SOLUTIONS MANUAL for University Physics with Modern Physics 15th Edition by Hugh Young and Roger Fre by welldoneassistant

1-2

Chapter 1 EXECUTE: Using the components from Ex. 1.7 we have Sx = 38.37 m – (–46.36 m) + 0.00 m = 84.73 m Sy = 61.40 m – (–33.68 m) + (–17.80 m) = 77.28 m

S = S x2 + S y2 =

θ = arctan

Sy Sx

(84.73 m)2 + (77.28 m)2 = 115 m

= arctan[(77.28 m)/(84.73 m)] = 42°

  Since both components of S are positive, S points into the first quadrant. Therefore it makes an angle of 42° with the +x-axis. EVALUATE:

Figure VP1.7.2

VP1.7.3.

The graphical solution shown in Fig. VP1.7.2 shows that our results are reasonable.     IDENTIFY: We know three vectors A , B , and C and we want to find the sum T where     T = A + B + 2C .    SET UP: Find the components of vectors A , B , and C and use them to find the magnitude and    direction of T . The components of 2 C are twice those of C . EXECUTE: S x = Ax + Bx + 2C x and S y = Ay + By + 2C y (a) Using the components from Ex. 1.7 gives Tx = 38.37 m + (–46.36 m) + 2(0.00 m) = –7.99 m Ty = 61.40 m + (–33.68 m) + 2(–17.80 m) = –7.88 m (b) T = Tx2 + Ty2 =

(–7.99 m)2 + (–7.88 m)2 = 11.2 m

θ = arctan y = arctan[(–7.88 m)/(–7.99 m)] = 45° Tx

 Both components of T are negative, so it points into the third quadrant, making an angle of 45° below the –x-axis or 45° + 180° = 225° counterclockwise with the +x-axis, in the third quadrant.

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Units, Physical Quantities, and Vectors

1-3

EVALUATE:

Figure VP1.7.3

The graphical solution shown in Fig. VP1.7.3 shows that this result is reasonable. VP1.7.4.

IDENTIFY: The hiker makes two displacements. We know the first one and their resultant, and we want to find the second displacement.    SET UP: Calling A the known displacement, R the known resultant, and D the unknown vector, we     know that A + D = R . We also know that R = 38.0 m and R makes an angle θ R = 37.0° + 90° =

127° with the +x-axis. Fig. VP1.7.4 shows a sketch of these vectors.

Figure VP1.7.4

 EXECUTE: From Ex. 1.7 we have Ax = 38.37 m and Ay = 61.40 m. The components of R are Rx = R cos 127.0° = (38.0 m) cos 127.0° = –22.87 m Ry = R sin 38.0° = (38.0 m) sin 127.0° = 30.35 m Rx = Ax + Dx and Ry = Ay + Dy  Using these components, we find the components of D . → Dx = –22.87 m 38.37 m + Dx = –22.87 m → Dy = –31.05 m 61.40 m + Dy = –31.05 m

D = Dx2 + Dy2 =

θ = arctan

Dy D

(–61.24 m)2 + (–31.05 m) 2 = 68.7 m

= arctan[(–31.05 m)/(–61.24 m)] = 27°

1-4

Chapter 1

 Both components of D are negative, so it points into the third quadrant, making an angle of 27° + 180° = 207° with the +x-axis. EVALUATE: A graphical solution will confirm these results. VP1.10.1.

IDENTIFY: We know the magnitude and direction of two vectors. We want to use these to find their components and their scalar product. SET UP: Ax = A cos θ A , Ay = A sin θ A , Bx = B cos θ B , By = B sin θ B . We can find the scalar product

using the vector components or using their magnitudes and the angle between them.     A ⋅ B = Ax Bx + Ay By and A ⋅ B = AB cos φ . Which form you use depends on the information you have. EXECUTE: (a) Ax = A cos θ A = (5.00) cos(360° – 36.9°) = 4.00

Ay = A sin θ A = (5.00) sin(360° – 36.9°) = –3.00 Bx = (6.40) cos(90° + 20.0°) = –2.19 By = (6.40)sin(90° + 20.0°) = 6.01 (b) Using components gives   A ⋅ B = Ax Bx + Ay By = (4.00)(–2.19) + (–3.00)(6.01) = –26.8   EVALUATE: We check by using A ⋅ B = AB cos φ .   A ⋅ B = AB cos φ = (5.00)(6.40) cos(20.0° + 90° + 36.9°) = –26.8 This agrees with our result in part (b). VP1.10.2.

IDENTIFY: We know the magnitude and direction of one vector and the components of another vector. We want to use these to find their scalar product and the angle between them.     SET UP: The scalar product can be expressed as A ⋅ B = Ax Bx + Ay By and A ⋅ B = AB cos φ . Which

form you use depends on the information you have. EXECUTE: (a) Cx = C cos θC = (6.50) cos 55.0° = 3.728

Cy = C sin θC = (6.50) sin 55.0° = 5.324 Dx = 4.80 and Dy = –8.40   Using components gives C ⋅ D = C x Dx + C y Dy = (3.728)(4.80) + (5.324)(–8.40) = –26.8 (b) D = Dx2 + Dy2 = (4.80) 2 +(–8.40) 2 = 9.675     C ⋅ D = CD cos φ , so cos φ = C ⋅ D /CD = (–26.8)/[(6.50)(9.67)] = –0.426. φ = 115°.  EVALUATE: Find the angle that D makes with the +x-axis.

Figure VP1.10.2

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Units, Physical Quantities, and Vectors

θ D = arctan

Dy Dx

1-5

= arctan[8.40/(–4.80] = –60.3°, which is 60.3° below the +x-axis. From Fig. VP1.10.2,

  we can easily seed that the angle between C and D is φ = 60.3° + 55.0° = 115°, as we found in (b). VP1.10.3.

IDENTIFY: We know the components of two vectors and want to find the angle between them.   SET UP: The scalar product A ⋅ B = AB cos φ involves the angle between two vectors. We can find   this product using components from A ⋅ B = Ax Bx + Ay By . From this result we can find the angle φ . EXECUTE: First find the magnitudes of the two vectors.

A = Ax2 + Ay2 + Az2 =

(–5.00) 2 + (3.00) 2 + 02 = 5.83

B = Bx2 + By2 + Bz2 = (2.50)2 + (4.00)2 + (–1.50) 2 = 4.95   Now use A ⋅ B = AB cos φ = Ax Bx + Ay By and solve for φ . (5.83)(4.95) cos φ = (–5.00)(2.50) + (3.00)(4.00) + (0)(–1.50) → φ = 91°. EVALUATE: The scalar product is positive, so φ must be between 90° and 180°, which agrees with our result. VP1.10.4.

IDENTIFY: We know the scalar product of two vectors. We also know both components of one of them and the x-component of the other one. We want to find the y-component of the other one and the angle between the two vectors. The scalar product involves the angle between two vectors.     SET UP: We use A ⋅ B = Ax Bx + Ay By and A ⋅ B = AB cos φ .   EXECUTE: (a) Use F ⋅ s = Fx sx + Fy s y to find Fy.

→ Fy = 14.8 N. 26.0 N ⋅ m = (–12.0 N)(4.00 m) + Fy(5.00 m)   2 2 (b) Use F ⋅ s = Fs cos φ and A = Ax + Ay to find the magnitudes of the two vectors. (–12.0 N)2 + (14.8 N) 2

(4.00 m)2 + (5.00 m)2 cos φ = 26.0 N ⋅ m → φ = 77.7°.   EVALUATE: The work is positive, so the angle between F and s must be between 0° and 90°, which agrees with our result in part (b). 1.1.

IDENTIFY: Convert units from mi to km and from km to ft. SET UP: 1 in. = 2.54 cm, 1 km = 1000 m, 12 in. = 1 ft, 1 mi = 5280 ft.  5280 ft  12 in.  2.54 cm  1 m  1 km  EXECUTE: (a) 1.00 mi = (1.00 mi)     2  3  = 1.61 km  1 mi  1 ft  1 in.  10 cm  10 m   103 m  102 cm   1 in.  1 ft  3 (b) 1.00 km = (1.00 km)     = 3.28 × 10 ft    2.54 cm  1 km 1 m 12 in .     EVALUATE: A mile is a greater distance than a kilometer. There are 5280 ft in a mile but only 3280 ft in a km.

1.2.

IDENTIFY: Convert volume units from L to in.3. SET UP: 1 L = 1000 cm3. 1 in. = 2.54 cm

 1000 cm3   1 in. 3 3 EXECUTE: 0.473 L ×   ×   = 28.9 in. .  1 L   2.54 cm  EVALUATE: 1 in.3 is greater than 1 cm3 , so the volume in in.3 is a smaller number than the volume in

cm3 , which is 473 cm3. 1.3.

IDENTIFY: We know the speed of light in m/s. t = d /v. Convert 1.00 ft to m and t from s to ns. SET UP: The speed of light is v = 3.00 × 108 m/s. 1 ft = 0.3048 m. 1 s = 109 ns.

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1-6

Chapter 1 EXECUTE: t =

0.3048 m = 1.02 × 10−9 s = 1.02 ns 3.00 × 108 m/s

EVALUATE: In 1.00 s light travels 3.00 × 108 m = 3.00 × 105 km = 1.86 × 105 mi. 1.4.

IDENTIFY: Convert the units from g to kg and from cm3 to m3. SET UP: 1 kg = 1000 g. 1 m = 100 cm. 3

EXECUTE: 19.3

g  1 kg   100 cm  4 kg × ×  = 1.93 × 10 cm3  1000 g   1 m  m3

EVALUATE: The ratio that converts cm to m is cubed, because we need to convert cm3 to m3. 1.5.

IDENTIFY: Convert seconds to years. 1 gigasecond is a billion seconds. SET UP: 1 gigasecond = 1 × 109 s. 1 day = 24 h. 1 h = 3600 s.

 1 h  1 day   1 y  EXECUTE: 1.00 gigasecond = (1.00 × 109 s)   = 31.7 y.    3600 s  24 h   365 days  EVALUATE: The conversion 1 y = 3.156 × 107 s assumes 1 y = 365.24 d, which is the average for one

extra day every four years, in leap years. The problem says instead to assume a 365-day year. 1.6.

IDENTIFY: Convert units. SET UP: Use the unit conversions given in the problem. Also, 100 cm = 1 m and 1000 g = 1 kg.

ft  mi   1 h  5280 ft  EXECUTE: (a)  60     = 88 h   3600 s  1 mi  s 

m  ft   30.48 cm   1 m  (b)  32 2     = 9.8 2 s  s   1 ft   100 cm  3

g  100 cm   1 kg   3 kg (c) 1.0 3   = 10 3   m  cm  1 m   1000 g 

EVALUATE: The relations 60 mi/h = 88 ft/s and 1 g/cm3 = 103 kg/m3 are exact. The relation

32 ft/s 2 = 9.8 m/s 2 is accurate to only two significant figures. 1.7.

IDENTIFY: Convert miles/gallon to km/L. SET UP: 1 mi = 1.609 km. 1 gallon = 3.788 L.  1.609 km  1 gallon  EXECUTE: (a) 55.0 miles/gallon = (55.0 miles/gallon)    = 23.4 km/L.  1 mi  3.788 L  1500 km 64.1 L = 64.1 L. = 1.4 tanks. 23.4 km/L 45 L/tank EVALUATE: 1 mi/gal = 0.425 km/L. A km is very roughly half a mile and there are roughly 4 liters in

(b) The volume of gas required is

a gallon, so 1 mi/gal ∼ 24 km/L, which is roughly our result. 1.8.

IDENTIFY: Convert units. SET UP: We know the equalities 1 mg = 10−3 g, 1 µg 10−6 g, and 1 kg = 103 g.

 10−3 g  1 μ g  5 EXECUTE: (a) (410 mg/day)   10−6 g  = 4.10 × 10 μ g/day. 1 mg     10−3 g  (b) (12 mg/kg)(75 kg) = (900 mg)   = 0.900 g.  1 mg 

Units, Physical Quantities, and Vectors

1-7

 10−3 g  −3 (c) The mass of each tablet is (2.0 mg)   = 2.0 × 10 g. The number of tablets required each day  1 mg  is the number of grams recommended per day divided by the number of grams per tablet: 0.0030 g/day = 1.5 tablet/day. Take 2 tablets each day. 2.0 × 10−3 g/tablet  1 mg  (d) (0.000070 g/day)  −3  = 0.070 mg/day.  10 g  EVALUATE: Quantities in medicine and nutrition are frequently expressed in a wide variety of units. 1.9.

IDENTIFY: We know the density and mass; thus we can find the volume using the relation density = mass/volume = m /V . The radius is then found from the volume equation for a sphere and the

result for the volume. SET UP: Density = 19.5 g/cm3 and mcritical = 60.0 kg. For a sphere V = 43 π r 3.

 60.0 kg  1000 g  3 EXECUTE: V = mcritical /density =   = 3080 cm . 3   19.5 g/cm   1.0 kg  r=3

3V 3 3 (3080 cm3 ) = 9.0 cm. = 4π 4π

EVALUATE: The density is very large, so the 130-pound sphere is small in size. 1.10.

IDENTIFY: Model the bacteria as spheres. Use the diameter to find the radius, then find the volume and surface area using the radius. SET UP: From Appendix B, the volume V of a sphere in terms of its radius is V = 43 π r 3 while its

surface area A is A = 4π r 2 . The radius is one-half the diameter or r = d /2 = 1.0 μ m. Finally, the

necessary equalities for this problem are: 1 μ m = 10−6 m; 1 cm = 10−2 m; and 1 mm = 10−3 m. 3

 10−6 m   1 cm 3 3 −12 EXECUTE: V = 43 π r 3 = 43 π (1.0 μ m)3    −2  = 4.2 × 10 cm and μ 1 m 10 m     2

 10−6 m   1 mm 2 −5 2 A = 4π r 2 = 4π (1.0 μ m) 2    10−3 m  = 1.3 × 10 mm μ 1 m   EVALUATE: On a human scale, the results are extremely small. This is reasonable because bacteria are not visible without a microscope. 1.11.

IDENTIFY: When numbers are multiplied or divided, the number of significant figures in the result can be no greater than in the factor with the fewest significant figures. When we add or subtract numbers it is the location of the decimal that matters. SET UP: 12 mm has two significant figures and 5.98 mm has three significant figures. EXECUTE: (a) (12 mm) × (5.98 mm) = 72 mm 2 (two significant figures) 5.98 mm = 0.50 (also two significant figures) 12 mm (c) 36 mm (to the nearest millimeter) (d) 6 mm (e) 2.0 (two significant figures) EVALUATE: The length of the rectangle is known only to the nearest mm, so the answers in parts (c) and (d) are known only to the nearest mm.

(b) s

1-8

Chapter 1

1.12.

IDENTIFY: This is a problem in conversion of units. SET UP: 10 mm = 1 cm, V = πr2h. EXECUTE: V = π(0.036 cm)2(12.1 cm) = 0.049 cm3. Now convert to mm3. 3

 10 mm  3 0.049 cm3   = 49 mm .  1 cm  EVALUATE: The answer has only 2 significant figures. Even though π and h have more than that, r has only 2 which limits the answer. 1.13.

IDENTIFY: Use your calculator to display π × 107. Compare that number to the number of seconds in a year. SET UP: 1 yr = 365.24 days, 1 day = 24 h, and 1 h = 3600 s.

 24 h   3600 s  7 7 7 EXECUTE: (365.24 days/1 yr)    = 3.15567…× 10 s; π × 10 s = 3.14159…× 10 s 1 day 1 h    The approximate expression is accurate to two significant figures. The percent error is 0.45%. EVALUATE: The close agreement is a numerical accident. 1.14.

IDENTIFY: To asses the accuracy of the approximations, we must convert them to decimals. SET UP: Use a calculator to calculate the decimal equivalent of each fraction and then round the numeral to the specified number of significant figures. Compare to π rounded to the same number of significant figures. EXECUTE: (a) 22/7 = 3.14286 (b) 355/113 = 3.14159 (c) The exact value of π rounded to six significant figures is 3.14159. EVALUATE: We see that 355/113 is a much better approximation to π than is 22/7.

1.15.

IDENTIFY: Express 200 kg in pounds. Express each of 200 m, 200 cm and 200 mm in inches. Express 200 months in years. SET UP: A mass of 1 kg is equivalent to a weight of about 2.2 lbs.1 in. = 2.54 cm. 1 y = 12 months. EXECUTE: (a) 200 kg is a weight of 440 lb. This is much larger than the typical weight of a man.  1 in.  3 (b) 200 m = (2.00 × 104 cm)   = 7.9 × 10 inches. This is much greater than the height of a  2.54 cm  person. (c) 200 cm = 2.00 m = 79 inches = 6.6 ft. Some people are this tall, but not an ordinary man. (d) 200 mm = 0.200 m = 7.9 inches. This is much too short.  1y  (e) 200 months = (200 mon)   = 17 y. This is the age of a teenager; a middle-aged man is much  12 mon 

older than this. EVALUATE: None are plausible. When specifying the value of a measured quantity it is essential to give the units in which it is being expressed. 1.16.

IDENTIFY: Estimate the number of people and then use the estimates given in the problem to calculate the number of gallons. SET UP: Estimate 3 × 108 people, so 2 × 108 cars. EXECUTE: (Number of cars × miles/car day)/(mi/gal) = gallons/day

(2 × 108 cars × 10000 mi/yr/car × 1 yr/365 days)/(20 mi/gal) = 3 × 108 gal/day EVALUATE: The number of gallons of gas used each day approximately equals the population of the U.S.

Units, Physical Quantities, and Vectors

1.17.

1-9

IDENTIFY: Estimation problem. SET UP: Estimate that the pile is 18 in.× 18 in.× 5 ft 8 in.. Use the density of gold to calculate the mass of gold in the pile and from this calculate the dollar value. EXECUTE: The volume of gold in the pile is V = 18 in.× 18 in.× 68 in. = 22,000 in.3. First convert to

cm3: V = 22,000 in.3 (1000 cm3 /61.02 in.3 ) = 3.6 × 105 cm3 .

The density of gold is 19.3 g/cm3 , so the mass of this volume of gold is m = (19.3 g/cm3 )(3.6 × 105 cm3 ) = 6.95 × 106 g.

The monetary value of one gram is $40, so the gold has a value of ($40 /gram) (6.95 × 106 grams) = $2.8 × 108 or about $300 × 106 (three hundred million dollars). EVALUATE: This is quite a large pile of gold, so such a large monetary value is reasonable. 1.18.

IDENTIFY: Approximate the number of breaths per minute. Convert minutes to years and cm3 to m3 to

find the volume in m3 breathed in a year.  24 h  60 min  5 2 SET UP: Assume 10 breaths/min. 1 y = (365 d)    = 5.3 × 10 min. 10 cm = 1 m so 1 d 1 h   

106 cm3 = 1 m3. The volume of a sphere is V = 43 π r 3 = 16 π d 3 , where r is the radius and d is the diameter. Don’t forget to account for four astronauts.  5.3 × 105 min  4 3 EXECUTE: (a) The volume is (4)(10 breaths/min)(500 × 10−6 m3 )   = 1 × 10 m /yr. 1y   1/3

 6V  (b) d =    π 

1/3

 6[1 × 104 m3 ]  =   π  

= 27 m

EVALUATE: Our estimate assumes that each cm3 of air is breathed in only once, where in reality not all the oxygen is absorbed from the air in each breath. Therefore, a somewhat smaller volume would actually be required. 1.19.

IDENTIFY: Estimate the diameter of a drop and from that calculate the volume of a drop, in m3.

Convert m3 to L. SET UP: Estimate the diameter of a drop to be d = 2 mm. The volume of a spherical drop is V = 43 π r 3 = 16 π d 3 . 103 cm3 = 1 L. EXECUTE: V = 16 π (0.2 cm)3 = 4 × 10−3 cm3. The number of drops in 1.0 L is

1000 cm3 = 2 × 105 4 × 10−3 cm3

EVALUATE: Since V ∼ d 3 , if our estimate of the diameter of a drop is off by a factor of 2 then our

estimate of the number of drops is off by a factor of 8. 1.20.

