , the car cannot make it through all three lights without stopping. The car has to travel another 51.43 m to clear the third intersection, and is traveling at a speed of 11.11 m/s. Thus the front of the car would clear the intersection a time 51.43 m x t= = = 4.6 s after the light turns red. 11.11m s v 82. (a) Choose downward to be the positive direction, and y0 = 0 to be the level from which the car was dropped. The initial velocity is v0 = 0, the final location is y = H , and the acceleration is a = g . Find the final velocity from Eq. 2–12c, replacing x with y.

v 2 = v02 + 2a ( y − y0 ) → v =  v02 + 2a ( y − y0 ) =  2 gH . The speed is the magnitude of the velocity, v =

2gH .

(b) Solving the above equation for the height, we have that H =

v2 2g

. Thus for a collision of

v = 35 km h , the corresponding height is as follows.

v2 2g

  1m s   ( 35 km h )    3.6 km h   

(

2 9.80 m s 2

= 4.823 m  4.8 m

)

  1m s   ( 95 km h )   2 v  3.6 km h    H= = = 35.53 m  36 m 2 2g

(

2 9.80 m s

)

83. Choose downward to be the positive direction, and y0 = 0 to be at the roof from which the stones are dropped. The first stone has an initial velocity of v 0 = 0 and an acceleration of a = g . Eqs. 2–12a and 2–12b (with x replaced by y) give the velocity and location, respectively, of the first stone as a function of time. v = v0 + at → v1 = gt1 y = y0 + v0t + 12 at 2 → y1 = 12 gt12 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

The second stone has the same initial conditions, but its elapsed time is t2 = t1 − 1.50 s, and so has velocity and location equations as follows. 2 v2 = g ( t1 − 1.50s ) y2 = 12 g ( t1 − 1.50s ) The second stone reaches a speed of v 2 = 12.0 m s at a time given by the following.

t2 =

v2 g

= t1 − 1.50 s → t1 = 1.50 s +

v2 g

= 1.50 s +

12.0 m s 9.80 m s

(

The location of the first stone at that time is y1 = 12 gt12 = 12 9.80 m s

= 2.724 s 2

) ( 2.724 s) = 36.36 m. 2

The location of the second stone at that time is 2 2 2 y2 = 12 g ( t1 − 1.50s ) = 12 9.80 m s ( 2.724 − 1.50s ) = 7.34 m.

(

)

Thus the distance between the two stones is y1 − y2 = 36.36 m − 7.34 m = 29.0 m . 84. For the motion in the air, choose downward to be the positive direction, and y0 = 0 to be at the height of the diving board. Then diver has v0 = 0, (assuming the diver does not jump upward or downward), a = g = 9.80 m s 2, and y = 4.0 m when reaching the surface of the water. Find the diver’s speed at the water’s surface from Eq. 2–12c, with x replaced by y.

v 2 = v02 + 2a ( y − y0 ) → v =  v02 + 2a ( y − y0 ) = 0 + 2 ( 9.80 m s 2 ) ( 4.0 m ) = 8.85 m s

For the motion in the water, again choose down to be positive, but redefine y0 = 0 to be at the surface of the water. For this motion, v0 = 8.85 m s , v = 0, and y − y0 = 1.8 m . Find the acceleration from Eq. 2–12c, with x replaced by y.

v 2 = v02 + 2a ( y − y0 ) → a =

v 2 − v02

0 − ( 8.85 m s )

= −22 m s 2

2 ( y − y0 ) 2 (1.8 m ) The negative sign indicates that the acceleration is directed upwards.

85. Take the origin to be the location where the speeder passes the police car. The speeder’s constant  1m s  speed is vspeeder = (140 km h )   = 38.89 m s , and the location of the speeder as a function  3.6 km h  of time is given by xspeeder = vspeeder tspeeder = ( 38.89 m s ) tspeeder. The police car has an initial velocity of v 0 = 0 m s and a constant acceleration of apolice. The location of the police car as a function of time is given by Eq. 2–12b. 2 xpolice = v0t + 12 at 2 = 12 apolicetpolice . (a) The position vs. time graphs would qualitatively look like the graph shown here.

Speeder Police car

t1 t (b) The time to overtake the speeder occurs when the speeder has gone a distance of 850 m. The time is found using the speeder’s equation from above. 850 m 850 m = ( 38.89 m s ) tspeeder → tspeeder = = 21.86 s  22 s 38.89 m s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(c) The police car’s acceleration can be calculated knowing that the police car also had gone a distance of 850 m in a time of 21.86 s. 850 m = 12 a p ( 21.86 s )

→ ap =

2 ( 850 m )

( 21.86 s )

= 3.558 m s 2  3.6 m s 2

(d) The speed of the police car at the overtaking point can be found from Eq. 2–12a. v = v0 + at = 0 + ( 3.558 m s 2 ) ( 21.86 s ) = 77.78 m s  78 m s Note that this is exactly twice the speed of the speeder, and so is 280 km/h. (e) The police officer would not want to pass the car at that high speed, because then he would go past the car for a long distance before being able to pull the speeder over. The police officer would not have a constant acceleration, but instead, once he gets close to the speeder, will start to slow down so as to match the speeder’s velocity and pull in close behind the speeder. 86. Choose downward to be the positive direction, and y0 = 0 to be at the height of the bridge. Agent Bond has an initial velocity of v0 = 0, an acceleration of a = g , and will have a displacement of y = 15 m − 3.5 m = 11.5 m. Find the time of fall from Eq. 2–12b with x replaced by y.

y = y0 + v0t + 12 at 2 → t =

2y a

2 (11.5 m ) 9.80 m s 2

= 1.532 s

If the truck is approaching with v = 25 m s , then he needs to jump when the truck is a distance away given by d = v t = ( 25 m s )(1.532 s ) = 38.3 m. Convert this distance into “poles.” d = ( 38.3 m )(1 pole 25 m ) = 1.53 poles

So he should jump when the truck is about 1.5 poles away from the bridge. 87. Assume that y0 = 0 for each child is the level at which the child loses contact with the trampoline surface. Choose upward to be the positive direction. (a) The second child has v02 = 4.0 m s, a = − g = −9.80 m s 2 , and v = 0 m s at the maximum height position. Find the child’s maximum height from Eq. 2–12c, with x replaced by y. v 2 = v022 + 2a ( y2 − y0 ) → y 2 = y0 +

v 2 − v022 2a

= 0+

0 − ( 4.0 m s )

(

2 −9.80 m s 2

)

= 0.8163 m  0.82 m

(b) Since the first child can bounce up to one-and-a-half times higher than the second child, the first child can bounce up to a height of 1.5 ( 0.8163 m ) = 1.224 m = y1 − y0 . Eq. 2–12c is again used to find the initial speed of the first child. v 2 = v012 + 2a ( y1 − y0 ) →

v01 =  v 2 − 2a ( y1 − y0 ) = 0 − 2 ( −9.80 m s 2 ) (1.224 m ) = 4.898 m s  4.9 m s The positive root was chosen since the child was initially moving upward. (c) To find the time that the first child was in the air, use Eq. 2–12b with a total displacement of 0, since the child returns to the original position. y = y0 + v01t1 + 12 at12 → 0 = ( 4.898 m s ) t1 + 12 ( −9.80 m s 2 ) t12 → t1 = 0 s, 0.9996 s The time of 0 s corresponds to the time the child started the jump, so the correct answer is 1.0 s . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

88. First consider the “uphill lie,” in which the ball is being putted down the hill. Choose x0 = 0 to be the ball’s original location, and the direction of the ball’s travel as the positive direction. The final velocity of the ball is v = 0 m s , the acceleration of the ball is a = −1.8 m s 2 , and the displacement of the ball will be x − x0 = 6.0 m for the first case and x − x0 = 8.0 m for the second case. Find the initial velocity of the ball from Eq. 2–12c.

v = v + 2a ( x − x0 ) 2

2 0

 0 − 2 ( −1.8 m s 2 ) ( 6.0 m ) = 4.648 m s  → v0 = v − 2a ( x − x0 ) =  2   0 − 2 ( −1.8 m s ) ( 8.0 m ) = 5.367 m s 2

The range of acceptable velocities for the uphill lie is 4.6 m s to 5.4 m s , a spread of 0.719 m/s if more digits are kept. Now consider the “downhill lie,” in which the ball is being putted up the hill. Use a very similar setup for the problem, with the basic difference being that the acceleration of the ball is now a = −2.6 m s 2. Find the initial velocity of the ball from Eq. 2–12c.

v = v + 2a ( x − x0 ) 2

2 0

 0 − 2 ( −2.6 m s 2 ) ( 6.0 m ) = 5.586 m s  → v0 = v − 2a ( x − x0 ) =  2   0 − 2 ( −2.6 m s ) ( 8.0 m ) = 6.450 m s 2

The range of acceptable velocities for the downhill lie is 5.6 m s to 6.5 m s , a spread of 0.864 m/s if more digits are kept. Because the range of acceptable velocities is smaller for putting down the hill, more control in putting is necessary, and so putting the ball downhill (the “uphill lie”) is more difficult.

 1m s   = 9.722 m s.  3.6 km h 

89. The car’s initial speed is v0 = ( 35 km h ) 

Case I: trying to stop. The constraint is, with the braking deceleration of the car ( a = −5.8 m s 2 ) , can the car stop in a 28 m displacement? The 2.0 seconds has no relation to this part of the problem. Using Eq. 2–12c, the distance traveled during braking is as follows.

( x − x0 ) =

v 2 − v02 2a

0 − ( 9.722 m s )

(

2 −5.8 m s 2

)

= 8.14 m She can stop the car in time.

Case II: crossing the intersection. The constraint is, with the acceleration of the car

  65 km h − 45 km h   1m s   = 0.9259 m s 2  , can she get through the intersection   a =  6.0 s   3.6 km h     (travel 43 meters) in the 2.0 seconds before the light turns red? Using Eq. 2–12b, the distance traveled during the 2.0 sec is 2 ( x − x0 ) = v0t + 12 at 2 = ( 9.722 m s )( 2.0 s ) + 12 0.9259 m s 2 ( 2.0 s ) = 21.3 m.

(

)

She should stop.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

90. The critical condition is that the total distance covered by the passing car and the approaching car must be less than 500 m so that they do not collide. The passing car has a total displacement composed of several individual parts. These are: i) the 10 m of clear room at the rear of the truck; ii) the 20 m length of the truck; iii) the 10 m of clear room at the front of the truck; and iv) the distance the truck travels. Since the truck travels at a speed of v = 18 m s , the truck will have a displacement of xtruck = (18 m s ) t. Thus the total displacement of the car during passing is xpassing = 40 m + (18 m s ) t. car

To express the motion of the car, we choose the origin to be at the location of the passing car when the decision to pass is made. For the passing car, we have an initial velocity of v0 = 18 m s and an acceleration of a = 0.60 m s 2 . Find xpassing from Eq. 2–12b. car

(

)

xpassing = xc − x0 = v0t + at = (18 m s ) t + 12 0.60 m s 2 t 2 1 2

car

Set the two expressions for xpassing equal to each other in order to find the time required to pass. car

(

)

(

)

2 2 40 m + (18 m s ) tpass = (18 m s ) tpass + 12 0.60 m s 2 tpass → 40 m = 12 0.60 m s 2 tpass →

tpass =

80 2 s = 11.55s 0.60

Calculate the displacements of the two cars during this time. xpassing = 40 m + (18 m s )(11.55s ) = 247.9 m car

xapproaching = vapproaching t = ( 25 m s )(11.55s ) = 288.75 m car

car

Thus the two cars together have covered a total distance of 247.9 m + 288.75 m = 536.65 m, which is more than allowed. The car should not pass. 91. For the falling rock, choose downward to be the positive direction, and y0 = 0 to be the height from which the stone is dropped. The initial velocity is v0 = 0 m s, the acceleration is a = g , the final position is y = H , and the time of fall is t1 . Using Eq. 2–12b with y substituting for x, we have H = y0 + v0t + 12 t 2 = 0 + 0 + 12 gt12 . For the sound wave, use the constant speed equation that

vsnd =

x t

H T − t1

, which can be rearranged to give t1 = T −

H vsnd

, where T = 4.1s is the total time

elapsed from dropping the rock to hearing the sound. Insert this expression for t1 into the equation for H from the stone, and solve for H.

 H  H = g T −  vsnd   1 2

−5

→

g 2v

2 snd

 gT

H2 −

 vsnd



+ 1 H + 12 gT 2 = 0 →



4.2388  10 H − 1.1182 H + 82.369 = 0 → H = 73.9 m, 2.63 10 4 m 2

Chapter 2

Describing Motion: Kinematics in One Dimension

If the larger answer is used in t1 = T −

H vs

, a negative time of fall results, and so the physically

correct answer is the smaller one: H = 74 m . 92. The speed of the conveyor belt is found from Eq. 2–2 for average velocity. 1.2 m x x = vt → v = = = 0.4286 m min  0.43 m min t 2.8 min The rate of burger production, assuming the spacing given is center to center, can be found as follows.

 1 burger   0.4286 m  burgers  0.25 m   1 min  = 1.7 min    93. (a) Choose up to be the positive direction. Let the throwing height of both objects be the y = 0 location, and so y0 = 0 for both objects. The acceleration of both objects is a = − g . The equation of motion for the rock, using Eq. 2–12b, is yrock = y0 + v0 rock t + 12 at 2 = v0 rock t − 12 gt 2 , where t is the time elapsed from the throwing of the rock. The equation of motion for the ball, 2 being thrown 1.00 s later, is yball = y0 + v0 ball ( t − 1.00s ) + 12 a ( t − 1.00s ) =

v0 ball ( t − 1.00s ) − 12 g ( t − 1.00s ) . Set the two equations equal (meaning the two objects are at the same place) and solve for the time of the collision. 2

yrock = yball → v0 rock t − 12 gt 2 = v0 ball ( t − 1.00 s ) − 12 g ( t − 1.00 s )

→

(15.0 m s ) t − 12 ( 9.80 m s 2 ) t 2 = ( 22.0 m s )( t − 1.00 s ) − 12 ( 9.80 m s 2 ) ( t − 1.00 s ) ( −16.8 ) t = −26.9 (16.8 m s ) t = ( 26.9 m ) → t = 1.60 s after the rock was thrown

→

(b) Use the time for the collision to find the position of either object.

(

)

yrock = v0 rock t − 12 gt 2 = (15.0 m s )(1.60 s ) − 12 9.80 m s 2 (1.60 s ) = 11.4 6 m  11.5 m 2

(

)

yball = v0 ball ( t − 1.00 s ) − 12 g ( t − 1.00 s ) = ( 22.0 m s )( 0.60 s ) − 12 9.80 m s 2 ( 0.60 s ) 2

= 11.44 m (c) Now the ball is thrown first, and so yball = v0 ballt − 12 gt 2 and yrock = v0 rock ( t − 1.00s ) − 12 g ( t − 1.00s ) . Again set the two equations equal to find the time of collision. 2

yball = yrock → v0 ball t − 12 gt 2 = v0 rock ( t − 1.00 s ) − 12 g ( t − 1.00 s )

→

( 22.0 m s ) t − 12 ( 9.80 m s 2 ) t 2 = (15.0 m s )( t − 1.00 s ) − 12 ( 9.80 m s 2 ) ( t − 1.00 s ) ( 22.0 m s ) t = (15.0 m s ) t − 15 + 9.8t − 4.9 ( 2.80 m s ) t = 19.9 m → t = 7.11s after the ball was thrown

→

But this answer can be deceptive. Where do the objects collide?

