Solutions Manual for Physics for Scientists and Engineers with Modern Physics, 10 Edition. Raymond S

Physics for Scientists and Engineers with Modern Physics, 10e Raymond Serway, John Jewett (Solutions Manual All Chapters, 100% original verified, A+ Grade)

All Chapters Download Link At The End Of This File

1 Physics and Measurement CHAPTER OUTLINE 1.1

Standards of Length, Mass, and Time

1.2

Modeling and Alternative Representations

1.3

Dimensional Analysis

1.4

Conversion of Units

1.5

Estimates and Order-of-Magnitude Calculations

1.6

Significant Figures

* An asterisk indicates a question or problem new to this edition.

SOLUTIONS TO THINK-PAIR-SHARE AND ACTIVITIES TP1.1

(a) The fourth experimental point from the top is a circle: this point lies just above the best-fit curve that passes through the point (400 cm2, 0.20 g). The interval between horizontal grid lines is 1 space = 0.05 g.

An estimate from the graph shows that the circle has a vertical separation of 0.3 spaces = 0.015 g above the best-fit curve.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2

Physics and Measurement (b) The best-fit curve passes through 0.20 g, so the percentage difference is

⎛ 0.015 g ⎞ ⎜⎝ 0.20 g ⎟⎠ × 100 = 7.5% (c) The best-fit curve passes through the origin and the point (600 cm2, 0.32 g). Therefore, the slope of the best-fit curve is

slope =

0.32 g − 0 = 5.3 × 10 −4 g/ cm 2 = 5.3 g/ m 2 2 600 cm − 0

(d) For shapes cut from this copy paper, the mass of the cutout is proportional to its area: m = aA. The proportionality constant a is 5.3 g/m2. (e) This result is to be expected if the paper has thickness and density that are uniform within the experimental uncertainty. (f) The slope is the areal density of the paper, its mass per unit area.]

Answers: Answers will vary, depending on how the graph is read *TP1.2

Answer: All results should be close to 2.54, representing the conversion factor 2.54 cm/in.

*TP1.3

Solution: The difference is due to the average density, which is related to the composition of the penny. Before 1982, U.S. pennies were 95% copper and 5% zinc. After that date, they are 97.5% zinc, with a coating of 2.5% copper. Both copper and zinc pennies were produced in 1982.

Perhaps a measurement of the mass of a sample of 1982 pennies would be interesting.

Answer: Pre-1982 pennies have an average mass of 3.1 g, while post1982 pennies have a lower average mass of 2.5 g

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1

3

SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 1.1 P1.1

(a)

Standards of Length, Mass, and Time Modeling the Earth as a sphere, we find its volume as

Its density is then

(b)

This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2000 to 3000 kg/m3. The average density of the Earth is significantly higher, so higher-density material must be down below the surface.

P1.2

(a)

where d is the diameter.

Then

(b)

P1.3

For sphere the volume is

and the mass is

We divide this equation for the larger sphere by the same equation for the smaller:

Then

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4

Physics and Measurement

P1.4

The volume of a spherical shell can be calculated from

From the definition of density,

*P1.5

, so

Let us find the angle subtended by the width of the Great Wall at the height of the spacecraft orbit. From the description of a subtended angle in the problem statement, we obtain

θ=

width of Great Wall 7m = = 3.5 × 10 −5 rad distance to Great Wall 200 000 m

The angle subtended by the width of the Great Wall at a height of 200 km is 3.5 × 10–5 rad, which is smaller than the normal visual acuity of the eye by about a factor of ten. Therefore, despite its great length, its width cannot be seen. In the same way, a single human hair cannot be seen from several meters away, despite its length. Your argument should be based on this calculation.]

Answer: The angle subtended by the Great Wall is less than the visual acuity of the eye.

Section 1.2 P 1.6

Matter and Model Building

Figure P1.6 suggests a right triangle where, relative to angle θ, its adjacent side has length d and its opposite side is equal to width of the river, y; thus,

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1

5

y = (100 m)tan(35.0°) = 70.0 m The width of the river is P1.7

.

From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane.

This diagonal distance may be obtained from the Pythagorean theorem,

Thus, since the atoms are separated by a

distance L = 0.200 nm, the diagonal planes are separated by

Section 1.3 P1.8

Dimensional Analysis

The term x has dimensions of L, a has dimensions of

and t has

has dimensions of

dimensions of T. Therefore, the equation

The powers of L and T must be the same on each side of the equation.

Therefore, and Likewise, equating terms in T, we see that n – 2m must equal 0. Thus, . The value of k, a dimensionless constant, .

P1.9

(a)

Write out dimensions for each quantity in the equation vf = vi + ax

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

6

Physics and Measurement The variables vf and vi are expressed in units of m/s, so [vf] = [vi] = LT –1 The variable a is expressed in units of m/s2;

[a] = LT –2

The variable x is expressed in meters. Therefore, [ax] = L2 T –2 Consider the right-hand member (RHM) of equation (a): [RHM] = LT –1+L2 T –2 Quantities to be added must have the same dimensions.

Therefore, (b)

Write out dimensions for each quantity in the equation y = (2 m) cos (kx) For y,

[y] = L

for 2 m,

[2 m] = L

and for (kx), Therefore we can think of the quantity kx as an angle in radians, and we can take its cosine. The cosine itself will be a pure number with no dimensions. For the left-hand member (LHM) and the right-hand member (RHM) of the equation we have

These are the same, so P1.10

Summed terms must have the same dimensions. (a)

[X] = [At3] + [Bt]

(b)

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1

Section 1.4 P1.11

7

Conversion of Units

From Table 14.1, the density of lead is 1.13 × 104 kg/m3, so we should expect our calculated value to be close to this value. The density of water is 1.00 × 103 kg/m3, so we see that lead is about 11 times denser than water, which agrees with our experience that lead sinks.

Density is defined as

We must convert to SI units in the

calculation.

Observe how we set up the unit conversion fractions to divide out the units of grams and cubic centimeters, and to make the answer come out in kilograms per cubic meter. At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference from the tabulated values is possibly due to measurement uncertainty and does not indicate a discrepancy.

P1.12

The area of the four walls is (3.6 + 3.8 + 3.6 + 3.8) m × (2.5 m) = 37 m2.

Each sheet in the book has area (0.21 m)(0.28 m) = 0.059 m2. The number of sheets required for wallpaper is 37 m2/0.059 m2 = 629 sheets = 629 sheets (2 pages/1 sheet) = 1260 pages.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

8

Physics and Measurement

P1.13

The aluminum sphere must be larger in volume to compensate for its lower density. We require equal masses:

then use the volume of a sphere. By substitution,

Now solving for the unknown,

Taking the cube root, The aluminum sphere is 43% larger than the iron one in radius, diameter, and circumference. Volume is proportional to the cube of the linear dimension, so this excess in linear size gives it the (1.43)(1.43)(1.43) = 2.92 times larger volume it needs for equal mass.

P1.14

The mass of each sphere is and

Setting these masses equal,

The resulting expression shows that the radius of the aluminum sphere is directly proportional to the radius of the balancing iron sphere. The aluminum sphere is 43% larger than the iron one in radius, diameter, and circumference. Volume is proportional to the cube of the linear

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1

9

dimension, so this excess in linear size gives it the (1.43)3 = 2.92 times larger volume it needs for equal mass.

P1.15

We assume the paint keeps the same volume in the can and on the wall, and model the film on the wall as a rectangular solid, with its volume given by its “footprint” area, which is the area of the wall, multiplied by its thickness t perpendicular to this area and assumed to be uniform. Then,

The thickness of 1.5 tenths of a millimeter is comparable to the thickness of a sheet of paper, so this answer is reasonable. The film is many molecules thick.

P1.16

(a)

To obtain the volume, we multiply the length, width, and height of the room, and use the conversion 1 m = 3.281 ft.

(b)

The mass of the air is

The student must look up the definition of weight in the index to find

where the unit of N of force (weight) is newtons.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10

Physics and Measurement Converting newtons to pounds,

Section 1.5 P1.17

(a)

Estimates and Order-of-Magnitude Calculations We estimate the mass of the water in the bathtub. Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then V = (0.5)(1.3)(0.5)(0.3) = 0.10 m3 The mass of this volume of water is

(b)

Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is

P1.18

Don’t reach for the telephone book or do a Google search! Think. Each full-time piano tuner must keep busy enough to earn a living. Assume a total population of 107 people. Also, let us estimate that one person in one hundred owns a piano. Assume that in one year a single piano tuner can service about 1 000 pianos (about 4 per day for 250 weekdays), and that each piano is tuned once per year.

Therefore, the number of tuners

If you did reach for an Internet directory, you would have to count.

Instead, have faith in your estimate. Fermi’s own ability in making an order-of-magnitude estimate is exemplified by his measurement of the © 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1

11

energy output of the first nuclear bomb (the Trinity test at Alamogordo, New Mexico) by observing the fall of bits of paper as the blast wave swept past his station, 14 km away from ground zero. *P1.19 Solution Imagine movies, television shows, or theme park rides where you may have seen a spacecraft traveling through a crowded asteroid field. In reality, in such a crowded field, the asteroids would be colliding and grinding themselves down into small particles.

We will assume that the distribution of the asteroids is uniform, which is not necessarily true but will allow us to make an estimate. With this assumption, the problem can be solved with a geometric model of a doughnut.

Let us first set up the geometric model of the doughnut-shaped asteroid belt. The diagram below shows this model.

The dashed line is at the average position of the inner and outer radii and represents the path through the centers of all circular cross sections of the doughnut, which have radius r. The volume of the thin slice of the doughnut shown at the top of the figure is the product of © 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

12

Physics and Measurement the area of the circular cross section and the average width of the slice, which we take to be the width at the dashed line. Therefore, (1)

Integrate Equation (1) around the doughnut:

From the geometry, we see that (3)

Substituting Equation (3) into Equation (2),

Divide the number N of asteroids in this region by this volume to find the number of asteroids per unit volume:

Substitute numerical values:

Convert to metric units:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1

13

Taking the reciprocal, we find that, on average, the volume associated with one asteroid is

Taking a cube root of this result, we see that an average asteroid occupies a volume equivalent to a cube with side length 4.01 × 108 m, or 401 000 km, which is more than 30 times the diameter of the earth.

This is an enormous volume compared to any conceivable spacecraft.

There is very little chance that you would be near an asteroid of radius 100 m or more, not to mention fighting your way through a crowded field of them. Despite the fact that there is a large number of asteroids, they are distributed through a tremendous volume of space.]

Answer: The average distance between asteroids in the asteroid belt is about 400 000 km.

Section 1.6 P1.20

(a)

Significant Figures The ± 0.2 following the 78.9 expresses uncertainty in the last digit.

Therefore, there are

(b)

significant figures in 78.9 ± 0.2.

Scientific notation is often used to remove the ambiguity of the number of significant figures in a number. Therefore, all the digits in 3.788 are significant, and 3.788 × 109 has

significant

figures. (c)

Similarly, 2.46 has three significant figures, therefore 2.46 × 10–6 has

significant figures.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

14

Physics and Measurement (d) Zeros used to position the decimal point are not significant.

Therefore 0.005 3 has

significant figures.

Uncertainty in a measurement can be the result of a number of factors, including the skill of the person doing the measurements, the precision and the quality of the instrument used, and the number of measurements made.