IDENTIFY: Estimate the number of beats per minute and the duration of a lifetime. The volume of blood pumped during this interval is then the volume per beat multiplied by the total beats. SET UP: An average middle-aged (40 year-old) adult at rest has a heart rate of roughly 75 beats per minute. To calculate the number of beats in a lifetime, use the current average lifespan of 80 years.  60 min   24 h  365 days  80 yr  9 EXECUTE: N beats = (75 beats/min)     = 3 × 10 beats/lifespan  yr lifespan  1 h   1 day    9  1 L  1 gal   3 × 10 beats  7 Vblood = (50 cm3 /beat)    = 4 × 10 gal/lifespan  3  Browsegrades.net  1000 cm  3.788 L   lifespan 

1-10

Chapter 1 EVALUATE: This is a very large volume. 1.21.

IDENTIFY: Draw each subsequent displacement tail to head with the previous displacement. The resultant displacement is the single vector that points from the starting point to the stopping point.     SET UP: Call the three displacements A, B, and C . The resultant displacement R is given by     R = A + B + C.  EXECUTE: The vector addition diagram is given in Figure 1.21. Careful measurement gives that R is 7.8 km, 38 north of east. EVALUATE: The magnitude of the resultant displacement, 7.8 km, is less than the sum of the magnitudes of the individual displacements, 2.6 km + 4.0 km + 3.1 km.

Figure 1.21 1.22.

IDENTIFY: Draw the vector addition diagram to scale.   SET UP: The two vectors A and B are specified in the figure that accompanies the problem.    EXECUTE: (a) The diagram for R = A + B is given in Figure 1.22a. Measuring the length and angle of  R gives R = 9.0 m and an angle of θ = 34°.     (b) The diagram for E = A − B is given in Figure 1.22b. Measuring the length and angle of E gives D = 22 m and an angle of θ = 250°.       (c) − A − B = −( A + B ), so − A − B has a magnitude of 9.0 m (the same as A + B ) and an angle with the   + x axis of 214° (opposite to the direction of A + B ).       (d) B − A = −( A − B ), so B − A has a magnitude of 22 m and an angle with the + x axis of 70°   (opposite to the direction of A − B ).   EVALUATE: The vector − A is equal in magnitude and opposite in direction to the vector A.

Figure 1.22

Units, Physical Quantities, and Vectors 1.23.

1-11

IDENTIFY: Since she returns to the starting point, the vector sum of the four displacements must be zero.     SET UP: Call the three given displacements A, B, and C , and call the fourth displacement D .     A + B + C + D = 0.  EXECUTE: The vector addition diagram is sketched in Figure 1.23. Careful measurement gives that D is144 m, 41° south of west.     EVALUATE: D is equal in magnitude and opposite in direction to the sum A + B + C .

Figure 1.23 1.24.

IDENTIFY: tan θ =

Ay Ax

, for θ measured counterclockwise from the + x -axis.

  SET UP: A sketch of Ax , Ay and A tells us the quadrant in which A lies. EXECUTE:

1.25.

(a) tan θ =

(b) tan θ =

(d) tan θ =

Ax Ax Ax Ax

−1.00 m = −0.500. θ = tan −1 (−0.500) = 360° − 26.6° = 333°. 2.00 m

1.00 m = 0.500. θ = tan −1 (0.500) = 26.6°. 2.00 m

1.00 m = −0.500. θ = tan −1 (−0.500) = 180° − 26.6° = 153°. −2.00 m

−1.00 m = 0.500. θ = tan −1 (0.500) = 180° + 26.6° = 207° −2.00 m

EVALUATE: The angles 26.6° and 207° have the same tangent. Our sketch tells us which is the correct value of θ .   IDENTIFY: For each vector V , use that Vx = V cosθ and Vy = V sin θ , when θ is the angle V makes

with the + x axis, measured counterclockwise from the axis.     SET UP: For A, θ = 270.0°. For B, θ = 60.0°. For C , θ = 205.0°. For D, θ = 143.0°. EXECUTE: Ax = 0, Ay = −8.00 m. Bx = 7.50 m, By = 13.0 m. C x = −10.9 m, C y = −5.07 m.

Dx = −7.99 m, Dy = 6.02 m. EVALUATE: The signs of the components correspond to the quadrant in which the vector lies. 1.26.

IDENTIFY: Given the direction and one component of a vector, find the other component and the magnitude. SET UP: Use the tangent of the given angle and the definition of vector magnitude. A EXECUTE: (a) tan 34.0° = x Ay

1-12

Chapter 1

Ay =

Ax tan 34.0°

16.0 m = 23.72 m tan 34.0°

Ay = −23.7 m. (b) A = Ax2 + Ay2 = 28.6 m. EVALUATE: The magnitude is greater than either of the components. 1.27.

Ax = (9.60 m)tan 32.0° = 6.00 m. Ax = −6.00 m. (b) A = Ax2 + Ay2 = 11.3 m. EVALUATE: The magnitude is greater than either of the components. 1.28.

IDENTIFY: Find the vector sum of the three given displacements. SET UP: Use coordinates for which + x is east and + y is north. The driver’s vector displacements are:    A = 2.6 km, 0° of north; B = 4.0 km, 0° of east; C = 3.1 km, 45° north of east. EXECUTE: Rx = Ax + Bx + C x = 0 + 4.0 km + (3.1 km)cos(45°) = 6.2 km; Ry = Ay + By + C y = 2.6 km + 0 + (3.1 km)(sin 45°) = 4.8 km; R = Rx2 + Ry2 = 7.8 km; θ = tan −1[(4.8 km)/(6.2 km)] = 38°;  R = 7.8 km, 38° north of east. This result is confirmed by the sketch in Figure 1.28.  EVALUATE: Both Rx and Ry are positive and R is in the first quadrant.

Figure 1.28 1.29.

   IDENTIFY: If C = A + B, then C x = Ax + Bx and C y = Ay + B y . Use C x and C y to find the magnitude  and direction of C . SET UP: From Figure E1.30 in the textbook, Ax = 0, Ay = −8.00 m and Bx = + B sin 30.0° = 7.50 m,

By = + B cos30.0° = 13.0 m.    EXECUTE: (a) C = A + B so C x = Ax + Bx = 7.50 m and C y = Ay + By = +5.00 m. C = 9.01 m. 5.00 m = and θ = 33.7°. C x 7.50 m       (b) B + A = A + B, so B + A has magnitude 9.01 m and direction specified by 33.7°. tan θ =

Units, Physical Quantities, and Vectors

1-13

   (c) D = A − B so Dx = Ax − Bx = −7.50 m and Dy = Ay − By = −21.0 m. D = 22.3 m. tan φ =

Dy Dx

 −21.0 m and φ = 70.3°. D is in the 3rd quadrant and the angle θ counterclockwise from −7.50 m

the + x axis is 180° + 70.3° = 250.3°.       (d) B − A = −( A − B ), so B − A has magnitude 22.3 m and direction specified by θ = 70.3°. EVALUATE: These results agree with those calculated from a scale drawing in Problem 1.22. 1.30.

IDENTIFY: Use A = Ax2 + Ay2 and tan θ =

Ay Ax

to calculate the magnitude and direction of each of the

given vectors.   SET UP: A sketch of Ax , Ay and A tells us the quadrant in which A lies. EXECUTE: (a) (b)

 5.20  (−8.60 cm) 2 + (5.20 cm) 2 = 10.0 cm, arctan   = 148.8° (which is180° − 31.2° ).  −8.60 

 −2.45  (−9.7 m)2 + (−2.45 m) 2 = 10.0 m, arctan   = 14° + 180° = 194°.  −9.7 

 −2.7  (7.75 km) 2 + (−2.70 km) 2 = 8.21 km, arctan   = 340.8° (which is 360° − 19.2° ).  7.75  EVALUATE: In each case the angle is measured counterclockwise from the + x axis. Our results for θ agree with our sketches. (c)

1.31.

IDENTIFY: Vector addition problem. We are given the magnitude and direction of three vectors and are asked to find their sum. SET UP:

A = 3.25 km B = 2.20 km C = 1.50 km

Figure 1.31a

   Select a coordinate system where + x is east and + y is north. Let A, B, and C be the three      displacements of the professor. Then the resultant displacement R is given by R = A + B + C . By the method of components, Rx = Ax + Bx + C x and Ry = Ay + By + C y . Find the x and y components of each vector; add them to find the components of the resultant. Then the magnitude and direction of the resultant can be found from its x and y components that we have calculated. As always it is essential to draw a sketch. EXECUTE:

1-14

Chapter 1

Ax = 0, Ay = +3.25 km Bx = −2.20 km, B y = 0 C x = 0, C y = −1.50 km

Rx = Ax + Bx + C x Rx = 0 − 2.20 km + 0 = −2.20 km

Ry = Ay + By + C y Ry = 3.25 km + 0 − 1.50 km = 1.75 km

Figure 1.31b

R = Rx2 + Ry2 = (−2.20 km) 2 + (1.75 km) 2 R = 2.81 km Ry 1.75 km = = −0.800 tan θ = Rx −2.20 km

θ = 141.5° Figure 1.31c

The angle θ measured counterclockwise from the + x-axis. In terms of compass directions, the resultant displacement is 38.5° N of W.  EVALUATE: Rx < 0 and Ry > 0, so R is in the 2nd quadrant. This agrees with the vector addition diagram. 1.32.

IDENTIFY: This problem involves vector addition. We know one vector and the resultant of that vector with a second vector, and we want to find the magnitude and direction of the second vector.       SET UP: A + B = R . We know A and R and want to find B . Use Ax + Bx = Rx and  By to Ay + By = Ry to find the components of B , then use B = Bx2 + By2 to find B and θ = arctan Bx

find its direction. First do a graphical sum, as shown in Fig. 1.32.

Figure 1.32

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Units, Physical Quantities, and Vectors

1-15

EXECUTE: First find the components. Ax = A cos 127° = (8.00 m) cos 127° = –4.185 m, Rx = 0, Ay = A sin 127° = (8.00 m) sin 127° = 6.389 m, Ry = –12.0 m. Now use Ax + Bx = Rx and Ay + By = Ry to find  the components of B . –4.185 m + Bx = 0 → Bx = 4.185 m. 6.389 m + By = –12.0 m → By = –18.39 m.

B = Bx2 + By2 =

  −18.39 m  = arctan   = −75.3°. From Fig. 1.32 we can see that B must point below the Bx  4.185 m   x-axis. This tells us that B makes an angle of 75.3° clockwise below the +x-axis, which we can also express as 360° – 75.3° = 284.7° counterclockwise with the +x-axis. EVALUATE: Our vector sum in Fig. 1.32 agrees with our calculations.     IDENTIFY: Vector addition problem. A − B = A + (− B ).   SET UP: Find the x- and y-components of A and B. Then the x- and y-components of the vector sum   are calculated from the x- and y-components of A and B. EXECUTE:

θ = arctan

1.33.

(4.185 m)2 + (–18.39 m) 2 = 19.0 m.

Ax = A cos(60.0°) Ax = (2.80 cm)cos(60.0°) = +1.40 cm Ay = A sin (60.0°) Ay = (2.80 cm)sin (60.0°) = +2.425 cm Bx = B cos( −60.0°) Bx = (1.90 cm)cos( −60.0°) = +0.95 cm

By = B sin (−60.0°) By = (1.90 cm)sin (−60.0°) = −1.645 cm Note that the signs of the components correspond to the directions of the component vectors. Figure 1.33a

   (a) Now let R = A + B. Rx = Ax + Bx = +1.40 cm + 0.95 cm = +2.35 cm.

Ry = Ay + By = +2.425 cm − 1.645 cm = +0.78 cm. R = Rx2 + Ry2 = (2.35 cm) 2 + (0.78 cm) 2 R = 2.48 cm +0.78 cm = = +0.3319 tan θ = Rx +2.35 cm Ry

θ = 18.4° Figure 1.33b

   EVALUATE: The vector addition diagram for R = A + B is

1-16

Chapter 1

 R is in the 1st quadrant, with | Ry | < | Rx | , in agreement with our calculation.

Figure 1.33c

   (b) EXECUTE: Now let R = A − B. Rx = Ax − Bx = +1.40 cm − 0.95 cm = +0.45 cm.

R y = Ay − By = +2.425 cm + 1.645 cm = +4.070 cm. R = Rx2 + Ry2 = (0.45 cm) 2 + (4.070 cm) 2 R = 4.09 cm 4.070 cm = = +9.044 tan θ = Rx 0.45 cm Ry

θ = 83.7°

Figure 1.33d

   EVALUATE: The vector addition diagram for R = A + (− B ) is

 R is in the 1st quadrant, with | Rx | < | Ry |, in agreement with our calculation.

Figure 1.33e (c) EXECUTE:

Units, Physical Quantities, and Vectors

1-17

    B − A = −( A − B )     B − A and A − B are equal in magnitude and opposite in direction. R = 4.09 cm and θ = 83.7° + 180° = 264°

Figure 1.33f

   EVALUATE: The vector addition diagram for R = B + ( − A) is  R is in the 3rd quadrant, with | Rx | < | Ry |, in agreement with our calculation.

Figure 1.33g 1.34.

IDENTIFY: The general expression for a vector written in terms of components and unit vectors is  A = Ax iˆ + Ay ˆj.    SET UP: 5.0 B = 5.0(4iˆ − 6 ˆj ) = 20i − 30 j EXECUTE: (a) Ax = 5.0, Ay = −6.3 (b) Ax = 11.2, Ay = −9.91 (c) Ax = −15.0, Ay = 22.4 (d) Ax = 20, Ay = −30 EVALUATE: The components are signed scalars.

1.35.

 IDENTIFY: Find the components of each vector and then use the general equation A = Ax iˆ + Ay ˆj for a vector in terms of its components and unit vectors. SET UP: Ax = 0, Ay = −8.00 m. Bx = 7.50 m, By = 13.0 m. C x = −10.9 m, C y = −5.07 m.

Dx = −7.99 m, Dy = 6.02 m.    EXECUTE: A = (−8.00 m) ˆj; B = (7.50 m) iˆ + (13.0 m) ˆj; C = (−10.9 m) iˆ + (−5.07 m) ˆj;  D = ( −7.99 m) iˆ + (6.02 m) ˆj. EVALUATE: All these vectors lie in the xy-plane and have no z-component. 1.36.

IDENTIFY: Find A and B. Find the vector difference using components. SET UP: Identify the x- and y-components and use A = Ax2 + Ay2 .  EXECUTE: (a) A = 4.00iˆ + 7.00 ˆj; Ax = +4.00; Ay = +7.00.

1-18

Chapter 1

 A = Ax2 + Ay2 = (4.00) 2 + (7.00) 2 = 8.06. B = 5.00iˆ − 2.00 ˆj; Bx = +5.00; By = −2.00;

B = Bx2 + By2 = (5.00) 2 + ( −2.00) 2 = 5.39.   EVALUATE: Note that the magnitudes of A and B are each larger than either of their components.   EXECUTE: (b) A − B = 4.00iˆ + 7.00 ˆj − (5.00iˆ − 2.00 ˆj ) = (4.00 − 5.00) iˆ + (7.00 + 2.00) ˆj.   A − B = −1.00iˆ + 9.00 ˆj    (c) Let R = A − B = −1.00iˆ + 9.00 ˆj. Then Rx = −1.00, Ry = 9.00. R = Rx2 + Ry2

R = ( −1.00) 2 + (9.00) 2 = 9.06. tan θ =

Ry Rx

9.00 = −9.00 −1.00

θ = −83.6° + 180° = 96.3°.

Figure 1.36

 EVALUATE: Rx < 0 and Ry > 0, so R is in the 2nd quadrant. 1.37.

IDENTIFY: Use trigonometry to find the components of each vector. Use Rx = Ax + Bx +  and  Ry = Ay + By +  to find the components of the vector sum. The equation A = Ax iˆ + Ay ˆj expresses a

vector in terms of its components. SET UP: Use the coordinates in the figure that accompanies the problem.  EXECUTE: (a) A = (3.60 m)cos 70.0°iˆ + (3.60 m)sin 70.0° ˆj = (1.23 m)iˆ + (3.38 m) ˆj  B = −(2.40 m)cos30.0°iˆ − (2.40 m)sin 30.0° ˆj = ( −2.08 m)iˆ + (−1.20 m) ˆj    (b) C = (3.00) A − (4.00) B = (3.00)(1.23 m)iˆ + (3.00)(3.38 m) ˆj − (4.00)(−2.08 m) iˆ − (4.00)( −1.20 m) ˆj  C = (12.01 m)iˆ + (14.94 m) ˆj (c) From A = Ax2 + Ay2 and tan θ =

Ay Ax

 14.94 m  C = (12.01 m) 2 + (14.94 m) 2 = 19.17 m, arctan   = 51.2°  12.01 m  EVALUATE: C x and C y are both positive, so θ is in the first quadrant. 1.38.

IDENTIFY: We use the vector components and trigonometry to find the angles. SET UP: Use the fact that tan θ = Ay / Ax . EXECUTE: (a) tan θ = Ay / Ax = (b) tan θ = By / Bx =

6.00 . θ = 117° with the +x-axis. −3.00

2.00 . θ = 15.9°. 7.00

Units, Physical Quantities, and Vectors

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 (c) First find the components of C . Cx = Ax + Bx = –3.00 + 7.00 = 4.00,

1.39.

Cy = Ay + By = 6.00 + 2.00 = 8.00 8.00 = 2.00 . θ = 63.4° tan θ = C y / C x = 4.00 EVALUATE: Sketching each of the three vectors to scale will show that the answers are reasonable.     IDENTIFY: A and B are given in unit vector form. Find A, B and the vector difference A − B.         SET UP: A = −2.00i + 3.00 j + 4.00k , B = 3.00i + 1.00 j − 3.00k Use A = Ax2 + Ay2 + Az2 to find the magnitudes of the vectors. EXECUTE: (a) A = Ax2 + Ay2 + Az2 = (−2.00) 2 + (3.00)2 + (4.00) 2 = 5.38

B = Bx2 + By2 + Bz2 = (3.00) 2 + (1.00) 2 + ( −3.00) 2 = 4.36   (b) A − B = ( −2.00iˆ + 3.00 ˆj + 4.00kˆ ) − (3.00iˆ + 1.00 ˆj − 3.00kˆ )   A − B = ( −2.00 − 3.00) iˆ + (3.00 − 1.00) ˆj + (4.00 − ( −3.00))kˆ = −5.00iˆ + 2.00 ˆj + 7.00kˆ.    (c) Let C = A − B, so C x = −5.00, C y = +2.00, C z = +7.00 C = C x2 + C y2 + C z2 = (−5.00) 2 + (2.00)2 + (7.00) 2 = 8.83         B − A = −( A − B ), so A − B and B − A have the same magnitude but opposite directions. 1.40.

1.41.

EVALUATE: A, B, and C are each larger than any of their components.   IDENTIFY: Target variables are A ⋅ B and the angle φ between the two vectors.   SET UP: We are given A and B in unit vector form and can take the scalar product using     A ⋅ B = Ax Bx + Ay By + Az Bz . The angle φ can then be found from A ⋅ B = AB cos φ .   EXECUTE: (a) A = 4.00iˆ + 7.00 ˆj , B = 5.00iˆ − 2.00 ˆj; A = 8.06, B = 5.39.   A ⋅ B = (4.00iˆ + 7.00 ˆj ) ⋅ (5.00iˆ − 2.00 ˆj ) = (4.00)(5.00) + (7.00)(−2.00) = 20.0 − 14.0 = +6.00.   A⋅ B 6.00 (b) cos φ = = = 0.1382; φ = 82.1°. AB (8.06)(5.39)    EVALUATE: The component of B along A is in the same direction as A, so the scalar product is positive and the angle φ is less than 90°.   IDENTIFY: A ⋅ B = AB cos φ       SET UP: For A and B, φ = 150.0°. For B and C , φ = 145.0°. For A and C , φ = 65.0°.   EXECUTE: (a) A ⋅ B = (8.00 m)(15.0 m)cos150.0° = −104 m 2   (b) B ⋅ C = (15.0 m)(12.0 m)cos145.0° = −148 m 2   (c) A ⋅ C = (8.00 m)(12.0 m)cos65.0° = 40.6 m 2 EVALUATE: When φ < 90° the scalar product is positive and when φ > 90° the scalar product is

negative. 1.42.