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

Instructor Solutions Manual

)

yball = v0 ball t − 12 gt 2 = ( 22.0 m s )( 7.11s ) − 12 9.80 m s 2 ( 7.11s ) = −91.3 m 2

(

)

yrock = v0 rock ( t − 1.00 s ) − 12 g ( t − 1.00 s ) = (15.0 m s )( 6.11s ) − 12 9.80 m s 2 ( 6.11s ) 2

= −91.3 m Thus, assuming they were thrown from ground level, they collide below ground level, which cannot happen. Thus they never collide . Or, they had to be more than 91.3 m above the

ground at their launch point. 94. (a) During the interval from A to B, it is moving in the negative direction , because its displacement is negative. (b) During the interval from A to B, it is speeding up , because the magnitude of its slope is increasing (changing from less steep to more steep). (c) During the interval from A to B, the acceleration is negative , because the graph is concave downward, indicating that the slope is getting more negative, and thus the acceleration is negative. (d) During the interval from D to E, it is moving in the positive direction , because the displacement is positive. (e) During the interval from D to E, it is speeding up , because the magnitude of its slope is (f)

increasing (changing from less steep to more steep). During the interval from D to E, the acceleration is positive , because the graph is

concave upward, indicating the slope is getting more positive, and thus the acceleration is positive. (g) During the interval from C to D, the object is not moving in either direction . The velocity and acceleration are both 0 .

95. This problem can be analyzed as a series of three one-dimensional motions; the acceleration phase, the constant speed phase, and the deceleration phase. The maximum speed of the train is:  1m s  ( 95 km h )   = 26.39 m s  3.6 km h  In the acceleration phase, the initial velocity is v0 = 0 m s , the acceleration is a = 1.1m s 2 , and the final velocity is v = 26.39 m s. Find the elapsed time for the acceleration phase from Eq. 2–12a. v − v0 26.39 m s − 0 v = v0 + at → tacc = = = 23.99 s a 1.1m s 2 Find the displacement during the acceleration phase from Eq. 2–12b. 2 ( x − x0 )acc = v0t + 12 at 2 = 0 + 12 (1.1m s 2 ) ( 23.99 s ) = 316.5 m In the deceleration phase, the initial velocity is v0 = 26.39 m s , the acceleration is a = −2.0 m s 2 , and the final velocity is v = 0 m s. Find the elapsed time for the deceleration phase from Eq. 2–12a. v − v0 0 − 26.39 m s = = 13.20 s v = v0 + at → tdec = a −2.0 m s 2 © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 2

Describing Motion: Kinematics in One Dimension

Find the distance traveled during the deceleration phase from Eq. 2–12b. 2 ( x − x0 )dec = v0t + 12 at 2 = ( 26.39 m s )(13.20 s ) + 12 −2.0 m s2 (13.20 s ) = 174.1m

(

)

The total elapsed time and distance traveled for the acceleration / deceleration phases are: tacc + tdec = 23.99 s + 13.20s = 37.19 s

( x − x0 )acc + ( x − x0 )dec = 316.5 m + 174.1m = 491m 15000 m

= 5 inter3000 m station segments. A train making the entire trip would thus have a total of 5 inter-station segments and 4 stops of 22 s each at the intermediate stations. Since 491 m is traveled during acceleration and deceleration, 3000 m − 491m = 2509 m of each segment is traveled at an

(a) If the stations are spaced 3.0 km = 3000 m apart, then there is a total of

average speed of v = 26.39 m s . The time for that 2509 m is given by x = vt → 2509 m x tconstant = = = 95.07 s. Thus a total inter-station segment will take 37.19 s + 26.39 m s v speed 95.07 s = 132.26 s. With 5 inter-station segments of 132.26 s each, and 4 stops of 22 s each, the total time is given by t3.0 km = 5 (132.26 s ) + 4 ( 22 s ) = 749 s = 12.5 min . 15000 m

= 3 inter5000 m station segments. A train making the entire trip would thus have a total of 3 inter-station segments and 2 stops of 22 s each at the intermediate stations. Since 491 m is traveled during acceleration and deceleration, 5000 m − 491m = 4509 m of each segment is traveled at an

(b) If the stations are spaced 5.0 km = 5000 m apart, then there is a total of

average speed of v = 26.39 m s. The time for that 4509 m is given by d = v t → 4509 m d t= = = 170.86 s. Thus a total inter-station segment will take 37.19 s + 170.86 s = v 26.39 m s 208.05 s. With 3 inter-station segments of 208.05 s each, and 2 stops of 22 s each, the total time is given by t5.0 km = 3 ( 208.05s ) + 2 ( 22 s ) = 668s = 11.1min . 96. To find the average speed for the entire race, we must take the total distance divided by the total time. If one lap is a distance of L, then the total distance will be 10 L. The time elapsed at a given 9L constant speed is given by t = d v , so the time for the first 9 laps would be t1 = , and 196.0 km h the time for the last lap would be t2 = L v2 , where v 2 is the average speed for the last lap. Write an expression for the average speed for the entire race, and then solve for v 2. v=

d total t1 + t2

10 L 9L

196.0 km h

v2 =

= 200.0 km h →

1 10 200.0 km h

−

= 245.0 km h

196.0 km h

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

97. Since all of the motion and most of the acceleration is downward, we will call downward the positive direction. Let the parachutist’s initial position be y 0 = 0 m, and their initial speed is v0 = 0 m s. We break the motion up into two constant-acceleration segments. The final position, duration, velocity, and acceleration for the first segment are represented by y1 , t1 , v1 , and a1, respectively. Note that the final velocity for the first segment is the initial velocity for the second segment. First segment:

2 y1

y 1 = y 0 + v0t1 + 21 a1t12 → t1 =

(

)

2 ( 75 m )

9.8 m s 2

= 3.912 s

v1 = a1t1 = 9.8 m s 2 ( 3.912 s ) = 38.34 m s

v 22 − v12

v22 = v12 + 2a2 y2 → y2 = Second segment:

v2 = v1 + a2t2 → t2 =

2 a2

v 2 − v1 a2

( 1.5 m s ) − ( 38.34 m s ) 2

(

2 −1.5 m s 2

1.5 m s − 38.34 m s −1.5 m s 2

)

= 489.2 m

= 24.56 s

(a) Total height: 75 m + 489.2 m = 564.2 m  560 m (b) Total time:

3.912 s + 24.56 s = 28.472 s  28s

98. As shown in Example 2–18, the speed with which the ball was thrown upward is the same as its speed on returning to the ground. From the symmetry of the two motions (both motions have speed = 0 at top, have same distance traveled, and have same acceleration), the time for the ball to rise is the same as the time for the ball to fall, 1.4 s. Choose upward to be the positive direction, and the origin to be at the level where the ball was thrown. For the ball, v = 0 at the top of the motion, and a = − g . Find the initial velocity from Eq. 2–12a.

v = v0 + at → v0 = v − at = 0 − ( −9.80 m s 2 ) (1.4 s ) = 13.72 m s  14 m s 99. To find the distance, we divide the motion of the robot into three segments. First, the initial acceleration from rest; second, motion at constant speed; and third, deceleration back to rest.

(

)

d1 = v0 t + 12 a1t12 = 0 + 12 0.20 m s 2 ( 4.5s ) = 2.025 m 2

v1 = a1t1 = ( 0.20 m s 2 ) ( 4.5s ) = 0.90 m s d 2 = v1t2 = ( 0.90 m s )( 68s ) = 61.2 m

v2 = v1 = 0.90 m s

(

)

d3 = v2t3 + 12 a1t12 = ( 0.90 m s )( 2.5s ) + 12 −0.40 m s 2 ( 2.5s ) = 1.0 m 2

d = d1 + d 2 + d 3 = 2.025 m + 61.2 m + 1.0 m = 64.225 m  64 m

100. Choose upward to be the positive direction, and the origin to be at the level where the ball was thrown. The velocity at the top is v = 0, and the ball will have an acceleration of a = − g . If the maximum height that the ball reaches is y = H , then the relationship between the initial velocity and the maximum height can be found from Eq. 2–12c, with x replaced by y. v 2 = v02 + 2a ( y − y0 ) → 0 = v02 + 2 ( − g ) H → H = v02 2 g It is given that v0 Bill = 1.5v0 Joe

( v ) 2 g ( v0 Bill ) , so = 0 Bill 2 = = 1.52 = 2.25  2.3 . 2 H Joe ( v0 Joe ) 2 g ( v0 Joe ) 2

H Bill

Chapter 2

Describing Motion: Kinematics in One Dimension

101. (a) Multiply the reading rate times the bit density to find the bit reading rate. 1.2 m 1bit N=  = 4.3  106 bits s −6 1s 0.28  10 m (b) The number of excess bits is N − N 0 . N − N 0 = 4.3  106 bits s − 1.4  106 bits s = 2.9  106 bits s

N − N0 N

2.9  106 bits s 4.3  106 bits s

= 0.67 = 67%

102. (a) The two bicycles will have the same velocity at any A time when the instantaneous slopes of their x vs. t B graphs are the same. That occurs near the time t1 x as marked on the graph. (b) Bicycle A has the larger acceleration, because its graph is concave upward, indicating a positive acceleration. Bicycle B has no acceleration because t1 t its graph has a constant slope. (c) The bicycles are passing each other at the times when the two graphs cross, because they both have the same position at that time. The graph with the steepest slope is the faster bicycle, and so is the one that is passing at that instant. So at the first crossing, bicycle B is passing bicycle A. At the second crossing, bicycle A is passing bicycle B. (d) Bicycle B has the highest instantaneous velocity at all times until the time t1, where both graphs have the same slope. For all times after t1 , bicycle A has the highest instantaneous velocity. The largest instantaneous velocity is for bicycle A at the latest time shown on the graph. (e) The bicycles appear to have the same average velocity. If the starting point of the graph for a particular bicycle is connected to the ending point with a straight line, the slope of that line is the average velocity. Both appear to have the same slope for that “average” line. 103. In this problem, note that a  0 and x  0. Take your starting position as 0. Then your position is given by Eq. 2–12b, x1 = vM t + 12 at 2, and the other car’s position is given by x2 = x + v At. Set the two positions equal to each other and solve for the time of collision. If this time is negative or imaginary, then there will be no collision. x1 = x2 → v M t + 12 at 2 = x + v At → 12 at 2 + ( v M − v A ) t − x = 0

( v A − vM )  ( vM − v A ) − 4 12 a ( − x ) 2

2 12 a

No collision: ( v M − v A ) − 4 a ( − x )  0 → 2

1 2

x

( vM − v A )

−2a

CHAPTER 3: Kinematics in Two or Three Dimensions; Vectors Responses to Questions 1.

No, the two velocities are not equal. Velocity is a vector quantity, with a magnitude and direction. If two vectors have different directions, they cannot be equal.

No. The car may be traveling at a constant speed of 60 km/h and going around a curve, in which case it would be accelerating.

(i)

During one year, the Earth travels a distance equal to the circumference of its orbit, but has a displacement of 0 relative to the Sun. (ii) Any kind of cross country “round trip” air travel would result in a large distance traveled, but a displacement of 0. (iii) The displacement for a race car from the start to the finish of the Indianapolis 500 auto race is 0, even though the car has traveled 500 miles.

The length of the displacement vector is the straight-line distance between the beginning point and the ending point of the trip and therefore the shortest distance between the two points. If the path is a straight line, then the length of the displacement vector is the same as the length of the path. If the path is curved or consists of different straight line segments, then the distance from beginning to end will be less than the path length. Therefore, the displacement vector can never be longer than the length of the path traveled, but it can be shorter.

Since both the batter and the ball started their motion at the same location (where the ball was hit) and ended their motion at the same location (where the ball was caught), the displacement of both was the same.

V is the magnitude of the vector V ; it is not necessarily larger than the magnitudes V1 and/or V2. For instance, if V1 and V2 have the same magnitude as each other and are in opposite directions, then V is zero. The magnitude of the sum is dependent on not just the magnitudes V1 and V2 , but also depends on the angle between V1 and V2.

If the two vectors are in the same direction, the magnitude of their sum will be a maximum, and will be 9.5 km. If the two vectors are in the opposite direction, the magnitude of their sum will be a minimum, and will be 0.5 km. If the two vectors are oriented in any other configuration, the magnitude of their sum will be between 0.5 km and 9.5 km.

Two vectors of unequal magnitude can never add to give the zero vector. The only way that two vectors can add up to give the zero vector is if they have the same magnitude and point in exactly opposite directions. However, three vectors of unequal magnitude can add to give the zero vector. As a one-dimensional example, a vector 10 units long in the positive x direction added to two vectors of 4 and 6 units each in the negative x direction will result in the zero vector. In two dimensions, if their sum using the tail-to-tip method gives a closed triangle, then the vector sum will be zero. See the diagram, in which A + B + C = 0

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

(a) The magnitude of a vector can equal the length of one of its components if the other components of the vector are all 0; i.e., if the vector lies along one of the coordinate axes. (b) The magnitude of a vector can never be less than one of its components, because each component contributes a positive amount to the overall magnitude, through the Pythagorean relationship. The square root of a sum of squares is never less than the absolute value of any individual term.

10. The odometer and the speedometer of a car both measure scalar quantities (distance and speed, respectively). 11. To find the initial speed, use the slingshot to shoot the rock directly horizontally (no initial vertical speed) from a height of 1 meter (measured with the meter stick). The vertical displacement of the rock can be related to the time of flight by Eq. 2–12b. Take downward to be positive.

(

)

y = y0 + v y 0t + 12 at 2 → 1 m = 12 gt 2 → t = 2 (1 m ) 9.8 m s 2 = 0.45 s. Measure the horizontal range R of the rock with the meter stick. Then, if we measure the horizontal range R, we know that R = v x t = v x ( 0.45 s ) , and so v x = R 0.45 s, which is the speed the slingshot imparts to the rock. The only measurements are the height of fall and the range, both of which can be measured by a meter stick. 12. The arrow should be aimed above the target, because gravity will deflect the arrow downward from a horizontal or “line of sight” flight path. The angle of aim (above the horizontal) should increase as the distance from the target increases, because gravity will have more time to act in deflecting the arrow from a straight-line path. If we assume that the arrow was shot from the same height as the target, then the “level horizontal range” formula is applicable: R = v02 sin 2 0 g →  = 12 sin −1 ( Rg v02 ). As the range and hence the argument of the inverse sine function increases, the angle increases. 13. (a) In Fig. 3–22, the object is always moving. It never stops, not even instantaneously. Thus, there is no location where v = 0. (b) v y = 0 at the highest point of the motion. (c) The object has a constant horizontal velocity. Thus, there is no location where v x = 0. 14. If the bullet was fired from the ground, then the y-component of its velocity slowed considerably by the time it reached an altitude of 2.0 km, because of both the downward acceleration due to gravity and air resistance. The x-component of its velocity would have slowed due to air resistance as well. Therefore, the bullet could have been traveling slowly enough to be caught. 15. The balloons will hit each other, although not along the line of sight from you to your friend. If there were no acceleration due to gravity, the balloons would hit each other along the line of sight. Gravity causes each balloon to accelerate downward from that “line of sight” path, each with the same acceleration. Thus each balloon falls below the line of sight by the same amount at every instant along their flight, and so they collide. This situation is similar to Conceptual Example 3–9, and to Problem 58. 16. The horizontal component of the velocity stays constant in projectile motion, assuming that air resistance is negligible. Thus the horizontal component of velocity 1.0 seconds after launch will be the same as the horizontal component of velocity 2.0 seconds after launch. In both cases the horizontal velocity will be given by v x = v0 cos  = ( 30 m s )( cos 30 ) = 26 m s . © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

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17. A projectile has the least speed at the top of its path. At that point the vertical speed is zero. The horizontal speed remains constant throughout the flight, if we neglect the effects of air resistance. 18. (a) Cannonball A, with the larger angle, will reach a higher elevation. It has a larger initial vertical velocity, and so by Eq. 2–12c it will rise higher before the vertical component of velocity is 0. (b) Cannonball A, with the larger angle, will stay in the air longer. It has a larger initial vertical velocity, and so takes more time to decelerate to 0 and start to fall. (c) The cannonball with a launch angle closest to 45 will travel the farthest. The range is a maximum for a launch angle of 45, and decreases for angles either larger or smaller than 45. 19. (a) The ball lands at the same point from which it was thrown inside the train car – back in the thrower’s hand. (b) If the car accelerates, the ball will land behind the thrower’s hand. (c) If the car decelerates, the ball will land in front of the thrower’s hand. (d) If the car rounds a curve (assume it curves to the right), then the ball will land to the left of the thrower’s hand. (e) The ball will be slowed by air resistance, and so will land behind the thrower’s hand. 20. Your reference frame is that of the train you are riding. If you are traveling with a relatively constant velocity (not over a hill or around a curve or drastically changing speed), then you will interpret your reference frame as being at rest. Since you are moving forward faster than the other train, the other train is moving backwards relative to you. Seeing the other train go past your window from front to rear makes it look like the other train is going backwards. 21. Both rowers need to cover the same “cross river” distance. The rower with the greatest speed in the “cross river” direction will be the one that reaches the other side first. The current has no bearing on the time to cross the river because the current doesn’t help either of the boats move across the river. Thus the rower heading straight across will reach the other side first. All of his “rowing effort” has gone into crossing the river. For the upstream rower, some of his “rowing effort” goes into battling the current, and so his “cross river” speed will be only a fraction of his rowing speed. 22. When you stand still under the umbrella in a vertical rain, you are in a cylinder-shaped volume in which there is no rain. The rain has no horizontal component of velocity, and so the rain cannot move from outside that cylinder into it. You stay dry. But as you run, you have a forward horizontal velocity relative to the rain, and so the rain has a backwards horizontal velocity relative to you. It is the same as if you were standing still under the umbrella but the rain had some horizontal component of velocity towards you. The perfectly vertical umbrella would not completely shield you.