P1.21

We work to nine significant digits:

P1.22

We are given the ratio of the masses and radii of the planets Uranus and Neptune:

The definition of density is

for a

sphere, and we assume the planets have a spherical shape.

We know

Compare densities:

which gives

P1.23

Let o represent the number of ordinary cars and s the number of sport utility vehicles. We know o = s + 0.947s = 1.947s, and o = s + 18.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1

15

We eliminate o by substitution:

P1.24

We require

ANS. FIG. P1.24 For tan–1(–3) = arctan(–3), your calculator may return –71.6°, but this angle is not between 0° and 360° as the problem requires. The tangent function is negative in the second quadrant (between 90° and 180°) and in the fourth quadrant (from 270° to 360°). The solutions to the equation are then

P1.25

Let s represent the number of sparrows and m the number of more interesting birds. We know s/m = 2.25 and s + m = 91.

We eliminate m by substitution:

P1.26

For those who are not familiar with solving equations numerically, we provide a detailed solution. It goes beyond proving that the suggested answer works.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Get complete Order files download link below htps://www.mediafire.com/file/rus03sp9cw3psp7/SM +Physics+for+Scien�sts+and+Engineers+with+Modern+ Physics,+10e+Raymond+Serway,+John+Jewet.zip/file

If The link does not work with a click, then copy the complete Download link in internet explorer/firefox/google chrome and get all files download successfully.

16

Physics and Measurement The equation 2x4 – 3x3 + 5x – 70 = 0 is quartic, so we do not attempt to solve it with algebra. To find how many real solutions the equation has and to estimate them, we graph the expression:

ANS. FIG. P1.26 x

–3

–2

–1

0

1

2

3

4

y = 2x4 – 3x3 + 5x – 70

158

–24

–70

–70

–66

–52

26

270

We see that the equation y = 0 has two roots, one around x = –2.2 and the other near x = +2.7. To home in on the first of these solutions we compute in sequence: When x = –2.2, y = –2.20. The root must be between x = –2.2 and x = –3.

When x = –2.3, y = 11.0. The root is between x = –2.2 and x = –2.3. When x = –2.23, y = 1.58. The root is between x = –2.20 and x = –2.23. When x = –2.22, y = 0.301. The root is between x = –2.20 and –2.22. When x = – 2.215, y = –0.331. The root is between x = –2.215 and –2.22. We could next try x = –2.218, but we already know to three-digit precision that the root is x = –2.22.

P1.27

Use substitution to solve simultaneous equations. We substitute p = 3q into each of the other two equations to eliminate p:

These simplify to © 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1

17

We substitute the upper relation into the lower equation to eliminate s:

We now have the ratio of t to P1.28

First, solve the given equation for

(a)

Making d three times larger with d2 in the bottom of the fraction makes

(b) (c) (d)

From the last version of the equation, the slope is Note that this quantity is constant as both ∆t and d vary.

Additional Problems P1.29

It is desired to find the distance x such that

(i.e., such that x is the same multiple of 100 m as the multiple that 1 000 m is of x). Thus, it is seen that x2 = (100 m)(1 000 m) = 1.00 × 105 m2 and therefore

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

18 P1.30

Physics and Measurement (a)

A Google search yields the following dimensions of the intestinal tract: small intestines: length ≅ 20 ft ≅ 6 m, diameter ≅ 1.5 in ≅ 4 cm large intestines: length ≅ 5 ft ≅ 1.5 m, diameter ≅ 2.5 in ≅ 6 cm Treat the intestines as two cylinders: the volume of a cylinder of diameter d and length L is The volume of the intestinal tract is

Assuming 1% of this volume is occupied by bacteria, the volume of bacteria is

Treating a bacterium as a cube of side L = 10–6 m, the volume of one bacterium is about L3 = 10–18 m3. The number of bacteria in the intestinal tract is about

(b)

The large number of bacteria suggests they must be otherwise the body would have developed methods a long time ago to reduce their number. It is well known that certain types of bacteria in the intestinal tract are beneficial: they aid digestion, as well as prevent dangerous bacteria from flourishing in the intestines.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1 P1.31

19

The volume of the galaxy is

If the distance between stars is 4 × 1016, then there is one star in a volume on the order of

The number of stars is about P1.32

Assume the winner counts one dollar per second, and the winner tries to maintain the count without stopping. The time interval required for the task would be

P1.33

Answers may vary depending on assumptions: typical length of bacterium: L = 10–6 m typical volume of bacterium: L3 = 10–18 m3 surface area of Earth: (a)

If we assume the bacteria are found to a depth d = 1000 m below Earth’s surface, the volume of Earth containing bacteria is about

If we assume an average of 1000 bacteria in every 1 mm3 of volume, then the number of bacteria is

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

20

Physics and Measurement (b)

Assuming a bacterium is basically composed of water, the total mass is

P1.34

(a)

The mass is equal to the mass of a sphere of radius 2.6 cm and density 4.7 g/cm3, minus the mass of a sphere of radius a and density 4.7 g/cm3, plus the mass of a sphere of radius a and density 1.23 g/cm3.

(b)

The mass is maximum for

.

(c) (d)

This is the mass of the uniform sphere we considered in the first term of the calculation.

(e) P1.35

The rate of volume increase is

(a) (b)

The rate of increase of the balloon’s radius is

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1

21

(c)

P1.36

The table below shows α in degrees, α in radians, tan(α), and sin(α) for angles from 15.0° to 31.1°: difference between

α′ (deg)

α (rad)

tan(α)

sin(α)

α and tan α

15.0

0.262

0.268

0.259

2.30%

20.0

0.349

0.364

0.342

4.09%

30.0

0.524

0.577

0.500

9.32%

33.0

0.576

0.649

0.545

11.3%

31.0

0.541

0.601

0.515

9.95%

31.1

0.543

0.603

0.516

10.02%

We see that α in radians, tan(α), and sin(α) start out together from zero and diverge only slightly in value for small angles. Thus

is the

largest angle for which P1.37

We write “millions of cubic feet” as 106 ft3, and use the given units of time and volume to assign units to the equation.

To convert the units to seconds, use

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

22

Physics and Measurement to obtain

or

where V is in cubic feet and t is in seconds. The coefficient of the first term is the volume rate of flow of gas at the beginning of the month.

The second term’s coefficient is related to how much the rate of flow increases every second.

P1.38

(a) and (b), the two triangles are shown.

ANS. FIG. P1.38(a) (c)

ANS. FIG. P1.38(b)

From the triangles,

and (d)

Equating the two expressions for y, we solve to find

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1

23

Challenge Problems P1.39

The geometry of the problem is shown below.

ANS. FIG. P1.39 From the triangles in ANS. FIG. P1.39,

and

Equate these two expressions for y and solve for x:

Take the expression for x and substitute it into either expression for y:

ANSWERS TO QUICK_QUIZZES 1. (a) 2. False 3. (b)

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

24

Physics and Measurement 4.

ANSWERS TO EVEN-NUMBERED PROBLEMS P1.2

(a) 2.3 × 1017 kg/m3; (b) 1.0 × 1013 times the density of osmium

P1.4 P1.6

70.0m

P1.8

The value of k, a dimensionless constant cannot be obtained by dimensional analysis

P1.10

(a) [A] = L/T3 and [B] = L/T; (b) L/T

P1.12

The number of pages in Volume 1 is sufficient

P1.14

rFe(1.43)

P1.16

(a) 3.39 × 105 ft3; (b) 2.54 × 104 lb

P1.18

100 tuners

P1.20

(a) 3; (b) 4; (c) 3; (d) 2

P1.22

(a)

P1.24

288°; 108°

P1.26

See P1.26 for complete description.

P1.28

(a) nine times smaller; (b) Δt is inversely proportional to the square of

(b)

(c) 1.03 h

d; (c) Plot Δt on the vertical axis and 1/d2 on the horizontal axis; (d) P1.30

(a) 1014 bacteria; (b) beneficial

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Chapter 1 P1.32

25

The scenario has the contestants succeeding on the whole. But the calculation shows that is impossible. It just takes too long!

P1.34

(a) m = 346 g − (14.5 g/cm3)a3; (b) a = 0; (c) 346 g; (d) yes; (e) no change

P1.36

31.0°

P1.38

(a-b) see ANS. FIG. P1.38 (a) and P1.38 (b); (c) y = x tan12.0° and y = (x − 1.00 km) tan14.0°; (d) y = 1.44 km

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2 Motion in One Dimension CHAPTER OUTLINE 2.1

Position, Velocity, and Speed of a Particle

2.2

Instantaneous Velocity and Speed

2.3

Analysis Model: Particle Under Constant Velocity

2.4

The Analysis Model Approach to Problem Solving

2.5

Acceleration

2.6

Motion Diagrams

2.7

Analysis Model: Particle Under Constant Acceleration

2.8

Freely Falling Objects

2.9

Kinematic Equations Derived from Calculus

* An asterisk indicates a question or problem new to this edition.

SOLUTIONS TO THINK-PAIR-SHARE AND ACTIVITIES *TP2.1

Conceptualize The photo in Figure TP2.1 helps you to visualize the situation.

In the first demonstrations, the object, which appears as a small silver-colored slider at the left end of the red lever in the photograph, rises up and just touches the bell at the top of the column. We can approximate this situation by saying that the slider just reaches zero velocity at the position of the bell.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

In the final demonstration, the hitting of the hammer on the apparatus has caused the bell to shift to the side, so that the slider can fly vertically off the track when it is projected with a higher initial velocity. It looks like we have only two pieces of information: the height h of the object in the first situation and the fact that the initial velocity of the object in the second situation is twice that in the first. But we have two more pieces of information that can be gleaned from the description: in the first situation, the velocity of the object goes to zero just as its vertical position reaches height h; in the second situation, the initial and final vertical positions are the same.

Categorize The slider is an object in free fall, so we use the particle under constant acceleration model for its motion.

Analyze For the first situation, write an equation from the model relating velocity and position, using subscripts 1 to describe this situation. Evaluate the final values in the equation for the instant at which the object just touches the bell and momentarily comes to rest:

(

v 2f 1 = vi12 − 2g y f 1 − yi1

) → 0 = v − 2g ( h − 0) → v = 2gh (1) 2 i1

i1

For the second situation, write an equation from the model relating velocity and time, using subscripts 2. Solve for the time:

v f 2 = vi 2 − gt → t =

vi 2 − v f 2 g

(2)

Now evaluate the time at which the object reaches its highest position at the top of its motion:

ttop =

vi 2 − 0 vi 2 = g g

(3)

Recognize that the initial velocity in the second situation is twice that of the first, and combine Equations (1) and (2):

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

ttop =

2vi1 2 = g

( 2gh ) = 2 2h g

g

Finally, recognize by symmetry that the time at which the object lands back on the ground is twice that at which it is at the top, and substitute numerical values:

tground = 2ttop = 4

2 ( 4.50 m ) 2h =4 = 3.83 s g 9.80 m/ s 2

Finalize Does this time interval for the flight of the object seem reasonable?

Discuss with your partners the following. Is this time just twice the time for the object to fall back to the lever in the first situation? In the second situation, does the object rise to a maximum height of just twice that in the first situation? Does the fact that the object begins its upward motion a few centimeters above the ground make any significant difference in our calculation of the final time in the second situation?