  IDENTIFY: Target variable is the vector A × B expressed in terms of unit vectors.   SET UP: We are given A and B in unit vector form and can take the vector product using iˆ × iˆ = ˆj × ˆj = 0 , iˆ × ˆj = kˆ , and ˆj × iˆ = − kˆ .   EXECUTE: A = 4.00iˆ + 7.00 ˆj , B = 5.00iˆ − 2.00 ˆj.

1-20

Chapter 1

  A × B = (4.00iˆ + 7.00 ˆj ) × (5.00iˆ − 2.00 ˆj ) = 20.0iˆ × iˆ − 8.00iˆ × ˆj + 35.0 ˆj × iˆ − 14.0 ˆj × ˆj. But   iˆ × iˆ = ˆj × ˆj = 0 and iˆ × ˆj = kˆ , ˆj × iˆ = −kˆ , so A × B = −8.00kˆ + 35.0(− kˆ ) = −43.0kˆ. The magnitude of   A × B is 43.0.   EVALUATE: Sketch the vectors A and B in a coordinate system where the xy-plane is in the plane of   the paper and the z-axis is directed out toward you. By the right-hand rule A × B is directed into the plane of the paper, in the − z -direction. This agrees with the above calculation that used unit vectors.

Figure 1.42 1.43.

  IDENTIFY: For all of these pairs of vectors, the angle is found from combining A ⋅ B = AB cos φ and      A⋅ B   Ax Bx + Ay By  A ⋅ B = Ax Bx + Ay By + Az Bz , to give the angle φ as φ = arccos   = arccos  . AB    AB    SET UP: A ⋅ B = Ax Bx + Ay By + Az Bz shows how to obtain the components for a vector written in terms of unit vectors.    −22  EXECUTE: (a) A ⋅ B = −22, A = 40, B = 13, and so φ = arccos   = 165°.  40 13    60   (b) A ⋅ B = 60, A = 34, B = 136, φ = arccos   = 28°. 34 136     (c) A ⋅ B = 0 and φ = 90°.       EVALUATE: If A ⋅ B > 0, 0 ≤ φ < 90°. If A ⋅ B < 0, 90° < φ ≤ 180°. If A ⋅ B = 0, φ = 90° and the two

vectors are perpendicular. 1.44.

  IDENTIFY: The right-hand rule gives the direction and |A × B| = AB sin φ gives the magnitude. SET UP: φ = 120.0°.   EXECUTE: (a) The direction of A× B is into the page (the − z -direction ). The magnitude of the vector product is AB sin φ = (2.80 cm)(1.90 cm)sin120° = 4.61 cm 2 .       (b) Rather than repeat the calculations, B × A = – A× B may be used to see that B × A has magnitude 4.61 cm 2 and is in the + z -direction (out of the page). EVALUATE: For part (a) we could use the components of the cross product and note that the only nonvanishing component is Cz = Ax By − Ay Bx = (2.80 cm)cos60.0°( −1.90 cm)sin 60° − (2.80 cm)sin 60.0°(1.90 cm)cos 60.0° = −4.61 cm 2 .

1.45.

This gives the same result.   IDENTIFY: A × D has magnitude AD sin φ . Its direction is given by the right-hand rule. SET UP: φ = 180° − 53° = 127°     EXECUTE: (a) | A × D| = (8.00 m)(10.0 m)sin127° = 63.9 m 2 . The right-hand rule says A × D is in the − z -direction (into the page).     (b) D × A has the same magnitude as A × D and is in the opposite direction.

Units, Physical Quantities, and Vectors

1-21

  EVALUATE: The component of D perpendicular to A is D⊥ = D sin 53.0° = 7.99 m.   | A × D| = AD⊥ = 63.9 m 2 , which agrees with our previous result. 1.46.

IDENTIFY: Apply Eqs. (1.16) and (1.20). SET UP: The angle between the vectors is 20° + 90° + 30° = 140°.     EXECUTE: (a) A ⋅ B = AB cos φ gives A ⋅ B = (3.60 m)(2.40 m)cos140° = −6.62 m 2 .   (b) From |A × B| = AB sin φ , the magnitude of the cross product is (3.60 m)(2.40 m)sin140° = 5.55 m 2

and the direction, from the right-hand rule, is out of the page (the + z -direction ).   EVALUATE: We could also use A ⋅ B = Ax Bx + Ay By + Az Bz and the cross product, with the   components of A and B . 1.47.

IDENTIFY: This problem involves the vector product of two vectors.     SET UP: The magnitude is A × B = AB sin φ and the right-hand rule gives the direction. Since A × B   is in the +z direction, both A and B must lie in the xy-plane.   EXECUTE: A has no y-component and B has no x-component, so they must be perpendicular to each    other. Since A × B is in the +z direction, the right-hand rule tells us that B must point in the –y   direction. A × B = AB sin φ = (8.0 m)B sin 90° = 16.0 m2, so B = 2.0 m.  EVALUATE: In unit vector notation, B = –2.0 m ĵ .

1.48.

IDENTIFY: This problem involves the vector product and the scalar of two vectors.     SET UP: The scalar product is A ⋅ B = AB cos φ and the magnitude of the vector product is A × B =

AB sin φ .

  EXECUTE: (a) Calculate both products. A × B = AB sin φ = AB sin 30.0° = 0.500 AB and   A ⋅ B = AB cos φ = AB cos 30.0° = 0.866 AB. Therefore the scalar product has the greater magnitude. (b) Equate the magnitudes. AB sin φ = AB cos φ → tan φ = 1 → φ = 45° or 135°. At 45° both

products are positive, but at 135° the scalar product is negative. However in both cases the magnitudes are the same. EVALUATE: Note that the problem says that the magnitudes of the products are equal. We cannot say     that the products are equal because A ⋅ B is a scalar but A × B is a vector. 1.49.

IDENTIFY: We model the earth, white dwarf, and neutron star as spheres. Density is mass divided by volume. SET UP: We know that density = mass/volume = m/V where V = 43 π r 3 for a sphere. From Appendix B,

the earth has mass of m = 5.97 × 1024 kg and a radius of r = 6.37 × 106 m whereas for the sun at the end of its lifetime, m = 1.99 × 1030 kg and r = 7500 km = 7.5 × 106 m. The star possesses a radius of r = 10 km = 1.0 × 104 m and a mass of m = 1.99 × 1030 kg. EXECUTE: (a) The earth has volume V = 43 π r 3 = 43 π (6.37 × 106 m)3 = 1.0827 × 1021 m3 . Its density is

density =

3 3 m 5.97 × 1024 kg 3 3  10 g   1 m  3 = = . × (5 51 10 kg/m )    1 kg   102 cm  = 5.51 g/cm V 1.0827 × 1021 m3  

(b) V = 43 π r 3 = 43 π (7.5 × 106 m)3 = 1.77 × 1021 m3

density =

 1 g/cm3  m 1.99 × 1030 kg 6 3 = = (1.1 × 109 kg/m3 )  21 3 3  = 1.1 × 10 g/cm V 1.77 × 10 m 1000 kg/m  

1-22

Chapter 1 (c) V = 43 π r 3 = 43 π (1.0 × 104 m)3 = 4.19 × 1012 m3

density =

 1 g/cm3  m 1.99 × 1030 kg 14 3 = = (4.7 × 1017 kg/m3 )  12 3 3  = 4.7 × 10 g/cm V 4.19 × 10 m 1000 kg/m  

EVALUATE: For a fixed mass, the density scales as 1/r 3 . Thus, the answer to (c) can also be obtained from (b) as 3

 7.50 × 106 m  14 3 (1.1 × 10 g/cm )   = 4.7 × 10 g/cm . 4  1.0 × 10 m  6

1.50.

IDENTIFY and SET UP: Unit conversion. EXECUTE: (a) f = 1.420 × 109 cycles/s, so (b)

1 s = 7.04 × 10−10 s for one cycle. 1.420 × 109

3600 s/h = 5.11 × 1012 cycles/h 7.04 × 10−10 s/cycle

cycle: (4.6 × 109 y)(3.156 × 107 s/y) = 2.1 × 1026 cycles 7.04 × 10−10 s/cycle (d) The clock is off by 1 s in 100,000 y = 1 × 105 y, so in 4.60 × 109 y it is off by

 4.60 × 109  4 (1 s)   = 4.6 × 10 s (about 13 h). 5  1 10 ×   EVALUATE: In each case the units in the calculation combine algebraically to give the correct units for the answer. 1.51.

IDENTIFY: The density relates mass and volume. Use the given mass and density to find the volume and from this the radius. SET UP: The earth has mass mE = 5.97 × 1024 kg and radius rE = 6.37 × 106 m. The volume of a sphere

is V = 43 π r 3 . ρ = 1.76 g/cm 3 = 1760 km/m3. EXECUTE: (a) The planet has mass m = 5.5mE = 3.28 × 1025 kg.

3.28 × 1025 kg = 1.86 × 1022 m3 . 1760 kg/m3 1/3

 3V  r =   4π 

1/3

 3[1.86 × 1022 m3 ]  =   4π  

= 1.64 × 107 m = 1.64 × 104 km

(b) r = 2.57 rE EVALUATE: Volume V is proportional to mass and radius r is proportional to V 1/3 , so r is proportional

to m1/3 . If the planet and earth had the same density its radius would be (5.5)1/3 rE = 1.8rE . The radius of the planet is greater than this, so its density must be less than that of the earth. 1.52.

IDENTIFY: Use the extreme values in the piece’s length and width to find the uncertainty in the area. SET UP: The length could be as large as 7.61 cm and the width could be as large as 1.91 cm. EXECUTE: (a) The area is 14.44 ± 0.095 cm2.

Units, Physical Quantities, and Vectors

1-23

0.095 cm 2 = 0.66%, and the fractional uncertainties in the 14.44 cm 2 0.01 cm 0.01 cm = 0.13% and = 0.53%. The sum of these fractional uncertainties length and width are 7.61 cm 1.9 cm is 0.13% + 0.53% = 0.66%, in agreement with the fractional uncertainty in the area. (b) The fractional uncertainty in the area is

EVALUATE: The fractional uncertainty in a product of numbers is greater than the fractional uncertainty in any of the individual numbers. 1.53.

IDENTIFY: The number of atoms is your mass divided by the mass of one atom. SET UP: Assume a 70-kg person and that the human body is mostly water. Use Appendix D to find the

mass of one H 2O molecule: 18.015 u × 1.661 × 10−27 kg/u = 2.992 × 10−26 kg/molecule. EXECUTE: (70 kg)/(2.992 × 10−26 kg/molecule) = 2.34 × 1027 molecules. Each H 2O molecule has

3 atoms, so there are about 6 × 1027 atoms. EVALUATE: Assuming carbon to be the most common atom gives 3 × 1027 molecules, which is a result of the same order of magnitude. 1.54.

IDENTIFY: Estimate the volume of each object. The mass m is the density times the volume. SET UP: The volume of a sphere of radius r is V = 43 π r 3 . The volume of a cylinder of radius r and

length l is V = π r 2l. The density of water is 1000 kg/m3. EXECUTE: (a) Estimate the volume as that of a sphere of diameter 10 cm: V = 5.2 × 10−4 m3.

m = (0.98)(1000 kg / m3 )(5.2 × 10−4 m3 ) = 0.5 kg. (b) Approximate as a sphere of radius r = 0.25μ m (probably an overestimate): V = 6.5 × 10−20 m3. m = (0.98)(1000 kg / m 3 )(6.5 × 10−20 m 3 ) = 6 × 10−17 kg = 6 × 10 −14 g.

m = (0.98)(1000 kg/m3 )(2.8 × 10−7 m3 ) = 3 × 10−4 kg = 0.3 g. EVALUATE: The mass is directly proportional to the volume. 1.55.

IDENTIFY: We are dealing with unit vectors, which must have magnitude 1. We will need to use the scalar product and to express vectors using the unit vectors.   SET UP: A = Ax2 + Ay2 + Az2 , A ⋅ B = Ax Bx + Ay By + Az Bz . If two vectors are perpendicular, their

scalar product is zero. EXECUTE: (a) If we divide a vector by its magnitude, the result will have magnitude 1 but still point in the same direction as the original vector, so it will be a unit vector. First find the magnitude of the given 3.0iˆ − 4.0kˆ = 0.60 iˆ – 0.80 k̂ is a unit vector. A = Ax2 + Az2 = (3.0) 2 + (–4.0) 2 = 5.0. Therefore 5  vector that is parallel to A .  (b) Reversing the direction of the unit vector in (a) will make it antiparallel to A , so the unit vector is – 0.60 iˆ + 0.80 k̂ .  (c) Call B the unknown unit vector. Since it has no y-component, we can express it as      B = Bx iˆ + Bz k̂ . Since A and B are perpendicular, A ⋅ B = 0, so Ax Bx + Ay By + Az Bz = 0. This gives  2 (3.0)Bx – (4.0)Bz = 0 → Bz = 0.75 Bx. Since B is a unit vector, we have Bx2 + Bz2 = Bx2 + ( 0.75 Bx ) =  1. Solving gives Bx = ±0.80. Therefore Bz = ±(0.75 Bx) = ±(0.75)(0.80) = ±0.60. Therefore B = ±(0.80 iˆ + 0.60 k̂ ), so the two unit vectors are

1-24

Chapter 1

  B+ = 0.80 iˆ + 0.60 k̂ and B– = –0.80 iˆ – 0.60 k̂ .     EVALUATE: A B+ = (3.0)(0.80) + (–4.0)(6.0) = 0 and A ⋅ B– = (3.0)(–0.80) + (–4.0)(–6.0) = 0, so the  two vectors are perpendicular to A . Their magnitudes are (±0.80)2 + (±0.60)2 = 1, so they are unit vectors. 1.56.

          IDENTIFY: Let D be the fourth force. Find D such that A + B + C + D = 0, so D = −( A + B + C ).  SET UP: Use components and solve for the components Dx and D y of D. EXECUTE:

Ax = + A cos30.0° = +86.6 N, Ay = + A sin 30.0° = +50.00 N.

Bx = − B sin 30.0° = −40.00 N, By = + B cos30.0° = +69.28 N. C x = −C cos53.0° = −24.07 N, C y = −C sin 53.0° = −31.90 N.

Then Dx = −22.53 N, Dy = −87.34 N and D = Dx2 + Dy2 = 90.2 N. tan α = | Dy /Dx | = 87.34/22.53.

α = 75.54°. φ = 180° + α = 256°, counterclockwise from the + x-axis.

 EVALUATE: As shown in Figure 1.56, since Dx and D y are both negative, D must lie in the third quadrant.

: dO Figure 1.56 1.57.

IDENTIFY: Vector addition. Target variable is the 4th displacement. SET UP: Use a coordinate system where east is in the + x -direction and north is in the + y -direction.     Let A, B, and C be the three displacements that are given and let D be the fourth unmeasured      displacement. Then the resultant displacement is R = A + B + C + D. And since she ends up back where  she started, R = 0.         0 = A + B + C + D, so D = −( A + B + C )

Dx = −( Ax + Bx + Cx ) and Dy = −( Ay + By + C y ) EXECUTE:

Ax = −180 m, Ay = 0 Bx = B cos315° = (210 m)cos315° = +148.5 m B y = B sin 315° = (210 m)sin 315° = −148.5 m C x = C cos60° = (280 m)cos60° = +140 m

C y = C sin 60° = (280 m)sin 60° = +242.5 m

Figure 1.57a

Dx = −( Ax + Bx + C x ) = −(−180 m + 148.5 m + 140 m) = −108.5 m

Browsegrades.net D y = −( Ay + By + C y ) = −(0 − 148.5 m + 242 .5 m) = −94.0 m

Units, Physical Quantities, and Vectors

1-25

D = Dx2 + Dy2 D = (−108.5 m)2 + (−94.0 m)2 = 144 m

tan θ =

Dy Dx

−94.0 m = 0.8664 −108.5 m

θ = 180° + 40.9° = 220.9°  ( D is in the third quadrant since both Dx and D y are negative.) Figure 1.57b

  The direction of D can also be specified in terms of φ = θ − 180° = 40.9°; D is 41° south of west. EVALUATE: The vector addition diagram, approximately to scale, is

 Vector D in this diagram agrees qualitatively with our calculation using components.

Figure 1.57c 1.58.

IDENTIFY: Find the vector sum of the two displacements.      SET UP: Call the two displacements A and B, where A = 170 km and B = 230 km. A + B = R.   A and B are as shown in Figure 1.58. EXECUTE: Rx = Ax + Bx = (170 km)sin 68° + (230 km)cos36° = 343.7 km.

Ry = Ay + By = (170 km)cos68° − (230 km)sin 36° = −71.5 km.

a R = Rx2 + Ry2 = (343.7 km) 2 + ( −71.5 km) 2 = 351 km. tanθ R = |

Ry Rx

71.5 km = 0.208. 343.7 km

θ R = 11.8° south of east.

 EVALUATE: Our calculation using components agrees with R shown in the vector addition diagram, Figure 1.58.

Figure 1.58

1-26

Chapter 1

1.59.

IDENTIFY: This problem requires vector addition. We can find the components of the given vectors and then use them to find the magnitude and direction of the resultant vector. Ay SET UP: Ax = A cos θ , Ay = A sin θ , θ = arctan , A = Ax2 + Ay2 + Az2 , Rx = Ax + Bx , and Ry Ax

= Ay + By . Sketch the given vectors to help find the components (see Fig. 1.59).

Figure 1.59 EXECUTE: From Fig. 1.59 we can see that the component s are E1x = 0 and E1y = 450 N/C E2x = E2 cos θ = (600 N/C) cos 233.1° = –360.25 N/C E2y = E2 sin θ = (600 N/C) sin 233.1° = –479.81 N/C. Now find the components of the resultant field: Ex = E1x + E2x = 0 + (–360.25 N/C) = –360.25 N/C Ey = E1y + E2y = 450 N/C + (–479.81 N/C) = –29.81 N/C  Now find the magnitude and direction of E :

Ex2 + E y2 =

θ = arctan

(–360.25 N/C) 2 + (–29.81 N/C)2 = 361 N/C.

  –29.81 N/C  = θ = arctan   = 4.73°. Both components of E are negative, so it must Ax  –360.25 N/C 

point into the third quadrant. Therefore the angle below the –x-axis is 4.73°. The angle with the +x-axis is 180° + 4.73° = 184.73°. EVALUATE: Make a careful graphical sum to check your answer. 1.60.

IDENTIFY: Solve for one of the vectors in the vector sum. Use components. SET UP: Use coordinates for which + x is east and + y is north. The vector displacements are:    A = 2.00 km, 0°of east; B = 3.50 m, 45° south of east; and R = 5.80 m, 0° east EXECUTE: C x = Rx − Ax − Bx = 5.80 km − (2.00 km) − (3.50 km)(cos 45°) = 1.33 km;

C y = Ry − Ay − By = 0 km − 0 km − ( −3.50 km)(sin 45°) = 2.47 km; C = (1.33 km) 2 + (2.47 km)2 = 2.81 km;

θ = tan −1[(2.47 km)/(1.33 km)] = 61.7° north of east. The vector addition diagram in Figure 1.60 shows good qualitative agreement with these values.