Responses to MisConceptual Questions 1.

(c) The shortest possible resultant will be 20 units, which occurs when the vectors point in opposite directions. Since 0 units and 18 units are less than 20 units, (a) and (b) cannot be correct answers. The largest possible result will be 60 units, which occurs when the vectors point in the same direction. Since 64 units and 100 units are greater than 60 units, (d) and (e) cannot be correct answers. Answer (c) is the only choice that falls between the minimum and maximum vector lengths.

(a) The components of a vector make up the two legs of a right triangle when the vector is the hypotenuse. The legs of a right triangle cannot be longer than the hypotenuse, therefore (c) and (d) cannot be correct answers. Only when the vector is parallel to the component, is the

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

magnitude of the vector equal to the magnitude of the component, as in (b). For all other vectors the magnitude of the component is less than the magnitude of the vector. 3.

(b) If you turned 90, as in (a), your path would be that of a right triangle. The distance back would be the hypotenuse of that triangle, which would be longer than 100 m. If you turned by only 30, as in (c), your path would form an obtuse triangle, the distance back would have to be greater than if you had turned 90, and therefore it too would be greater than 100 m. If you turned 180, as in (d), you would end up back at your starting point, not 100 m away. Three equal distances of 100 m would form an equilateral triangle, so (b) is the correct answer.

(b) B and A share a common origin point, and B and C share a common terminal point. Thus B = A + C.

(c) The displacement is not 0, so the magnitude of the average velocity must be greater than 0, and so (a) is incorrect. It is impossible for the magnitude of the average velocity to be larger than the average speed of 10 m/s, and so (d) is incorrect. The only way for (b) to be correct would be if the object was traveling in a straight line. Thus (c) must be the correct answer. The displacement along a quarter of a circle is shorter than the actual distance traveled (the chord of the circle is less than the arc length), and so the average velocity is less than the average speed.

(a) The bullet falls due to the influence of gravity, not due to air resistance. Therefore (b) and (c) are incorrect. Inside the rifle the barrel prevents the bullet from falling, so the bullet does not begin to fall until it leaves the barrel.

(b) Assuming that we ignore air resistance, the ball is in free-fall after it leaves the bat. If the answer were (a) the ball would continue to accelerate forward and would not return to the ground. If the answer were (c) the ball would slow to a stop and return backwards toward the bat.

(b) If we ignore air friction, the horizontal and vertical components of the velocity are independent of each other. The vertical components of the two balls will be equal when the balls reach ground level. The ball thrown horizontally will have a horizontal component of velocity in addition to the vertical component. Therefore, it will have the greater speed.

(c) Both will hit the floor at the same time. Both were launched at the same time, neither had an initial vertical velocity, and both have the same vertical acceleration. The time of flight is determined by the vertical motion, and so is the same for both.

10. (c) Both you and the ball have the same constant horizontal velocity. Therefore in the time it takes the ball to travel up to its highest point and return to ground level, your hand and the ball have traveled the same horizontal distance, and the ball will land back in your hand. 11. (d) The time of flight and the maximum height are both determined by the vertical component of the initial velocity. Since all three kicks reach the same maximum height, they must also have the same time of flight. The horizontal components of the initial velocity are different, which accounts for them traveling different distances. 12. (c) The baseball is in projectile motion during the entire flight. In projectile motion the acceleration is downward and constant; it is never zero. Therefore, (a) is incorrect. Since the ball was hit high and far, it must have had an initial horizontal component of velocity. In projectile motion the horizontal component of velocity is constant, so at the highest point the © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

magnitude of the velocity cannot be zero, and thus (b) is incorrect. However, at the highest point, the vertical component of velocity is zero, so the magnitude of the velocity has a minimum at the highest point. So, (c) is the correct answer. 13. (b) Both the monkey and bullet fall at the same rate due to gravity. If the gun was pointed directly at the monkey and gravity did not act upon either the monkey or bullet, the bullet would hit the monkey. Since both start falling at the same time and fall at the same rate, the bullet will still hit the monkey. 14. (b, e) In projectile motion the acceleration is vertical, so the x velocity is constant throughout the motion, so (a) is valid. The acceleration is that of gravity, which, when up is positive, is a constant negative value, so (b) is not valid and (c) is valid. At the highest point in the trajectory the vertical velocity is changing from a positive to a negative value. At this point the y component of velocity is zero, so (d) is valid. However, the x component of the velocity is constant, but not necessarily zero, so (e) is not valid. 15. (a) The maximum relative speed between the two cars occurs when they are traveling in opposite directions. This maximum speed would be the sum of their speeds relative to the ground or 20 m/s. Since the two cars are traveling perpendicular to each other (not in opposite directions) their relative speed must be less than the maximum relative speed.

Solutions to Problems 1.

The resultant vector displacement of the car is given by D R = D west + Dsouth- . The westward displacement is west

245km + (118 km ) cos 45.0 = 328.4 km and the southward



Dsouth-

Dwest DR

west

displacement is (118 km ) sin 45.0 = 83.4 km. The resultant displacement has a magnitude of

328.42 + 83.42 = 339 km . The direction is  = tan −1 83.4 328.4 = 14.3 south of west . 2.

The truck has a displacement of 21 + ( −26 ) = −5 blocks north and 16 blocks east. The resultant has a magnitude of

Deast

( −5) + 16 = 16.76 blocks 2

Dnorth

Dsouth

−1

 17 blocks and a direction of tan 5 16 = 17 sorth of east . DR 3.

Given that Vx = 9.40 units and Vy = −6.80 units, the magnitude of V is given by V = V + V 2 x

2 y

direction is given by  = tan −1

9.40 + ( −6.80 ) = 11.6 units . The 2

−6.80 9.40

= −35.9 , or 35.9 below the



positive x-axis.

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

The vectors for the problem are drawn approximately to scale. The resultant has a length of 17.5 m and a direction 19 north of east. If calculations are done, the actual resultant should be 17 m at 23o north of east. Keeping one more significant figure would give 17.4 m at 22.5o.

(a) See the accompanying diagram. (b) Vx = −21.8cos 23.4 = −20.0 units (c) V = Vx2 + Vy2 =

 = tan −1 6.

8.66 20.0

Vy = 21.8sin 23.4 = 8.66 units

( −20.0)2 + (8.66)2 = 21.8 units

= 23.4above the − x axis

(a) V1 x = −6.2 units

V1 y = 0 units

V2 x = 8.1cos 55 = 4.646 units  4.6 units

V2 y = 8.1sin 55 = 6.635 units  6.6 units

(b) To find the components of the sum, add the components of the individual vectors. V1 + V2 = (V1 x + V2 x ,V1 y + V2 y ) = ( −1.554, 6.635) V1 + V2 =

( −1.554 )2 + ( 6.635)2 = 6.815 units  6.8 units  = tan −1

6.635 1.554

= 77

The sum has a magnitude of 6.8 units , and is 77o clockwise from the negative x-axis , or 103o counterclockwise from the positive x-axis. 7.

We see from the diagram that A = 6.8î and B = −5.5î. (a) C = A + B = 6.8î + ( −5.5) î = 1.3î . The magnitude is 1.3 units , and the direction is +x . (b) C = A − B = 6.8î − ( −5.5) î = 12.3î . The magnitude is 12.3 units , and the direction is +x . (c)

C = B − A = ( −5.5) î − 6.8î = −12.3î . The magnitude is 12.3 units , and the direction is –x .

(a) vnorth = ( 815 km h )( cos 41.5 ) = 610.4 km h  6.10  102 km h

vwest = ( 815 km h )( sin 41.5 ) = 540.0 km h  5.40 10 2 km h (b) d north = vnorth t = ( 610.4 km h )(1.75 h ) = 1068.2 km  1070 km d west = v west t = ( 540.0 km h )(1.75 h ) = 945 km

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

The x component is negative and the y component is positive, since the summit is to the west of north. The angle measured counterclockwise from the positive x axis would be 128.4o. Thus the components are found to be as follows. x = 4580 cos128.4 = −2844.9 m y = 4580 sin128.4 = 3589.3 m z = 2450 m r = −2840 m ˆi + 3590 m ˆj + 2450 m kˆ

10. Ax = 42.0cos 28.0 = 37.08

r =

( −2844.9 ) + ( 3589.3) + ( 2450 ) = 5190 m 2

Ay = 42.0sin 28.0 = 19.72

Bx = −29.7 cos 56.0 = −16.61

B y = 29.7 sin 56.0 = 24.62

C x = 31.0 cos 270 = 0.0

C y = 31.0sin 270 = −31.0

(a)

( A + B + C ) = 37.08 + ( −16.61) + 0.0 = 20.47  20.5 ( A + B + C ) = 19.72 + 24.62 + ( −31.0) = 13.34  13.3 x

A + B + C = 20.5ˆi + 13.3ˆj (b)

A+B+C =

( 20.47 )2 + (13.34 )2 = 24.43  24.4

 = tan −1

11. Ax = 42.0cos 28.0 = 37.08

Ay = 42.0sin 28.0 = 19.72

Bx = −29.7 cos 56.0 = −16.61

By = 29.7 sin 56.0 = 24.62

(a)

( B − A ) = ( −16.61) − 37.08 = −53.69

13.34 20.47

= 33.1

( B − A ) = 24.62 − 19.72 = 4.90

Since the x component < 0 and the y component > 0, the vector is in the 2nd quadrant. 2 2 B − A = −53.7ˆi + 4.9ˆj ; B − A = ( −53.69 ) + ( 4.90) = 53.9

 B − A = tan −1

4.90

= 5.2 above − x axis or 174.8 −53.69 (b) ( A − B ) x = 37.08 − ( −16.61) = 53.69 ( A − B ) y = 19.72 − 24.62 = −4.90

Since the x component > 0 and the y component < 0, the vector is in the 4th quadrant. 2 2 A − B = 53.7ˆi − 4.9ˆj A − B = ( 53.69 ) + ( −4.90 ) = 53.9

 = tan −1

−4.90

= 5.2 below + x axis or 354.8 53.69 (c) Comparing results shows that B − A = − ( A − B ) . The two results are opposite of each other.

12. Ax = 42.0cos 28.0 = 37.08

Ay = 42.0sin 28.0 = 19.72

Cx = 31.0 cos 270 = 0.0

C y = 31.0sin 270 = −31.0

( A − C ) = 37.08 − 0.0 = 37.08 x

( A − C ) = 19.72 − ( −31.0 ) = 50.72 y

A − C = 37.1ˆi + 50.7ˆj A−C =

( 37.08)2 + ( 50.72 )2 = 62.8

 = tan −1

50.72 37.08

= 53.8

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

13. Ax = 42.0cos 28.0 = 37.08

Ay = 42.0sin 28.0 = 19.72

Bx = −29.7 cos 56.0 = −16.61

B y = 29.7 sin 56.0 = 24.62

C x = 31.0 cos 270 = 0.0

C y = 31.0sin 270 = −31.0

(a)

( B − 3A ) = −16.61 − 3 ( 37.08) = −127.85 ( B − 3A ) = 24.62 − 3 (19.72 ) = −34.54 x

Note that since both components are negative, the vector is in the 3rd quadrant. B − 3A = −127.9ˆi − 34.5ˆj B − 3A =

( −127.85)2 + ( −34.54 )2 = 132.4

 = tan −1

−34.54

−127.85 (b) ( 2A − 3B + 2C ) x = 2 ( 37.08) − 3 ( −16.61) + 2 ( 0.0 ) = 123.99

= 15.1below − x axis

( 2A − 3B + 2C ) = 2 (19.72 ) − 3 ( 24.62 ) + 2 ( −31.0) = −96.42 y

Note that since the x component is positive and the y component is negative, the vector is in the 4th quadrant. 2 A − 3B + 2C = 124.0ˆi − 96.4ˆj

(123.99 )2 + ( −96.42 )2 = 157.1

2 A − 3B + 2C =

(c)

 = tan −1

( C − A − B ) = 0.0 − 37.08 − ( −16.61) = −20.47 ( C − A − B ) = −31.0 − 19.72 − 24.62 = −75.34

−96.42 123.99

= 37.9below + x axis

Note that since both components are negative, the vector is in the 3rd quadrant. C − A − B = −20.5ˆi − 75.3ˆj C−A−B =

( −20.47 )2 + ( −75.34 )2 = 78.1

 = tan −1

−75.34 −20.47

14. (a) V1 = −6.0ˆi + 8.0ˆj

V1 =

( −6.0)2 + (8.0)2 = 10  = tan −1

(b) V2 = −4.5ˆi − 5.0ˆj

V2 =

( −4.5)2 + ( −5.0)2 = 6.7

(c)

V1 + V2 = −10.5ˆi + 3.0ˆj

 = tan −1

3.0 −10.5

(d) V1 − V2 = −1.5ˆi + 13.0ˆj

 = tan −1

(

V1 + V2 =

13.0 −1.5

= 74.8 below − x axis

8.0

= 127 −6.0 −5.0 = 228  = tan −1 −4.5

( −10.5)2 + ( 3.0)2 = 10.9  11

= 164

V1 − V2 =

( −1.5)2 + (13.0)2 = 13.1  13

= 96.6  97

) (

)

15. (a) V1 + V2 + V3 = 4.0î − 8.0ˆj + 1.0î + 1.0ˆj + −2.0î + 4.0ˆj = 3.0î − 3.0ˆj

V1 + V2 + V3 =

3.02 + ( −3.0 ) = 4.2 2

 = tan −1

−3.0 3.0

= 315

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

) (

Instructor Solutions Manual

)

(b) V1 − V2 + V3 = 4.0î − 8.0ˆj − 1.0î + 1.0ˆj + −2.0î + 4.0ˆj = 1.0î − 5.0ˆj

V1 − V2 + V3 = 1.02 + ( −5.0 ) = 5.1 2

 = tan −1

−5.0 1.0

= 280

16. If the angle is in the first quadrant, then Vx = V sin  and Vy = V cos  . See the first diagram. If the angle is in the second quadrant, then Vx = V sin  and Vy = −V cos  . See the second diagram. Vx



17. (a) For the magnitudes to add linearly, the two vectors must be parallel. V1 V2 (b) For the magnitudes to add according to the Pythagorean theorem, the two vectors must be at right angles to each other. V1 ⊥ V2 (c) The magnitude of V2 vector 2 must be 0. V2 = 0 18. (a) Use the Pythagorean theorem to find the possible x components. 2 95.02 = x 2 + ( −60.0 ) → x 2 = 5425 → x = 73.7 units (b) Express each vector in component form, with V the vector to be determined. 73.7î − 60.0ˆj + V î + V ˆj = −80.0î + 0.0ˆj →

(

) (

)

Vx = ( −80.0 − 73.7 ) = −153.7

Vy = 60.0

V = −153.7ˆi + 60.0ˆj

( −153.7 ) + ( 60.0 ) = 165.0 ; V = tan −1 2

V =

60.0 −153.7

= 159

19. Differentiate the position vector in order to determine the velocity, and differentiate the velocity in order to determine the acceleration. dr r = 9.60t ˆi + 6.45 ˆj − 1.50t 2 kˆ m → v = = 9.60 ˆi − 3.00t kˆ m s → dt dv a= = −3.00 kˆ m s 2 dt