Answer: 3.83 s *TP2.2

Conceptualize Be sure you understand the setup of the problem. One person drops a rock from rest. The second person waits for a time interval and then throws the second rock with just the right speed so that it catches up to the first rock just as the two rocks enter the water.

Categorize The rocks are in free fall, so we use the particle under constant acceleration model for their motion.

Analyze Let us first look at the fall of the first rock through the vertical distance h. Using Equation 2.16, rewritten for the vertical direction, we find

y f = yi + v yit + 12 ayt 2

→

− h = 0 + 0 − 12 gt 2

→ t=

2h g

(1)

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

This is the time at which the dropped rock strikes the water. Now consider the fall of the second rock. The thrower waits for a time interval ∆t after the first rock is dropped before throwing his rock downward. Let us denote time t’ as the time at which the second rock reaches the water, with t’ = 0 at the instant the second rock is thrown downward. Using Equation 2.16, rewritten for the vertical direction, we find

y f = yi + v yit ′ + 12 ay ( t ′ )

2

h − 12 g ( t ′ ) v yi = t′

2

→

− h = 0 − v yit ′ − 12 g ( t ′ )

→

2

(2)

In order for there to be a single splash, the time t at which the first rock strikes the water must be the same as the sum of the waiting time ∆t and the time t’ for the second rock to fall:

t = Δt + t ′ → t ′ = t − Δt

(3)

Substitute Equation (3) into Equation (2) and then substitute Equation (1) into Equation (2):

h − 12 g ( t − Δt ) v yi = = t − Δt 2

⎛ 2h ⎞ h − 12 g ⎜ − Δt ⎟ ⎝ g ⎠ 2h − Δt g

2

=

Δt 2gh − 12 g ( Δt )

2

2h − Δt g

(4)

(a) Substitute numerical values:

(1.00 s) 2 ( 9.80 m/ s ) ( 75.0 m ) − 12 ( 9.80 m/ s ) (1.00 s) v = = 11.5 m/ s 2 ( 75.0 m ) − 1.00 s 2

2

2

yi

( 9.80 m/ s ) 2

(b) One way to solve this problem is to solve Equation (4) for ∆t. This leads to a quadratic equation whose appropriate solution is

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Δt =

v yi + 2gh − v yi2 + 2gh

(5)

g

Substitute numerical values:

Δt =

(

)

40.0 m/ s + 2 9.80 m/ s 2 ( 75.0 m ) −

( 40.0 m/ s) + 2 ( 9.80 m/ s ) ( 75.0 m ) 2

2

9.80 m/ s 2

= 2.34 s Another way to solve the problem is to set up a spreadsheet using Equation (4) and generate a list of velocities vyi for increasing values of ∆t. You can then look in the list for the speed for 40.0 m/s and read the approximate value of ∆t. Then the spreadsheet can be modified by looking at just the short range of ∆t that gives a speed of about 40.0 m/s and using a smaller increment for ∆t.

By narrowing in using this technique, you find again that the time interval is 2.34 s. (c) The second rock must be thrown before the first rock hits the water.

Therefore, the longest time interval is given by Equation (1):

Δtmax = t =

2h g

Substitute numerical values:

Δtmax =

2 ( 75.0 m ) = 3.91 s 9.80 m/ s 2

Finalize If the second person waited to throw the second rock for the time interval in part (c), it would have to be thrown with infinite speed. (We will find in Chapter 38 that this is not possible.) If you have set up the spreadsheet for part (b), you will see the values for the speed vyi become very large as you approach 3.91 s and then become negative for larger values of the time interval.

Answers: (a) 11.5 m/s (b) 2.34 s (c) 3.91 s

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

*TP2.3

Conceptualize Think about the falling ruler. When it begins to fall, your eyes see the motion and send a signal to your brain. Your brain has to process the information and send a signal to your hand to grab the ruler. The muscles in your hand then have to respond to this signal and close your fingers onto the ruler. The time intervals for all of these things to happen are added to equal the reaction time.

Categorize The ruler falls with an acceleration due to gravity, so we use the particle under constant acceleration model for its motion.

Analyze There are no numbers given in the problem, so let’s make one up.

Suppose your fingers close on the ruler at the 8.0-inch mark. Then, using Equation 2.16 in the vertical direction,

y f = yi + v yit + 12 ayt 2

→

− h = 0 + 0 − 12 gt 2

→ t=

2h g

Substitute the value of h:

t=

2 ( 8.0 in ) ⎛ 0.025 4 m ⎞ = 0.20 s 9.80 m/ s 2 ⎜⎝ 1 in ⎟⎠

Finalize This value is a typical human reaction time.

Answer: Answers will vary depending on your individual reaction time. A typical human reaction time is about 0.2 s.

*TP2.4

Conceptualize From the graphical representation, generate a mental representation by running the motion in your mind. (a) From –50 s to 0, the Acela is cruising at a constant positive velocity in the +x direction. Between 0 and 50 s, the constant velocity is maintained for a while and then the train begins to speed up. Between 50s and 100s, the train speeds up; notice that the graph is linear in this segment. Between 100 s and 200 s, the train reaches its

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

maximum speed of about 175 mi/h. Between 200 s and 300 s, the engineer applies the brakes: the train is slowing down. The train is moving very slowly between 300 s and 350 s and then stops at 350 s. After 350 s, the train reverses direction: the velocity is negative. The train is backing up faster and faster until the data ends at 400 s.

Categorize Much of the motion of the train does not fit into a simple analysis model. However, between 50 s and 100 s, the train is well modeled as a particle under constant acceleration, since the v–t graph is linear.

Analyze (b) Draw a best-fit line between 50 s and 100 s on the graph, taking advantage of the full vertical extent of the graph:

Now, find the slope of the green line, using its topmost and bottommost points, reading the coordinates from the graph:

a = slope =

Δv 200 mi/ h − ( −100 mi/ h ) = = 1.9 mi/ h · s 125 s − ( −30 s ) Δt

This result can be converted to metric units:

⎛ 1 609 m ⎞ ⎛ 1 h ⎞ a = 1.9 mi/ h · s ⎜ = 0.87 m/ s2 ⎜ ⎟ ⎟ ⎝ 1 mi ⎠ ⎝ 3 600 s ⎠

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(c) To find the displacement, we need to use the notion in Section 2.9 that the displacement of a particle is the area under the v–t curve. Because we do not have a functional description of the velocity of the train, we will need to approximate the displacement. Let’s approximate the area under the curve between 0 and 200 s with the following triangle (yellow) and two rectangles (green):

Find the area of each shape: Lower rectangle:

Δx1 = ( 40 mi/ h ) ( 200 s) = 8 000 mi · s/ h Upper rectangle:

Δx2 = (170 mi/ h − 40 mi/ h ) (100 s) = 13 000 mi · s/ h Triangle:

Δx3 = 12 (110 mi/ h ) ( 55 s) = 3 025 mi · s/ h Therefore, the total displacement is

Δxtotal = Δx1 + Δx2 + Δx3 = 8 000 mi · s/ h + 13 000 mi · s/ h + 3 025 mi · s/ h = 2.4 × 10 4 mi · s/ h

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

This is not a very useful unit! So let’s reconcile the hours and seconds:

⎛ 1h ⎞ Δxtotal = 2.4 × 10 4 mi · s/ h ⎜ = 6.7 mi ⎝ 3 600 s ⎟⎠ Finalize There are a variety of ways that the area under the curve in part (c) could be approximated, so the results may vary a bit depending on how this approximation is made.

Answers: (a) From –50 s to 0, the Acela is cruising at a constant positive velocity in the +x direction. Between 0 and 50 s, the constant velocity is maintained for a while and then the train begins to speed up. Between 50s and 100s, the train speeds up; notice that the graph is linear in this segment.

Between 100 s and 200 s, the train reaches its maximum speed of about 175 mi/h. Between 200 s and 300 s, the engineer applies the brakes: the train is slowing down. The train is moving very slowly between 300 s and 350 s and then stops at 350 s. After 350 s, the train reverses direction: the velocity is negative. The train is backing up faster and faster until the data ends at 400 s. (b) 0.87 m/s2 (c) 6.7 mi (answers may vary, depending on estimation from the graph.)

SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 2.1 P2.1

Position, Velocity, and Speed

We assume that you are approximately 2 m tall and that the nerve impulse travels at uniform speed. The elapsed time is then

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P2.2

We substitute for t in x = 10t2, then use the definition of average velocity: t (s)

2.00

2.10

3.00

x (m)

40.0

44.1

90.0

(a) (b) P2.3

We read the data from the table provided, assume three significant figures of precision for all the numbers, and use Equation 2.2 for the definition of average velocity. (a) (b) (c)

Section 2.2 P2.4

Instantaneous Velocity and Speed

We use the definition of average velocity. (a)

(b) (c)

To find the average velocity for the round trip, we add the displacement and time for each of the two halves of the swim:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(d) The average speed of the round trip is the total distance the athlete travels divided by the total time for the trip:

P2.5

For average velocity, we find the slope of a secant line running across the graph between the 1.5-s and 4-s points. Then for instantaneous velocities we think of slopes of tangent lines, which means the slope of the graph itself at a point.

We place two points on the curve: Point A, at t = 1.5 s, and Point B, at t = 4.0 s, and read the corresponding values of x. (a)

At ti = 1.5 s, xi = 8.0 m (Point A) At tf = 4.0 s, xf = 2.0 m (Point B)

ANS. FIG. P2.5

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b)

The slope of the tangent line can be found from points C and D. (tC = 1.0 s, xC = 9.5 m) and (tD = 3.5 s, xD = 0),

The negative sign shows that the direction of vxis along the negative x direction. (c)

The velocity will be zero when the slope of the tangent line is zero. This occurs for the point on the graph where x has its minimum value. This is at

Section 2.3 P2.6

.

Analysis Model: Particle Under Constant Velocity

The trip has two parts: first the car travels at constant speed v1 for distance d, then it travels at constant speed v2 for distance d. The first part takes the time interval (a)

= d/v1, and the second part takes the time interval ∆t2 = d/v2.

By definition, the average velocity for the entire trip is where

and

Putting

these together, we have

We know vavg = 30 mi/h and v1 = 60 mi/h.

Solving for v2 gives

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b)

The average velocity for this trip is

where

so, vavg= (c)

The average speed for this trip is d + d = 2d and

where d = d1 + d2 = so, the average speed is the

same as in part (a): vavg = P2.7

(a)

The total time for the trip is ttotal = t1 + 22.0 min = t1 + 0.367 h, where t1 is the time spent traveling at v1 = 89.5 km/h. Thus, the distance traveled is which gives

or from which, t1= 2.44 h, for a total time of

(b)

Section 2.5 P2.8

The distance traveled during the trip is

giving

Acceleration

The acceleration is zero whenever the marble is on a horizontal section. The acceleration has a constant positive value when the marble is rolling on the 20-to-40-cm section and has a constant negative value when it is rolling on the second sloping section. The position graph is a straight sloping line whenever the speed is constant and a section of a parabola when the speed changes.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

ANS. FIG. P2.8 P2.9

(a)

In the interval ti = 0 s and tf= 6.00 s, the motorcyclist’s velocity changes from vi = 0 to vf = 8.00 m/s. Then,

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b)

Maximum positive acceleration occurs when the slope of the velocitytime curve is greatest, at t = 3 s, and is equal to the slope of the graph, approximately (6 m/s – 2 m/s)/(4 s − 2 s) =

(c)

The acceleration a = 0 when the slope of the velocity-time graph is zero, which occurs at

, and also for

(d) Maximum negative acceleration occurs when the velocity-time graph has its maximum negative slope, at t = 8 s, and is equal to the slope of the graph, approximately P2.10(a) The graph is shown in ANS. FIG. P2.10 below.