Browsegrades.net

Units, Physical Quantities, and Vectors

1-27

EVALUATE: The third leg lies in the first quadrant since its x and y components are both positive.

Figure 1.60 1.61.

IDENTIFY: We know the resultant of two forces of known equal magnitudes and want to find that magnitude (the target variable). SET UP: Use coordinates having a horizontal + x axis and an upward + y axis. Then Ax + Bx = Rx and

Rx = 12.8 N. SOLVE: Ax + Bx = Rx and A cos32° + B sin 32° = Rx . Since A = B,

2 A cos32° = Rx , so A =

1.62.

Rx = 7.55 N. (2)(cos32°)

EVALUATE: The magnitude of the x component of each pull is 6.40 N, so the magnitude of each pull (7.55 N) is greater than its x component, as it should be.       IDENTIFY: The four displacements return her to her starting point, so D = −( A + B + C ), where A, B,   and C are in the three given displacements and D is the displacement for her return. SET UP: Let + x be east and + y be north. EXECUTE: (a) Dx = −[(147 km)sin85° + (106 km)sin167° + (166 km)sin 235°] = −34.3 km.

Dy = −[(147 km)cos85° + (106 km)cos167° + (166 km)cos 235°] = +185.7 km. D = (−34.3 km)2 + (185.7 km) 2 = 189 km.  34.3 km  (b) The direction relative to north is φ = arctan   = 10.5°. Since Dx < 0 and Dy > 0, the  185.7 km   direction of D is 10.5° west of north. EVALUATE: The four displacements add to zero. 1.63.

IDENTIFY: We have two known vectors and a third unknown vector, and we know the resultant of these three vectors. SET UP: Use coordinates for which + x is east and + y is north. The vector displacements are:     A = 23.0 km at 34.0° south of east; B = 46.0 km due north; R = 32.0 km due west ; C is unknown. EXECUTE: C x = Rx − Ax − Bx = −32.0 km − (23.0 km)cos34.0° − 0 = −51.07 km;

C y = Ry − Ay − By = 0 − ( −23.0 km)sin34.0° − 46.0 km = −33.14 km; C = C x2 + C y2 = 60.9 km  Calling θ the angle that C makes with the –x-axis (the westward direction), we have 33.14 ; θ = 33.0° south of west. tan θ = C y / C x = 51.07 EVALUATE: A graphical vector sum will confirm this result.

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Chapter 1

1.64.

   IDENTIFY: Let the three given displacements be A, B and C , where A = 40 steps, B = 80 steps and       C = 50 steps. R = A + B + C . The displacement C that will return him to his hut is − R. SET UP: Let the east direction be the + x -direction and the north direction be the + y -direction. EXECUTE: (a) The three displacements and their resultant are sketched in Figure 1.64. (b) Rx = (40)cos 45° − (80)cos 60° = −11.7 and R y = (40)sin 45° + (80)sin 60° − 50 = 47.6.

The magnitude and direction of the resultant are

 47.6  (−11.7) 2 + (47.6) 2 = 49, acrtan   = 76°, north  11.7 

 of west. We know that R is in the second quadrant because Rx < 0, Ry > 0. To return to the hut, the explorer must take 49 steps in a direction 76° south of east, which is 14° east of south.  EVALUATE: It is useful to show Rx , Ry , and R on a sketch, so we can specify what angle we are computing.

Figure 1.64 1.65.

    IDENTIFY: We want to find the resultant of three known displacement vectors: R = A + B + C . SET UP: Let + x be east and + y be north and find the components of the vectors.

EXECUTE: The magnitudes are A = 20.8 m, B = 38.0 m, C = 18.0 m. The components are Ax = 0, Ay = 28.0 m, Bx = 38.0 m, By = 0, Cx = –(18.0 m)(sin33.0°) = –9.804 m, Cy = –(18.0 m)(cos33.0°) = –15.10 m Rx = Ax + Bx + Cx = 0 + 38.0 m + (–9.80 m) = 28.2 m Ry = Ay + By + Cy = 20.8 m + 0 + (–15.10 m) = 5.70 m

R = Rx2 + Ry2 = 28.8 m is the distance you must run. Calling θ R the angle the resultant makes with the +x-axis (the easterly direction), we have tan θ R = Ry/Rx = (5.70 km)/(28.2 km); θ R = 11.4° north of east. EVALUATE: A graphical sketch will confirm this result. 1.66.

IDENTIFY: This is a problem in vector addition. We want the magnitude of the resultant of four known displacement vectors in three dimensions. We can use components to do this. SET UP: Rx = Ax + Bx + C x + Dx , and likewise for the other components. The magnitude is

R = Rx2 + Ry2 + Rz2 . Call the +x-axis toward the east, the +y-axis toward the north, and the +z-axis vertically upward. First find the components of the four displacements, then use them to find the magnitude of the resultant displacement. EXECUTE: Finding the components in the order in which the displacements are listed in the problem, the components of the four displacement vectors are Ax = –14.0 m, Bz = +22.0 m, Cy = +12.0 m, and Dx  = 6.0 m. All the other components are zero. Now find the components of R .

Units, Physical Quantities, and Vectors

1-29

Rx = –14.0 m + 6.0 m = –8.0 m, Ry = 12.0 m, Rz = 22.0 m. Now find the resultant displacement using R = Rx2 + Ry2 + Rz2 . R =

(–8.0 m)2 + (12.0 m)2 + (22.0 m)2 = 26 m.

EVALUATE: This is one case where a graphical solution would not be useful as a check since threedimensional drawings are very difficult to visualize. Note that the answer has only 2 significant figures even though all the given numbers have 3 significant figures. The reason for this is that in the subtraction to find Rx we lost one significant figure because –14.0 + 6.0 = 8.0, which has only 2 significant figures. 1.67.

IDENTIFY: We know the resultant of two vectors and one of the vectors, and we want to find the second vector. SET UP: Let the westerly direction be the + x-direction and the northerly direction be the + y -direction.     We also know that R = A + B where R is the vector from you to the truck. Your GPS tells you that you are 122.0 m from the truck in a direction of 58.0° east of south, so a vector from the truck to you is 122.0 m at 58.0° east of south. Therefore the vector from you to the truck is 122.0 m at 58.0° west of   north. Thus R = 122.0 m at 58.0° west of north and A is 72.0 m due west. We want to find the  magnitude and direction of vector B . EXECUTE: Bx = Rx – Ax = (122.0 m)(sin 58.0°) – 72.0 m = 31.462 m

By = Ry – Ay = (122.0 m)(cos 58.0°) – 0 = 64.450 m; B = Bx2 + By2 = 71.9 m . 64.650 m = 2.05486 ; θ B = 64.1° north of west. 31.462 m EVALUATE: A graphical sum will show that the results are reasonable. tan θ B = By / Bx =

1.68.

IDENTIFY: We are dealing with vector addition. We know the resultant vector is the same for both trips, but we take different displacements on two different days. It is best to use components. SET UP: First make a clear sketch showing the displacement vectors on the two different days (see Fig.

1.68). Use Rx = Ax + Bx and Ry = Ay + By and A = Ax2 + Ay2 for the magnitude of a vector. Let the xaxis be eastward the y-axis be toward the north.

Figure 1.68 EXECUTE: (a) The magnitude R of the resultant is the distance to the store from your apartment. Using      the Saturday trip, let A be the first drive and B be the second drive, so A + B = R . First find the  components of R .

1-30

Chapter 1

Rx = Ax + Bx = 0 + (1.40 km) cos 150.0° = –1.212 km Ry = Ay + By = 0.600 km + (1.40 km) sin 150.0° = 1.30 km. Now find R: R =

Rx2 + R y2 =

(–1.212 km) 2 + (1.30 km) 2 = 1.78 km.

(b) The distance traveled each day is not the magnitude of the resultant. Rather, it is the sum of the magnitudes of both displacement vectors for each trip. We know all of them except the second drive on   the Sunday trip. Call C the first drive on Sunday and D the second Sunday drive. The resultant of  these is the same as for the Saturday trip, so we can find the components of D . Cx + Dx = Rx: (0.80 km) cos 130° + Dx = –1.212 km → Dx = –0.6978 km → Dy = 0.6872 km. Cy + Dy = Ry: (0.80 km) sin 130° + Dy = 1.30 km  2 2 Now find the magnitude of D : D = Dx + Dy = (–0.6978 km) 2 + (0.6872 km)2 = 0.9793 km.

The distances driven on the two days are Saturday: 0.60 km + 1.40 km = 2.00 km Sunday: 0.80 km + 0.9793 km = 1.7793 km The difference in distance is 2.00 km – 1.7793 km = 0.22 km. On Saturday you drove 0.22 km farther than on Sunday. EVALUATE: Even though the resultant displacement was the same on both days, you drove different distances on the two days because you took different paths. 1.69.

IDENTIFY: The sum of the four displacements must be zero. Use components.      SET UP: Call the displacements A, B, C , and D, where D is the final unknown displacement for    the return from the treasure to the oak tree. Vectors A, B, and C are sketched in Figure 1.69a.     A + B + C + D = 0 says Ax + Bx + C x + Dx = 0 and Ay + By + C y + Dy = 0. A = 825 m, B = 1250 m,

and C = 1000 m. Let + x be eastward and + y be north. EXECUTE: (a) Ax + Bx + C x + Dx = 0 gives

Dx = −( Ax + Bx + C x ) = −[0 − (1250 m)sin 30.0° + (1000 m)cos32.0°] = −223.0 m. Ay + By + C y + Dy = 0 gives Dy = −( Ay + By + C y ) = −[−825 m + (1250 m)cos30.0° + (1000 m)sin 32.0°] = −787.4 m. The  fourth displacement D and its components are sketched in Figure 1.69b. D = Dx2 + Dy2 = 818.4 m. | Dx | 223.0 m = and φ = 15.8°. You should head 15.8° west of south and must walk 818 m. | Dy | 787.4 m  (b) The vector diagram is sketched in Figure 1.69c. The final displacement D from this diagram agrees  with the vector D calculated in part (a) using components.     EVALUATE: Note that D is the negative of the sum of A, B, and C , as it should be. tan φ =

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1-31

Figure 1.69 1.70.

IDENTIFY: The displacements are vectors in which we want to find the magnitude of the resultant and know the other vectors.   SET UP: Calling A the vector from you to the first post, B the vector from you to the second post, and     C the vector from the first post to the second post, we have A + C = B. We want to find the magnitude   of vector B . We use components and the magnitude of C . Let +x be toward the east and +y be toward the north. EXECUTE: Bx = 0 and By is unknown. Cx = –Ax = –(52.0 m)(cos 37.0°) = –41.529 m Ax = 41.53 m

C = 68.0 m, so C y = ± C 2 − C x2 = –53.8455 m. We use the minus sign because the second post is south of the first post. By = Ay + Cy = (52.0 m)(sin 37°) + (–53.8455 m) = –22.551 m. Therefore you are 22.6 m from the second post. EVALUATE: By is negative since post is south of you (in the negative y direction), but the distance to you is positive. 1.71.

IDENTIFY: We are given the resultant of three vectors, two of which we know, and want to find the magnitude and direction of the third vector.        SET UP: Calling C the unknown vector and A and B the known vectors, we have A + B + C = R. The components are Ax + Bx + C x = Rx and Ay + By + C y = Ry . EXECUTE: The components of the known vectors are Ax = 12.0 m, Ay = 0,

Bx = − B sin 50.0° = −21.45 m, By = B cos50.0° = +18.00 m, Rx = 0, and Ry = −10.0 m. Therefore the  components of C are C x = Rx − Ax − Bx = 0 − 12.0 m − ( −21.45 m) = 9.45 m and

C y = Ry − Ay − By = −10.0 m − 0 − 18.0 m = −28.0 m.  9.45 Using these components to find the magnitude and direction of C gives C = 29.6 m and tan θ = 28.0 and θ = 18.6° east of south. EVALUATE: A graphical sketch shows that this answer is reasonable.

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Chapter 1

1.72.

IDENTIFY: The displacements are vectors in which we know the magnitude of the resultant and want to find the magnitude of one of the other vectors.   SET UP: Calling A the vector of Ricardo’s displacement from the tree, B the vector of Jane’s     displacement from the tree, and C the vector from Ricardo to Jane, we have A + C = B. Let the +x-axis be to the east and the +y-axis be to the north. Solving using components we have Ax + C x = Bx and

Ay + C y = By .

  EXECUTE: (a) The components of A and B are Ax = −(26.0 m)sin 60.0° = −22.52 m, Ay = (26.0 m)cos60.0° = +13.0 m, Bx = −(16.0 m)cos30.0° = −13.86 m,

By = −(16.0 m)sin 30.0° = −8.00 m, C x = Bx − Ax = −13.86 m − ( −22.52 m) = +8.66 m, C y = By − Ay = −8.00 m − (13.0 m) = −21.0 m

1.73.

Finding the magnitude from the components gives C = 22.7 m. 8.66 and θ = 22.4°, east of south. (b) Finding the direction from the components gives tan θ = 21.0 EVALUATE: A graphical sketch confirms that this answer is reasonable.   IDENTIFY: If the vector from your tent to Joe’s is A and from your tent to Karl’s is B, then the vector   from Karl’s tent to Joe’s tent is A − B . SET UP: Take your tent’s position as the origin. Let + x be east and + y be north. EXECUTE: The position vector for Joe’s tent is [(21.0 m)cos 23°] iˆ − [(21.0 m)sin 23°] ˆj = (19.33 m) iˆ − (8.205 m) ˆj.

The position vector for Karl’s tent is [(32.0 m)cos 37°]iˆ + [(32.0 m)sin 37°] ˆj = (25.56 m) iˆ + (19.26 m) ˆj. The difference between the two positions is (19.33 m − 25.56 m)iˆ + (−8.205 m − 19.25 m) ˆj = −(6.23 m) iˆ − (27.46 m) ˆj. The magnitude of this vector is the distance between the two tents: D = (−6.23 m) 2 + (−27.46 m)2 = 28.2 m . EVALUATE: If both tents were due east of yours, the distance between them would be 32.0 m − 21.0 m = 11.0 m. If Joe’s was due north of yours and Karl’s was due south of yours, then the distance between them would be 32.0 m + 21.0 m = 53.0 m. The actual distance between them lies between these limiting values. 1.74.

IDENTIFY: Calculate the scalar product and use Eq. (1.16) to determine φ . SET UP: The unit vectors are perpendicular to each other. EXECUTE: The direction vectors each have magnitude 3, and their scalar product is (1)(1) + (1)(−1) + (1)( −1) = −1, so from Eq. (1.16) the angle between the bonds is  −1   1 arccos   = arccos  − 3  = 109°. 3 3     EVALUATE: The angle between the two vectors in the bond directions is greater than 90°.

1.75.

IDENTIFY: This problem involves the scalar product of two vectors.   SET UP: W = F ⋅ s = Fs cos φ = Fx sx + Fy s y .   EXECUTE: Since F is at 60° above the –x-axis and s is along the +x-axis, the angle between them is 120°. The work is W = Fs cos φ = (5.00 N)(0.800 m) cos 120° = –2.00 J. EVALUATE: Use W = Fx sx + Fy s y to check.

W = (5.00 N cos 120°)(0.800 m) + (5.00 N sin 120°)(0) = –2.00 J, which agrees with our result. 1.76.

IDENTIFY: This problem involves the vector product of two vectors.    SET UP: The magnetic force is F = qv ×Browsegrades.net B , F = qvB sin φ .

Units, Physical Quantities, and Vectors

1-33

EXECUTE: F = qvB sin φ gives the magnitude of a vector, so it must be positive. Therefore we only

need to use the sign of q, so F = (8.00 × 10–6 C)(3.00 × 104 m/s)(5.00 T) sin 90° = 1.20 N.       Since v is in the +x direction and B is in the –y direction, v × B is in the –z direction. But qv × B is in the +z direction because q is negative, so the force is in the –z direction. EVALUATE: Careful! The quantity qvB sin φ cannot be negative since it is the magnitude of a vector. Both v and B are vector magnitudes, so they are always positive, and sin φ is positive because 0 ≤ φ ≤ 120°. Only q could be negative, but when using qvB sin φ , we must use only the magnitude of q. When    using F = qv × B , we do use the minus sign for q because it affects the direction of the force. 1.77.

IDENTIFY: We know the scalar product and the magnitude of the vector product of two vectors and want to know the angle between them.     SET UP: The scalar product is A ⋅ B = AB cosθ and the vector product is A × B = AB sin θ .     EXECUTE: A ⋅ B = AB cosθ = −6.00 and A × B = AB sin θ = +9.00. Taking the ratio gives

9.00 , so θ = 124°. −6.00 EVALUATE: Since the scalar product is negative, the angle must be between 90° and 180°. tan θ =

1.78.

IDENTIFY: This problem involves the vector product of two vectors.   SET UP: The torque is r × F , so its magnitude is rF sin φ .   EXECUTE: We know that r makes a 36° angle counterclockwise from the +y-axis and F points in the –y direction. Therefore the angle between these two vectors is 180° – 36° = 144°. So the magnitude   of the torque is r × F = rF sin φ = (4.0 m)(22.0 N) sin 144° = 52 N ⋅ m . The direction of the torque is   in the direction of r × F . By the right-hand rule, this is in the +z direction. EVALUATE: The torque vector could point along a negative axis (such as –z), but it would still always have a positive magnitude.

1.79.

IDENTIFY: This problem involves the vector product and the scalar product of two vectors. It is best to use components.     SET UP: A ⋅ B = Ax Bx + Ay By + Az Bz , the components of A × B are shown in Eq. 1.25 in the text.   EXECUTE: (a) A ⋅ B = Ax Bx + Ay By + Az Bz = a(0) + (0)(–c) + (–b)(d) = –bd.   Realizing that Ay = 0 and Bx = 0, Eq. 1.25 gives the components of A × B .   ( A × B ) x = –AzBy = –(–b)(–c) = –bc   ( A × B ) y = –AxBz = –(a)(d) = –ad   ( A × B ) z = AxBy = (a)(–c) = –ac   A × B = –bc iˆ – ad ĵ – ac k̂ .       (b) If c = 0, A ⋅ B = –bd and A × B = –ad ĵ . The magnitude of A × B is ad and its direction is − ˆj   (which is in the –y direction). Figure 1.79 shows a sketch of A and B in the xy plane. In this figure, the   +y axis would point into the paper. By the right-hand rule, A × B points out of the paper, which is in the –y direction (or the – ĵ direction), which agrees with our results.

1-34

Chapter 1

Figure 1.79

  From Fig. 1.79 we see that the component of A that is parallel to B is –b. So the product of B with the   component of A that is parallel to B is d(–b) = –bd, which agrees with our result. From the same   figure we see that the component of A that is perpendicular to B is a. So the product of B and the   component of A that is perpendicular to B is da, which is the magnitude of the vector product we found above.       EVALUATE: The geometric interpretations of A ⋅ B and A × B can be reversed in the sense that A ⋅ B     equals A times the component of B that is parallel to A and A × B equals A times the component of   B that is perpendicular to A . 1.80.

IDENTIFY: We are dealing with the scalar product and the vector product of two vectors.     SET UP: A ⋅ B = AB cos φ and A × B = AB sin φ . EXECUTE: (a) For the notation in the problem, AB cos θ has its maximum value when θ = 0°. In that   case A × B = 0 because sin 0° = 0.   (b) A × B = AB sin θ , so its value occurs when θ = 90°. The scalar product is zero at that angle

because cos 90° = 0.     (c) A ⋅ B = 2 A × B , so AB cos θ = 2AB sin θ → tan θ = ½ → θ = 26.6°. EVALUATE: It might appear that a second solution is θ = 180° – 26.6° = 153.4°, but that is not true     because in that case A ⋅ B = –2 A × B . 1.81.