(

)

(

)

20. The average velocity is found from the displacement at the two times. r ( t2 ) − r ( t1 ) r ( 3.00s ) − r (1.00s ) v avg = = t2 − t1 3.00s − 1.00s

(

)

  9.60 ( 3.00 ) î + 6.45 ˆj − (1.50 ) ( 3.00 ) 2 kˆ m   19.20î − 12.00kˆ    = 2.00s =  2 2.00s −  9.60 (1.00 ) î + 6.45 ˆj − (1.50 )(1.00 ) kˆ m      

(

)

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

(

)

= 9.60 ˆi − 6.00 kˆ m s

The magnitude of the instantaneous velocity is found from the velocity vector. dr v= = 9.60 î − 3.00t kˆ m s dt v ( 2.00 ) = 9.60 î − ( 3.00 )( 2.00 ) kˆ m s = 9.60 î − 6.00 kˆ m s →

( (

)

(

)

( 9.60 ) + ( 6.00 ) m s = 11.3 m s 2

Note that, since the acceleration of this object is constant, the average velocity over the time interval is equal to the instantaneous velocity at the midpoint of the time interval. 21. (a) Average velocity is displacement divided by elapsed time. Since the displacement is not known, the average velocity cannot be determined . A special case exists in the case of constant acceleration, where the average velocity is the numeric average of the initial and final velocities. But this is not specified as motion with constant acceleration, and so that special case cannot be assumed. (b) Define east as the positive x-direction, and north as the positive y-direction. The average acceleration is the change in velocity divided by the elapsed time. v 25.7ˆi m s − ( −16.0ˆj m s ) = = 3.21ˆi m s 2 + 2.00ˆj m s 2 aavg = 8.00s t aavg =

( 3.21m s ) + ( 2.00 m s ) = 3.78 m s 2

 = tan −1

2.00

= 31.9 3.21 (c) Average speed is distance traveled divided by elapsed time. Since the distance traveled is not known, the average speed cannot be determined . 2

22. Note that the acceleration vector is constant, and so Eqs. 3–13a and 3–13b are applicable. Also v 0 = 0 and r0 = 0.

(

)

(a) v = v 0 + at = 4.0t ˆi + 3.0t ˆj m s → (b) v = v x2 + v 2y =

v x = 4.0t m s, v y = 3.0t m s

( 4.0t m s ) + ( 3.0t m s ) = 5.0t m s 2

(

)

(c)

r = r0 + v 0t + 12 at 2 = 2.0t 2 ˆi + 1.5t 2 ˆj m

(d)

v x ( 2.0 ) = 8.0 m s, v y ( 2.0 ) = 6.0 m s, v ( 2.0 ) = 10.0 m s, r ( 2.0 ) = 8.0 ˆi + 6.0 ˆj m

(

)

23. Choose downward to be the positive y-direction for this problem. Her acceleration is directed along the slope. (a) The vertical component of her acceleration is directed downward, and its magnitude will be

(

)

given by a y = a sin  = 1.80 m s2 sin 30.0o = 0.900 m s2 . (b) The time to reach the bottom of the hill is calculated from Eq. 2–12b, with a y displacement of 125 m, v y 0 = 0, and a y = 0.900 m s 2 .

(

)

y = y0 + v y 0t + 12 a y t 2 → 125 m = 0 + 0 + 12 0.900 m s 2 ( t ) t=

2 (125 m )

( 0.900 m s ) 2

→

= 16.7 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

24. The average speed is the distance traveled divided by the elapsed time, Eq. 2–1. distance traveled 11.5 km average speed = = = 2.1km h time elapsed 5.5 h The average velocity is the displacement divided by the elapsed time. Let north be the y-direction and up be the z-direction. 8.0 km ˆj + 0.85 km kˆ = 1.455 ˆj + 0.1545 kˆ km h v avg = 5.5 h

(

vavg =

)

(1.455 km h ) + ( 0.1545 km h ) = 1.46 km h  1.5 km h 2

0.1545

 = tan −1

1.455

= 6.1elevation above straight north

25. The three displacements for the ant are shown in the diagram, along with the net displacement. In x and y components, they are +10.0 cm ˆi , 10.0 cos 40.0 ˆi + 10.0 sin 40.0 ˆj cm, and

(

)

(10.0 cos110.0 î + 10.0 sin110.0 ˆj) cm. To find the average velocity, divide the net displacement by the elapsed time. (a) r = +10.0 cm î + 10.0 cos 40.0î + 10.0sin 40.0ˆj cm

(

)

+ 10.0 cos110 ˆi + 10.0sin110 ˆj cm

(

)

= 14.24 ˆi + 15.82 ˆj cm v avg = (b)

v avg =

r t

(14.24 ˆi + 15.82 ˆj) cm = 2.48 ˆi + 2.75ˆj cm s ( ) 2.40s + 1.80s + 1.55s

( 2.48cm s )2 + ( 2.75cm s )2 = 3.70cm s

 = tan −1

vy vx

= tan −1

2.75 2.48

= 48.0

26. (a) Differentiate the position vector, r = ( 3.0 t 2 î − 6.0t 3ˆj) m , with respect to time in order to determine the velocity, and differentiate the velocity to determine the acceleration. dr dv v= = 6.0 t î − 18.0t 2 ˆj m s a= = 6.0 î − 36.0t ˆj m s 2 dt dt 2 3 (b) r ( 3.5s ) = 3.0 ( 3.5) î − 6.0 ( 3.5) ˆj m = 37 î − 260ˆj m

(

)

(



(



)

(

)

2 v ( 3.5s ) = 6.0 ( 3.5) ˆi − 18.0 ( 3.5 ) ˆj m s = 21 ˆi − 220ˆj m s





27. Find the position at t = 5.0 s, and then subtract the initial point from that new location. 2 3 r ( 5.0) = 5.0 ( 5.0) + 6.0 ( 5.0 )  m ˆi + 7.0 − 3.0 (5.0 )  m ˆj = 175m ˆi − 368 m ˆj

(



) (





)



r = 175 m î − 368 m ˆj − 2.0 m î + 7.0 m ˆj = 173m î − 375 m ˆj

r =

(173m )2 + ( −375 m )2 = 413m  410 m

 = tan −1

−375 173

= −65.2  −65

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

 1m s   = 30.6 m s ,  3.6 km h 

28. The deceleration is along a straight line. The starting velocity is 110 km h 

and the ending velocity is 0 m/s. The acceleration is found from Eq. 2–12a. 30.6 m s v = v0 + at → 0 = 30.6 m s + a ( 7.0 s ) → a = − = −4.37 m s 2 7.0 s The horizontal acceleration is ahoriz = a cos  = −4.37 m s2 ( cos 26 ) = −3.9 m s2 . The vertical acceleration is avert = a sin  = −4.37 m s2 ( sin 26 ) = −1.9 m s 2 . The horizontal acceleration is to the left in Fig, 3–43, and the vertical acceleration is down. 29. Call east the positive x-direction and north the positive y-direction. Then this relative velocity relationship follows (see the accompanying diagram). v plane rel. = v plane + v air rel. ground

rel. air

15

v plane rel.

ground

Equate the x components of the velocity vectors. The magnitude of v plane rel.

v plane

ground

is given as

135 km 1.25 h

ground

rel. air

v air rel.

= 108 km h.

 ground

(108 km h ) sin15.0 = 0 + vwind x

→

v wind x = 27.95 km h

From the y components of the relative velocity equation, we find vwind y.

−108cos15.0 = −185 + vwind y → vwind y = 185 − 108cos15.0 = 80.68 km h The magnitude of the wind velocity is as follows. 2 2 vwind = vwind + vwind = x y

( 27.95 km h ) + (80.68 km h ) = 85.38 km h  85 km h 2

The direction of the wind is  = tan

−1

v wind-y

= tan −1

v wind-x

80.68 27.95

= 70.89  71 north of east .

30. The velocity vector can be found from Eq. 3–13a, and the position vector can be found from Eq. 3–13b, since the acceleration vector is constant. The time at which the object comes to rest is found by setting the velocity vector equal to 0. Both components of the velocity must be 0 at the same time for the object to be at rest. v = v 0 + at = −14 î − 7.0ˆj m s + 6.0t î + 3.0t ˆj m s = ( −14 + 6.0t ) î + ( −7.0 + 3.0t ) ˆj m s

(

)

(

)

v rest = 0.0 ˆi + 0.0ˆj m s =  ( −14 + 6.0t ) ˆi + ( −7.0 + 3.0t ) ˆj m s →

( v x ) rest = 0.0 = −14 + 6.0t → t =

(v ) y

= 0.0 = −7.0 + 3.0t → t =

s = 73 s

6.0 7.0

s = 73 s

3.0 Since both components of velocity are 0 at t = 73 s, the object is at rest at that time. rest

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

(

)

(

)

Instructor Solutions Manual

(

)

r = r0 + v 0t + 12 at 2 = 0.0 î + 0.0ˆj m + −14t î − 7.0t ˆj m + 12 6.0t 2 î + 3.0t 2 ˆj m

( ) = ( −14 ( ) + 6.0 ( ) ) ˆi m + ( −7.0 ( ) + 3.0 ( ) ) ˆj m (

)

2 2 = −14 ( 73 ) ˆi − 7.0 ( 73 ) ˆj m + 12 6.0 ( 73 ) ˆi + 3.0 ( 73 ) ˆj m

7 3

7 2 3

1 2

(

)

7 3

(

7 2 3

1 2

)

= −16.3î − 8.16ˆj m  −16.3î − 8.2ˆj m 31. Since the acceleration vector is constant, Eqs. 3–13a and 3–13b are applicable. The particle reaches its maximum x coordinate when the x velocity is 0. Note that v 0 = 5.0 m s î and r0 = 0.

(

)

v = v 0 + at = 5.0 ˆi m s + −3.0t ˆi + 4.5t ˆj m s v x = ( 5.0 − 3.0t ) m s → v x = 0 = ( 5.0 − 3.0t x − max ) m s → t x − max =

5.0 m s 3.0 m s 2

= 1.67 s

v ( t x − max ) = 5.0 ˆi m s +  −3.0 (1.67 ) ˆi + 4.5 (1.67 ) t ˆj m s = 7.5 m s ˆj

(

)

r = r0 + v 0 t + 12 at 2 = 5.0 t î m + 12 −3.0t 2 î + 4.5t 2 ˆj m 2 2 r ( t x − max ) = 5.0 (1.67 ) î m s + 12  −3.0 (1.67 ) î + 4.5 (1.67 ) ˆj m = 4.2î m + 6.3ˆj m





32. Choose downward to be the positive y-direction. The origin will be at the point where the tiger leaps from the rock. In the horizontal direction, v x 0 = 3.0 m s and ax = 0. In the vertical direction,

v y 0 = 0, a y = 9.80 m s , y0 = 0, and the final location is y = 7.5 m. The time for the tiger to reach 2

the ground is found from applying Eq. 2–12b to the vertical motion.

(

)

y = y0 + v y 0 t + 12 a y t 2 → 7.5m = 0 + 0 + 12 9.80 m s 2 t 2 → t =

2 ( 7.5m ) 9.80 m s 2

= 1.237 sec

The horizontal displacement is calculated from the constant horizontal velocity. x = v x t = ( 3.0 m s )(1.237 sec ) = 3.7 m 33. Choose downward to be the positive y-direction. The origin will be at the point where the diver dives from the cliff. In the horizontal direction, v x 0 = 2.5 m s and a x = 0. In the vertical direction, v y 0 = 0, a y = 9.80 m s2 , y0 = 0, and the time of flight is t = 3.5 s. The height of the cliff is found

from applying Eq. 2–12b to the vertical motion.

(

)

y = y0 + v y 0t + 12 a y t 2 → y = 0 + 0 + 12 9.80 m s 2 ( 3.5s ) = 6.0 101 m 2

The distance from the base of the cliff to where the diver hits the water is found from the horizontal motion at constant velocity: x = v x t = ( 2.5 m s )( 3.5s ) = 8.75 m  8.8 m 34. Choose downward to be the positive y-direction. The origin will be at the point where the ball is thrown from the roof of the building. In the vertical direction, v y 0 = 0, a y = 9.80 m s 2 , y0 = 0, and the displacement is 7.5 m. The time of flight is found from applying Eq. 2–12b to the vertical motion. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

(

)

y = y0 + v y 0t + 12 a y t 2 → 7.5 m = 12 9.80 m s 2 t 2 → t =

2 ( 9.0 m ) 9.80 m s 2

= 1.237 sec

The horizontal speed (which is the initial speed) is found from the horizontal motion at constant velocity. x = v x t → v x = x t = 9.5 m 1.237 s = 7.7 m s 35. Choose downward to be the positive y-direction. The origin is the point where the ball is thrown from the roof of the building. In the vertical direction v y 0 = 0, y0 = 0, and a y = 9.80 m s 2. The initial horizontal velocity is 10.8 m/s and the horizontal range is 21.0 m. The time of flight is found from the horizontal motion at constant velocity. x = v x t → t = x v x = ( 21.0 m ) (10.8 m s ) = 1.944 s The vertical displacement, which is the height of the building, is found by applying Eq. 2–12b to the vertical motion. 2 y = y0 + v y 0 t + 12 a y t 2 → y = 0 + 0 + 12 ( 9.80 m s 2 ) (1.944 s ) = 18.5 m 36. Choose the point at which the football is kicked as the origin, and choose upward to be the positive y-direction. When the football reaches the ground again, the y-displacement is 0. For the football, 2 v y 0 = (18.0sin 31.0 ) m s, a y = −9.80 m s , and the final y-velocity will be the opposite of the starting y-velocity. Use Eq. 2–12a to find the time of flight. v y − v y 0 ( −18.0 sin 31.0 ) m s − (18.0 sin 31.0 ) m s = = 1.89 s v y = v y 0 + at → t = a −9.80 m s 2 37. Apply the “level horizontal range” formula. v 2 sin 2 0 R= 0 → g

( 2.5 m ) ( 9.80 m s 2 ) = 0.5799 sin 2 0 = 2 = 2 v0 ( 6.5 m s ) Rg

2 0 = sin −1 0.5799 →  0 = 18, 72

There are two angles because each angle gives the same range. If one angle is  = 45 +  , then  = 45 −  is also a solution. The two paths are shown in the graph. 38. When shooting the gun vertically, half the time of flight is spent moving upwards. Thus the upwards flight takes 1.7 seconds. Choose upward as the positive y-direction. Since at the top of the flight, the vertical velocity is zero, find the launching velocity from Eq. 2–12a. v y = v y 0 + at → v y 0 = v y − at = 0 = ( 9.80 m s 2 ) (1.7 s ) = 16.66 m s Using this initial velocity and an angle of 45o in the level horizontal range formula derived in the text will give the maximum range for the gun. R=

v02 sin 2 0 g

(16.66 m s ) sin ( 2  45o ) 2

9.80 m s 2

= 28 m

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

39. Choose the origin to be where the projectile is launched, and upwards to be the positive y-direction. The initial velocity of the projectile is v 0 , the launching angle is  0 , a y = − g, and v y 0 = v0 sin 0 . (a) The maximum height is found from Eq. 2–12c with v y = 0 at the maximum height.

ymax = 0 +

v y2 − v y20 2a y

−v02 sin 2  0 −2 g

v02 sin 2  0 2g

( 38.8 m s ) sin 2 42.2 2

(

2 9.80 m s 2

)

= 34.7 m

(b) The total time in the air is found from Eq. 2–12b, with a total vertical displacement of 0 for the ball to reach the ground. y = y0 + v y 0t + 12 a y t 2 → 0 = v0 sin  0t − 12 gt 2 →

2v0 sin  0

2 ( 38.8 m s ) sin 42.2

= 5.3189 s  5.32 s and t = 0 9.80 m s 2 g The time of 0 represents the launching of the ball. (c) The total horizontal distance covered is found from the horizontal motion at constant velocity.

x = v x t = ( v0 cos  0 ) t = ( 38.8 m s )( cos 42.2 )( 5.3189 s ) = 153 m

(d) The velocity of the projectile 1.50 s after firing is found as the vector sum of the horizontal and vertical velocities at that time. The horizontal velocity is a constant v 0 cos  0 =