ANS. FIG. P2.10 (b)

At t = 5.0 s, the slope is

.

At t = 4.0 s, the slope is

.

At t = 3.0 s, the slope is

.

At t = 2.0 s, the slope is

.

(c) (d) The initial velocity of the car was

.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Get complete Order files download link below htps://www.mediafire.com/file/rus03sp9cw3psp7/SM +Physics+for+Scien�sts+and+Engineers+with+Modern+ Physics,+10e+Raymond+Serway,+John+Jewet.zip/file

If The link does not work with a click, then copy the complete Download link in internet explorer/firefox/google chrome and get all files download successfully.

P2.11

(a)

The area under a graph of a vs. t is equal to the change in velocity, ∆v.

We can use Figure P2.11 to find the change in velocity during specific time intervals.

The area under the curve for the time interval 0 to 10 s has the shape of a rectangle. Its area is = (2 m/s2)(10 s) = 20 m/s The particle starts from rest, v0 = 0, so its velocity at the end of the 10-s time interval is v = v0 +

= 0 + 20 m/s =

Between t = 10 s and t = 15 s, the area is zero:

= 0 m/s.

Between t = 15 s and t = 20 s, the area is a rectangle:

=

(−3 m/s2)(5 s) = −15 m/s.

So, between t = 0 s and t = 20 s, the total area is

= (20 m/s) +

(0 m/s) + (−15 m/s) = 5 m/s, and the velocity at t = 20 s is (b)

We can use the information we derived in part (a) to construct a graph of x vs. t; the area under such a graph is equal to the displacement, of the particle.

From (a), we have these points (t, v) = (0 s, 0 m/s), (10 s, 20 m/s), (15 s, 20 m/s), and (20 s, 5 m/s). The graph appears below.

The displacements are: 0 to 10 s (area of triangle):

= (1/2)(20 m/s)(10 s) = 100 m

10 to 15 s (area of rectangle):

= (20 m/s)(5 s) = 100 m

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

15 to 20 s (area of triangle and rectangle): = (1/2)[(20 – 5) m/s](5 s) + (5 m/s)(5 s) = 37.5 m + 25 m = 62.5 m Total displacement over the first 20.0 s: = 100 m + 100 m + 62.5 m = 262.5 m =

Section 2.6 P2.12

Motion Diagrams

(a) (b) (c) (d) (e)

(f)

One way of phrasing the answer: The spacing of the successive positions would change with Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the acceleration vectors would vary in magnitude and direction.

P2.13

(a)

The motion is fast at first but slowing until the speed is constant. We assume the acceleration is constant as the object slows.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b)

The motion is constant in speed.

(c)

The motion is speeding up, and we suppose the acceleration is constant.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Section 2.7 P2.14

Analysis Model: Particle Under Constant Acceleration and x f − xi = 1.50 × 10

We have

−2

m.

(a)

:

(b)

P2.15

In parts (a) – (c), we use Equation 2.13 to determine the velocity at the times indicated. (a)

The time given is 1.00 s after 10:05:00 a.m., so vf = vi + at = 13.0 m/s + (–4.00 m/s2)(1.00 s) =

(b)

The time given is 4.00 s after 10:05:00 a.m., so vf = vi + at = 13.0 m/s + (–4.00 m/s2)(4.00 s) =

(c)

The time given is 1.00 s before 10:05:00 a.m., so vf = vi + at = 13.0 m/s + (–4.00 m/s2)(–1.00 s) =

(d)

(e)

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P2.16

We think of the plane moving with maximum-size backward acceleration throughout the landing, so the acceleration is constant, the stopping time a minimum, and the stopping distance as short as it can be. The negative acceleration of the plane as it lands can be called deceleration, but it is simpler to use the single general term acceleration for all rates of velocity change. (a)

The plane can be modeled as a particle under constant acceleration, with Given

we use the equation

and solve for t:

(b)

Find the required stopping distance and compare this to the length of the runway. Taking xi to be zero, we get

or (c)

P2.17

The stopping distance is greater than the length of the runway;

The velocity is always changing; there is always nonzero acceleration and the problem says it is constant. So we can use one of the set of equations describing constant-acceleration motion. Take the initial point to be the moment when xi= 3.00 cmand vxi = 12.0 cm/s. Also, at t = 2.00 s, xf = –5.00 cm.

Once you have classified the object as a particle moving with constant acceleration and have the standard set of four equations in front of you, how do you choose which equation to use? Make a list of all of the six symbols in the equations: xi , xf, vxi , vxf, ax, and t. On the list fill in values as above,

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

showing that xi, xf, vxi, and t are known. Identify ax as the unknown. Choose an equation involving only one unknown and the knowns. That is, choose an equation not involving vxf. Thus we choose the kinematic equation

and solve for ax:

We substitute:

ax =

2[ −5.00 cm − 3.00 cm −(12.0 cm/s)(2.00 s)] (2.00 s)2

= −16.0 cm/s 2 P2.18

As in the algebraic solution to Example 2.8, we let t represent the time the trooper has been moving. We graph xcar = 45+ 45t and

xtrooper = 1.5t2

They intersect at

ANS. FIG. P2.18 P2.19

Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its motion from entry to exit,

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(a)

The speed halfway through the photogate in space is given by

and this is (b)

The speed halfway through the photogate in time is given by and this is

P2.20

as determined above.

We ask whether the constant acceleration of the rhinoceros from rest over a period of 10.0 s can result in a final velocity of 8.00 m/s and a displacement of 50.0 m? To check, we solve for the acceleration in two ways.

P2.21

1)

ti = 0, vi = 0; t = 10.0 s, vf = 8.00 m/s:

2)

ti = 0, xi = 0, vi = 0; t = 10.0 s, xf = 50.0 m:

(a)

Let a stopwatch start from t = 0 as the front end of the glider passes point A. The average speed of the glider over the interval between t = 0 and

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

t = 0.628 s is 12.4 cm/(0.628 s) =

, and this is the instantaneous

speed halfway through the time interval, at t = 0.314 s. (b)

The average speed of the glider over the time interval between 0.628 + 1.39 = 2.02 s and 0.628 + 1.39 + 0.431 = 2.45 s is 12.4 cm/(0.431 s) = 28.8 cm/s and this is the instantaneous speed at the instant t = (2.02 + 2.45)/2 = 2.23 s.

Now we know the velocities at two instants, so the acceleration is found from

(c)

P2.22

Take any two of the standard four equations, such as

Solve one for vxi, and substitute into the other: vxi = vxf – axt

Thus

We note that the equation is dimensionally correct. The units are units of length in each term. Like the standard equation

this

equation represents that displacement is a quadratic function of time.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P2.23

(a)

For the first car, the speed as a function of time is

For the second car, the speed is

Setting the two expressions equal gives

Solving for t gives

(b)

The first car then has speed

and this is also the constant speed of the second car. (c)

For the first car, the position as a function of time is

For the second car, the position is

At the point where the cars pass one another, their positions are equal:

rearranging gives

We solve this with the quadratic formula. Suppressing units,

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(d) At t = 0.604 s, the second and also the first car’s position is

At t = 6.90 s, both are at position

(e)

*P2.24

Conceptualize A diagram of the information provided will be helpful. The diagram below shows three positions of the car and the times at which the car is at those positions.

Categorize The phrase “slowing down the car uniformly” tells us to use the particle under constant acceleration analysis model for this problem.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Analyze We have no information about the velocity of the car, so that suggests we focus on the equation in the particle under constant acceleration model involving position: x f = xi + vit +

1 2 at . (a) Write the position equation 2

twice, once for the car at pole #2 and once at pole #3, each referencing pole #1 as the initial situation:

x2 = x1 + v1t2 + 12 at22 = 0 + v1t2 + 12 at22 = v1t2 + 12 at22

(1)

x3 = x1 + v1t3 + 12 at32 = 0 + v1t3 + 12 at32 = v1t3 + 12 at32

(2)

Eliminate v1 between Equations (1) and (2) and solve for the acceleration a:

a=

2 ( x3t2 − x2t3 )

(3)

t32t2 − t22t3

Substitute numerical values:

2 ⎡( 80.0 m ) (10.0 s) − ( 40.0 m ) ( 25.0 s ) ⎤⎦ a= ⎣ = −0.107 m/ s2 2 2 ( 25.0 s) (10.0 s) − (10.0 s) ( 25.0 s) (b) Solve Equation (1) for the initial speed of the car:

x2 = v1t2 + 12 at22

x2 − 12 at22 v1 = t2

→

(4)

Substitute numerical values:

v1 =

(

)

40.0 m − 12 −0.107 m/ s 2 (10.0 s)

(10.0 s)

2

= 4.53 m/ s

(c) Using Equation 2.17 in the particle under constant acceleration model relating velocity and position, solve for the position at which the velocity goes to zero:

v = v + 2ax f 2 f

2 1

→

xf =

v 2f − v12 2a

=

0 − ( 4.53 m/ s)

(

2

2 −0.107 m/ s 2

)

= 96.3 m

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Because the fourth pole would be at x = 120 m, the last pole passed is pole #3.

Finalize Part (c) can also be solved less directly by finding the time at which the velocity goes to zero and then substituting into the position equation to find the position at that time. In doing so, you find the time at which the car stops to be 42.5 s. It might be surprising that the car travels for an additional 42.5 – 25.0 = 17.5 s beyond pole #3 (longer than between poles #1 and #2, and between poles #2 and #3), but only moves 96.3 – 80.0 m = 16.3 m beyond that pole.

Answers: (a)–0.107 m/s2(b) 4.53 m/s (c) pole #3 (stops at x = 96.3 m at t = 42.4 s

Section 2.8 P2.25

Freely Falling Objects

The bill starts from rest, vi = 0, and falls with a downward acceleration of 9.80 m/s2 (due to gravity). For an average human reaction time of about 0.20 s, we can find the distance the bill will fall:

The bill falls about 20 cm—this distance is about twice the distance between the center of the bill and its top edge, about 8 cm. Thus

P2.26

We can solve (a) and (b) at the same time by assuming the rock passes the top of the wall and finding its speed there. If the speed comes out imaginary, the rock will not reach this elevation.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

which gives . (c)

The rock travels from yi = 3.65 m to yf = 1.55 m. We find the final speed of the rock thrown down:

which gives The change in speed of the rock thrown down is

(d) The magnitude of the speed change of the rock thrown up is This

with 2.39 m/s.

(e)

P2.27

We are given the height of the helicopter: y = h = 3.00t3.

At t = 2.00 s, y = 3.00(2.00 s)3 = 24.0 m and

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

If the helicopter releases a small mailbag at this time, the mailbag starts its free fall with velocity 36.0 m/s upward. The equation of motion of the mailbag is

Setting yf = 0, dropping units, and rearranging the equation, we have 4.90t2 – 36.0t – 24.0 = 0 We solve for t using the quadratic formula:

Since only positive values of t count, we find P2.28

The falling ball moves a distance of (15 m – h) before they meet, where h is the height above the ground where they meet. We apply

to the falling ball to obtain

or

Applying

[1]

to the rising ball gives

[2]

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Combining equations [1] and [2] gives

or P2.29

We model the keys as a particle under the constant free-fall acceleration. Take the first student’s position to be

and the second student’s position to be

We are given that the time of flight of the keys is t = 1.50 s, and . (a)

We choose the equation

to connect the data and the

unknown.