IDENTIFY: We know the magnitude of two vectors and their scalar product and want to find the magnitude of their vector product.     SET UP: The scalar product is A ⋅ B = AB cos φ and the vector product is |A × B| = AB sin φ .

  112.0 m 2 112.0 m 2 A ⋅ B = AB cos φ = 90.0 m2, which gives cos φ = = = 0.5833, so (12.0 m)(16.0 m) AB   φ = 54.31°. Therefore A × B = AB sin φ = (12.0 m)(16.0 m)(sin 54.31°) = 156 m 2 .

EXECUTE:

EVALUATE: The magnitude of the vector product is greater than the scalar product because the angle between the vectors is greater than 45º. 1.82.

IDENTIFY: We are dealing with the scalar product and the vector product of two vectors.     SET UP: A ⋅ B = AB cos φ and A × B = AB sin φ .

Browsegrades.net

Units, Physical Quantities, and Vectors

1-35

  EXECUTE: (a) In order to have the maximum positive z-component, A × B should have its maximum  magnitude (which is AB) and it should all point in the +z direction. Thus B should be perpendicular to    A and have a direction so that A × B points in the +z direction, as shown in Fig. 1.82. As you can see  in this figure, the direction of B is at an angle of 53.0° + 90° = 143.0° with the +x-axis.

Figure 1.82

   (b) In this case, A × B must point in the –z direction, so B must be the reverse of what we found in part (a). Therefore its angle with the +x-axis is 90° – 53.0° = 37.0° clockwise as shown in Fig. 1.82. This angle is 323° counterclockwise with the +x-axis.     (c) A × B = 0 when the angle φ between A and B is 0° or 180°. When φ = 0°, the vectors point in the same direction, so they are parallel. When φ = 180°, they point in opposite directions, so they are    antiparallel. When B is parallel to A , B makes an angle of 53.0° counterclockwise from the +x-axis.    When B is antiparallel to A , B makes an angle of 53° + 180° = 233° with the +x-axis. EVALUATE: When calculating the work done by a force, we frequently encounter parallel and antiparallel vectors. 1.83.

1.84.

IDENTIFY: We know the scalar product of two vectors, both their directions, and the magnitude of one of them, and we want to find the magnitude of the other vector.   SET UP: A ⋅ B = AB cos φ . Since we know the direction of each vector, we can find the angle between

them.   EXECUTE: The angle between the vectors is θ = 79.0°. Since A ⋅ B = AB cos φ , we have   A ⋅B 48.0 m 2 = = 28.0 m. B= A cos φ (9.00 m)cos79.0°   EVALUATE: Vector B has the same units as vector A.     IDENTIFY: The cross product A × B is perpendicular to both A and B.   SET UP: Use Eq. (1.23) to calculate the components of A × B. EXECUTE: The cross product is   6.00  ˆ 11.00 ˆ  (−13.00) iˆ + (6.00) ˆj + (−11.00) kˆ = 13  −(1.00) iˆ +  k  . The magnitude of the vector in  j−  13.00  13.00   square brackets is

1.93, and so a unit vector in this direction is  −(1.00) iˆ + (6.00/13.00) ˆj − (11.00/13.00) kˆ   . 1.93  

The negative of this vector,

1-36

Chapter 1

1.85.

 (1.00) iˆ − (6.00/13.00) ˆj + (11.00/13.00) kˆ   , 1.93     is also a unit vector perpendicular to A and B. EVALUATE: Any two vectors that are not parallel or antiparallel form a plane and a vector perpendicular to both vectors is perpendicular to this plane.    IDENTIFY and SET UP: The target variables are the components of C . We are given A and B. We     also know A ⋅ C and B ⋅ C , and this gives us two equations in the two unknowns C x and C y .     EXECUTE: A and C are perpendicular, so A ⋅ C = 0. AxC x + Ay C y = 0, which gives

5.0Cx − 6.5C y = 0.   B ⋅ C = 15.0, so 3.5C x − 7.0C y = 15.0

1.86.

We have two equations in two unknowns C x and C y . Solving gives C x = −8.0 and C y = −6.1.  EVALUATE: We can check that our result does give us a vector C that satisfies the two equations     A ⋅ C = 0 and B ⋅ C = 15.0.   IDENTIFY: Calculate the magnitude of the vector product and then use | A × B| = AB sin θ . SET UP: The magnitude of a vector is related to its components by A = Ax2 + Ay2 .     (−5.00) 2 + (2.00) 2 | A × B| EXECUTE: | A × B| = AB sin θ . sin θ = = = 0.5984 and AB (3.00)(3.00)

θ = sin −1 (0.5984) = 36.8°.

  EVALUATE: We haven’t found A and B, just the angle between them. 1.87.

IDENTIFY: Express all the densities in the same units to make a comparison. SET UP: Density ρ is mass divided by volume. Use the numbers given in the table in the problem and

convert all the densities to kg/m3.  1 kg  8.00 g    1000 g  = 4790 kg/m3 EXECUTE: Sample A: ρ A = 1.67 × 10-6 m3

 1 kg  6.00 × 10-6 g    1000 g  = 640 kg/m3 Sample B: ρ B = 3 -6 6 3  10 m  9.38 × 10 µm    1 µm   1 kg  8.00 × 10-3 g    1000 g  = 3200 kg/m3 Sample C: ρ C = 3 –3 3 1 m  2.50 × 10 cm    100 cm 

Sample D: ρ D =

9.00 × 10-4 kg  1m  2.81 × 103 mm3    1000 mm 

= 320 kg/m3

 1 g  1 kg  9.00 × 104 ng  9    10 ng   1000 g  = 6380 kg/m3 Sample E: ρ E = 3  1m  1.41 × 10 –2 mm3   mm   1000Browsegrades.net © Copyright 2020 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Units, Physical Quantities, and Vectors

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 1 kg  6.00 × 10-5 g    1000 g  = 480 kg/m3 Sample F: ρ F = 3  1m  1.25 × 108 µm3  6   10 µm  EVALUATE: In order of increasing density, the samples are D, F, B, C, A, E. 1.88.

IDENTIFY: We know the magnitude of the resultant of two vectors at four known angles between them, and we want to find out the magnitude of each of these two vectors. SET UP: Use the information in the table in the problem for θ = 0.0° and 90.0°. Call A and B the magnitudes of the vectors. EXECUTE: (a) At 0°: The vectors point in the same direction, so A + B = 8.00 N. At 90.0°: The vectors are perpendicular to each other, so A2 + B2 = R2 = (5.83 N)2 = 33.99 N2. Solving these two equations simultaneously gives B = 8.00 N – A A2 + (8.00 N – A)2 = 33.99 N2 A2 + 64.00 N2 – 16.00 N A + A2 = 33.99 N2 The quadratic formula gives two solutions: A = 5.00 N and B = 3.00 N or A = 3.00 N and B = 5.00 N. In either case, the larger force has magnitude 5.00 N. (b) Let A = 5.00 N and B = 3.00 N, with the larger vector along the x-axis and the smaller one making an angle of +30.0° with the +x-axis in the first quadrant. The components of the resultant are Rx = Ax + Bx = 5.00 N + (3.00 N)(cos 30.0°) = 7.598 N Ry = Ay + By = 0 + (3.00 N)(sin 30.0°) = 1.500 N

R = Rx2 + Ry2 = 7.74 N EVALUATE: To check our answer, we could use the other resultants and angles given in the table with the problem. 1.89.

IDENTIFY: Use the x and y coordinates for each object to find the vector from one object to the other; the distance between two objects is the magnitude of this vector. Use the scalar product to find the angle between two vectors.  SET UP: If object A has coordinates ( x A , y A ) and object B has coordinates ( xB , yB ), the vector rAB from

A to B has x-component xB − x A and y-component yB − y A . EXECUTE: (a) The diagram is sketched in Figure 1.89. (b) (i) In AU,

(0.3182) 2 + (0.9329) 2 = 0.9857.

(ii) In AU, (1.3087)2 + (−0.4423) 2 + (−0.0414) 2 = 1.3820. (iii) In AU, (0.3182 − 1.3087) 2 + (0.9329 − ( −0.4423))2 + (0.0414) 2 = 1.695. (c) The angle between the directions from the Earth to the Sun and to Mars is obtained from the dot product. Combining Eqs. (1.16) and (1.19),  (−0.3182)(1.3087 − 0.3182) + ( −0.9329)(−0.4423 − 0.9329) + (0)  φ = arccos   = 54.6°. (0.9857)(1.695)   (d) Mars could not have been visible at midnight, because the Sun-Mars angle is less than 90°. EVALUATE: Our calculations correctly give that Mars is farther from the Sun than the earth is. Note that on this date Mars was farther from the earth than it is from the Sun.

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Chapter 1

Figure 1.89 1.90.

IDENTIFY: Add the vector displacements of the receiver and then find the vector from the quarterback to the receiver. SET UP: Add the x-components and the y-components. EXECUTE: The receiver’s position is [( +1.0 + 9.0 − 6.0 + 12.0)yd]iˆ + [(−5.0 + 11.0 + 4.0 + 18.0) yd] ˆj = (16.0 yd) iˆ + (28.0 yd) ˆj.

The vector from the quarterback to the receiver is the receiver’s position minus the quarterback’s position, or (16.0 yd)iˆ + (35.0 yd) ˆj , a vector with magnitude (16.0 yd) 2 + (35.0 yd) 2 = 38.5 yd. The  16.0  angle is arctan   = 24.6° to the right of downfield.  35.0  EVALUATE: The vector from the quarterback to receiver has positive x-component and positive y-component.

1.91.

IDENTIFY: Draw the vector addition diagram for the position vectors.  SET UP: Use coordinates in which the Sun to Merak line lies along the x-axis. Let A be the position   vector of Alkaid relative to the Sun, M is the position vector of Merak relative to the Sun, and R is the position vector for Alkaid relative to Merak. A = 138 ly and M = 77 ly.    EXECUTE: The relative positions are shown in Figure 1.91. M + R = A. Ax = M x + Rx so

Rx = Ax − M x = (138 ly)cos 25.6° − 77 ly = 47.5 ly. Ry = Ay − M y = (138 ly)sin 25.6° − 0 = 59.6 ly. R = 76.2 ly is the distance between Alkaid and Merak. (b) The angle is angle φ in Figure 1.91. cosθ =

Rx 47.5 ly = and θ = 51.4°. Then φ = 180° − θ = 129°. R 76.2 ly

EVALUATE: The concepts of vector addition and components make these calculations very simple.

Units, Physical Quantities, and Vectors

1-39

Figure 1.91 1.92.

IDENTIFY: The total volume of the gas-exchanging region of the lungs must be at least as great as the total volume of all the alveoli, which is the product of the volume per alveoli times the number of alveoli. SET UP: V = NValv, and we use the numbers given in the introduction to the problem. EXECUTE: V = NValv = (480 × 106)(4.2 × 106 µm3) = 2.02 × 1015 µm3. Converting to liters gives 3

 1m  V = 2.02 × 1015 m3  6  = 2.02 L ≈ 2.0 L. Therefore choice (c) is correct.  10 µm  EVALUATE: A volume of 2 L is reasonable for the lungs. 1.93.

IDENTIFY: We know the volume and want to find the diameter of a typical alveolus, assuming it to be a sphere. SET UP: The volume of a sphere of radius r is V = 4/3 πr3 and its diameter is D = 2r. EXECUTE: Solving for the radius in terms of the volume gives r = (3V/4π)1/3, so the diameter is

(

)

1/3

 3 4.2 × 106 µm3  1/3  = 200 µm. Converting to mm gives D = 2r = 2(3V/4π) = 2    4π   D = (200 µm)[(1 mm)/(1000 µm)] = 0.20 mm, so choice (a) is correct. EVALUATE: A sphere that is 0.20 mm in diameter should be visible to the naked eye for someone with good eyesight. 1.94.

IDENTIFY: Draw conclusions from a given graph. SET UP: The dots lie more-or-less along a horizontal line, which means that the average alveolar volume does not vary significantly as the lung volume increases. EXECUTE: The volume of individual alveoli does not vary (as stated in the introduction). The graph shows that the volume occupied by alveoli stays constant for higher and higher lung volumes, so there must be more of them, which makes choice (c) the correct one. EVALUATE: It is reasonable that a large lung would need more alveoli than a small lung because a large lung probably belongs to a larger person than a small lung.

MOTION ALONG A STRAIGHT LINE

VP2.5.1.

IDENTIFY: The bus and the car leave the same point at the same time. The bus has constant velocity, but the car starts from rest with constant acceleration. So the constant-acceleration formulas apply. We want to know how long it takes for the car to catch up to the bus and how far they both travel during that time. SET UP: When they meet, x is the same for both of them and they have traveled for the same time. The 1 formulas x = x0 + v0 xt + axt 2 and vx = v0 x + axt both apply. 2 EXECUTE: (a) When the car and bus meet, they have traveled the same distance in the same time. We 1 apply the formula x = x0 + v0 xt + axt 2 to each of them, with the origin at their starting point, which 2 makes x0 = 0 for both of them. The bus has no acceleration and the car has no initial velocity. The 1 → t = 2vbus/ acar. equation reduces to acar t 2 = vbust 2 t = 2(18 m/s)/(8.0 m/s2) = 4.5 s. (b) The bus has zero acceleration, so vx = v0 x + axt reduces to xbus = vbust

xbus = (18 m/s)(4.5 s) = 81 m. EVALUATE: To check, use the car’s motion to find the distance. 1 1 xcar = acar t 2 = (8.0 m/s2)(4.5 s)2 = 81 m, which agrees with our result in part (b). 2 2 VP2.5.2.

IDENTIFY: This is very similar to VP2.5.1 and VP2.5.2. The motorcycle and the SUV leave the same point at the same time. The motorcycle has a constant velocity, but the SUV has an initial velocity and a constant acceleration. So the constant-acceleration formulas apply. SET UP: When they meet, x is the same for both of them and they have traveled for the same time. The 1 formulas x = x0 + v0 xt + axt 2 and vx = v0 x + axt both apply. 2 EXECUTE: (a) When the SUV and motorcycle meet, they have traveled the same distance in the same 1 time. We apply the formula x = x0 + v0 xt + axt 2 to each of them, with the origin at their starting point, 2 which makes x0 = 0 for both of them. The motorcycle has no acceleration and the SUV has an initial velocity and an acceleration. The acceleration is opposite to the velocity of the SUV. If we take the xaxis to be in the direction of motion, aSUV is negative. The equation reduces to 1 vmotorcyclet = vSUVt + aSUVt 2 . We want the time. Putting in the numbers gives 2 1 (20.0 m/s)t = (30.0 m/s)t + (–1.80 m/s2)t2 2 Browsegrades.net

2-1

2-2

Chapter 2

t = 0 s and t = 11.1 s. The t = 0 s solution is when they both of them leave the same point, and the t = 11.1 s is the time when the motorcycle passes the SUV. (b) Both have traveled the same distance when they meet. For the motorcycle this gives xmotorcycle = vmotorcyclet = (20.0 m/s)(11.1 s) = 222 m. (c) The equation vx = v0 x + axt gives the speed of the SUV when they meet in 11.1 s. vx = 30.0 m/s + (–1.80 m/s2)(11.1 s) = 10.0 s. EVALUATE: Use x = vavt to find the distance the SUV has traveled in 11.1 s. For constant acceleration, the average velocity is vav = (v1 + v2)/2, which gives us x = [(30.0 m/s + 10.0 m/s)/2](11.1 s) = 222 m, which agrees with our previous result. VP2.5.4.

IDENTIFY: The truck and car have constant (but different) accelerations and the car has an initial velocity but the truck starts from rest. They leave from the same place at the same time and the truck 1 eventually passes the car. The constant-acceleration equation x = x0 + v0 xt + axt 2 applies. 2 SET UP: (a) The truck and car have traveled the same distance in the same time when the truck reaches the car to pass. They start at the same place so x0 is the same for both and the truck has no initial velocity. 1 EXECUTE: Use the equation x = x0 + v0 xt + axt 2 for each of them, which gives 2 1 2 1 2v C . aTt = vCt − aCt 2 → t= aT + aC 2 2 2

(b) Looking at the truck gives xT =

1 2 1  2vC  2aT vC2 . aTt = aT   = 2 2  aT + aC  ( aT + aC )2

1 EVALUATE: We can calculate the distance the car travels using xC = vCt − aCt 2 and the value of t we 2 found in part (a). Doing this and simplifying the result gives the same answer as in part (b). VP2.7.1.

IDENTIFY: The ball is in freefall so its acceleration is g downward and the constant-acceleration equations apply. 1 SET UP: Calling the y-axis vertical, the formulas y = y0 + v0 yt + a y t 2 and v y = v0 y + a yt apply to the 2 motion of the ball. We know that ay = 9.80 m/s2 downward and v0y = 12.0 m/s upward. 1 EXECUTE: (a) At time t = 0.300 s, the vertical coordinate of the ball is given by y = y0 + v0 yt + a y t 2 , 2 where y0 = 0 at the location of the hand. 1 y = 0 + (12.0 m/s)(0.300 s) + (–9.80 m/s2)(0.300 s)2 = 3.16 m. Since y is positive, the ball is above the 2 hand. The vertical velocity is given by v y = v0 y + a yt .

vy = 12.0 m/s + (–9.80 m/s2)(0.300 s) = 9.06 m/s. Since vy is positive, the ball is moving upward. 1 (b) At t = 2.60 s, y = (12.0 m/s)(2.60 s) + (–9.80 m/s2)(2.60 s)2 = –1.92 m. Since y is negative, the 2 ball is now below the hand. The ball must be moving downward since it is now below the hand. EVALUATE: Check with vy: v y = v0 y + a yt = 12.0 m/s + (–9.80 m/s2)(2.60 s) = –13.5 m/s. Since vy is negative, the ball is moving downward, as we saw above.

Motion Along a Straight Line VP2.7.2.

2-3

IDENTIFY: The stone is in freefall so its acceleration is g downward and the constant-acceleration equations apply. 1 SET UP: Calling the y-axis vertical, the formulas y = y0 + v0 y t + a y t 2 , v y = v0 y + a yt , and 2

v 2y = v02 y + 2a y ( y − y0 ) apply to the motion of the stone. We know that ay = 9.80 m/s2 downward, v0y =

8.00 m/s downward, and y0 = 0. 1 EXECUTE: (a) The equation y = y0 + v0 yt + a yt 2 gives 2 1 y = 0 + (–8.00 m/s)(1.50 s) + (–9.80 m/s2)(1.50 s)2 = –23.0 m. The minus sign means that the stone is 2 below your hand. The velocity of the stone is given by v y = v0 y + a yt = –8.00 m/s + (–9.80 m/s2)(1.50 s) = –22.7 m/s. The minus sign tells us it is moving

downward. (b) We know the stone’s position and acceleration and want its velocity. The equation v 2y = v02 y + 2a y ( y − y0 ) gives v 2y = (–8.00 m/s)2 + 2(–9.80 m/s2)(–8.00 m) vy = ±14.9 m/s. The stone must be moving downward, so vy = –14.9 m/s. EVALUATE: When the stone returned to the level of your hand, its speed was the same as its initial speed of 8.00 m/s. But as the stone has continued to accelerate downward since then, its speed must be greater than its initial speed, which is what we found. VP2.7.3.

IDENTIFY: The football is in freefall so its acceleration is g downward and the constant-acceleration equations apply. 1 SET UP: Calling the y-axis vertical, the formulas y = y0 + v0 y t + a y t 2 , v y = v0 y + a yt , and 2

v 2y = v02 y + 2a y ( y − y0 ) apply. We know that ay = 9.80 m/s2 downward, vy = 0.500 m/s upward when y =

4.00 m, and y0 = 0. EXECUTE: (a) We know the speed, acceleration, and position of the ball and want its initial speed, so we use the equation v 2y = v02 y + 2a y ( y − y0 ) to find its initial speed v0y. (0.500 m/s)2 = v02 y + 2(–9.80 m/s2)(4.00 m) →

v0y = 8.87 m/s.