( 38.8 m s )( cos 42.2 ) = 28.74 m s. The vertical velocity is found from Eq. 2–12a. v y = v y 0 + at = v0 sin  0 − gt = ( 38.8 m s ) sin 42.2 − ( 9.80 m s 2 ) (1.50 s ) = 11.36 m s

Thus the speed of the projectile is as follows.

v x2 + v y2 =

( 28.74 m s ) + (11.36 m s ) = 30.9 m s 2

40. First, we find the vertical component of the launch speed using the fact that she rises 1.2 m above the launch point, and that her vertical speed at that height is 0. Use Eq. 2–12c, and take upwards to be positive. v y2 − v y2 0 → v y 0 = v y2 − 2a y ( ymax − y0 ) = 2 9.80 m s 2 (1.2 m ) = 4.85 m s ymax − y0 = 2a y

(

)

Next, find the time of flight knowing the starting vertical velocity and the total vertical displacement, using Eq. 2–12b. Take the water surface to be y = 0. y = y0 + v y 0 t + 12 a y t 2 → 0 = 2.0 + 4.85t + 12 −9.80 m s 2 t 2 → 4.90t 2 − 4.85t − 2.0 = 0

(

4.85 

)

( 4.85 ) + 4 ( 4.90 )( 2.0 ) 2

= 1.30 s or − 0.313s 9.80 We use the positive time. Now find the horizontal component of the launch speed from the time and the horizontal distance traveled. x 2.2 m x = v x t → v x = = = 1.69 m s t 1.30 s Now that we have both components of the launch velocity, we can find the magnitude and direction.

v0 =

v x2 + v y2 0 =

 0 = tan −1

vy0 vx

(1.69 m s ) + ( 4.85 m s ) = 5.136 m s  5.1m s

= tan −1

4.85 m s 1.69 m s

= 70.7  71 above the horizontal

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

41, First use the level horizontal range formula to find the launching speed for the maximum range, using 45. R=

v02 sin 2 0 g

→ v0 =

Rmax g

sin  2 ( 45 )

Rmax g

Then use the level horizontal range formula again to find the launching angle for the speed found above, with the given range. v 2 sin 2 0 Rg Rg 68 m R= 0 → sin 2 0 = 2 = = = 0.7158 → 2 0 = 45.71, 134.29 g Rmax g 95 m v0

 0 = 22.86, 67.15  23, 67 42. Choose the origin to be at ground level, under the place where the projectile is launched, and upwards to be the positive y direction. For the projectile, v0 = 62.0 m s,  0 = 35.0, a y = − g , y0 = 125 m, and v y 0 = v0 sin 0 .

(a) The time taken to reach the ground is found from Eq. 2–12b, with a final height of 0. y = y0 + v y 0t + 12 a y t 2 → 0 = y0 + v0 sin 0t − 12 gt 2 → t=

−v0 sin  0  v02 sin 2  0 − 4 ( − 12 g ) y0 2 ( − 12 g )

−35.56 m s 

( −35.56 m s ) + 2450 m 2 s 2 2

−9.80 m s 2

= 9.848s, − 2.590 s = 9.85s

Choose the positive time since the projectile was launched at time t = 0. (b) The horizontal range is found from the horizontal motion at constant velocity. x = v x t = ( v0 cos  0 ) t = ( 62.0 m s ) ( cos 35.0 ) ( 9.848 s ) = 5.00  10 2 m

(c) At the instant just before the particle reaches the ground, the horizontal component of its velocity is the constant v x = v0 cos  0 = ( 62.0 m s ) cos 35.0 = 50.8 m s . The vertical component is found from Eq. 2–12a. v y = v y 0 + at = v0 sin  0 − gt = ( 62.0 m s ) sin 35.0 − ( 9.80 m s 2 ) ( 9.848s ) = −60.9 m s

(d) The magnitude of the velocity is found from the x and y components calculated in part (c) above.

v = v x2 + v y2 =

( 50.8 m s ) + ( −60.9 m s ) = 79.3m s 2

(e) The direction of the velocity is  = tan −1

(f)

vy vx

= tan −1

−60.9 50.8

= −50.2, and so the object is

moving 50.2 below the horizon . The maximum height above the cliff top reached by the projectile will occur when the y-velocity is 0, and is found from Eq. 2–12c. v y2 = v y2 0 + 2a y ( y − y0 ) → 0 = v02 sin 2  0 − 2 gymax ymax =

v02 sin 2  0 2g

( 62.0 m s ) sin 2 35.0 2

(

2 9.80 m s 2

)

= 64.5 m

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

43. (a) Use the level horizontal range formula derived in the text. R=

v02 sin 2 0

→ v0 =

sin 2 0

( 7.80 m ) ( 9.80 m s 2 ) sin 54.0

= 9.72 m s

(b) Now increase the speed by 5.0% and calculate the new range. The new speed would be 9.72 m s (1.05 ) = 10.21m s and the new range would be as follows.

(10.21m s ) sin 54 2

v02 sin 2 0

9.80 m s 2

= 8.606 m

This is an increase of 0.81m (10% increase ) . 44. We choose the origin at the same place as in Example 3–11. With the new definition of the coordinate axes, we have the following data: y0 = 0, y = +1.00 m, v y 0 = −12.0 m s, v x 0 = −16.0 m s, a = 9.80 m s 2 . Use Eq. 2–12b to solve for the time. y = y0 + v y 0t + 12 gt 2 → 1.00 m = 0 − (12.0 m s ) t + 4.90 m s 2 t 2 →

(

)

( 4.90 m s ) t − (12.0 m s ) t − (1.00 m ) = 0 2

This is the same equation as in Example 3–11, and so we know the appropriate solution is t = 2.53s. We use that time to calculate the horizontal distance the ball travels. x = v x 0 t = ( −16.0 m s )( 2.53s ) = −40.5 m Since the x-direction is now positive to the left, the negative value means that the ball lands 40.5 m to the right of where it departed the punter’s foot. 45. (a) Choose the origin to be the location where the car leaves the ramp, and choose upward to be the positive y-direction. At the end of its flight over the 8 cars, the car must be at y = −1.5m. Also for the car, v y 0 = 0, a y = − g , v x = v0 , and x = 22 m. The time of flight is found from the horizontal motion at constant velocity: x = v x t → t = x v0 . That expression for the time is used in Eq. 2–12b for the vertical motion. 2 y = y0 + v y 0t + 12 a y t 2 → y = 0 + 0 + 12 ( − g )( x v0 ) → v0 =

− g ( x )

(

2( y)

)

− 9.80 m s 2 ( 22 m )

2 ( −1.5 m )

= 39.76 m s  4.0  101 m s

(b) Again choose the origin to be the location where the car leaves the ramp, and choose upward to be the positive y-direction. The y-displacement of the car at the end of its flight over the 8 cars must again be y = −1.5 m. For the car, v y 0 = v0 sin 0 , a y = − g , v x = v0 cos  0 , and

x = 22 m. The launch angle is  0 = 9.0. The time of flight is found from the horizontal motion at constant velocity. x x = v x t → t = v0 cos  0 That expression for the time is used in Eq. 2–12b for the vertical motion. y = y0 + v y 0 t + a y t 1 2

 x  → y = v0 sin  0 + (−g )  v0 cos  0  v0 cos  0  x

1 2

→

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

g ( x )

v0 =

( 9.80 m s ) ( 22 m ) 2

2 ( x tan  0 − y ) cos 2  0

2 ( ( 22 m ) tan 9.0 + 1.5 m ) cos 2 9.0

= 22 m s

46. Choose the origin to be the point on the ground directly below the point where the baseball was hit. Choose upward to be the positive y-direction. Then y0 = 1.2 m, y = 12.5 m at the end of the motion, v y 0 = ( 27.0sin 45.0 ) m s = 19.09 m s , and a y = −9.80 m s 2 . Use Eq. 2–12b to find the time of flight. y = y0 + v y 0 t + 12 a y t 2 t=

→

1 2

a y t 2 + v y 0 t + ( y0 − y ) = 0

−v y 0  v y2 0 − 4 ( 12 a y ) ( y0 − y ) 2 12 a y

−19.09 

→

(19.09 ) − 2 ( −9.80 )( −11.3 ) 2

−9.80

= 0.728 s, 3.168 s The smaller time is the time the baseball reached the building’s height on the way up, and the larger time is the time the baseball reached the building’s height on the way down. We must choose the larger result, because the baseball cannot land on the roof on the way up. Now calculate the horizontal distance traveled using the horizontal motion at constant velocity. x = v x t = ( 27.0 cos 45.0 ) m s  ( 3.168s ) = 60.5 m

47. Choose the origin to be the location on the ground directly below the airplane at the time the supplies are dropped, and choose upward as the positive y-direction. For the supplies, y0 = 265 m, v y 0 = 0, a y = − g, and the final y-location is y = 0 m. The initial (and constant) x-velocity of the supplies is

 1h  1000 m    = 34.72 m s. The time for the supplies to reach the ground is  3600 s  1km 

v x = 125 km h 

found from Eq. 2–12b. y = y0 + v y 0t + 12 a y t 2

→

0 = y0 + 0 + 12 at 2 → t =

−2 y0

−2 ( 265 m )

( −9.80 m s ) 2

Then the horizontal distance of travel for the package is found from the horizontal constant velocity. x = v x t = ( 34.72 m s )

−2 ( 265 m )

( −9.80 m s ) 2

= 255 m

48. Choose the origin to be the point of launch, and upwards to be the positive y-direction. The initial velocity is v0 , the launching angle is  0 , a y = − g , y0 = 0, and v y 0 = v0 sin 0 . Eq. 2–12a is used to find the time required to reach the highest point, at which v y = 0. v y = v y 0 + atup

→

tup =

vy − vy0

0 − v0 sin  0

a −g Eq. 2–12c is used to find the height at this highest point. v y2 = v y2 0 + 2a y ( ymax − y0 )

→

ymax = y0 +

v y2 − v y2 0 2a y

v0 sin  0 g = 0+

−v02 sin 2  0 −2 g

v02 sin 2  0 2g

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Eq. 2–12b is used to find the time for the object to fall the same distance with a starting velocity of 0. v 2 sin 2  0 v sin  0 2 + 0tdown − 12 gtdown → tdown = 0 y = yo + v y 0t + 12 a y t 2 → 0 = 0 2g g A comparison shows that tup = tdown . 49. The constant acceleration of the projectile is given by a = −9.80 m s 2 ˆj. We use Eq. 3–13a with the given velocity, the acceleration, and the time to find the initial velocity. v = v 0 + at → v 0 = v − at = ( 7.8 ˆi + 5.2 ˆj) m s − ( −9.80 m s 2 ˆj) ( 3.0s ) = ( 7.8 ˆi + 34.6 ˆj) m s

( 7.8 m s ) + ( 34.6 m s ) = 35.47 m s , and the original launch direction is 2

The initial speed is v0 = given by  0 = tan −1

34.6 m s 7.8 m s

= 77.30. Use this information with the level horizontal range formula

to find the range. (a) R =

v02 sin 2 0

( 35.47 m s ) ( sin154.6 ) 2

= = 55 m g 9.80 m s 2 (b) We use the vertical information to find the maximum height. The initial vertical velocity is 34.6 m/s, and the vertical acceleration is −9.80 m s 2 . The vertical velocity at the maximum height is 0, and the initial height is 0. Use Eq. 2–12c. v y2 = v y2 0 + 2a y ( ymax − y0 ) →

ymax = y0 +

v y2 − v y2 0 2a y

−v02 y 2a y

− ( 34.6 m s )

(

2 −9.80 m s 2

)

= 61m

(c) From the information above and the symmetry of projectile motion, we know that the final speed just before the projectile hits the ground is the same as the initial speed, and the angle is the same as the launching angle, but below the horizontal. So vfinal = 35 m s and

final = 77 below the horizontal . 50. Choose the origin to be where the projectile is launched, and upwards to be the positive y-direction. The initial velocity of the projectile is v0 , the launching angle is  0 , a y = − g , and v y 0 = v0 sin 0 . The range of the projectile is given by the level horizontal range formula, R =

v02 sin 2 0

. The g maximum height of the projectile will occur when its vertical speed is 0. Apply Eq. 2–12c. v02 sin 2  0 2 2 2 2 v y = v y 0 + 2a y ( y − y0 ) → 0 = v0 sin  0 − 2 gymax → ymax = 2g Now find the angle for which R = ymax . R = ymax

→

v02 sin 2 0 g

v02 sin 2  0 2g

→

sin 2 0 = 12 sin 2  0 →

2sin 0 cos 0 = 12 sin 0 → 4 cos 0 = sin 0 → tan 0 = 4 → 0 = tan −1 4 = 76 2

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

51. The angle is in the direction of the velocity, so find the components of the velocity, and use them to define the angle. Let the positive y-direction be down. vy gt v x = v0 v y = v y 0 + a y t = gt  = tan −1 = tan −1 vx v0 52. Choose upward to be the positive y-direction. The origin is the point from which the pebbles are released. In the vertical direction, a y = −9.80 m s 2 , the velocity at the window is v y = 0, and the vertical displacement is 8.0 m. The initial y-velocity is found from Eq. 2–12c. v y2 = v y2 0 + 2a y ( y − y0 ) →

vy0 =

v y2 − 2a y ( y − y0 ) =

(

)

0 − 2 −9.80 m s 2 ( 8.0 m ) = 12.5 m s

Find the time for the pebbles to travel to the window from Eq. 2–12a. v y − v y 0 0 − 12.5 m s = = 1.28s v y = v y 0 + at → t = a −9.80 m s 2 Find the horizontal speed from the horizontal motion at constant velocity. x = v x t → v x = x t = 8.5 m 1.28 s = 6.6 m s This is the speed of the pebbles when they hit the window. 53. (a) Use the level horizontal range formula to find her takeoff speed.

v02 sin 2 0 g

→ v0 =

sin 2 0

( 9.80 m s ) (8.0 m ) = 8.854 m s  8.9 m s 2

sin 90

(b) Let the launch point be at the y = 0 level, and choose upward to be positive. Use Eq. 2–12b to solve for the time to fall to 2.5 meters below the starting height, and then calculate the horizontal distance traveled. y = y0 + v0 y t + 12 a y t 2 → − 2.5 m = ( 8.854 m s ) sin 5t + 12 ( −9.80 m s 2 ) t 2

4.9t 2 − 6.261t − 2.5 m = 0 → t=

6.261 

( 6.261) − 4 ( 4.9 )( −2.5) 6.261  9.391 = = −0.319 s, 1.597 s 2 ( 4.9 ) 2 ( 4.9 ) 2

Use the positive time to find the horizontal displacement during the jump. x = v0 x t = v0 cos 45t = ( 8.854 m s ) cos 45 (1.597 s ) = 10.0 m She will land exactly on the opposite bank, neither long nor short. 54. The horizontal component of the speed does not change during the course of the motion, and so v x = v x 0. The net vertical displacement is 0 if the firing level equals the landing level, and so y − y0 = 0. Eq. 2–12c then gives v y = v y 0 + 2a y ( y − y0 ) = v y 0 . Thus v y = v y 0 , and from the 2

horizontal v x2 = v x20. The initial speed is v0 = v x20 + v y20 . The final speed is v = v x2 + v y2

= v x20 + v y20 = v0 . Thus v = v0 .