We solve:

and substitute:

(b)

The velocity at any time t > 0 is given by vyf = vyi + ayt.

Therefore, at t =1.50 s,

The negative sign means that the keys are moving downward just before they are caught.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P2.30

(a)

The keys, moving freely under the influence of gravity (a = −g), undergo a vertical displacement of ∆y = +h in time t. We use

to find

the initial velocity as

(b)

We find the velocity of the keys just before they were caught (at time t) using v = vi + at:

*P2.31

Conceptualize This is a simple situation in which an object is thrown straight upward and is in free fall after being thrown. We wish to relate the speed of the throw and the height of the projectile.

Categorize We recognize a simple one-dimensional free-fall problem, in which we use the particle under constant acceleration model for the motion of the box.

Analyze (a) From the particle under constant acceleration model, chose the equation that relates position and equation (Equation 2.17), noting that the motion is in the y direction:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(1)

Consider the point at which the box reaches its highest position. Note that the velocity of the box at that position is vyf = 0, and define the point at which the box is thrown as the origin, so that yi = 0. Also assign the acceleration to be that due to gravity, so that Equation (1) becomes (2)

Find the highest position for the box if thrown upward at the speed of 20.0 m/s found in the demonstration:

Because this position is higher than the bottom of the window in which the alleged accomplice caught the box, the action of throwing the box to the accomplice is possible. (b) The defense attorney could argue as follows: The demonstration involved throwing a baseball horizontally. It is more difficult to throw something upward, so the throwing speed could be in error. You might counter-argue that the throwing speed could actually be higher because the defendant was attempting to minimize his throwing speed during the demonstration in order to show that the feat could not be done.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Answers: (a) The box could reach the window according to the data provided.(b) Answers will vary.

Section 2.9 P2.32

(a)

Kinematic Equations Derived from Calculus See the x vs. t graph on the top panel of ANS. FIG. P2.32.Choosex = 0 at t

= 0.

(b)

See the a vs. t graph at the bottom right.

For 3 <t< 5 s, a = 0.

ANS. FIG. P2.32

At the points of inflection, t = 3 and 5 s, the slope of the velocity curve changes abruptly, so the acceleration is not defined. (c)

For 5 s <t< 9 s,

(d) The average velocity between t = 5 and 7 s is vavg = (8 m/s +0)/2 = 4 m/s At

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(e)

The average velocity between t = 5 and 9 s is vavg = [(8 m/s) + (−8 m/s)]/2 = 0 m/s At t = 9 s, x = 28 m + (0 m/s)(1 s) =

P2.33

This is a derivation problem. We start from basic definitions. We are given J = dax/dt= constant, so we know that dax = Jdt. (a)

Integrating from the ‘initial’ moment when we know the acceleration to any later moment,

Therefore, From ax = dvx/dt,

dvx = axdt.

Integration between the same two points tells us the velocity as a function of time:

From vx = dx/dt, dx = vxdt. Integrating a third time gives us x(t):

and (b)

Squaring the acceleration,

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Rearranging,

The expression for vxwas

So and by substitution

Additional Problems P2.34

(a)

Area A1 is a rectangle. Thus, A1 = hw = vxit.

Area A2 is triangular. Therefore, The total area under the curve is

and since vx – vxi = axt,

(b)

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P2.35

(a)

From

the insect’s velocity after straightening its legs is

(b)

The time to reach this velocity is

(c)

The upward displacement of the insect between when its feet leave the ground and its speed is momentarily zero is

P2.36

Take downward as the positive y direction. (a)

While the woman was in free fall,

and we take

Thus,

giving

(b)

Her velocity just before impact is:

While crushing the box,

,

, and

Therefore,

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

or (c)

Time to crush box:

or P2.37

(a)

In order for the trailing athlete to be able to catch the leader, his speed (v1) must be greater than that of the leading athlete (v2), and the distance between the leading athlete and the finish line must be great enough to give the trailing athlete sufficient time to make up the deficient distance, d.

(b)

During a time interval t the leading athlete will travel a distance d2 = v2t and the trailing athlete will travel a distance d1 = v1t. Only when d1 = d2 + d (where d is the initial distance the trailing athlete was behind the leader) will the trailing athlete have caught the leader. Requiring that this condition be satisfied gives the elapsed time required for the second athlete to overtake the first: d1 = d2 + d

or

v1t = v2t + d

giving

(c)

In order for the trailing athlete to be able to at least tie for first place, the initial distance D between the leader and the finish line must be greater than or equal to the distance the leader can travel in the time t calculated above (i.e., the time required to overtake the leader). That is, we must

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

require that

P2.38

For the collision not to occur, the front of the passenger train must not have a position that is equal to or greater than the position of the back of the freight train at any time. We can write expressions of position to see whether the front of the passenger car (P) meets the back of the freight car (F) at some time.

Assume at t = 0, the coordinate of the front of the passenger car is xPi = 0; and the coordinate of the back of the freight car is xFi = 58.5 m. At later time t, the coordinate of the front of the passenger car is

and the coordinate of the back of the freight car is

Setting these expression equal to each other gives

or after simplifying and suppressing units.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

We do not have to solve this equation, we just want to check if a solution exists; if a solution does exist, then the trains collide. A solution does exist:

P2.39

We have constant-acceleration equations to apply to the two cars separately. (a)

Let the times of travel for Hannah and Stan be tKand tS, where tS = tK + 1.00 s Both start from rest (vxi,K= vxi,S = 0), so the expressions for the distances traveled are

and When Hannah overtakes Stan, the two distances will be equal. Setting xK = xSgives

This we simplify and write in the standard form of a quadratic as

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

We solve using the quadratic formula

, suppressing

units, to find

Only the positive root makes sense physically, because the overtake point must be after the starting point in time.

P2.40

(b)

Use the equation from part (a) for distance of travel,

(c)

Remembering that vxi,K= vxi,S= 0, the final velocities will be:

We translate from a pictorial representation through a geometric model to a mathematical representation by observing that the distances x and y are always related by x2+y2 = L2 . (a)

Differentiating this equation with respect to time, we have

Now the unknown velocity of B is

and

so the differentiated equation becomes

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

But

so

(b)

P2.41

The average speed of every point on the train as the first car passes Lisa is given by:

The train has this as its instantaneous speed halfway through the 1.50-s time.

Similarly, halfway through the next 1.10 s, the speed of the train is . The time required for the speed to change from 5.73 m/s to 7.82 m/s is

so the acceleration is:

Challenge Problems P2.42

(a)

The factors to consider are as follows. The red bead falls through a greater distance with a downward acceleration of g. The blue bead travels a shorter distance, but with acceleration of A first guess would be that the blue bead “wins,” but not by much. We do note, however, that points with

(b)

,

, and

are the vertices of a right triangle

as the hypotenuse.

The red bead is a particle under constant acceleration. Taking downward as the positive direction, we can write

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

as

which gives (c)

The blue bead is a particle under constant acceleration, with Taking the direction along L as the positive direction, we can write

as which gives (d) For the two beads to reach point

simultaneously, tR = tB. Then,

Squaring both sides and cross-multiplying gives

or We note that the angle between chords

(e)

P2.43

and

that the angle between chords

and

and the beads arrive at point

simultaneously.

is is

so

Then,

Once we recognize that the two rods form one side and the hypotenuse of a right triangle with θ as its smallest angle, then the result becomes obvious.

Consider the runners in general. Each completes the race in a total time so each interval T. Each runs at constant acceleration a for a time interval covers a distance (displacement) final speed (velocity)

where they eventually reach a

after which they run at this constant speed for

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

the remaining time

until the end of the race, covering distance The total distance (displacement) each covers is

the same:

so

where (a)

and T = 10.4 s.

For Laura (runner 1),

= 2.00 s:

a1 = (100 m)/(18.8 s2) = For Healan (runner 2),

= 3.00 s:

a2 = (100 m)/(26.7 s2) = (b)

Laura (runner 1): v1 = a1 Healan (runner 2): v2 = a2

(c)

= =

The 6.00-s mark occurs after either time interval above, each has covered the distance

From the reasoning

where t = 6.00 s.

Laura (runner 1):

= 53.19 m

Healan (runner 2):

= 50.56 m

(d) Laura accelerates at the greater rate, so she will be ahead of Healen at, and immediately after, the 2.00-s mark. After the 3.00-s mark, Healan is travelling faster than Laura, so the distance between them will shrink. In the time interval

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

During that time interval, the distance between them (the position of Laura relative to Healan) is

because Laura has ceased to accelerate but Healan is still accelerating.

Differentiating with respect to time, (and doing some simplification), we can solve for the time t when D is an maximum:

which gives

Substituting this time back into the expression for D, we find that D = 4.47 m, that is, Laura ahead of Healan by *P2.44

Conceptualize You may have had this experience when driving. A bicycle can exhibit a high acceleration, so it can pull ahead of the car when leaving a traffic light. But the maximum speed of the bicycle is less than that of the car, so the car will eventually catch up and pass the bicycle. A graphical representation may be of some help here:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Because of the higher acceleration of the bicycle, the line representing its velocity is steeper in slope than that of the car. But it soon reaches its maximum speed of 20.0 mi/h, and the line for its velocity changes to horizontal. The slope of the line representing the acceleration of your car is less steep, but the acceleration lasts for a longer time, so that eventually your car is traveling faster than the bicycle.

Categorize The phrase “constant acceleration” tells us to use the particle under constant acceleration analysis model for this problem. Once the vehicles reach the

indicated speeds, they travel at constant velocity, so we will also need to employ the particle under constant velocity analysis model.

Analyze Assuming that your car catches up with the bicycle before reaching its

maximum speed, the motion of your car is described completely during the time interval of interest by the particle under constant acceleration model. Therefore, its position is given by

xcar = xcar ,i + vcar ,it + 12 acart 2 = 12 acart 2 (1) where we have incorporated the fact that the car begins from rest (vcar,i = 0), and have defined the position of the traffic light as the origin (xcar,i = 0). The position of the bicycle is described by the particle under constant acceleration model until it reaches its maximum speed and by the particle under constant velocity model thereafter.