(b) Use the result from (a) in the equation v y = v0 y + a yt to find the time. 0.500 m/s = 8.87 m/s + (–9.80 m/s2)t → t = 0.854 s. EVALUATE: Calculate y using the time from (b) and compare it with the given value of 4.00 m. 1 1 y = y0 + v0 yt + a y t 2 = 0 + (8.87 m/s)(0.854 s) + (–9.80 m/s2)(0.854 s)2 = 4.00 m, which agrees with 2 2 the given value. VP2.7.4.

IDENTIFY: The tennis ball is in freefall so its acceleration is g downward and the constant-acceleration equations apply. SET UP: When the ball is at its highest point, its vertical velocity is zero. The equation v 2y = v02 y + 2a y ( y − y0 ) applies. EXECUTE: (a) At the highest point, vy = 0. Use v 2y = v02 y + 2a y ( y − y0 ) to find v0y.

0 = v02 y + 2(–g)(H)

→

v0y =

2gH

(b) We now know H, v0y and want vy. The same equation gives v 2y = v02 y + 2( − g )( H / 2) = 2gH – 2gH/2 = 2gH/2

Browsegrades.net

2-4

Chapter 2

2 gH v 2 gH = = 0 . 2 2 2 (c) We want y – y0, we know v0, a, and v. The same equation gives v 2y = v02 y + 2a y ( y − y0 ) vy =

 v0  2   = v0 + 2(− g )( y − y0 ) 2 y − y0 =

3v02 3(2 gH ) 3H = = . 8g 8g 4

EVALUATE: When the ball is half way to the top, v ≠ v0/2 because the motion equations involve the squares of quanties such as v2 and t2. VP2.8.1.

IDENTIFY: The rock is in freefall so its acceleration is g downward and the constant-acceleration equations apply. 1 SET UP: The equation y = y0 + v0 yt + a y t 2 applies. 2 EXECUTE: (a) With the origin at the hand and the y-axis positive upward, y = 4.00 m and y0 = 0. We 1 want the time at which this occurs. The equation y = y0 + v0 y t + a y t 2 gives 2 1 2 2 4.00 m = (12.0 m/s)t + (–9.80 m/s )t . Solving this quadratic equation for t gives two answers: t = 2 0.398 s and t = 2.05 s. (b) Use the same procedure as in (a) except that y = –4.00 m. This gives 1 –4.00 m = (12.0 m/s)t + (–9.80 m/s2)t2. This quadratic equation has two solutions, t = 2.75 s and t = – 2 0.297 s. The negative answer is not physical, so t = 2.75 s. EVALUATE: The ball is at 4.00 m above your hand twice, when it is going up and when it is going down, so we get two answers. It is at 4.00 m below the hand only once, when it is going down, so we have just one answer.

VP2.8.2.

IDENTIFY: The ball is in freefall so its acceleration is g downward and the constant-acceleration equations apply. 1 SET UP: Calling the y-axis vertical with the origin at the hand, the formulas y = y0 + v0 yt + a y t 2 , 2

v y = v0 y + a yt , and v 2y = v02 y + 2a y ( y − y0 ) apply. We know that ay = 9.80 m/s2 downward and vy is initially 9.00 m/s downward. Since all the quantities are downward, it is convenient to call the +y-axis downward. EXECUTE: First find vy when y = 5.00 m. Then use this result to find the time for the ball to reach this height. v 2y = v02 y + 2a y ( y − y0 ) = (9.00 m/s)2 + 2(9.80 m/s2)(5.00 m) vy = 13.38 m/s Now use v y = v0 y + a y t to find the time t. 13.38 m/s = 9.00 m/s + (9.80 m/s2)t

→

t = 0.447 s. 1 EVALUATE: Check using the equation y = y0 + v0 y t + a y t 2 when t = 0.447 s. 2 1 y = (9.00 m/s)(0.447 s) + (9.80 m/s2)(0.447 s)2 = 5.00 m, which agrees with the given value. 2

Motion Along a Straight Line VP2.8.3.

2-5

IDENTIFY: The apple is in freefall so its acceleration is g downward and the constant-acceleration equations apply. SET UP: Calling the y-axis vertically upward with the origin at the hand, the formulas 1 y = y0 + v0 y t + a y t 2 , v y = v0 y + a yt , and v 2y = v02 y + 2a y ( y − y0 ) apply. We know that ay = 9.80 m/s2 2 downward and vy is initially 5.50 m/s upward. 1 EXECUTE: (a) Using y = y0 + v0 y t + a y t 2 gives 2 1 1.30 m = (5.50 m/s)t + (–9.80 m/s2)(0.447 s)2 2 Solving using the quadratic formula gives t = 0.338 s and t = 0.784 s. The apple passes through this point twice: going up at 0.338 s and going down at 0784 s. (b) Use the same approach as in (a). 1 1.80 m = (5.50 m/s)t + (–9.80 m/s2)t2 2 This equation has no real solutions, so the apple never reaches a height of 1.80 m. EVALUATE: The highest point the apple reaches is when vy = 0. Use v 2y = v02 y + 2a y ( y − y0 ) to find the

maximum height. 0 = (5.50 m/s)2 + 2(–9.80 m/s2)(y – y0) y – y0 = 1.54 m, which is less than 1.80 m. This is why we had no solutions to the quadratic equation in part (b). VP2.8.4.

IDENTIFY: The orange is in freefall so its acceleration is g downward and the constant-acceleration equations apply. SET UP: Calling the y-axis vertically upward with the origin at the hand, the formula 1 y = y0 + v0 y t + a y t 2 applies. We know that ay = g downward and vy is initially v0 upward. 2 EXECUTE: We want the time when y =

v02 1 , so we use y = y0 + v0 yt + a y t 2 . 2g 2

v02 1 = v0t – gt2. Solving this quadratic equation gives t = v0/g. There is only one solution, so the 2g 2 orange reaches this height only once. 1 (b) Use the same approach as in part (a). y = y0 + v0 yt + a y t 2 gives 2 3v02 1 = v0t – gt2. The quadratic formula gives two solutions: t = v0/2g and t = 3v0/2g. 2 8g The orange is going up at the smaller solution and going down at the larger solution. v2 EVALUATE: We got only one solution for part (a) because y = 0 is the highest point the orange 2g 3v 2 reaches, and that occurs only once because the orange stops there. In (b) the height 0 is less than the 8g maximum height, so the orange reaches this height twice, once going up and once going down. 2.1.

IDENTIFY: Δx = vav-x Δt SET UP: We know the average velocity is 6.25 m/s. EXECUTE: Δx = vav-x Δt = 25.0 m EVALUATE: In round numbers, 6 m/s × 4 s = 24 m ≈ 25 m, so the answer is reasonable. Browsegrades.net

2-6

Chapter 2

2.2.

IDENTIFY: vav-x =

Δx Δt

SET UP: 13.5 days = 1.166 × 106 s. At the release point, x = +5.150 × 106 m.

x2 − x1 −5.150 × 106 m = = −4.42 m/s. Δt 1.166 × 106 s (b) For the round trip, x2 = x1 and Δx = 0. The average velocity is zero. EXECUTE: (a) vav-x =

EVALUATE: The average velocity for the trip from the nest to the release point is positive. 2.3.

IDENTIFY: Target variable is the time Δt it takes to make the trip in heavy traffic. Use Eq. (2.2) that relates the average velocity to the displacement and average time. Δx Δx SET UP: vav-x = so Δx = vav-x Δt and Δt = . Δt vav-x EXECUTE: Use the information given for normal driving conditions to calculate the distance between the two cities, where the time is 1 h and 50 min, which is 110 min: Δx = vav-x Δt = (105 km/h)(1 h/60 min)(110 min) = 192.5 km.

Now use vav-x for heavy traffic to calculate Δt ; Δx is the same as before: Δt =

Δx 192.5 km = = 2.75 h = 2 h and 45 min. vav-x 70 km/h

The additional time is (2 h and 45 min) – (1 h and 50 min) = (1 h and 105 min) – (1 h and 50 min) = 55 min. EVALUATE: At the normal speed of 105 km/s the trip takes 110 min, but at the reduced speed of 70 km/h it takes 165 min. So decreasing your average speed by about 30% adds 55 min to the time, which is 50% of 110 min. Thus a 30% reduction in speed leads to a 50% increase in travel time. This result (perhaps surprising) occurs because the time interval is inversely proportional to the average speed, not directly proportional to it. 2.4.

2.5.

Δx . Use the average speed for each segment to find the time Δt traveled in that segment. The average speed is the distance traveled divided by the time. SET UP: The post is 80 m west of the pillar. The total distance traveled is 200 m + 280 m = 480 m. 200 m = 40.0 s and the westward run takes EXECUTE: (a) The eastward run takes time 5.0 m/s 280 m 480 m = 70.0 s. The average speed for the entire trip is = 4.4 m/s. 4.0 m/s 110.0 s Δx −80 m = = −0.73 m/s. The average velocity is directed westward. (b) vav-x = Δt 110.0 s EVALUATE: The displacement is much less than the distance traveled, and the magnitude of the average velocity is much less than the average speed. The average speed for the entire trip has a value that lies between the average speed for the two segments. IDENTIFY: The average velocity is vav-x =

IDENTIFY: Given two displacements, we want the average velocity and the average speed. Δx SET UP: The average velocity is vav-x = and the average speed is just the total distance walked Δt divided by the total time to walk this distance.

Motion Along a Straight Line

2-7

Figure 2.5 EXECUTE: (a) Let +x be eastward with the origin at the front door. The trip begins at the front door and ends at the bench as shown in Fig. 2.5. Therefore x1 = 0.00 m and x2 = 20.0 m. Δx = x2 − x1 = 20.0 m − 0.00 m = 20.0 m . The total time is Δt = 28.0 s + 36.0 s = 64.0 s. So Δx 20.0 m = = 0.313 m/s. Δt 64.0 s 60.0 m + 40.0 m = 1.56 m/s . (b) Average speed = 64.0 s EVALUATE: The average speed is much greater than the average velocity because the total distance walked is much greater than the magnitude of the displacement vector.

vav-x =

2.6.

Δx . Use x (t ) to find x for each t. Δt SET UP: x (0) = 0, x(2.00 s) = 5.60 m, and x (4.00 s) = 20.8 m

IDENTIFY: The average velocity is vav-x =

EXECUTE: (a) vav-x =

5.60 m − 0 = +2.80 m/s 2.00 s

20.8 m − 0 = +5.20 m/s 4.00 s 20.8 m − 5.60 m (c) vav-x = = +7.60 m/s 2.00 s EVALUATE: The average velocity depends on the time interval being considered. (b) vav-x =

2.7.

(a) IDENTIFY: Calculate the average velocity using vav-x =

Δx . Δt

Δx so use x (t ) to find the displacement Δx for this time interval. Δt EXECUTE: t = 0 : x = 0

SET UP: vav-x =

t = 10.0 s: x = (2.40 m/s 2 )(10.0 s)2 − (0.120 m/s3 )(10.0 s)3 = 240 m − 120 m = 120 m. Δx 120 m = = 12.0 m/s. Δt 10.0 s dx (b) IDENTIFY: Use vx = to calculate vx (t ) and evaluate this expression at each specified t. dt dx SET UP: vx = = 2bt − 3ct 2 . dt EXECUTE: (i) t = 0 : vx = 0

Then vav-x =

(ii) t = 5.0 s: vx = 2(2.40 m/s 2 )(5.0 s) − 3(0.120 m/s3 )(5.0 s) 2 = 24.0 m/s − 9.0 m/s = 15.0 m/s. (iii) t = 10.0 s: vx = 2(2.40 m/s 2 )(10.0 s) − 3(0.120 m/s3 )(10.0 s) 2 = 48.0 m/s − 36.0 m/s = 12.0 m/s. (c) IDENTIFY: Find the value of t when vx (t ) from part (b) is zero. SET UP: vx = 2bt − 3ct 2

vx = 0 at t = 0. vx = 0 next when 2bt − 3ct 2 =Browsegrades.net 0 © Copyright 2020 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2-8

Chapter 2

EXECUTE: 2b = 3ct so t =

2b 2(2.40 m/s 2 ) = = 13.3 s 3c 3(0.120 m/s3 )

EVALUATE: vx (t ) for this motion says the car starts from rest, speeds up, and then slows down again. 2.8.

2.9.

IDENTIFY: We know the position x(t) of the bird as a function of time and want to find its instantaneous velocity at a particular time. 3 3 dx d  28.0 m + (12.4 m/s)t − (0.0450 m/s )t  SET UP: The instantaneous velocity is vx (t ) = . = dt dt dx EXECUTE: vx (t ) = = 12.4 m/s − (0.135 m/s3 )t 2 . Evaluating this at t = 8.0 s gives vx = 3.76 m/s. dt EVALUATE: The acceleration is not constant in this case.

Δx . We can find the displacement Δt for each Δt constant velocity time interval. The average speed is the distance traveled divided by the time. SET UP: For t = 0 to t = 2.0 s, vx = 2.0 m/s. For t = 2.0 s to t = 3.0 s, vx = 3.0 m/s. In part (b), IDENTIFY: The average velocity is given by vav-x =

vx = −3.0 m/s for t = 2.0 s to t = 3.0 s. When the velocity is constant, Δx = vx Δt. EXECUTE: (a) For t = 0 to t = 2.0 s, Δx = (2.0 m/s)(2.0 s) = 4.0 m. For t = 2.0 s to t = 3.0 s, Δx = (3.0 m/s)(1.0 s) = 3.0 m. For the first 3.0 s, Δx = 4.0 m + 3.0 m = 7.0 m. The distance traveled is Δx 7.0 m = = 2.33 m/s. The average speed is also 2.33 m/s. Δt 3.0 s (b) For t = 2.0 s to 3.0 s, Δx = ( −3.0 m/s)(1.0 s) = −3.0 m. For the first 3.0 s,

also 7.0 m. The average velocity is vav-x =

Δx = 4.0 m + (−3.0 m) = +1.0 m. The ball travels 4.0 m in the +x-direction and then 3.0 m in the −xdirection, so the distance traveled is still 7.0 m. vav-x =

Δx 1.0 m = = 0.33 m/s. The average speed is Δt 3.0 s

7.00 m = 2.33 m/s. 3.00 s EVALUATE: When the motion is always in the same direction, the displacement and the distance traveled are equal and the average velocity has the same magnitude as the average speed. When the motion changes direction during the time interval, those quantities are different. 2.10.

IDENTIFY and SET UP: The instantaneous velocity is the slope of the tangent to the x versus t graph. EXECUTE: (a) The velocity is zero where the graph is horizontal; point IV. (b) The velocity is constant and positive where the graph is a straight line with positive slope; point I. (c) The velocity is constant and negative where the graph is a straight line with negative slope; point V. (d) The slope is positive and increasing at point II. (e) The slope is positive and decreasing at point III. EVALUATE: The sign of the velocity indicates its direction.

2.11.

IDENTIFY: Find the instantaneous velocity of a car using a graph of its position as a function of time. SET UP: The instantaneous velocity at any point is the slope of the x versus t graph at that point. Estimate the slope from the graph. EXECUTE: A: vx = 6.7 m/s; B: vx = 6.7 m/s; C: vx = 0; D: vx = − 40.0 m/s; E: vx = − 40.0 m/s;

F: vx = −40.0 m/s; G: vx = 0. EVALUATE: The sign of vx shows the direction the car is moving. vx is constant when x versus t is a

straight line. 2.12.

Δvx . a x (t ) is the slope of the vx versus t graph. Δt Browsegrades.net SET UP: 60 km/h = 16.7 m/s

IDENTIFY: aav-x =

Motion Along a Straight Line

2-9

16.7 m/s − 0 0 − 16.7 m/s = 1.7 m/s 2 . (ii) aav-x = = −1.7 m/s 2 . 10 s 10 s (iii) Δvx = 0 and aav-x = 0. (iv) Δvx = 0 and aav-x = 0. EXECUTE: (a) (i) aav-x =

(b) At t = 20 s, vx is constant and a x = 0. At t = 35 s, the graph of vx versus t is a straight line and a x = aav-x = −1.7 m/s 2 .

EVALUATE: When aav-x and vx have the same sign the speed is increasing. When they have opposite

signs, the speed is decreasing. 2.13.

IDENTIFY and SET UP: Use vx = EXECUTE: vx =

ax =

dx dv and a x = x to calculate vx (t ) and a x (t ). dt dt

dx = 2.00 cm/s − (0.125 cm/s 2 )t dt

dvx = −0.125 cm/s 2 dt

(a) At t = 0, x = 50.0 cm, vx = 2.00 cm/s, a x = −0.125 cm/s 2 . (b) Set vx = 0 and solve for t: t = 16.0 s. (c) Set x = 50.0 cm and solve for t. This gives t = 0 and t = 32.0 s. The turtle returns to the starting point after 32.0 s. (d) The turtle is 10.0 cm from starting point when x = 60.0 cm or x = 40.0 cm. Set x = 60.0 cm and solve for t: t = 6.20 s and t = 25.8 s. At t = 6.20 s, vx = +1.23 cm/s.

At t = 25.8 s, vx = −1.23 cm/s. Set x = 40.0 cm and solve for t: t = 36.4 s (other root to the quadratic equation is negative and hence nonphysical). At t = 36.4 s, vx = −2.55 cm/s. (e) The graphs are sketched in Figure 2.13.

Figure 2.13 EVALUATE: The acceleration is constant and negative. vx is linear in time. It is initially positive,

decreases to zero, and then becomes negative with increasing magnitude. The turtle initially moves farther away from the origin but then stops and moves in the − x -direction. 2.14.

IDENTIFY: We know the velocity v(t) of the car as a function of time and want to find its acceleration at the instant that its velocity is 12.0 m/s. d (0.860 m/s3 )t 2  dv . SET UP: We know that vx(t) = (0.860 m/s3)t2 and that a x (t ) = x =  dt dt dv EXECUTE: a x (t ) = x = (1.72 m/s3 )t. When vx = 12.0 m/s, (0.860 m/s3)t2 = 12.0 m/s, which gives dt

t = 3.735 s. At this time, a x = 6.42 m/s 2 . EVALUATE: The acceleration of this car is not constant.

2-10

Chapter 2

2.15.

IDENTIFY: The average acceleration is aav-x =

Δvx . Use vx (t ) to find vx at each t. The instantaneous Δt

dvx . dt SET UP: vx (0) = 3.00 m/s and vx (5.00 s) = 5.50 m/s.

acceleration is a x =

EXECUTE: (a) aav-x =

Δvx 5.50 m/s − 3.00 m/s = = 0.500 m/s 2 Δt 5.00 s

dvx = (0.100 m/s3 )(2t ) = (0.200 m/s3 )t. At t = 0, a x = 0. At t = 5.00 s, a x = 1.00 m/s 2 . dt (c) Graphs of vx (t ) and a x (t ) are given in Figure 2.15. (b) a x =

EVALUATE: a x (t ) is the slope of vx (t ) and increases as t increases. The average acceleration for t = 0 to t = 5.00 s equals the instantaneous acceleration at the midpoint of the time interval, t = 2.50 s, since ax (t ) is a linear function of t.

Figure 2.15 2.16.

Δvx , with Δt = 10 s in all cases. Δt SET UP: vx is negative if the motion is to the left.

IDENTIFY: Use aav-x =

EXECUTE: (a) [(5.0 m/s) − (15.0 m/s)]/(10 s) = −1.0 m/s 2 (b) [( −15.0 m/s) − ( −5.0 m/s)]/(10 s) = −1.0 m/s 2 (c) [( −15.0 m/s) − (+15.0 m/s)]/(10 s) = −3.0 m/s 2 EVALUATE: In all cases, the negative acceleration indicates an acceleration to the left. 2.17.