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

55. (a) Use the level horizontal range formula to find the two angles. 2 v 2 sin 2 0 Rg ( 20.0 m ) 9.80 m s R= 0 → sin 2 0 = 2 = = 0.8711 → 2 g v0 (15.0 m s )

(

)

2 0 = 60.59, 119.41 →  0 = 30.29, 59.71  30.3, 59.7

(b) The maximum height can be found from the y-component of the velocity and Eq. 2–12c, using the fact that the vertical velocity is 0 at the top of the path. Call upwards the positive direction. v y2 = v y2 0 + 2a y ( y − y0 ) →

y − y0 = y − y0 =

v y2 − v y2 0 2a y v y2 − v y2 0 2a y

v − ( v0 sin 0 )

(

2 −9.80 m s 2 v − ( v0 sin 0 )

(

2 −9.80 m s 2

) 2

)

(15.0 m s )( sin 30.29 ) = 2.92 m = 2

(

2 9.80 m s 2

)

(15.0 m s )( sin 59.71 ) = 8.56 m = 2

(

2 9.80 m s 2

)

(c) The time of flight can be found Eq. 2–12a and using the fact that the initial and final vertical velocities are opposite of each other. v y − v y 0 −2v y 0 −2v0 sin  0 2v0 sin  0 v y = v y 0 + ayt → t = = = = ay ay g −g t1 =

2 (15.0 m s )( sin 30.29 ) 9.80 m s

= 1.54 s

t2 =

2 (15.0 m s )( sin 59.71 ) 9.80 m s 2

= 2.64 s

56. Choose the origin to be the location on the ground directly underneath the ball when served, and choose upward as the positive y-direction. Then for the ball, y0 = 2.30 m, v y 0 = 0, a y = − g , and the y-location when the ball just clears the net is y = 0.90 m. The time for the ball to reach the net is calculated from Eq. 2–12b. y = y0 + v y 0 t + 12 a y t 2 → 0.90 m = 2.30 m + 0 + 12 −9.80 m s 2 t 2 →

(

t to =

)

2 ( −1.40 m )

= 0.5345s −9.80 m s 2 The x-velocity is found from the horizontal motion at constant velocity. 15.0 m x x = v x t → v x = = = 28.06  28.1m s t 0.5345 s This is the minimum speed required to clear the net. net

To find the full time of flight of the ball, set the final y-location to be y = 0, and again use Eq. 2–12b. y = y0 + v y 0t + 12 a y t 2 → 0.0 m = 2.30 m + 12 −9.80 m s 2 t 2 →

(

t total =

)

2 ( −2.30 m )

= 0.6851  0.685s −9.80 m s 2 The horizontal position where the ball lands is found from the horizontal motion at constant velocity. x = v x t = ( 28.06 m s )( 0.6851s ) = 19.22  19.2 m Since this is between 15.0 and 22.0 m, the ball lands in the “good” region .

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

57. The time of flight is found from the constant velocity relationship for horizontal motion.

x = v x t → t = x v x = 8.0 m 9.1m s = 0.88s The y-motion is symmetric in time – it takes half the time of flight to rise, and half to fall. Thus the time for the jumper to fall from his highest point to the ground is 0.44 sec. His vertical speed is zero at the highest point. From this time, starting vertical speed, and the acceleration of gravity, the maximum height can be found. Call upward the positive y-direction. The point of maximum height is the starting position y0 , the ending position is y = 0, the starting vertical speed is 0, and a = − g . Use Eq. 2–12b to find the height. 2 y = y0 + v y 0 t + 12 a y t 2 → 0 = y0 + 0 − 12 ( 9.8 m s 2 ) ( 0.44 s ) → y0 = 0.95 m . 58. Choose the origin to be the location from which the balloon is fired, and choose upward as the positive y-direction. Assume the boy in the tree is a distance H up from the point at which the balloon is fired, and that the tree is a distance d horizontally from the point at which the balloon is fired. The equations of motion for the balloon and boy are as follows, using constant acceleration relationships.

 d

xballoon = v0 cos 0t

yballoon = 0 + v0 sin 0t − 12 gt 2

yboy = H − 12 gt 2

Use the horizontal motion at constant velocity to find the elapsed time after the balloon has traveled d to the right. d d = v0 cos  0 t D tD = → v0 cos  0 Where is the balloon vertically at that time? 2

    d d 1 yballoon = v0 sin  0 t D − gt = v0 sin  0 v − g  = d tan  0 − 2 g   v0 cos  0  v0 cos  0   v0 cos  0  Where is the boy vertically at that time? Note that H = d tan  0 . d

2 D

1 2

    d d 1 yboy = H − gt = H − g   = d tan  0 − 2 g    v0 cos  0   v0 cos  0  Note that yballoon = yboy , and so the boy and the balloon are at the same height and the same 1 2

2 D

1 2

horizontal location at the same time. Thus they collide! 59. Choose upward to be the positive y-direction. The origin is the point from which the football is kicked. The initial speed of the football is v0 = 20.0 m s . We have v y 0 = v0 sin 37.0 = 12.04 m s , y0 = 0, and a y = −9.80 m s2 . In the horizontal direction, v x = v0 cos 37.0 = 15.97 m s , and x = 36.0 m. The time of flight to reach the goalposts is found from the horizontal motion at constant speed. x = v x t → t = x v x = 36.0 m 15.97 m s = 2.254 s Now use this time with the vertical motion data and Eq. 2–12b to find the height of the football when it reaches the horizontal location of the goalposts. 2 y = y0 + v y 0t + 12 a y t 2 = 0 + (12.04 m s )( 2.254 s ) + 12 −9.80 m s 2 ( 2.254 s ) = 2.24 m

(

)

Since the ball’s height is less than 3.05 m, the football does not clear the bar . It is 0.81 m too low when it reaches the horizontal location of the goalposts. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

To find the distances from which a score can be made, redo the problem (with the same initial conditions) to find the times at which the ball is exactly 3.05 m above the ground. Those times would correspond with the maximum and minimum distances for making the score. Use Eq. 2–12b. y = y0 + v y 0 t + 12 a y t 2 → 3.05 = 0 + (12.04 m s ) t + 12 −9.80 m s 2 t 2 →

(

)

(12.04 ) − 4 ( 4.90 )( 3.05) 4.90t − 12.04t + 3.05 = 0 → t = = 2.1703s, 0.2868s 2 ( 4.90 ) x = v t = 15.97 m s ( 0.2868 s ) = 4.580 m ; x = v t = 15.97 m s ( 2.1703s ) = 34.660 m 1

12.04 

So the kick must be made in the range from 4.6 m to 34.7 m . 60. Choose the origin to be at the bottom of the hill, just where the incline starts. The equation of the line describing the hill is y2 = x tan  . The equations of the motion of the object are y1 = v0 y t + 12 a y t 2 and x = v 0 x t , with v0 x = v0 cos  and v0 y = v0 sin  . Solve the horizontal

equation for the time of flight, and insert that into the vertical projectile motion equation. t=

 x  gx 2 x tan → y1 = v0 sin  − g =  −  2v02 cos 2  v0 cos   v0 cos  

= v0 x v0 cos  Equate the y-expressions for the line and the parabola to find the location where the two x-coordinates intersect. gx 2 gx → tan  − tan  = 2 → x tan  = x tan  − 2 2 2v0 cos  2v0 cos 2 

( tan  − tan  )

1 2

2v02 cos 2 

g This intersection x-coordinate is related to the desired quantity d by x = d cos . d cos  = ( tan  − tan  )

2v02 cos 2 

2v02

(

)

sin  cos  − tan  cos 2  g g cos  To maximize the distance, set the derivative of d with respect to  equal to 0, and solve for  . d (d ) 2v02 d sin  cos  − tan  cos 2  = d g cos  d → d=

(

= =

2v02 g cos 

)

sin  ( − sin  ) + cos  ( cos  ) − tan  ( 2 ) cos  ( − sin  )

2v02

2v 2

0  − sin 2  + cos 2  + 2 tan  cos  sin   = cos 2 + sin  tan   = 0 g cos   g cos 



1    tan   This expression can be confusing, because it would seem that a negative sign enters the solution. In order to get appropriate values, 180 or  radians must be added to the angle resulting from the inverse tangent operation, to have a positive angle. Thus a more appropriate expression would be the following: cos 2 + sin  tan  = 0 →  = 12 tan −1  −

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

  1    . This can be shown to be equivalent to  = + , because  = 12  + tan −1  −   2 4  tan       1   tan −1  − = tan −1 ( − cot  ) = cot −1 cot  − =  − .  2 2  tan   61. See the diagram. Solve for R, the horizontal range, which is the horizontal speed times the time of flight. R R = ( v0 cos 0 ) t → t = v0 cos   h = ( v0 sin  0 ) t − 12 gt 2 → R2 − R

2v02 cos 2  0 tan  g

R= =

1 2

gt 2 − ( v0 sin  0 ) t + h = 0 →

2hv02 cos 2  0 g

=0 2

 2v 2 cos 2  0 tan   2hv02 cos 2  0   0 − 4  g g   2

v0 cos  0 

v sin  0  v02 sin 2  0 − 2 gh 

 0

 g Which sign is to be used? We know the result if h = 0 from the level horizontal range formula. v cos  0 Substituting h = 0 gives R = 0  v0 sin  0  v0 sin  0 . To agree with the level horizontal g range formula, we must choose the + sign, and so R =

v0 cos  0  g

v sin  0 + v02 sin 2  0 − 2 gh  .

 0



We see from this result that if h > 0, the range will shorten, and if h < 0, the range will lengthen. 62. Call the direction of the boat relative to the water the positive direction. For the jogger moving towards the bow, we have the following: v =v +v = 2.5m s î + 8.8 m s î = 11.3m s î . jogger rel. water

jogger rel. boat

boat rel. water

For the jogger moving towards the stern, we have the following. v =v +v = −2.5m s î + 8.8 m s î = 6.3m s î jogger rel. water

jogger rel. boat

boat rel. water

63. Call the direction of the flow of the river the x-direction, and the direction of Huck walking relative to the raft the y-direction. v Huck = v Huck + v raft rel. = 0.70ˆj m s + 1.50ˆi m s rel. bank

rel. raft

bank

(

)

= 1.50ˆi + 0.70ˆj m s Magnitude: vHuck

= 1.502 + 0.702 = 1.66 m s

rel. bank

Direction:  = tan −1

0.70 1.50

= 25relative to river

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

64. From Fig. 3–33 in Example 3–13, it is seen that vboat rel. = vboat rel. cos  = (1.85 m s ) cos 40.4 = 1.41m s . shore

water

65. The boat is traveling directly across the stream, with a heading of  = 23.5 upstream, and speed of vboat rel. = 4.30 m s . water

(a) v water rel. = vboat rel. sin  = ( 4.30 m s ) sin 23.5 = 1.71m s water

shore

(b) vboat rel. = vboat rel. cos  = ( 4.30 m s ) cos 23.5 = 3.94 m s water

shore

66. Call the direction of the boat relative to the water the x-direction, and upward the y-direction. Also see the diagram. v passenger = v passenger + v boat rel. rel. water

rel. boat

water

( = ( 2.12ˆi + 0.42ˆj) m s

)

= 0.60 cos 45ˆi + 0.60 sin 45ˆj m s + 1.70ˆi m s

67. Call east the positive x-direction and north the positive y-direction. Then the following vector velocity relationship exists. (a) v plane rel. = v plane + v air rel. ground

rel. air

ground

(

)

= −688ˆj km h + 85.0 cos 45.0ˆi + 85.0 sin 45.0ˆj km h

(

)

= 60.1ˆi − 628ˆj km h v plane rel. =

( 60.1km h ) + ( −628 km h ) = 631km h 2

ground

 = tan −1

60.1

= −5.47 = 5.47 east of south −628 (b) The plane is away from its intended position by the distance the air has caused it to move. The wind speed is 85.0 km/h, so after 11.0 min the plane is off course by the following amount.  1h  d = vt = ( 85.0 km h )(11.0 min )   = 15.6 km .  60 min 

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

68. Call east the positive x-direction and north the positive y-direction. Then the following vector velocity relationship exists. v plane rel. = v plane + v air rel. → ground

rel. air

ground

(

)

−v plane rel. ˆj = −688sin  ˆi + 688 cos  ˆj km h ground

(

)

+ 85.0 cos 45.0ˆi + 85.0 sin 45.0ˆj km h Equate x components in the above equation. 0 = −688sin  + 85.0 cos 45.0 →

 = sin −1

85.0 cos 45.0 688

= 5.01, west of south

69. Choose the x-direction to be the direction of train travel (the direction the passenger is facing) and choose the y-direction to be up. This relationship exists among the velocities: v rain rel. = v rain rel. + v train rel. . From the diagram, find the ground

train

ground

expression for the speed of the raindrops. v train rel. vT v ground = → vrain rel. = T tan  = vrain rel. vrain rel. tan  ground ground

ground

70. From the diagram,  = tan −1 118 m 265 m = 24.0. Equate the vertical components of the velocities in the diagram to find the speed of the boat relative to the shore. v boat rel. cos  = vboat rel. sin 45.0 → shore

water

v boat rel. = ( 2.80 m s )

sin 45.0

= 2.17 m s cos 24.0 Equate the horizontal components of the velocities. v boat rel. sin  = v boat rel. cos 45.0 − v water → shore

shore

v water rel. shore

water

rel. shore

= v boat rel. cos 45.0 − v boat rel. sin  water

shore

= ( 2.80 m s ) cos 45.0 − ( 2.17 m s ) sin 24.0 = 1.10 m s 71. Call the direction of the flow of the river the x-direction, and the direction straight across the river the y-direction. Call the location of the swimmer’s starting point the origin. v swimmer = v swimmer + v water rel. = 0.60 m s ˆj + 0.50 m s ˆi rel. shore

rel. water

shore

(a) Since the swimmer starts from the origin, the distances covered in the x- and y-directions will be exactly proportional to the speeds in those directions. x v x t v x x 0.50 m s = = → = → x = 46 m y v y t v y 55 m 0.60 m s © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

(b) The time is found from the constant velocity relationship for either the x- or y-directions. y 55 m y = v y t → t = = = 92 s v y 0.60 m s 72. (a) Call the direction of the flow of the river the x-direction, and the direction straight across the river the y-direction. v water rel. 0.50 m s 0.50 shore sin  = = →  = sin −1 = 56.44  56 0.60 vswimmer 0.60 m s rel. water

(b) From the diagram her speed with respect to the shore is found as follows. vswimmer = vswimmer cos  = ( 0.60 m s ) cos 56.44 = 0.332 m s rel. shore

rel. water

The time to cross the river can be found from the constant velocity relationship. y 55 m y = v y t → t = = = 170 s = 2.8 min v y 0.332 m s 73. Call east the positive x-direction and north the positive y-direction. From the first diagram, this relative velocity relationship is seen. v car 1 rel. = v car 1 rel. + v car 2 rel. → street

car 2

street

(

)

v car 1 rel. = v car 1 rel. − v car 2 rel. = 35ˆj km h − 55ˆi km h = −55ˆi + 35ˆj km h car 2

street

For the other relative velocity relationship: v car 2 rel. = v car 2 rel. + v car 1 rel. → street

car 1

street

(

)

v car 2 rel. = v car 2 rel. − v car 1 rel. = 55ˆi km h − 35ˆj km h = 55ˆi − 35ˆj km h car 1

street

Notice that the two relative velocities are opposites of each other: v car 2 rel. = − v car 1 rel. . car 1

car 2

74. See the diagram for the orientation of the velocities. Apply the law of sines to the triangle formed by the three vectors. vplane vair rel. vair rel. rel. air ground ground = → sin  = sin128 → vplane sin128 sin rel. air

 vair rel.   82 km h  ground    = sin  sin128 = sin −1  sin128  = 6.6   560 km h   vplane  rel. air   −1

So the plane should head in a direction of 38.0 + 6.6 = 44.6north of east .

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

75. The river is flowing to the right in the diagram. From the diagram,  = 132.0. Use the law of sines to determine the angle  . v boat

v water rel.

rel. water

sin

shore

sin

→

 v water rel.   5.60 km h  shore  sin  = sin −1  sin132.0  =13.37  = sin v   18.0 km h   boat   rel. water  −1

(a) The boat should head at the angle of 42.0 + 13.37 = 55.4 upstream . (b) The across-river components of v boat rel. and v boat rel. are equal. Use that to solve for vboat rel. . shore

v boat v boat

shore

cos ( + 42.0 ) →

cos 42.0 = v boat

rel. shore

shore

rel. water

= (18.0 km h )

rel. shore

cos 55.37 cos 42.0

= 13.76 km h  13.8 km h

vboat

cos 42.0 =

rel. shore

1.45 km t

→ t =

1.45 km vboat

cos 42.0

1.45 km 13.76 km h ( cos 42.0 )

= 0.142 h

rel. shore

The time is also equal to 8.51 min. 76. Let east be the positive x-direction, north be the positive y-direction, and up be the positive z-direction. Then the plumber’s resultant displacement in component notation is D = 55m ˆi − 38 m ˆj − 12 m kˆ . Since this is a 3-



Dxy

dimensional problem, it requires 2 angles to determine his location (similar to latitude and longitude on the surface of the Earth). For the xy (horizontal) plane, see the first figure. D −38  = tan −1 y = tan −1 = −35 = 35 south of east Dx 55

Dxy =

Dx2 + Dy2 =

one leg, and the vertical displacement D z as the other leg. See the second figure, and the following calculations. −12 m D = −12.3 = 12 below the x -axis  2 = tan −1 z = tan −1 55 m Dx

Dxy2 + Dz2 =

( 55)2 + ( −38)2 = 66.9 m  67 m

For the vertical motion, consider another right triangle, made up of D x as

Dx2 + Dy2 + Dz2 =

Dx 

( 55)2 + ( −38)2 + ( −12 )2 = 68 m

The result is that the displacement is 68 m , at an angle of 35 south of east , and 12 below the x -axis .