Therefore, knowing that the bicycle also begins from the origin at rest,

where t1 is the time at which the bicycle reaches its maximum speed and vbicyle,max is that maximum speed. As noted, the bicycle maintains that speed thereafter. The time

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

at which the bicycle has a given speed is provided by the velocity equation in the particle under constant acceleration model:

vbicycle = vbicycle ,i + abicyclet = 0 + abicyclet

→ t=

(4)

vbicycle abicycle

Evaluate this time when the bicycle first reaches its maximum speed:

t1 =

(5)

vbicycle,max abicycle

(a) To find the time interval during which the bicycle is ahead of the car, we find the time at which the car and the bicycle are at the same position by using Equations (1) and (3):

xbicycle = xcar

→

(

1 a t 2 + vbicycle,max t − t1 2 bicycle 1

) = 12 a t

2

car

Rearranging this equation gives the standard form of a quadratic equation:

(

1 a t 2 − vbicycle,max t − 12 abicyclet12 − vbicycle,maxt1 2 car

) = 0 (6)

Applying the quadratic formula, we can solve for t as follows:

t=

=

vbicycle,max ±

(

−vbicycle,max

) − 4 ( a ) ⎡⎣− ( a 2

1 2

1 2

car

)

t − vbicycle,maxt1 ⎤ ⎦

2 bicycle 1

acar vbicycle,max ±

( −v

) + 2a ( a 2

bicycle,max

car

1 2

t − vbicycle,maxt1

2 bicycle 1

)

acar

Finally, substitute for t1:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(7)

vbicycle,max ±

( −v

bicycle,max

t=

=

)

2

2 ⎡ ⎛ ⎞ ⎛ vbicycle,max ⎞ ⎤ v bicycle,max 1 + 2acar ⎢ 2 abicycle ⎜ ⎟ − vbicycle,max ⎜ a ⎟⎥ ⎢ ⎝ abicycle ⎠ ⎝ bicycle ⎠ ⎥ ⎣ ⎦ acar

⎛ acar ⎞ 2 vbicycle,max ± vbicycle,max 1 − ⎜ abicycle ⎟⎠ ⎝ acar Substitute numerical values:

20.0 mi/ h ± t=

⎡

⎞⎤ ⎠⎦

⎛ ⎝

9.00 mi/ h/ s ( 20.0 mi/ h ) ⎢1− ⎜ 13.0 ⎥ mi/ h/ s ⎟ 2

⎣ 9.00 mi/ h/ s

= 3.45 s (The other root is t = 0.990 s, but this turns out to be less than t1, so it is not a realistic solution.) (b) The speed of the your car at any time is found from the particle under constant acceleration model:

vcar = vcar ,i + acart

(8)

The maximum separation occurs when the speeds of both vehicles are the same, because after that instant the car will be moving faster than the bicycle, so the separation distance will decrease. Therefore, set the speed of the car equal to the speed of the bicycle in Equation (8) and find the time t2 at which the two objects have the same speed:

vcar = vbicycle,max = vcar ,i + acart2

→ t2 =

vbicycle,max

(9)

acar

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

At any time after the bicycle reaches its maximum speed, the separation distance between the vehicles is found by subtracting Equation (1) from Equation (3):

xbicycle − xcar = 12 abicyclet12 + vbicycle,max ( t − t1 ) − 12 acart 2

(10)

Evaluate the separation distance at the time found in Equation (9):

xbicycle − xcar = 12 abicyclet12 + vbicycle,max ( t2 − t1 ) − 12 acart12 2

⎛ vbicycle,max ⎞ ⎛ vbicycle,max vbicycle,max ⎞ = 12 abicycle ⎜ + v − ⎟ bicycle,max ⎜ abicycle ⎟⎠ ⎝ abicycle ⎠ ⎝ acar ⎛ vbicycle,max ⎞ − 12 acar ⎜ ⎟⎠ ⎝ a

2

car

(

= vbicycle,max

)

2

⎡ 1 ⎛ 1 1 ⎞ 1 ⎤ +⎜ − ⎢ ⎥ ⎟− ⎢⎣ 2abicycle ⎝ acar abicycle ⎠ 2acar ⎥⎦

Substitute numerical values:

1 ⎡ ⎤ ⎢ 2 ( 13.0 mi/ h/ s ) ⎥ ⎢ ⎥ ⎢ ⎛ ⎞⎥ 1 1 2 xbicycle − xcar = ( 20.0 mi/ h ) ⎢ +⎜ − ⎥ ⎝ 9.00 mi/ h/ s 13.0 mi/ h/ s ⎟⎠ ⎥ ⎢ ⎢ ⎥ 1 − ⎢ ⎥ 2 ( 9.00 mi/ h/ s ) ⎣ ⎦ ⎛ 5 280 ft ⎞ ⎛ 1 h ⎞ = 10.0 ft = 6.84 mi · s/ h ⎜ ⎝ mi ⎟⎠ ⎜⎝ 3 600 s ⎟⎠ Finalize It would be helpful to graph the positions of the two vehicles against time as

shown below:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

This graph shows both the time at which the car passes the bicycle and the time at which the maximum separation occurs between the vehicles. Notice that the bicycle reaches its maximum speed at about 1.5 s, but the car is still accelerating at 4 seconds (the blue line is still curved). The black dot represents the instant that the bicycle stops accelerating. The corresponding black dot for the car would be to the right of the graph as drawn, at about 5.6 s.]

Answers: (a) 3.45 s (b) 10.0 ft

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Get complete Order files download link below htps://www.mediafire.com/file/rus03sp9cw3psp7/SM +Physics+for+Scien�sts+and+Engineers+with+Modern+ Physics,+10e+Raymond+Serway,+John+Jewet.zip/file

If The link does not work with a click, then copy the complete Download link in internet explorer/firefox/google chrome and get all files download successfully.

ANSWERS TO QUICK-QUIZZES 1. (b) 2. (c) 3. (b) 4. False.

5. (b) 6. (c) 7. (a)–(e), (b)–(d), (c)–(f) 8. (i) (e) (ii) (d)

ANSWERS TO EVEN-NUMBERED PROBLEMS P2.2

(a) 50.0 m/s; (b) 41.0 m/s

P2.4

(a) +L/t1; (b) –L/t2; (c) 0; (d) 2L/t1+ t2

P2.6

(a) 20 mi/h; (b) 0; (c) 30 mi/h

P2.8

See graphs in P2.8.

P2.10

(a) See ANS. FIG. P2.10; (b) 23 m/s, 18 m/s, 14 m/s, and 9.0 m/s; (c) 4.6 m/s2; (d) zero

P2.12

(a–e) See graphs in P2.12; (f) with less regularity

P2.14

(a) 4.98×10-19s

(b) 1.20×1015m/s2

P2.16

(a) 20.0s (b) 1000m

(c) the plane cannot land

P2.18

31s

P2.20

The accelerations do not match, therefore the situation is impossible.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P2.22

The equation represents that displacement is a quadratic function of time.

P2.24

(a)–0.107 m/s2(b) 4.53 m/s (c) pole #3 (stops at x = 96.3 m at t = 42.4 s

P2.26

(a and b) The rock does not reach the top of the wall with vf = 3.69 m/s; (c) 2.39m/s; (d) does not agree; (e) The average speed of the upward-moving rock is smaller than the downward moving rock.

P2.28

0.60 s

P2.30

(a)

P2.32

(a) See graphs in P2.32; (b) See graph in P2.32; (c) −4 m/s2; (d) 32 m; (e) 28 m

P2.34

(a)

P2.36

(a) 96.0 ft/s; (b)

P2.38

The trains do collide.

P2.40

(a) vB= (1/tan u)v (b) The velocity vB starts off larger than v for small angles

(b)

(b) The displacement is the same result for the total area. ; (c)

theta and then decreases, approaching zero as theta approaches 90°.

P2.42

(a) The red bead falls through a greater distance with a downward acceleration of g. The blue bead travels a shorter distance, but with acceleration of

A first guess would be that the blue bead “wins,” but

not by much. (b)

(c)

(d) the beads arrive at point ©

simultaneously; (e) Once we recognize that the two rods form one side and the hypotenuse of a right triangle with θ as its smallest angle, then the result becomes obvious.

P2.44

(a) 3.45 s; (b) 10.0 ft.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3 Vectors CHAPTER OUTLINE 3.1

Coordinate Systems

3.2

Vector and Scalar Quantities

3.3

Basic Vector Arithmetic

3.4

Components of a Vector and Unit Vectors

* An asterisk indicates a question or problem new to this edition.

SOLUTIONS TO THINK-PAIR-SHARE AND ACTIVITIES *TP3.1

Conceptualize Imagine your situation, staring at a radar screen with five ships indicated. You want to choose which ship to send, but how do you do it, especially when they all have different speeds? Should you simply send the ship that has the highest speed?

Categorize Based on the assumption in the problem statement, we can use the particle under constant velocity model for each ship.

Analyze First, find the x and y components of the four ships, as well as the sinking ship, using east as the +x direction and north as the +y direction:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Ship #

x (km)

y (km)

1

–24.2

26.8

2

–32.6

18.1

3

–6.00

8.25

4

–50.3

9.77

Sinking –30.1

41.4

Now find the differences in x components and the differences in y components between the sinking ship and each of the other ships: Ship #

Δx (km)

Δy (km)

1

5.94

–14.6

2

–2.53

–23.3

3

24.1

–33.2

4

–20.2

–31.7

Finally, use the Pythagorean theorem to find the distance from each ship to the sinking ship and use the speed, assumed constant, to find the time interval required for the ship to reach the sinking ship: Ship #

Distance from

Maximum

Time

Sinking Ship (km)

Speed (km/h)

Interval (h)

1

15.8

30.0

0.53

2

23.5

38.0

0.62

3

10.2

32.0

1.28

4

51.2

45.0

0.83

Finalize Well, the fastest ship would not get there first! In fact, you need to alert the slowest ship, because it is the closest and, despite its lower speed, it will arrive at the sinking ship first.

Answer: ship 1

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

*TP3.2

A typical drawing will look like that below.

Albuquerque (35°06′39″N, 106°36′36″W) and Memphis (35°07′03″N 89°58′16″W) are almost on the exact same latitude, so we can say that the vector from Albuquerque to Memphis is pointed due east. The comparison to the scale in the diagram shows a distance of 912 miles, so we can write the Albuquerque–Memphis vector as

The comparison to the scale for the Memphis–Chicago vector shows a distance of 473 miles. A protractor shows that the vector makes an angle of 80.7° with the latitude line running through Memphis.

Therefore, the Memphis–Chicago vector is

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

A vector drawn from Albuquerque to Chicago would be the sum of these two vectors:

Use the Pythagorean theorem to find the distance between Albuquerque and Chicago:

The direction of Chicago from Albuquerque is

⎛ 467 mi ⎞ = 25.3° ⎝ 988 mi ⎟⎠

θ = tan −1 ⎜ Answer: Answers will vary.

SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 3.1 P3.1

Coordinate Systems

(a)

The distance between the points is given by

(b)

To find the polar coordinates of each point, we measure the radial distance to that point and the angle it makes with the +x axis:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

measured from the +x axis.

P3.2

therefore,

(a)

x1 = (2.50 m) cos 30.0°, y1 = (2.50 m) sin 30.0°, and

x2 = (3.80 m) cos 120°, y2 = (3.80 m) sin 120°, and

(b) P3.3

For polar coordinates (r, θ), the Cartesian coordinates are (x = r cosθ, y = r sinθ), if the angle is measured relative to the +x axis. (a) (b) (c) (d)

P3.4

We have (a)

The radius for this new point is

and its angle is

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

This point is in the third quadrant if (x, y)

(b)

is in the first quadrant or in the fourth quadrant if (x, y) is in the second quadrant. It is at an angle of

.

This point is in the fourth quadrant if (x, y)

(c)

is in the first quadrant or in the third quadrant if (x, y) is in the second quadrant. It is at an angle of

.

Section 3.2 Vector and Scalar Quantities P3.5

In solving this problem we must contrast displacement with distance traveled. We draw a diagram of the skater’s path in ANS. FIG. P3.5, which is the view from a hovering helicopter so that we can see the circular path as circular in shape. To start with a concrete example, we have chosen to draw motion ABC around one half of a circle of radius 5 m.