IDENTIFY: vx (t ) = SET UP:

dx dv and ax (t ) = x dt dt

d n (t ) = nt n −1 for n ≥ 1. dt

EXECUTE: (a) vx (t ) = (9.60 m/s 2 )t − (0.600 m/s 6 )t 5 and a x (t ) = 9.60 m/s 2 − (3.00 m/s6 )t 4 . Setting

vx = 0 gives t = 0 and t = 2.00 s. At t = 0, x = 2.17 m and a x = 9.60 m/s 2 . At t = 2.00 s, x = 15.0 m and a x = −38.4 m/s 2 . (b) The graphs are given in Figure 2.17. EVALUATE: For the entire time interval from t = 0 to t = 2.00 s, the velocity vx is positive and x

increases. While a x is also positive the speed increases and while a x is negative the speed decreases.

Motion Along a Straight Line

2-11

Figure 2.17 2.18.

IDENTIFY: We have motion with constant acceleration, so the constant-acceleration equations apply. We want to determine the acceleration of a car on the entrance ramp of a highway. SET UP: Estimate: 300 ft on the entrance ramp. We know the initial and final velocities and the distance traveled, and we want to find the acceleration. So vx2 = v02x + 2ax ( x − x0 ) applies.

 1.466 ft/s   1.466 ft/s  EXECUTE: Convert units: 30 mph   = 44 ft/s ; 70 mph   = 103 ft/s.  1 mph   1 mph  Now use vx2 = v02x + 2ax ( x − x0 ) : (103 ft/s)2 = (44 ft/s)2 + 2ax(300 ft)

→

ax = 14 ft/s2.

 0.3048 m/s 2  2 In SI units, ax = 14 ft/s 2   = 4.4 m/s . 2  1 ft/s  EVALUATE: Compare with acceleration due to gravity: a/g = (14 ft/s2)/(32 ft/s2) = 0.45, so this acceleration is around 45% of g. This seems rather large for an ordinary car. Calculate the time to reach the 70 mph speed, using vx = v0 x + axt : 103 ft/s = 44 ft/s + (14 ft/s2)t → t = 4.2 s. This is too small to be reasonable for most cars. My estimate of the length of the on-ramp must be too small. 2.19.

IDENTIFY: Use the constant acceleration equations to find v0x and a x . (a) SET UP: The situation is sketched in Figure 2.19.

x − x0 = 70.0 m t = 6.00 s vx = 15.0 m/s v0 x = ? Figure 2.19

2( x − x0 ) 2(70.0 m) v +v  − vx = − 15.0 m/s = 8.33 m/s. EXECUTE: Use x − x0 =  0 x x  t , so v0 x = 2 t 6.00 s   v −v 15.0 m/s − 5.0 m/s (b) Use vx = v0 x + axt , so a x = x 0 x = = 1.11 m/s 2 . t 6.00 s EVALUATE: The average velocity is (70.0 m)/(6.00 s) = 11.7 m/s. The final velocity is larger than this, so the antelope must be speeding up during the time interval; v0x < vx and a x > 0.

2-12

Chapter 2

2.20.

IDENTIFY: For constant acceleration, the standard kinematics equations apply. SET UP: Assume the ball moves in the + x direction. EXECUTE: (a) vx = 73.14 m/s, v0 x = 0 and t = 30.0 ms. vx = v0 x + axt gives

ax =

vx − v0 x 73.14 m/s − 0 = = 2440 m/s 2 . t 30.0 × 10−3 s

 v + v   0 + 73.14 m/s  −3 (b) x − x0 =  0 x x  t =   (30.0 × 10 s) = 1.10 m . 2  2    EVALUATE: We could also use x − x0 = v0 xt + 12 axt 2 to calculate x − x0 :

x − x0 = 12 (2440 m/s 2 )(30.0 × 10−3 s)2 = 1.10 m, which agrees with our previous result. The acceleration of the ball is very large. 2.21.

IDENTIFY: For constant acceleration, the standard kinematics equations apply. SET UP: Assume the ball starts from rest and moves in the + x-direction. EXECUTE: (a) x − x0 = 1.50 m, vx = 45.0 m/s and v0 x = 0. vx2 = v02x + 2ax ( x − x0 ) gives

ax =

vx2 − v02x (45.0 m/s) 2 = = 675 m/s 2 . 2( x − x0 ) 2(1.50 m)

2( x − x0 ) 2(1.50 m) v +v  (b) x − x0 =  0 x x  t gives t = = = 0.0667 s v0 x + vx 45.0 m/s  2 

EVALUATE: We could also use vx = v0 x + axt to find t =

vx 45.0 m/s = = 0.0667 s which agrees with ax 675 m/s 2

our previous result. The acceleration of the ball is very large. 2.22.

IDENTIFY: A car is slowing down with uniform acceleration, so the constant-acceleration equations apply. We want to determine the acceleration of the car as it slows down. SET UP: Estimate: It takes 5.0 s to reduce the speed from 70 mph to 30 mph. vx = v0 x + axt ,

vx2 = v02x + 2ax ( x − x0 ) , and a x -av =

Δvx apply. Call the +x-axis the direction in which the car is Δt

moving. EXECUTE: (a) Δvx = 30 mph – 70 mph = –40 mph = –59 ft/s, so a x -av =

–59 ft/s = –12 ft/s2. The 5.0 s

magnitude is 12 ft/s2 and its direction is opposite to the velocity of the car. (b) The initial velocity is 70 mph = 103 ft/s and vx = 0 when the car stops, so v x = v0 x + a xt gives 0 = 103 ft/s + (–12 ft/s2)t → t = 8.6 s. This is the time from first hitting the brakes. The time to stop from 30 mph is 8.6 s – 5.0 s = 3.6 s. (c) Use vx2 = v02x + 2ax ( x − x0 ) since we know everything in it except the distance traveled. 0 = (103 ft/s)2 + 2(–12 ft/s2)(x – x0)

→ x – x0 = 440 ft. 1 2 EVALUATE: As a check, use x = x0 + v0 xt + axt to calculate the distance in (c). This gives 2 1 x = 0 + (103 ft/s)(8.6 s) + (–12 ft/s2)(8.6 s)2 = 440 ft, which agrees with our answer. Compare the 2 acceleration to g: axg = (12 ft/s2)/(32 ft/s2) = 0.38, so ax is about 38% of g. This is fairly large, but if you really slam on your brakes, it might be reasonable.

2.23.

IDENTIFY: Assume that the acceleration is constant and apply the constant acceleration kinematic equations. Set |a x | equal to its maximum allowed value. SET UP: Let + x be the direction of the initial velocity of the car. ax = −250 m/s 2 .

105 km/h = 29.17 m/s.

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Motion Along a Straight Line

2-13

EXECUTE: v0 x = 29.17 m/s. vx = 0. vx2 = v02x + 2ax ( x − x0 ) gives

x − x0 =

vx2 − v02x 0 − (29.17 m/s) 2 = = 1.70 m. 2a x 2( −250 m/s 2 )

EVALUATE: The car frame stops over a shorter distance and has a larger magnitude of acceleration. Part of your 1.70 m stopping distance is the stopping distance of the car and part is how far you move relative to the car while stopping. 2.24.

IDENTIFY: Apply constant acceleration equations to the motion of the car. SET UP: Let + x be the direction the car is moving. EXECUTE: (a) From vx2 = v02x + 2ax ( x − x0 ) , with v0 x = 0, ax = (b) Using x − x0 =

1 2

vx2 (20 m/s) 2 = = 1.67 m/s 2 . 2( x − x0 ) 2(120 m)

( v0 x + v x )t , we have t = 2( x − x0 )/vx = 2(120 m)/(20 m/s) = 12 s.

(c) (12 s)(20 m/s) = 240 m. EVALUATE: The average velocity of the car is half the constant speed of the traffic, so the traffic travels twice as far. 2.25.

IDENTIFY: If a person comes to a stop in 36 ms while slowing down with an acceleration of 60g, how far does he travel during this time? SET UP: Let + x be the direction the person travels. vx = 0 (he stops), ax is negative since it is

opposite to the direction of the motion, and t = 36 ms = 3.6 × 10−2 s. The equations vx = v0 x + axt and

x = x0 + v0 xt + 12 axt 2 both apply since the acceleration is constant. EXECUTE: Solving vx = v0 x + axt for v0x gives v0 x = − axt . Then x = x0 + v0 xt + 12 axt 2 gives

x = − 12 axt 2 = − 12 (−588 m/s 2 )(3.6 × 10−2 s) 2 = 38 cm. EVALUATE: Notice that we were not given the initial speed, but we could find it:

v0 x = − axt = − (−588 m/s 2 )(36 × 10−3 s) = 21 m/s = 47 mph. 2.26.

IDENTIFY: The acceleration ax is the slope of the graph of vx versus t. SET UP: The signs of vx and of ax indicate their directions. EXECUTE: (a) Reading from the graph, at t = 4.0 s, vx = 2.7 cm/s, to the right and at t = 7.0 s,

vx = 1.3 cm/s, to the left. (b) vx versus t is a straight line with slope −

8.0 cm/s = −1.3 cm/s 2 . The acceleration is constant and 6. 0 s

equal to 1.3 cm/s 2 , to the left. It has this value at all times. (c) Since the acceleration is constant, x − x0 = v0 xt + 12 axt 2 . Call x0 = 0 the cat’s position when t = 0.

During the first 4.5 s: x = (8.0 cm/s)(4.5 s) + 12 ( −1.3 cm/s 2 )(4.5 s)2 = 22.8 cm which rounds to 23 cm. This is the change in the cat’s position, but it is also the distance the cat walks. During the first 7.5 s: After 6.0 s, vx becomes negative so the cat is walking backward. However the distance it is moving does not become negative. The position of the car at t = 6.0 s is x = (8.0 cm/s)(6.0 s) + 12 (−1.3 cm/s 2 )(6.0 s) 2 = 24.6 cm. At the end of 7.5 s, the position is

x = (8.0 cm/s)(7.5 s) + 12 (−1.3 cm/s 2 )(7.5 s) 2 = 23.4 cm. Therefore from t = 6.0 s to t = 7.5 s, the cat has walked back a distance of 24.6 cm – 23.4 cm = 1.2 cm. During the first 7.5 s, the cat has walked a total distance of 24.6 cm + 1.2 cm = 25.8 cm which rounds to 26 cm. (d) The graphs of ax and x versus t are given in Figure 2.26.

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2-14

Chapter 2

v +v  EVALUATE: In part (c) we could have instead used x − x0 =  0 x x  t.  2 

Figure 2.26 2.27.

IDENTIFY: We know the initial and final velocities of the object, and the distance over which the velocity change occurs. From this we want to find the magnitude and duration of the acceleration of the object. SET UP: The constant-acceleration kinematics formulas apply. vx2 = v02x + 2ax ( x − x0 ), where

v0 x = 0, vx = 5.0 × 103 m/s, and x − x0 = 4.0 m. EXECUTE: (a) vx2 = v02x + 2ax ( x − x0 ) gives

ax =

vx2 − v02x (5.0 × 103 m/s) 2 = = 3.1 × 106 m/s 2 = 3.2 × 105 g. 2( x − x0 ) 2(4.0 m)

(b) vx = v0 x + axt gives t =

vx − v0 x 5.0 × 103 m/s = = 1.6 ms. ax 3.1 × 106 m/s 2

EVALUATE: (c) The calculated a is less than 450,000 g so the acceleration required doesn’t rule out this hypothesis. 2.28.

IDENTIFY: vx (t ) is the slope of the x versus t graph. Car B moves with constant speed and zero

acceleration. Car A moves with positive acceleration; assume the acceleration is constant. SET UP: For car B, vx is positive and ax = 0. For car A, ax is positive and vx increases with t. EXECUTE: (a) The motion diagrams for the cars are given in Figure 2.28a. (b) The two cars have the same position at times when their x-t graphs cross. The figure in the problem shows this occurs at approximately t = 1 s and t = 3 s. (c) The graphs of vx versus t for each car are sketched in Figure 2.28b. (d) The cars have the same velocity when their x-t graphs have the same slope. This occurs at approximately t = 2 s. (e) Car A passes car B when x A moves above xB in the x-t graph. This happens at t = 3 s. (f) Car B passes car A when xB moves above x A in the x-t graph. This happens at t = 1 s. EVALUATE: When ax = 0, the graph of vx versus t is a horizontal line. When ax is positive, the graph

of vx versus t is a straight line with positive slope.

Motion Along a Straight Line

2-15

Figure 2.28 2.29.

(a) IDENTIFY and SET UP: The acceleration ax at time t is the slope of the tangent to the vx versus t

curve at time t. EXECUTE: At t = 3 s, the vx versus t curve is a horizontal straight line, with zero slope. Thus ax = 0. At t = 7 s, the vx versus t curve is a straight-line segment with slope

45 m/s − 20 m/s = 6.3 m/s 2 . 9 s−5 s

Thus ax = 6.3 m/s 2 . At t = 11 s the curve is again a straight-line segment, now with slope

−0 − 45 m/s = −11.2 m/s 2 . 13 s − 9 s

Thus ax = −11.2 m/s 2 . EVALUATE: ax = 0 when vx is constant, ax > 0 when vx is positive and the speed is increasing, and

ax < 0 when vx is positive and the speed is decreasing. (b) IDENTIFY: Calculate the displacement during the specified time interval. SET UP: We can use the constant acceleration equations only for time intervals during which the acceleration is constant. If necessary, break the motion up into constant acceleration segments and apply the constant acceleration equations for each segment. For the time interval t = 0 to t = 5 s the acceleration is constant and equal to zero. For the time interval t = 5 s to t = 9 s the acceleration is

constant and equal to 6.25 m/s 2 . For the interval t = 9 s to t = 13 s the acceleration is constant and equal to −11.2 m/s 2 . EXECUTE: During the first 5 seconds the acceleration is constant, so the constant acceleration kinematic formulas can be used. v0 x = 20 m/s ax = 0 t = 5 s x − x0 = ?

x − x0 = v0 xt (ax = 0 so no 12 axt 2 term) x − x0 = (20 m/s)(5 s) = 100 m; this is the distance the officer travels in the first 5 seconds. During the interval t = 5 s to 9 s the acceleration is again constant. The constant acceleration formulas can be applied to this 4-second interval. It is convenient to restart our clock so the interval starts at time t = 0 and ends at time t = 4 s. (Note that the acceleration is not constant over the entire t = 0 to t = 9 s interval.) v0 x = 20 m/s ax = 6.25 m/s 2 t = 4 s x0 = 100 m x − x0 = ?

x − x0 = v0 xt + 12 axt 2 x − x0 = (20 m/s)(4 s) + 12 (6.25 m/s 2 )(4 s) 2 = 80 m + 50 m = 130 m. Thus x − x0 + 130 m = 100 m + 130 m = 230 m. At t = 9 s the officer is at x = 230 m, so she has traveled 230 m in the first 9 seconds.

2-16

Chapter 2

During the interval t = 9 s to t = 13 s the acceleration is again constant. The constant acceleration formulas can be applied for this 4-second interval but not for the whole t = 0 to t = 13 s interval. To use the equations restart our clock so this interval begins at time t = 0 and ends at time t = 4 s. v0 x = 45 m/s (at the start of this time interval)

ax = −11.2 m/s 2 t = 4 s x0 = 230 m x − x0 = ? x − x0 = v0 xt + 12 axt 2 x − x0 = (45 m/s)(4 s) + 12 (−11.2 m/s 2 )(4 s)2 = 180 m − 89.6 m = 90.4 m. Thus x = x0 + 90.4 m = 230 m + 90.4 m = 320 m. At t = 13 s the officer is at x = 320 m, so she has traveled 320 m in the first 13 seconds. EVALUATE: The velocity vx is always positive so the displacement is always positive and displacement and distance traveled are the same. The average velocity for time interval Δt is vav-x = Δx /Δt . For t = 0 to 5 s, vav-x = 20 m/s. For t = 0 to 9 s, vav-x = 26 m/s. For t = 0 to 13 s,

vav-x = 25 m/s. These results are consistent with the figure in the textbook. 2.30.

IDENTIFY: For constant acceleration, the kinematics formulas apply. We can use the total displacement and final velocity to calculate the acceleration and then use the acceleration and shorter distance to find the speed. SET UP: Take +x to be down the incline, so the motion is in the +x direction. The formula vx2 = v02x + 2a( x − x0 ) applies. EXECUTE: First look at the motion over 6.80 m. We use the following numbers: v0x = 0, x – x0 = 6.80 m, and vx = 3.80 /s. Solving the above equation for ax gives ax = 1.062 m/s2. Now look at the motion over the 3.40 m using v0x = 0, ax = 1.062 m/s2 and x – x0 = 3.40 m. Solving the same equation, but this time for vx, gives vx = 2.69 m/s. EVALUATE: Even though the block has traveled half way down the incline, its speed is not half of its speed at the bottom.

2.31.

IDENTIFY: Apply the constant acceleration equations to the motion of the flea. After the flea leaves the ground, a y = g , downward. Take the origin at the ground and the positive direction to be upward. (a) SET UP: At the maximum height v y = 0.

v y = 0 y − y0 = 0.440 m a y = −9.80 m/s 2 v0 y = ? v 2y = v02 y + 2a y ( y − y0 ) EXECUTE: v0 y = −2a y ( y − y0 ) = −2(−9.80 m/s2 )(0.440 m) = 2.94 m/s (b) SET UP: When the flea has returned to the ground y − y0 = 0.

y − y0 = 0 v0 y = +2.94 m/s a y = −9.80 m/s 2 t = ? y − y0 = v0 y t + 12 a y t 2 EXECUTE: With y − y0 = 0 this gives t = −

2v0 y

=−

2(2.94 m/s) = 0.600 s. −9.80 m/s 2

EVALUATE: We can use v y = v0 y + a y t to show that with v0 y = 2.94 m/s, v y = 0 after 0.300 s. 2.32.

IDENTIFY: The rock has a constant downward acceleration of 9.80 m/s2. We know its initial velocity and position and its final position. SET UP: We can use the kinematics formulas for constant acceleration.

Motion Along a Straight Line

2-17

EXECUTE: (a) y − y0 = −30 m, v0 y = 22.0 m/s, a y = −9.80 m/s 2 . The kinematics formulas give

v y = − v02 y + 2a y ( y − y0 ) = − (22.0 m/s) 2 + 2( −9.80 m/s 2 )(−30 m) = −32.74 m/s, so the speed is 32.7 m/s. (b) v y = v0 y + a yt and t =

v y − v0 y ay

−32.74 m/s − 22.0 m/s = 5.59 s. −9.80 m/s 2

EVALUATE: The vertical velocity in part (a) is negative because the rock is moving downward, but the speed is always positive. The 5.59 s is the total time in the air. 2.33.

IDENTIFY: The pin has a constant downward acceleration of 9.80 m/s2 and returns to its initial position. SET UP: We can use the kinematics formulas for constant acceleration. 1 EXECUTE: The kinematics formulas give y − y0 = v0 y t + a y t 2 . We know that y − y0 = 0, so 2 2v0 y 2(8.20 m/s) t=− =− = +1.67 s. ay −9.80 m/s 2 EVALUATE: It takes the pin half this time to reach its highest point and the remainder of the time to return.

2.34.

IDENTIFY: The putty has a constant downward acceleration of 9.80 m/s2. We know the initial velocity of the putty and the distance it travels. SET UP: We can use the kinematics formulas for constant acceleration. EXECUTE: (a) v0y = 9.50 m/s and y – y0 = 3.60 m, which gives

v y = v02 y + 2a y ( y − y0 ) = (9.50 m/s) 2 + 2(−9.80 m/s 2 )(3.60 m) = 4.44 m/s (b) t =

v y − v0 y ay

4.44 m/s − 9.50 m/s = 0.517 s −9.8 m/s 2

EVALUATE: The putty is stopped by the ceiling, not by gravity. 2.35.