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

77. Assume that the golf ball takes off and lands at the same height, so that the level horizontal range formula derived in the text can be applied. The only variable is to be the acceleration due to gravity. REarth = v02 sin 2 0 g Earth RMoon = v02 sin 2 0 g Moon REarth RMoon

v02 sin 2 0 g Earth v sin 2 0 g Moon 2 0

1 g Earth 1 g Moon

(

g Moon g Earth

32 m 180 m

= 0.1778

→

)

g Moon = 0.1778 g Earth = 0.1778 9.80 m s 2 = 1.742 m s 2  1.7 m s 2

78. Work in the frame of reference in which the train is at rest. Then, relative to the train, the car is moving at 20 km/h. The car has to travel 1.4 km in that frame of reference to pass the train, and so the time to pass can be found from the constant horizontal velocity relationship. x 1.40 km  3600 s  x = v x t → tsame = =   = 252 s  250 s 20 km h  1h  ( v x )same direction direction

The car travels 1.4 km in the frame of reference of the stationary train, but relative to the ground, the car is traveling at 95 km/h. Thus relative to the ground the car travels this distance:  1h  x = v x tsame = ( 95 km h )( 252 s )   = 6.65 km  6.7 km direction  3600 s  If the car and train are traveling in opposite directions, the velocity of the car relative to the train is 170 km/h. Thus the time to pass will be as follows. Note that the relative speed has 3 significant figures, due to addition. x 1.40 km  3600 s  topposite = = = 29.64 s  29.6 s ( v x )opposite 170 km h  1 h  direction direction

The distance traveled by the car relative to the ground is calculated.  1h  x = v x topposite = ( 95 km h ) ( 29.64s )   = 0.78 km direction  3600 s  79. The minimum speed will be that for which the ball just clears the fence; i.e., the ball has a height of 8.0 m when it is 98 m horizontally from home plate. The origin is at home plate, with upward as the positive y direction. For the ball, y0 = 1.0 m, y = 8.0 m, a y = − g , v y 0 = v0 sin 0 , v x = v0 cos  0 , and  0 = 36.

0 y0 = 1.0 m

y = 8.0 m

x = 98 m See the diagram (not to scale). For the constant-velocity horizontal x . For the vertical motion, apply Eq. 2–12b. motion, x = v x t = v0 cos  0 t , and so t = v0 cos  0

y = y0 + v y 0t + 12 a y t 2 = y0 + v0 ( sin 0 ) t − 12 gt 2

Substitute the value of the time of flight for the first occurrence only in the above equation, and then solve for the time. x − 12 gt 2 → y = y0 + v0 t sin  0 − 12 gt 2 → y = y0 + v0 sin  0 v0 cos  0

 1.0 m − 8.0 m + ( 98 m ) tan 36   y0 − y + x tan 0  = 2   = 3.620s g 9.80 m s 2    

2

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

Finally, use the time with the horizontal range to find the initial speed. 98 m x x = v0 cos  0 t → v0 = = = 33 m s t cos  0 ( 3.620 s ) cos 36 80. See the sketch of the geometry. We assume that the hill is sloping downward to the right. If we take the point where the child jumps as the origin, with the x-direction 1.3m positive to the right and the y-direction positive upwards, 12 then the equation for the hill is given by y = − x tan12. The path of the child (shown by the dashed line) is that of projectile motion. With the same origin and coordinate system, the horizontal motion of the child is given by x = v0 cos15 ( t ) , and the vertical motion of the child will be given by Eq. 2–12b, y = v0 sin15t − 12 gt 2. The landing point of the child is given by xlanding = 1.3cos12 and ylanding = −1.3sin12. Use the horizontal motion and landing point to find an expression for the time the child is in the air, and then use that time to find the initial speed. 1.3cos12 x , tlanding = x = v0 cos15 ( t ) → t = v0 cos15 v0 cos15 Equate the y-expressions, and use the landing time. We also use the trigonometric identity that sin12 cos15 + sin15 cos12 = sin (12 + 15 ) .

ylanding = yprojectile →

2 − 1.3sin12 = v0 sin15tlanding − 12 gtlanding →

 1.3cos12  −1.3sin12 = v0 sin15 − g  v0 cos15  v0 cos15  1.3cos12

v02 = 12 g

→

1 2

cos 2 12  1.3

   → v0 = 3.728 m s  3.7 m s sin 27  cos15 

81. Consider the downward vertical component of the motion, which will occur in half the total time. Take the starting position to be y = 0, and the positive direction to be downward. Use Eq. 2–12b with an initial vertical velocity of 0. 2

y = y0 + v 0 y t + a y t 1 2

→ h = 0 + 0 + gt 1 2

2 down

 t  9.80 2 = g  = t = 1.225t 2  1.2t 2 2 8   1 2

As can be seen from the equation, by starting the analysis from the “top” point of the motion, the initial vertical speed is 0. This eliminates the need to know the original launch speed or direction in that calculation. We then also realize that the time for the object to rise is the same as the time for it to fall, so we only have to analyze the downward motion. 82. Choose downward to be the positive y-direction. The origin is at the point from which the divers push off the cliff. In the vertical direction, the initial velocity is v y 0 = 0, the acceleration is

a y = 9.80 m s 2 , and the displacement is 35 m. The time of flight is found from Eq. 2–12b.

(

)

y = y0 + v y 0t + 12 a y t 2 → 35 m = 0 + 0 + 12 9.80 m s 2 t 2 → t =

2 ( 35 m ) 9.8 m s 2

= 2.7 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

The horizontal speed (which is the initial speed) is found from the horizontal motion at constant velocity. x = v x t → v x = x t = 5.0 m 2.7 s = 1.9 m s 83. The level horizontal range formula derived in the text can be used to find the launching velocity of the grasshopper.

v02 sin 2 0 g

→ v0 =

Rg sin 2 0

( 0.80 m ) ( 9.80 m s 2 ) sin 90

= 2.8 m s

Since there is no time between jumps, the horizontal velocity of the grasshopper is the horizontal component of the launching velocity. v x = v0 cos  0 = ( 2.8 m s ) cos 45o = 2.0 m s 84. Work in the frame of reference in which the car is at rest at ground level. In this reference frame, the  1m s  helicopter’s horizontal speed is 208 km h − 135 km h = 73km h   = 20.28 m s . For  3.6 km h  the vertical motion, choose the level of the helicopter to be the origin, and downward to be positive. Then the package’s y-displacement is y = 78.0 m, v y 0 = 0, and a y = g. The time for the package to fall is calculated from Eq. 2–12b. y = y0 + v y 0 t + 12 a y t 2

→

(

)

78.0 m = 12 9.80 m s 2 t 2

→ t=

2 ( 78.0 m ) 9.80 m s 2

= 3.99 sec

The horizontal distance that the package must move, relative to the “stationary” car, is found from the horizontal motion at constant velocity. x = v x t = ( 20.28 m s )( 3.99 s ) = 80.9 m Thus the angle under the horizontal for the package release will be as follows.  78.0 m   y   = tan −1   = tan −1   = 43.95  44  x   80.9 m  85. The proper initial speeds will be those for which the ball has traveled a horizontal distance somewhere between 10.78 m and 11.22 m while it changes height from 2.10 m to 3.05 m y = 0.95m 0 with an initial (shooting) angle of 38.0o. Choose the origin to be at the shooting location of the basketball, with upward x = 10.78 m − 11.22 m as the positive y-direction. Then the vertical displacement is 2 y = 0.95m, a y = −9.80 m s , v y 0 = v0 sin 0 , and the (constant) x-velocity is v x = v0 cos  0 . See the diagram (not to scale). For the constant-velocity horizontal motion, x = v x t = v0 cos  0 t and so t =

x v0 cos  0

. For the vertical motion, apply Eq. 2–12b.

y = y0 + v y 0t + 12 a y t 2 = ( v0 sin  ) t − 12 gt 2

Substitute the expression for the time of flight and solve for the initial velocity. g ( x )  x  y = ( v0 sin  ) t − gt = v0 sin  − g  = x tan  − 2 2v0 cos 2  0 v0 cos  0  v0 cos  0  1 2

x

1 2

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

v0 =

g ( x )

2 cos 2  0 ( − y + x tan  )

For x = 11.00 m − 0.22 m = 10.78 m, the shortest shot:

( 9.80 m s ) (10.78 m ) v = = 11.078 m s  11.1m s . 2 cos 38.0 ( −0.95 m + (10.78 m ) tan 38.0 ) 2

For x = 11.00 m + 0.22 m = 11.22 m, the longest shot:

( 9.80 m s ) (11.22 m ) v = = 11.274 m s  11.3 m s . 2 cos 38.0 ( −0.95 m + (11.22 m ) tan 38.0 ) 2

86. (a) Take the ground to be the y = 0 level, with upward as the positive direction. Use Eq. 2–12b to solve for the time, with an initial vertical velocity of 0. y = y0 + v0 y t + 12 a y t 2 → 150 m = 910 m + 12 ( −9.80 m s 2 ) t 2 →

2 (150 − 910 )

( −9.80 m s ) 2

= 12.45s  12 s

(b) The horizontal motion is at a constant speed, since air resistance is being ignored. x = v x t = ( 4.0 m s )(12.45s ) = 49.8 m  5.0  101 m

87. First we find the time of flight for the ball. From that time we can calculate the vertical speed of the ball. From that vertical speed we can calculate the total speed of the ball, and the % change in the speed. We chose the downward direction to be positive for vertical motion.  1m s  v x = v0 x = (130 km h )   = 36.11m s  3.6 km h  x = v x t

→

(

x vx

18 m 36.11m s

= 0.498 s

)

v y = v0 y + at = 9.80 m s 2 ( 0.498 s ) = 4.88 m s v=

v x2 + v y2 =

% change =

( 36.11m s ) + ( 4.88 m s ) = 36.44 m s

v − v0 v0

 100 =

36.44 − 36.11 36.11

 100 = 0.91%

88. Choose the origin to be the point from which the projectile is launched, and choose upward as the positive y-direction. The y-displacement of the projectile is 135 m and the horizontal range of the projectile is 195 m. The acceleration in the y-direction is a y = − g , and the time of flight is 6.6 s. The horizontal velocity is found from the horizontal motion at constant velocity. x 195 m x = v x t → v x = = = 29.55 m s t 6.6 s Calculate the initial y-velocity from the given data and Eq. 2–12b. 2 y = y0 + v y 0t + 12 a y t 2 → 135 m = v y 0 ( 6.6 s ) + 12 −9.80 m s 2 ( 6.6 s ) → v y 0 = 52.79 m s

(

)

Thus, the initial velocity and direction of the projectile are as follows. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

v0 = v x2 + v y2 0 =

 = tan −1

vy0 vx

Instructor Solutions Manual

( 29.55 m s ) + ( 52.79 m s ) = 60.4978 m s  6.0  101 m s 2

= tan −1

52.79 m s 29.55 m s

= 61

89. Choose the origin to be the water level directly underneath the diver when she left the board. Choose upward as the positive y-direction. For the diver, y0 = 3.5 m , the final y-position is y = 0.0 m (water level), a y = − g , the time of flight is t = 1.3 s, and the horizontal displacement is

x = 3.0 m. (a) The horizontal velocity is determined from the horizontal motion at constant velocity. x 3.0 m x = v x t → v x = = = 2.308 m s t 1.3s The initial y-velocity is found using Eq. 2–12b.

(

)

y = y0 + v y 0t + 12 a y t 2 → 0 m = 3.5 m + v y 0 (1.3s ) + 12 −9.80 m s 2 (1.3s )

→

v y 0 = 3.678 m s Thus, the velocity in both vector and magnitude / direction format are as follows. 2 2 v = v 2 + v 2 = ( 2.308 m s ) + ( 3.678 m s ) = 4.3 m s v = 2.3ˆi + 3.7 ˆj m s 0

(

 = tan −1

)

vy0 vx

= tan −1

3.678 m s 2.308 m s

= 57 above the horizontal

(b) The maximum height will be reached when v y = 0. Use Eq. 2–12c.

(

)

v y2 = v y2 0 + 2ay → 0 = ( 3.678 m s ) + 2 −9.80 m s 2 ( ymax − 3.5 m ) → 2

ymax = 4.2 m above the water. (c) To find the velocity when she enters the water, the horizontal velocity is the (constant) value of v x = 2.308 m s. The vertical velocity is found from Eq. 2–12a.

(

)

v y = v y 0 + at = 3.678 m s + −9.80 m s 2 (1.3s ) = −9.062 m s. The velocity is as follows. v = 2.3ˆi − 9.1ˆj m s f

(

)

vf = v x2 + v y2 =

 f = tan −1

vfy vfx

( 2.308 m s ) + ( −9.062 m s ) = 9.351m s  9.4 m s

= tan −1

−9.062 m s 2.308 m s

= −76 ( below the horizontal )

90. Find the time of flight from the vertical data, using Eq. 2–12b. Call the floor the y = 0 location, and choose upwards as positive. y = y0 + v0 y t + 12 a y t 2 → 3.05 m = 2.40 m + (12 m s ) sin 35t + 12 −9.80 m s 2 t 2

(

)

4.90t − 6.883t + 0.65 m = 0 → 2

6.883  6.8832 − 4 ( 4.90 )( 0.65 ) 2 ( 4.90 )

= 1.303s, 0.102 s

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

(a) Use the larger time for the time of flight. The shorter time is the time for the ball to rise to the basket height on the way up, while the longer time is the time for the ball to be at the basket height on the way down. x = v x t = v0 ( cos 35 ) t = (12 m s )( cos 35 )(1.303s ) = 12.81m  13 m (b) The angle to the horizontal is determined by the components of the velocity. v x = v0 cos  0 = 12 cos 35 = 9.830 m s v y = v y 0 + at = v0 sin  0 − gt = 12 sin 35 − 9.80 (1.303) = −5.886 m/s

 = tan −1

vy vx

= tan −1

−5.886 9.830

= −30.9  −31 , 31 below the horizontal.

91. (a) Choose downward to be the positive y-direction. The origin is the point where the arrow leaves the bow. In the vertical direction, v y 0 = 0, y0 = 0, and a y = 9.80 m s 2. In the horizontal direction, x = 38.0 m and v x = 21.3 m s. The time of flight is found from the horizontal motion at constant velocity. x = v x t → t = x v x = 38.0 m 21.3 m s = 1.784 s This time can now be used in Eq. 2–12b to find the vertical drop of the arrow. 2 y = y0 + v y 0t + 12 a y t 2 → y = 0 + 0 + 12 ( 9.80 m s 2 ) (1.784 s ) = 15.6 m (b) For the arrow to hit the target at the same level, the level horizontal range formula derived in the text applies. The range is 38.0 m, and the initial velocity is 21.3 m/s. Solving for the angle of launch results in the following. ( 38.0 m ) ( 9.80 m s 2 ) v02 sin 2 0 Rg −1 1 → sin 2 0 = 2 →  0 = 2 sin = 27.6 R= 2 g v0 ( 21.3 m s ) Because of the symmetry of the range formula, there is also an answer of the complement of the above answer, which would be 62.4. The larger angle is less reasonable – an archer rarely points the bow at such a steep angle. 92. (a) Call the upward direction positive for the vertical motion. Then the velocity of the ball relative to a person on the ground is the vector sum of the horizontal and vertical motions. The horizontal velocity is v x = 12.0 m s and the vertical velocity is v y = 3.0 m s . v = 12.0 m s ˆi + 3.0 m s ˆj → v =

 = tan −1

3.0 m s 12.0 m s

(12.0 m s ) + ( 3.0 m s ) = 12.4 m s 2

= 14 above the horizontal .