ANS. FIG. P3.5 The displacement, shown as

in the diagram, is the straight-line

change in position from starting point A to finish C. In the specific case we have chosen to draw, it lies along a diameter of the circle. Its magnitude is The distance skated is greater than the straight-line displacement. The distance follows the curved path of the semicircle (ABC). Its length is half of the circumference:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

A straight line is the shortest distance between two points. For any nonzero displacement, less or more than across a semicircle, the distance along the path will be greater than the displacement magnitude. Therefore:

Section 3.3 Basic Vector Arithmetic P3.6

We are given

. When two vectors are added graphically, the

second vector is positioned with its tail at the tip of the first vector. The resultant then runs from the tail of the first vector to the tip of the second vector. In this case, vector

will be positioned with its tail at the origin

and its tip at the point (0, 29). The resultant is then drawn, starting at the origin (tail of first vector) and going 14 units in the negative y direction to the point (0, −14). The second vector,

, must then start from the tip of

at point (0, 29) and end on the tip of

at point

ANS. FIG. P3.6 © 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(0, −14) as shown in the sketch at the right. From this, it is seen that

P3.7

We find the resultant head of

graphically by placing the tail of

. The resultant force vector

at the

is of magnitude

and at an angle of

ANS. FIG. P3.7 P3.8

(a)

The three diagrams are shown in ANS. FIG. P3.8 below.

ANS. FIG. P3.8a (b)

The diagrams in ANS. FIG. P3.8a represent the graphical solutions for the three vector sums: and

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P3.9

To find these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (Scale: 1 unit = 0.5 m) (a)

(c)

(b)

(d)

ANS. FIG. P3.9 P3.10

The scale drawing for the graphical solution should be similar to the figure to the right. The magnitude and direction of the final displacement from the starting point are obtained by measuring d and

θ on the drawing and applying the scale factor used in making the drawing. The results should be

ANS. FIG. P3.10

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Section 3.4 Components of a Vector and Unit Vectors P3.11

(a) (b)

Let v represent the speed of the camper. The northward component of its velocity is v cos 8.50°. To avoid crowding the minivan we require v cos 8.50° ≥ 28 m/s.

P3.12

The person would have to walk

and P3.13

We use the unit-vector addition method. It is just as easy to add three displacements as to add two. We take the direction east to be along The three displacements can be written as:

The resultant is

(or 9.20 m west and 2.30 m north).

The magnitude of the resultant displacement is

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

The direction of the resultant vector is given by

P3.14

(a)

See figure to the right.

(b)

ANS. FIG. P3.14

(c)

P3.15

(a)

The single force is obtained by summing the two forces:

We can also express this force in terms of its magnitude and direction:

(b)

A force equal and opposite the resultant force from part (a) is required for the total force to equal zero:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P3.16

We take the x axis along the slope downhill. (Students, get used to this choice!) The y axis is perpendicular to the slope, at 35.0° to the vertical.

Then the displacement of the snow makes an angle of 90.0° + 35.0° + 16.0° = 141° with the x axis.

ANS. FIG. P3.16

(a)

(b)

P3.17

P3.18

(a)

We add the components of the three vectors:

(b)

Again, using the components of the three vectors,

We are given

and Solving for

gives

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P3.19

Hold your fingertip at the center of the front edge of your study desk, defined as point O. Move your finger 8 cm to the right, then 12 cm vertically up, and then 4 cm horizontally away from you. Its location relative to the starting point represents position vector

Move three-

fourths of the way straight back toward O. Now your fingertip is at the location of

Now move your finger 50 cm straight through O,

through your left thigh, and down toward the floor. Its position vector now is We use unit-vector notation throughout. There is no adding to do here, but just multiplication of a vector by two different scalars. (a)

(b)

(c) P3.20

We carry out the prescribed mathematical operations using unit vectors.

(a)

(b)

P3.21

The component description of A is just restated to constitute the answer to part (a): Ax= −3.00, Ay = 2.00. (a)

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b)

θ is in the second quadrant, so (c) Bx = Rx – Ax = 0 – (–3.00) = 3.00, By = Ry – Ay= –4.00 – 2.00 = –6.00.

Therefore, P3.22

The given diagram shows the vectors individually, but not their addition. The second diagram represents a map view of the motion of the ball. According to the definition of a displacement, we ignore any departure from straightness of the actual path of the ball. We model each of the three motions as straight. The simplified problem is solved by straightforward application of the component method of vector addition. It works for adding two, three, or any number of vectors.

ANS. FIG. P3.22 (a)

We find the two components of each of the three vectors

and

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

and

and Now adding,

and so

R =

2

2

⎛ Ry ⎞ −1 ⎛ 27.1 ⎞ ⎟ = tan ⎜ ⎟ = 28.7° ⎝ 49.5 ⎠ ⎝ Rx ⎠

θ = tan −1 ⎜

(b) P3.23

( 49.5 ) + ( 27.1) = 56.4

(a)

Taking components along

and

we get two equations:

6.00a – 8.00b +26.0 = 0 and –8.00a + 3.00b + 19.0 = 0 Substituting a = 1.33b – 4.33 into the second equation, we find

and so a = 1.33(7.00) – 4.33 = 5.00.

Thus

Therefore,

(b)

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P3.24

We are given

The magnitude

of the vector is therefore

And the angle of the vector with the three coordinate axes is

P3.25

We use the numbers given in Problem 3.9:

Ax = A cos θA = 3.00 cos 30.0° = 2.60 m, Ay = A sin θA = 3.00 sin 30.0° = 1.50 m So

then P3.26

(a)

Her net x (east-west) displacement is –3.00 + 0 + 6.00 = +3.00 blocks, while her net y (north-south) displacement is 0 + 4.00 + 0 = +4.00 blocks.

The magnitude of the resultant displacement is

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

and the angle the resultant makes with the x axis (eastward direction) is

The resultant displacement is then (b) P3.27

The total distance traveled is 3.00 + 4.00 + 6.00 =

We will use the component method for a precise answer. We already know the total displacement, so the algebra of solving a vector equation will guide us to do a subtraction.

We have Ax = 150 cos 120° = −75.0 cm Ay = 150 sin 120° = 130 cm Rx = 140 cos 35.0° = 115 cm Ry = 140 sin 35.0° = 80.3 cm

ANS. FIG. P3.27

Therefore,

P3.28

First, we sum the components of the two vectors for the male:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

followed by the components of the two vectors for the female:

P3.29

The hurricane’s first displacement is

and its second displacement is

With

representing east and

representing north, its total

displacement is:

with magnitude P3.30

Let the positive x direction be eastward, the positive y direction be vertically upward, and the positive z direction be southward. The total displacement is then

(a)

The magnitude is

(b)

Its angle with the y axis follows from

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P3.31

The y coordinate of the airplane is constant and equal to 7.60 × 10 m 3

whereas the x coordinate is given by x = vit, where vi is the constant speed in the horizontal direction.

At t = 30.0 s we have x = 8.04 × 10 , so vi = 8 040 m/30 s = 268 m/s. The 3

position vector as a function of time is

At

The magnitude is

and the direction is

P3.32

Note that each shopper must make a choice whether to turn 90° to the left or right, each time he or she makes a turn. One set of such choices, following the rules in the problem, results in the shopper heading in the positive y direction and then again in the positive x direction.

Find the magnitude of the sum of the displacements:

P3.33

The displacement from the start to the finish is

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

The displacement from the starting point to A is (a)

The position vector of point A is

(b)

For f = 0 we have the position vector

meters.

(c) (d) For f = 1 = 100%, we have position vector

(e) *P3.34

Conceptualize A drawing is always a good way to help understand a vector problem. Below is a sketch of the situation.

The desired displacement from the position when the system comes up is vector

in the diagram. The desired heading angle is shown as the

angle θ.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Categorize This is a relatively simple problem in vector addition in two dimensions.

Analyze As noted in the diagram, we denote the early part of the displacement as vector

, the desired vector from the point at which

the navigation system comes up as originally desired displacement,

, and the resultant, which is the . Vector addition of

and

gives us (1) Separate the x and y components:

Ax î + Bx î = Rx î

→

Ax + Bx = Rx

Ay ĵ + By ĵ = R y ĵ →

A y + By = R y

(2)

Substitute numerical values:

25.0 km + Bx = 0 → Bx = −25.0 km 50.0 km + By = 200 km → By = 150 km Therefore, expressing the heading with respect to due north,

tan θ =

Bx 25.0 km = = 0.167 By 150 km

→ θ = 9.46°

Finalize You should advise the captain to set a heading of 9.46° west of north and maintain that heading to arrive at the destination port.

This is the vector

in the figure in the Conceptualize step.

Answer: 9.46° west of north

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Additional Problems P3.35

On our version of the diagram we have drawn in the resultant from the tail of the first arrow to the head of the last arrow. The resultant displacement

is equal

ANS. FIG. P3.35 to the sum of the four individual displacements, We translate from the pictorial representation to a mathematical representation by writing the individual displacements in unit-vector notation:

Summing the components together, we find

so altogether

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Its magnitude is

We calculate the angle The resultant points into the third quadrant instead of the first quadrant. The angle counter-clockwise from the +x axis is

P3.36

Let A represent the distance from island 2 to island 3. The Represent the displacement from 3 to 1

displacement is as

We have 4.76 km at

ANS. FIG. P3.36 For the x components:

For the y components:

(a)

We solve by eliminating B by substitution:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b) P3.37

Let θ represent the angle between the directions of and

have the same magnitudes,

,

, and

isosceles triangle in which the angles are magnitude of

is then

and

. Since

form an The

This can be seen from

applying the law of cosines to the isosceles triangle and using the fact that B = A.

ANS. FIG. P3.37 Again,

,

, and

form an isosceles triangle with apex

angle θ. Applying the law of cosines and the identity

gives the magnitude of

as

The problem requires that R = 100D. Thus,

This gives

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

P3.38

Let θ represent the angle between the directions of and

have the same magnitudes,

,

, and

and . Since form an

isosceles triangle in which the angles are magnitude of

is then

The

This can be seen by applying

the law of cosines to the isosceles triangle and using the fact that B = A. Again,

,

, and

form an isosceles triangle with apex

angle θ. Applying the law of cosines and the identity

ANS. FIG. P3.38

gives the magnitude of

as

The problem requires that R = nD or

The larger R is to be compared to D, the smaller the angle between and P3.39

(a)

becomes.

Take the x axis along the tail section of the snake. The displacement from tail to head is

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Its magnitude is From

the time for each child’s run is

Inge wins by

P3.40

(b)

Olaf must run the race in the same time:

(a)

The very small differences between the angles suggests we may consider this region of Earth to be small enough so that we may consider it to be flat (a plane); therefore, we may consider the lines of latitude and longitude to be parallel and perpendicular, so that we can use them as an xy coordinate system. Values of latitude, θ, increase as we travel north, so differences between latitudes can give the y coordinate. Values of longitude, φ, increase as we travel west, so differences between longitudes can give the x coordinate. Therefore, our coordinate system will have +y to the north and +x to the west.

ANS. FIG. P3.40 © 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Since we are near the equator, each line of latitude and longitude may be considered to form a circle with a radius equal to the radius of Earth, R = 6.36 × 10 m. Recall the length s of an arc of a 6

circle of radius R that subtends an angle (in radians) Δθ (or Δφ) is given by

We can use this equation to find

the components of the displacement from the starting point to the tree—these are parallel to the x and y coordinates axes. Therefore, we can regard the origin to be the starting point and the displacements as the x and y coordinates of the tree.

The angular difference Δφ for longitude values is (west being positive)

corresponding to the x coordinate (displacement west)

The angular difference Δθ for latitude values is (north being positive)

corresponding to the y coordinate (displacement north)

The distance to the tree is

The direction to the tree is

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b)

P3.41

(a) (b)

(c)

*P3.42

Conceptualize The aircraft are flying in three-dimensional space, so the distance between them is difficult to visualize easily or draw on paper based on the numbers provided. Your goal is to determine a straight-line distance between them to make sure they are far enough apart to be safe.

Categorize Despite the difficulty in visualizing the positions of the airplanes, this is a relatively simple substitution problem.

Write a vector expression for each airplane using the suggested axis orientations:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Find the difference vector between these two positions:

Use the Pythagorean theorem to find the distance between the planes:

The two airplanes are just slightly farther apart than the minimum separation distance.]

Answer: 2.28 km

P3.43

(a)

(b)

The position vector at t = 0 is

At t = 1 s, the position is

and so on. The object is moving straight downward at 2 m/s, so P3.44

Since

ANS. FIG. P3.44 we have

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

giving Because the vectors have the same magnitude and x components of equal magnitude but of opposite sign, the vectors are reflections of each other in the y axis, as shown in the diagram. Therefore, the two vectors have the same y components: Ay = By = (1/2)(6.00) = 3.00 Defining θ as the angle between either

or

and the y axis, it is seen

that

The angle between P3.45

(a)

and

is then

From the picture,

ANS. FIG. P3.45 (b) (c)

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Challenge Problems P3.46

(a)

You start at point The displacement to B is

You cover half of this,

to move to

Now the displacement from your current position to C is

You cover one-third, moving to

The displacement from where you are to D is

You traverse one-quarter of it, moving to

The displacement from your new location to E is

of which you cover one-fifth the distance, moving to

The treasure is at

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b)

Following the directions brings you to the average position of the trees. The steps we took numerically in part (a) bring you to

then to

then to

and last to

ANSWERS TO QUICK-QUIZZES

1. vectors: (b), (c); scalars: (a), (d), (e) 2. (c) 3. (b) and (c) 4. (b) 5. (c)

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1.

ANSWERS TO EVEN-NUMBERED PROBLEMS P3.2

(a) (2.17, 1.25) m, (−1.90, 3.29) m; (b) 4.55m

P3.4

(a) r, 180° – θ; (b) 180° + θ ; (c) –θ is 43 units in the negative y direction

P3.6 P3.8

(a) See ANS. FIG. P3.8; (b) The sum of a set of vectors is not affected by the order in which the vectors are added.

P3.10

approximately 420 ft at 23°

P3.12

1.31 km north and 2.81 km east

P3.14

(a) See ANS. FIG. P3.14; (b)

(c) 6.40 at 38.7°,

8.06 at 97.2° P3.16

(a) Its component parallel to the surface is (1.50 m) cos 141° = −1.17 m, or 1.17 m toward the top of the hill; (b) Its component perpendicular to the surface is (1.50 m) sin 141° = 0.944 m, or 0.944 m away from the snow.

P3.18 P3.20

(a)

(b)

P3.22

(a)

P3.24

59.2° with the x axis, 39.8° with the y axis, 67.4° with the z axis

P3.26

(a) 5.00 blocks at 53.1° N of E; (b) 13.00 blocks

P3.28

magnitude: 170.1 cm, direction: 57.2° above +x axis (first quadrant);

(b) 56.4, 28.7°

magnitude: 145.7 cm, direction: 58.6° above +x axis (first quadrant) P3.30

(a) 10.4 cm (b) u 5 35.5°

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Get complete Order files download link below htps://www.mediafire.com/file/rus03sp9cw3psp7/SM +Physics+for+Scien�sts+and+Engineers+with+Modern+ Physics,+10e+Raymond+Serway,+John+Jewet.zip/file

If The link does not work with a click, then copy the complete Download link in internet explorer/firefox/google chrome and get all files download successfully.

P3.32

Impossible because 12.4 m is greater than 5.00 m

P3.34

9.46° west of north

P3.36

(a) 7.17 km; (b) 6.15 km

P3.38 P3.40

(a) 496 m, 65.1° N of W; (b) The arguments are justified because the distances involved are small relative to the radius of the Earth.

P3.42

2.29 km

P3.44 P3.46

(a) (10.0 m, 16.0 m) (b) This center of mass of the tree distribution is the same location whatever order we take the trees in.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

4 Motion in Two Dimensions CHAPTER OUTLINE 4.1

The Position, Velocity, and Acceleration Vectors

4.2

Two-Dimensional Motion with Constant Acceleration

4.3

Projectile Motion

4.4

Analysis Model: Particle in Uniform Circular Motion

4.5

Tangential and Radial Acceleration

4.6

Relative Velocity and Relative Acceleration

* An asterisk indicates an item new to this edition.

SOLUTIONS TO THINK-PAIR-SHARE AND ACTIVITIES *TP4.1

Conceptualize

The diagram below (see Fig. ANS TP4.1) shows a

closeup of the stairs, with the appropriate length values indicated.

Imagine the marble rolling horizontally toward the right off the upper landing and flying into the air. If it is rolling slowly, it will fall off the upper landing and strike the first step. We want its trajectory to be such that it misses that step, which will require a certain minimum rolling speed.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Categorize This is a projectile problem, so we use the particle under constant acceleration model for the vertical motion of the ball and the particle under constant velocity model for its horizontal motion.

Analyze Let the marble leave the upper landing at time t = 0. At time t, the marble has position coordinates xf and yf. Therefore, from the particle under constant acceleration and particle under constant velocity models, substituting values for the initial coordinates and initial vertical velocity component,

x f = xi + vxit = 0 + vxit

→

x f = vxit

y f = yi + vyit − 12 gt 2 = 0 + 0 − 12 gt 2

→

y f = − 12 gt 2

(1) ( 2)

Solve Equation (1) for the time t and substitute into Equation (2) to find the horizontal projection velocity:

⎛ y f = − 12 g ⎜

xf ⎞ ⎝ v ⎟⎠

2

→

vxi = −

xi

gx 2f 2y f

(a) To find the minimum speed for the marble to miss the first step, evaluate the equation for the horizontal projection velocity using the coordinates of the edge of that step:

( 9.80 m/ s )( 0.300 m ) = 1.48 m/ s v = − 2

xi

2

2 ( −0.200 m )

(b) To find the minimum speed for the marble to miss the second step, evaluate the equation for the horizontal projection velocity using the coordinates of the edge of that step:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

( 9.80 m/ s ) ⎡⎣( 2 )( 0.300 m )⎤⎦ = 2.10 m/ s v = − 2

xi

2

2 ⎡⎣( 2 ) ( −0.200 m ) ⎤⎦

(c) To find the minimum speed for the marble to miss all the steps, evaluate the equation for the horizontal projection velocity using the coordinates of the edge of the eleventh step:

( 9.80 m/ s ) ⎡⎣(11)( 0.300 m )⎤⎦ = 4.92 m/ s v = − 2

xi

2

2 ⎡⎣( 11) ( −0.200 m ) ⎤⎦

This is not a particularly high speed, so it should be possible for your toddler nephew to propel the marble with this speed. (d) We can calculate the required speed for the marble to land on the sixth step as we have above for other steps. If the marble were projected horizontally from the sixth step at this same speed, it would clear the remaining six steps. When it bounces off the sixth step, however, it has this horizontal component of velocity along with an upward vertical component that will keep it in the air even longer and allow its horizontal displacement to increase more than it did for the flight from the upper landing. Therefore, the marble should hit the floor of the lower landing beyond the lowest step.

Finalize Is it clear why we substituted 11 in part (c) rather than 12?

Notice that the required projection speed does not increase linearly with the number of steps. From part (a) to part (c), it requires a little over three times as much speed to miss eleven times as many steps. Is that consistent with your conception of projectile motion?

Answers: (a) 1.48 m/s (b) 2.10 m/s (c) yes, at 4.92 m/s (d) The last step, the twelfth, is down to the landing. The speed of projection allows the ball to reach the sixth step and then another six steps to the landing, even without including its vertical motion after bouncing from the sixth step. Including the vertical motion for the bounce, the ball is in

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

the air after the sixth step for an even longer time, causing the ball to make its second bounce well beyond the last step. *TP4.2

The time interval should be zero, because the vertical falling motion is unaffected by the horizontal motion of the pennies. In the absence of an audio recording, students can just listen for the sound of the pennies hitting the floor. They should not hear two distinct sounds.

Answer: The time interval should be zero, because the vertical falling motion is unaffected by the horizontal motion of the pennies.

SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 4.1 The Position, Velocity,and Acceleration Vectors P4.1

(a) For the average velocity, we have

(b)

For the velocity components, we have

and Therefore,

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

and the speed is

P4.2

(a)

From

we determine the components of the

velocity by taking the time derivatives of x and y:

and At t = 0,

(b)

Acceleration is the time derivative of the velocity, so

and At t = 0,

(c)

(d)

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Section 4.2 Two-Dimensional Motion with Constant Acceleration P4.3

(a)

We differentiate the equation for the vector position of the particle with respect to time to obtain its velocity:

(b)

Differentiating the expression for velocity with respect to time gives the particle’s acceleration:

P4.4

(c)

By substitution, when t = 1.00 s,

(a)

For the xcomponent of the motion we have

We choose the + sign to represent the physical situation:

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Here

So, (b)

(c) (d) P4.5

The directions of the position, velocity, and acceleration vectors are given with respect to the x axis, and we know that the components of a vector with magnitude A and direction θare given by Ax = A cosθ and Ay = A sinθ; thus we have

where (a)

is in m,

in m/s,

2

in m/s , and t in s.

From

where

in m/s and t in s.

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b)

The car’s position vector is given by

where

Section 4.3 P4.6

is in m and t in s.

Projectile Motion

The mug is a projectile from just after leaving the counter until just before it reaches the floor. Taking the origin at the point where the mug leaves the bar, the coordinates of the mug at any time t are

and

(a)

When the mug reaches the floor, yf = h and xf = d, so

is the time of impact, and

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b)

Just before impact, the x component of velocity is still vxf = vxi while the y component is

Then the direction of motion just before impact is below the horizontal at an angle of

because the x component of velocity is positive (forward) and the y component is negative (downward).

P4.7

At the maximum height

and the time to reach this height is

found from

as

.

The vertical displacement that has occurred during this time is

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1m ⎞ Thus, if ( Δy )max = 12 ft ⎛⎜ ⎟ = 3.66 m, then ⎝ 3.281 ft ⎠

and if the angle of projection is

P4.8

, the launch speed is

We ignore the trivial case where the angle of projection equals zero degrees.

so

or

thus, P4.9

Consider the motion from original zero height to maximum height h: gives or Now consider the motion from the original point to half the maximum height: gives so

© 2019 Cengage Learning, Inc. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Get complete Order files download link below htps://www.mediafire.com/file/rus03sp9cw3psp7/SM +Physics+for+Scien�sts+and+Engineers+with+Modern+ Physics,+10e+Raymond+Serway,+John+Jewet.zip/file

If The link does not work with a click, then copy the complete Download link in internet explorer/firefox/google chrome and get all files download successfully.