(b) The only change is the initial vertical velocity, and so v y = −3.0 m s. v = 12.0 m s ˆi − 3.0 m s ˆj → v =

 = tan −1

−3.0 m s 12.0 m s

(12.0 m s ) + ( −3.0 m s ) = 12.4 m s 2

= 14 below the horizontal .

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

93. Let the launch point be the origin of coordinates, with right and upwards as the positive directions. The equation of the line representing the ground is ygnd = − x. The equations representing the motion of the rock are xrock = v0 t and yrock = − 12 gt 2 , which can be combined into yrock = − 12

2 xrock . v02 Find the intersection (the landing point of the rock) by equating the two expressions for y, and so finding where the rock meets the ground. 2v02 g 2 x 2v0 2 (15 m s ) 1 yrock = ygnd → − 2 2 x = − x → x = → t= = = = 3.1s v0 v0 9.80 m s 2 g g

94. Choose the origin to be the point at ground level directly below where the ball was hit. Call upwards the positive y-direction. For the ball, we have v0 = 28 m s,  0 = 64, a y = − g, y0 = 0.9 m, and y = 0.0 m. (a) To find the horizontal displacement of the ball, the horizontal velocity and the time of flight are needed. The (constant) horizontal velocity is given by v x = v0 cos  0. The time of flight is found from Eq. 2–12b. y = y0 + v y 0t + 12 a y t 2 → 0 = y0 + v0 sin  0t − 12 gt 2 →

−v0 sin  0  v02 sin 2  0 − 4 ( − 12 g ) y0 2 ( − 12 g )

− ( 28 m s ) sin 64 

( 28 m s ) sin 2 64 − 4 ( − 12 ) ( 9.80 m s 2 ) ( 0.9 m ) 2 ( − 12 ) ( 9.80 m s 2 ) 2

= 5.171s, − 0.0355s Choose the positive time, since the ball was hit at t = 0. The horizontal displacement of the ball will be found by the constant velocity relationship for horizontal motion.

x = v x t = v0 cos  0t = ( 28 m s )( cos 64 )( 5.171s ) = 63.47 m  63 m

(b) The center fielder catches the ball right at ground level. He ran 105 m – 63.47 m = 41.53 m to catch the ball, so his average running speed would be as follows. d 41.53 m vavg = = = 8.031m s  8.0 m s t 5.171s 95. Choose the origin to be the point at the top of the building from which the ball is shot, and call upwards the positive y-direction. The initial velocity is v0 = 24 m s at an angle of  0 = 42. The acceleration due to gravity is a y = − g. (a) v x = v0 cos  0 = ( 24 m s ) cos 42 = 17.84 m s  18 m s v y 0 = v0 sin  0 = ( 24 m s ) sin 42 = 16.06 m s  16 m s

(b) Since the horizontal velocity is known and the horizontal distance is known, the time of flight can be found from the constant velocity equation for horizontal motion. x 65 m x = v x t → t = = = 3.643s v x 17.84 m s With that time of flight, calculate the vertical position of the ball using Eq. 2–12b. © 2023 Pearson Education, Ltd. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

(

)

y = y0 + v y 0t + 12 a y t 2 = (16.06 m s )( 3.643s ) + 12 −9.80 m s 2 ( 3.643s )

= −6.52 m  −6.5 m So the ball will strike 6.5 m below the top of the building. 96. (a) For the upstream trip, the boat will cover a distance of D 2 with a net speed of v − u , so the time is t1 =

D 2 v −u

2 (v − u )

. For the downstream trip, the boat will cover a distance of D 2

with a net speed of v + u , so the time is t2 = round trip will be t = t1 + t2 =

D 2(v − u )

D 2 v+u

D 2(v + u )

D 2 (v + u )

. Thus the total time for the

(v − u ) 2

(b) For the boat to go directly across the river, it must be angled against the current in such a way that the net velocity is straight across the river, as in the diagram. This equation must be satisfied: v boat rel. = v boat rel. + v water rel. = v + u. shore

Thus vboat rel. =

water

shore

v water rel. = u shore

v boat rel. = v



v boat rel. shore

water

v 2 − u 2 , and the time to go a distance D 2 across

shore

the river is t1 =

D 2 v −u 2

D 2 v2 − u2

. The same relationship would be in effect for crossing

back, so the time to come back is given by t2 = t1 and the total time is t = t1 + t2 =

v − u2 The speed v must be greater than the speed u. The velocity of the boat relative to the shore when going upstream is v − u. If v  u , the boat will not move upstream at all, and so the first part of the trip would be impossible. Also, in part (b), we see that v is longer than u in the triangle, since v is the hypotenuse, and so we must have v  u. 2

97. Since the ball is being caught at the same height from which it was struck, use the level horizontal range formula to find the horizontal distance the ball travels. R=

v02 sin 2 0

( 28 m s ) sin ( 2  55 ) 2

= = 75.175 m g 9.80 m s 2 Then as seen from above, the location of home plate, the point where the ball must be caught, and the initial location of the outfielder are shown in the diagram. The dark arrow shows the direction in which the outfielder must run. The length of that distance is found from the law of cosines as applied to the triangle.

x= =

a 2 + b 2 − 2ab cos  75.1752 + 852 − 2 ( 75.175) ( 85) cos 22 = 32.048 m

The angle  at which the outfielder should run is found from the law of sines.

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

sin 22

sin 

Instructor Solutions Manual

 75.175 sin 22  = 61.49 or 118.51  = sin −1   32.048 m 75.175 m  32.048  2 2 2 Since 75.175  85 + 32.048 , the angle must be acute, so we choose  = 61.49. =

→

Now assume that the outfielder’s time for running is the same as the time of flight of the ball. The time of flight of the ball is found from the horizontal motion of the ball at constant velocity. R 75.175 m R = v x t = v0 cos  0 t → t = = = 4.681s v0 cos  0 ( 28 m s ) cos 55 Thus the average velocity of the outfielder must be vavg =

d t

32.048 m 4.681s

= 6.8 m s at an angle of

61 relative to the outfielder’s line of sight to home plate.

98. The lifeguard will be carried downstream at the same rate as the child. Thus only the horizontal 45 m = motion need be considered. To cover 45 meters horizontally at a rate of 2.0 m/s takes 2m s 22.5s  23s for the lifeguard to reach the child. During this time they would both be moving

downstream at 1.0 m/s, and so would travel (1.0 m s )( 22.5s ) = 22.5m  23m downstream. 99. The acceleration is the derivative of the velocity. dv a= = 4.5 m s2 ˆj dt Since the acceleration is constant, we can use Eq. 3–13b. r = r0 + v 0t + 12 at 2 = 2.5 ˆi − 3.1ˆj + −3.0 ˆi t + 12 4.5 ˆj t 2

( ) ( ) = ( 2.5 − 3.0t ) m ˆi + ( −3.1 + 2.25t ) m ˆj

(

)

The shape is parabolic , with the parabola opening in the y-direction. 100. We choose to initially point the boat downstream at an angle of  relative to straight across the river, because then all horizontal velocity components are in the same direction, and the algebraic signs might be less confusing. If the boat should in reality be pointed upstream, the solution will give a negative angle. We use v BW = 1.60 m s, the speed of the boat relative to the water (the rowing speed); v WS = 0.80 m s, the speed of the water relative to the shore (the current); and v R = 3.00 m s, his running speed. The width of the river is w = 1300 m, and the length traveled along the bank is l. The time spent in the water is tW , and the time running is tR. The actual vector velocity of the boat is v BS = v BW + v WS. That vector addition is illustrated on the diagram (not drawn to scale).

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

The distance straight across the river (w) is the velocity component across the river, times the time in the water. The distance along the bank (l) is the velocity component parallel to the river, times the time in the water. The distance along the bank (as he runs back to the point straight across the river) is also his running speed times the time running. These three distances are expressed below. w = ( v BW cos  ) t W ; l = ( v BW sin  + v WS ) t W ; l = v R tR The total time is t = t W + t R , and needs to be expressed as a function of  . Use the distance relations above to write this function.  ( v sin  + v WS )  l ( v sin  + v WS ) tW = t W + BW = t W 1 + BW t = tW + tR = t W +  vR vR vR  

vBW vR cos 

v + v R

+ vBW sin   =

w vBW vR

( v + v ) sec  + v R

To find the angle corresponding to the minimum time, we set

dt d

= =

dt d

tan  

= 0 and solve for the angle.

d 

 w ( vR + v WS ) sec  + vBW tan    d  v BW v R  w

( vR + v WS ) tan  sec  + v BW sec 2   = 0 →

v BW v R 

( v + v ) tan  + v R

sec   sec  = 0 → sec  = 0, sin  = −

v BW v R + v WS

The first answer is impossible, and so we must use the second solution. v BW 1.60 m s sin  = − =− = −0.421 →  = sin −1 ( −0.421) = −24.9 3.00 m s + 0.80 m s v R + v WS To know that this is really a minimum and not a maximum, some argument must be made. The maximum time would be infinity, if he pointed his boat either directly upstream or downstream. Thus, this angle should give a minimum. A second derivative test could be done, but that would be algebraically challenging. A graph of t vs.  could also be examined to see that the angle is a minimum. Here is a portion of such a graph, showing a minimum time of somewhat more than 900 seconds near  = −25. It is also interesting to note from the equations above that the angle for the minimum time is determined only by the velocities, not the actual distance across the river. The time he takes in getting to the final location can be calculated from the angle. 1300 m w tW = = = 895.77 s v BW cos  (1.60 m s ) cos ( −24.9 )

l = ( v BW sin  + v WS ) t W = (1.60 m s ) sin ( −24.9 ) + 0.80 m s  ( 895.77 s ) = 113.12 m tR =

l vR

113.12 m 3.00 m s

= 37.71s

t = t W + tR = 895.77 s + 37.71s = 933.48s  9.33  102 s

Physics for Scientists & Engineers with Modern Physics, 5e, Global Edition

Instructor Solutions Manual

Thus he must point the boat 24.9 upstream, taking 895.8 seconds to cross, and landing 113 m from the point directly across from his starting point. Then he runs the 113 m from his landing point to the point directly across from his starting point, in 37.7 seconds, for a total elapsed time of 933 seconds (about 15.6 minutes). 101. We have vcar rel. = 12 m s . Use the diagram, showing the snow falling ground

straight down relative to the ground and the car moving parallel to the ground, and illustrating v snow rel. = v snow rel. + v car rel. , to calculate the ground

car

ground

v snow rel. ground

v car rel.

→ vsnow rel. = (12 m s ) tan 9.0 = 1.901 m s  1.9 m s

ground

tan 9.0 =

car

ground

other speeds.

vsnow rel.

v snow rel.

9.0

vcar rel.

ground

102. First, we find the direction of the straight-line path that the boat must take to pass 150 m to the east of the buoy. See the first diagram (not to scale). We find the net displacement of the boat in the horizontal and vertical directions, and then calculate the angle. x = ( 2800 m ) sin 22.5 + 150 m = 1221.51m y = ( 2800m ) cos 22.5 = 2586.86 m

 = tan −1

y

= tan −1

buoy 150 m

2800 m

boat path

2586.86 m

= 64.7234 x 1221.51m This angle gives the direction that the boat must actually travel, so it is the direction of the velocity of the boat with respect to the shore, v boat rel.. So

(

22.5o



shore

)

v boat rel. = vboat rel. cos  ˆi + sin  ˆj . Then, using the second diagram (also not shore

shore

to scale), we can write the relative velocity equation relating the boat’s travel and the current. The relative velocity equation gives us the following. v boat rel. = v boat rel. + v water rel. → shore

water

(

shore

)

(

v water rel. shore

v boat rel.

)

vboat rel. cos  î + sin  ˆj = 2.1m s cos  î + sin  ˆj + 0.20 m s î → shore

vboat rel. cos  = ( 2.1m s ) cos  + 0.20 m s ; vboat rel. sin  = ( 2.1m s ) sin  shore

v boat rel.

water

shore





shore

These two component equations can then be solved for v boat rel. and  . One technique is to isolate the shore

terms with  in each equation, and then square those equations and add them. That gives a quadratic equation for vboat rel. . We will represent v boat rel. by “x” and drop the units in order to make shore

shore

the equations simpler to write. Once we have a value for v boat rel., we can solve for  . shore

Chapter 3

Kinematics in Two or Three Dimensions; Vectors

cos  =

x cos  − 0.20 2.1

; sin  =

x sin  2.1

→

 x cos  − 0.20   x sin   cos  + sin  =   +  =1 2.1    2.1  2

x 2 cos 2  − 2 ( 0.20 ) x cos  + ( 0.20 ) + x 2 sin 2  = ( 2.1) 2

x 2 − ( 0.40 cos  ) x − 4.37 = x 2 − 0.17079 x − 4.37 = 0 x=

0.17080 

( −0.17080 )2 − 4 (1)( −4.37 )

0.17080  4.1844

= 2.1776 m s , − 2.0068 m s 2 2 We must have v boat rel. > 0 since it is the magnitude of a vector, so vboat rel. = 2.1776 m s. shore

sin  =

x sin 

shore

( 2.1776 m s ) sin 64.723

= 0.93768 →  = sin −1 0.93768 = 69.7 N of E

2.1m s 2.1m s The final answer was very sensitive to the number of significant digits carried in the problem, so we kept more than usual.

CHAPTER 4: Dynamics: Newton’s Laws of Motion Responses to Questions 1.

When you give the wagon a sharp pull forward, the force of friction between the wagon and the child acts on the child to move her forward. But the force of friction acts at the contact point between the child and the wagon – we assume the child is sitting in the wagon. The lower part of the child begins to move forward, while the upper part, following Newton’s first law (the law of inertia), remains almost stationary, making it seem as if the child falls backward. The “backwards” motion is relative to the wagon, not to the ground.

(a) Francesca, standing on the ground beside the truck, will see the box remain motionless while the truck accelerates out from under it. Since there is no friction, there is no net horizontal force on the box and so the box will not gain any speed. Thus Francesca would describe the motion of the box in terms of Newton’s first law – there is no force on the box, so it does not accelerate. (b) Phil, riding on the truck, will see the box appear to accelerate backwards with respect to his frame of reference, which is not inertial. He might even say something about the box being “thrown” backwards from the truck, and try to invoke Newton’s second law to explain the motion of the box. But the source of the force would be impossible to specify. (Phil better hold on to the truck, though; if the truck bed is frictionless, he too will slide off if he is just standing!)

Yes, the net force can be zero on a moving object. If the net force is zero, then the object’s acceleration is zero, but its velocity is not necessarily zero. Instead of classifying objects as “moving” and “not moving,” Newtonian dynamics classifies them as “accelerating” and “not accelerating.” Both zero velocity and constant velocity fall in the “not accelerating” category.

If the acceleration of an object is zero, the vector sum of the forces acting on the object is zero (Newton’s second law), so there can be forces on an object that has no acceleration. For example, a book resting on a table is acted on by gravity and the normal force from the table, but it has zero acceleration, because the vector sum of the two forces on it is 0.

If only one force acts on an object, the net force cannot be zero, and so the object cannot have zero acceleration, by Newton’s second law. It is possible for the object to have zero velocity, but only for an instant. For example (if we neglect air resistance), a ball thrown up into the air has only the force of gravity acting on it. Its speed will decrease while it travels upward, stops, and then falls back to the ground. At the instant the ball is at its highest point, its velocity is zero. However, the ball has a non-zero net force and a non-zero acceleration throughout its flight, including at the top, where its velocity is zero.

(a) A force is needed to bounce the ball back up, because the ball changes direction, and so the ball accelerates. If the ball accelerates, there must be a force. (b) The pavement exerts the force on the golf ball.

As you take a step on the log, your foot exerts a force on the log in the direction opposite to the direction in which you want to move, which pushes the log “backwards.” (The log exerts an equal and opposite force forward on you, by Newton’s third law.) If the log had been on the ground, friction between the ground and the log could have kept the log from moving. However, the log is floating in water, which offers little resistance to the movement of the log as you push it backwards while walking forward.