IDENTIFY: A ball on Mars that is hit directly upward returns to the same level in 8.5 s with a constant downward acceleration of 0.379g. How high did it go and how fast was it initially traveling upward? SET UP: Take + y upward. v y = 0 at the maximum height. a y = − 0.379 g = − 3.71 m/s 2 . The

constant-acceleration formulas v y = v0 y + a y t and y = y0 + v0 y t + 12 a y t 2 both apply. EXECUTE: Consider the motion from the maximum height back to the initial level. For this motion v0 y = 0 and t = 4.25 s. y = y0 + v0 y t + 12 a y t 2 = 12 (−3.71 m/s 2 )(4.25 s)2 = −33.5 m. The ball went 33.5

m above its original position. (b) Consider the motion from just after it was hit to the maximum height. For this motion v y = 0 and

t = 4.25 s. v y = v0 y + a y t gives v0 y = − a yt = − ( −3.71 m/s 2 )(4.25 s) = 15.8 m/s. (c) The graphs are sketched in Figure 2.35.

Figure 2.35

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2-18

Chapter 2 EVALUATE: The answers can be checked several ways. For example, v y = 0, v0 y = 15.8 m/s, and

a y = − 3.71 m/s 2 in v y2 = v02 y + 2a y ( y − y0 ) gives y − y0 =

v y2 − v02 y 2a y

0 − (15.8 m/s) 2 = 33.6 m, 2( −3.71 m/s 2 )

which agrees with the height calculated in (a). 2.36.

IDENTIFY: A baseball is thrown upward, so its acceleration is downward and uniform. Therefore the constant-acceleration equations apply. SET UP: Estimate: It is not easy to throw a ball straight up, so estimate 25 ft for the maximum height. 1 Its acceleration is g = 32.2 ft/s2 downward, and the formulas y = y0 + v0 y t + a y t 2 and 2

v 2y = v02 y + 2a y ( y − y0 ) apply. Call the +y-axis upward, with the origin at the point where the ball leaves the hand; this makes y0 = 0 and ay = –32.2 ft/s2. At its maximum height, the ball stops moving, so vy = 0. EXECUTE: (a) First find the initial speed of the ball using v 2y = v02 y + 2a y ( y − y0 ) and vy = 0 at the maximum height. 0 = v02 y + 2(–32.2 ft/s2)(25 ft)

→

v0y = 40.1 ft/s. Now use

1 y = y0 + v0 yt + a y t 2 to get the time t to reach the maximum height.’ 2 1 0 = 0 + (40.1 ft/s)t + (–32.2 ft/s2)t2 → t = 2.5 s. 2 (b) Estimate: The ball moves about 2.5 ft while it is being thrown. We want the average accelerating during this time. The ball starts from rest and reaches a velocity of 40.1 ft/s while traveling 2.5 ft upward. So we use v 2y = v02 y + 2a y ( y − y0 ) . (40.1 ft/s)2 = 0 + 2ay(2.5 ft) → ay = 320 ft/s2 ≈ 10g. EVALUATE: The result in (b) seems rather large, so maybe the 2.5-ft estimate was too short, or maybe the maximum height of 25 ft was too large. 2.37.

IDENTIFY: A rock is thrown upward, so its acceleration is downward and uniform. Therefore the constant-acceleration equations apply. We want to know the rock’s velocity at times 1.0 s and 3.0 s after it is thrown. SET UP: The formula v y = v0 y + a yt applies. Call the +y-axis upward, with the origin at the point

where the rock leaves the hand; this makes y0 = 0, v0y = 24.0 m/s, and ay = –9.80 m/s2. EXECUTE: (a) At 1.0 s: v y = v0 y + a y t = 24.0 m/s + (–9.80 m/s2)(1.0 s) = +14.2 m/s. The acceleration is downward. The velocity is upward but the speed is decreasing because the acceleration is downward. (b) At 3.0 s: v y = v0 y + a yt = 24.0 m/s + (–9.80 m/s2)(3.0 s) = –5.40 m/s. The acceleration is downward. The velocity is downward and the speed is increasing because the acceleration is also downward. The rock has passed its highest point and is now coming down. EVALUATE: If only gravity acts on an object, its acceleration is always downward with a magnitude of 9.80 m/s2. 2.38.

IDENTIFY: Apply constant acceleration equations to the vertical motion of the brick. SET UP: Let + y be downward. a y = 9.80 m/s 2 EXECUTE: (a) v0 y = 0, t = 1.90 s, a y = 9.80 m/s 2 . y − y0 = v0 y t + 12 a y t 2 = 12 (9.80 m/s 2 )(1.90 s) 2 = 17.7 m.

The building is 17.7 m tall. (b) v y = v0 y + a y t = 0 + (9.80 m/s 2 )(1.90 s) = 18.6 m/s (c) The graphs of a y , v y and y versus t are given in Figure 2.38. Take y = 0 at the ground.

Motion Along a Straight Line

2-19

 v0 y + v y  2 2 EVALUATE: We could use either y − y0 =   t or v y = v0 y + 2a y ( y − y0 ) to check our results. 2  

Figure 2.38 2.39.

IDENTIFY: Apply constant acceleration equations to the motion of the meterstick. The time the meterstick falls is your reaction time. SET UP: Let + y be downward. The meter stick has v0 y = 0 and a y = 9.80 m/s 2 . Let d be the distance

the meterstick falls. EXECUTE: (a) y − y0 = v0 y t + 12 a y t 2 gives d = (4.90 m/s 2 )t 2 and t =

d . 4.90 m/s 2

0.176 m = 0.190 s 4.90 m/s 2 EVALUATE: The reaction time is proportional to the square of the distance the stick falls. (b) t =

2.40.

IDENTIFY: Apply constant acceleration equations to the motion of the lander. SET UP: Let + y be downward. Since the lander is in free-fall, a y = +1.6 m/s 2 . EXECUTE: v0 y = 0.8 m/s, y − y0 = 5.0 m, a y = +1.6 m/s 2 in v 2y = v02 y + 2a y ( y − y0 ) gives

v y = v02 y + 2a y ( y − y0 ) = (0.8 m/s) 2 + 2(1.6 m/s 2 )(5.0 m) = 4.1 m/s. EVALUATE: The same descent on earth would result in a final speed of 9.9 m/s, since the acceleration due to gravity on earth is much larger than on the moon. 2.41.

IDENTIFY: When the only force is gravity the acceleration is 9.80 m/s 2 , downward. There are two

intervals of constant acceleration and the constant acceleration equations apply during each of these intervals. SET UP: Let + y be upward. Let y = 0 at the launch pad. The final velocity for the first phase of the motion is the initial velocity for the free-fall phase. EXECUTE: (a) Find the velocity when the engines cut off. y − y0 = 525 m, a y = 2.25 m/s 2 , v0 y = 0.

v 2y = v02 y + 2a y ( y − y0 ) gives v y = 2(2.25 m/s 2 )(525 m) = 48.6 m/s. Now consider the motion from engine cut-off to maximum height: y0 = 525 m, v0 y = +48.6 m/s, v y = 0 (at the maximum height), a y = −9.80 m/s 2 . v 2y = v02 y + 2a y ( y − y0 ) gives y − y0 =

v 2y − v02 y 2a y

0 − (48.6 m/s) 2 = 121 m and y = 121 m + 525 m = 646 m. 2( −9.80 m/s 2 )

(b) Consider the motion from engine failure until just before the rocket strikes the ground: y − y0 = −525 m, a y = −9.80 m/s 2 , v0 y = +48.6 m/s. v 2y = v02 y + 2a y ( y − y0 ) gives

v y = − (48.6 m/s) 2 + 2(−9.80 m/s 2 )( −525 m) = −112 m/s. Then v y = v0 y + a y t gives t=

v y − v0 y ay

−112 m/s − 48.6 m/s = 16.4 s. −9.80 m/s 2Browsegrades.net

2-20

Chapter 2 (c) Find the time from blast-off until engine failure: y − y0 = 525 m, v0 y = 0, a y = +2.25 m/s 2 .

y − y0 = v0 y t + 12 a y t 2 gives t =

2( y − y0 ) 2(525 m) = = 21.6 s. The rocket strikes the launch pad ay 2.25 m/s 2

21.6 s + 16.4 s = 38.0 s after blast-off. The acceleration a y is +2.25 m/s 2 from t = 0 to t = 21.6 s. It is −9.80 m/s 2 from t = 21.6 s to 38.0 s. v y = v0 y + a y t applies during each constant acceleration segment, so the graph of v y versus t is a straight line with positive slope of 2.25 m/s 2 during the blastoff phase and with negative slope of −9.80 m/s 2 after engine failure. During each phase y − y0 = v0 y t + 12 a yt 2 . The sign of a y determines the curvature of y (t ). At t = 38.0 s the rocket has returned to y = 0. The graphs are sketched in Figure 2.41. EVALUATE: In part (b) we could have found the time from y − y0 = v0 y t + 12 a y t 2 , finding v y first

allows us to avoid solving for t from a quadratic equation.

Figure 2.41 2.42.

IDENTIFY: Apply constant acceleration equations to the vertical motion of the sandbag. SET UP: Take + y upward. a y = −9.80 m/s 2 . The initial velocity of the sandbag equals the velocity of

the balloon, so v0 y = +5.00 m/s. When the balloon reaches the ground, y − y0 = −40.0 m. At its maximum height the sandbag has v y = 0. EXECUTE: (a) t = 0.250 s: y − y0 = v0 y t + 12 a y t 2 = (5.00 m/s)(0.250 s) + 12 (−9.80 m/s 2 )(0.250 s) 2 = 0.94 m. The

sandbag is 40.9 m above the ground. v y = v0 y + a y t = +5.00 m/s + (−9.80 m/s 2 )(0.250 s) = 2.55 m/s. t = 1.00 s: y − y0 = (5.00 m/s)(1.00 s) + 12 ( −9.80 m/s 2 )(1.00 s)2 = 0.10 m. The sandbag is 40.1 m above the ground. v y = v0 y + a y t = +5.00 m/s + ( −9.80 m/s 2 )(1.00 s) = −4.80 m/s. (b) y − y0 = −40.0 m, v0 y = 5.00 m/s, a y = −9.80 m/s 2 . y − y0 = v0 y t + 12 a y t 2 gives −40.0 m = (5.00 m/s)t − (4.90 m/s 2 )t 2 . (4.90 m/s 2 )t 2 − (5.00 m/s)t − 40.0 m = 0 and

)

(

1 5.00 ± (−5.00)2 − 4(4.90)(−40.0) s = (0.51 ± 2.90) s. t must be positive, so t = 3.41 s. 9.80

(c) v y = v0 y + a y t = +5.00 m/s + (−9.80 m/s 2 )(3.41 s) = −28.4 m/s (d) v0 y = 5.00 m/s, a y = −9.80 m/s 2 , v y = 0. v 2y = v02 y + 2a y ( y − y0 ) gives

y − y0 =

v 2y − v02 y 2a y

0 − (5.00 m/s) 2 = 1.28 m. The maximum height is 41.3 m above the ground. 2(−9.80 m/s 2 )

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Motion Along a Straight Line

2-21

(e) The graphs of a y , v y , and y versus t are given in Figure 2.42. Take y = 0 at the ground. EVALUATE: The sandbag initially travels upward with decreasing velocity and then moves downward with increasing speed.

Figure 2.42 2.43.

IDENTIFY: You throw a rock thrown upward, so its acceleration is downward and uniform. Therefore the constant-acceleration equations apply. SET UP: We want to know how high the rock went if it returned to your hand 3.60 s after you threw it upward. With constant downward acceleration, the time it takes the rock to reach its highest point is the same as the time to fall back to your hand. Therefore it took 1.80 s to reach that highest point, and the rock’s velocity was zero at the point. We can use v y = v0 y + a y t to find the initial speed and then use

v 2y = v02 y + 2a y ( y − y0 ) to find how high it went. EXECUTE: First find v0y using v y = v0 y + a y t with vy = 0 at the highest point.

0 = v0y – gt, so v0y = (9.80 m/s2)(1.80 s) = 17.64 m/s. Now use v 2y = v02 y + 2a y ( y − y0 ) to find the maximum height the rock reached. 0 = (17.64 m/s)2 + 2(–9.80 m/s2)(y – y0) → y – y0 = 15.9 m. 1 EVALUATE: Check by using y = y0 + v0 yt + a y t 2 to calculate y at the maximum height. 2 1 2 y = 0 + (17.64 m/s)(1.80 s) + (–9.80 m/s )(1.80 s)2 = 15.9 m, which agrees with our result. 2 2.44.

IDENTIFY: Since air resistance is ignored, the egg is in free-fall and has a constant downward acceleration of magnitude 9.80 m/s 2 . Apply the constant acceleration equations to the motion of the egg. SET UP: Take + y to be upward. At the maximum height, v y = 0. EXECUTE: (a) y − y0 = −30.0 m, t = 5.00 s, a y = −9.80 m/s 2 . y − y0 = v0 y t + 12 a y t 2 gives v0 y =

y − y0 1 −30.0 m 1 − 2 a yt = − 2 (−9.80 m/s 2 )(5.00 s) = +18.5 m/s. t 5.00 s

(b) v0 y = +18.5 m/s, v y = 0 (at the maximum height), a y = −9.80 m/s 2 . v 2y = v02 y + 2a y ( y − y0 ) gives

y − y0 =

v 2y − v02 y 2a y

0 − (18.5 m/s) 2 = 17.5 m. 2( −9.80 m/s 2 )

(c) At the maximum height v y = 0. (d) The acceleration is constant and equal to 9.80 m/s 2 , downward, at all points in the motion,

including at the maximum height. (e) The graphs are sketched in Figure 2.44.

2-22

Chapter 2

EVALUATE: The time for the egg to reach its maximum height is t =

v y − v0 y ay

−18.5 m/s = 1.89 s. −9.8 m/s 2

The egg has returned to the level of the cornice after 3.78 s and after 5.00 s it has traveled downward from the cornice for 1.22 s.

Figure 2.44 2.45.

IDENTIFY: We can avoid solving for the common height by considering the relation between height, time of fall, and acceleration due to gravity, and setting up a ratio involving time of fall and acceleration due to gravity. SET UP: Let g En be the acceleration due to gravity on Enceladus and let g be this quantity on earth. Let h be the common height from which the object is dropped. Let + y be downward, so

y − y0 = h. v0 y = 0 2 y − y0 = v0 y t + 12 a y t 2 gives h = 12 gtE2 and h = 12 g En tEn . Combining these two equations

EXECUTE:

gives

2 gtE2 = g En tEn

t   1.75 s  2 and g En = g  E  = (9.80 m/s 2 )   = 0.0868 m/s . 18 6 s t .    En 

EVALUATE: The acceleration due to gravity is inversely proportional to the square of the time of fall. 2.46.

IDENTIFY: Since air resistance is ignored, the boulder is in free-fall and has a constant downward acceleration of magnitude 9.80 m/s 2 . Apply the constant acceleration equations to the motion of the boulder. SET UP: Take + y to be upward. EXECUTE: (a) v0 y = +40.0 m/s, v y = +20.0 m/s, a y = −9.80 m/s 2 . v y = v0 y + a y t gives

v y − v0 y ay

20.0 m/s − 40.0 m/s = +2.04 s. −9.80 m/s 2

(b) v y = −20.0 m/s. t =

v y − v0 y ay

−20.0 m/s − 40.0 m/s = +6.12 s. −9.80 m/s 2

t=−

2v0 y ay

=−

2(40.0 m/s) = +8.16 s. −9.80 m/s 2

(d) v y = 0, v0 y = +40.0 m/s, a y = −9.80 m/s 2 . v y = v0 y + a y t gives t =

v y − v0 y ay

0 − 40.0 m/s = 4.08 s. −9.80 m/s 2

(e) The acceleration is 9.80 m/s 2 , downward, at all points in the motion. (f) The graphs are sketched in Figure 2.46.

Motion Along a Straight Line

2-23

EVALUATE: v y = 0 at the maximum height. The time to reach the maximum height is half the total

time in the air, so the answer in part (d) is half the answer in part (c). Also note that 2.04 s < 4.08 s < 6.12 s. The boulder is going upward until it reaches its maximum height and after the maximum height it is traveling downward.

Figure 2.46 2.47.

IDENTIFY: The rock has a constant downward acceleration of 9.80 m/s2. The constant-acceleration kinematics formulas apply. SET UP: The formulas y = y0 + v0 yt + 12 a y t 2 and v y2 = v02 y + 2a y ( y − y0 ) both apply. Call +y

upward. First find the initial velocity and then the final speed. EXECUTE: (a) 6.00 s after it is thrown, the rock is back at its original height, so y = y0 at that instant. Using ay = –9.80 m/s2 and t = 6.00 s, the equation y = y0 + v0 yt + 12 a y t 2 gives v0y = 29.4 m/s. When the rock reaches the water, y – y0 = –28.0 m. The equation v y2 = v02 y + 2a y ( y − y0 ) gives vy = –37.6 m/s, so its speed is 37.6 m/s. EVALUATE: The final speed is greater than the initial speed because the rock accelerated on its way down below the bridge. 2.48.

IDENTIFY: We want to interpret a graph of vx versus t. SET UP: The area under a graph of vx versus t is equal to the change in position x2 – x1. The average Δx velocity is vx -av = . Δt 1 1 EXECUTE: (a) The area of a triangle is bh, which gives A = (6.0 s)(8.0 m/s) = 24 m. 2 2 (b) The area under a graph of vx versus t is equal to the change in position x2 – x1, which in this case is

the distance traveled, which is 24 m in 6.0 s. Thus vx -av =

Δx Δt

24 m 6.0 s

= 4.0 m/s.

(c) Δx = vx-av Δt = (4.0 m/s)(6.0 s) = 24 m, which is the same as the area in (a). v +v 8.0 m/s + 0 EVALUATE: For constant acceleration vx -av = x -0 x = = 4.0 m/s, which agrees with our 2 2 answer in (b). 2.49.

IDENTIFY: The acceleration is not constant, but we know how it varies with time. We can use the definitions of instantaneous velocity and position to find the rocket’s position and speed. t

SET UP: The basic definitions of velocity and position are v y (t ) = v0 y +  a y dt and y − y0 =  v y dt. t

EXECUTE: (a) v y (t ) =  a y dt =  (2.80 m/s3 )tdt = (1.40 m/s3 )t 2 t

y − y0 =  v y dt =  (1.40 m/s3 )t 2 dt = (0.4667 m/s3 )t 3. For t = 10.0 s, y − y0 = 467 m. (b) y − y0 = 325 m so (0.4667 m/s3 )t 3 = 325 m and t = 8.864 s. At this time

v y = (1.40 m/s3 )(8.864 s)2 = 110 m/s.

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2-24

Chapter 2 EVALUATE: The time in part (b) is less than 10.0 s, so the given formulas are valid. 2.50.

IDENTIFY: The acceleration is not constant, so we must use calculus instead of the standard kinematics formulas. t

SET UP: The general calculus formulas are vx = v0 x +  ax dt and x = x0 +  vx dt. First integrate ax to

find v(t), and then integrate that to find x(t). t

EXECUTE: Find v(t): vx (t ) = v0 x +  ax dt = v0 x +  −(0.0320 m/s3 )(15.0 s − t ) dt. Carrying out the

integral and putting in the numbers gives vx(t) = 8.00 m/s – (0.0320 m/s3)[(15.0 s)t – t2/2]. Now use this result to find x(t). t t 2 x = x0 +  vx dt = x0 +  8.00 m/s − (0.0320 m/s3 ) (15.0 s)t − t2  dt , which gives  0 0  3 2 3 x = x0 + (8.00 m/s)t – (0.0320 m/s )[(7.50 s)t – t /6)]. Using x0 = –14.0 m and t = 10.0 s, we get x = 47.3 m. EVALUATE: The standard kinematics formulas apply only when the acceleration is constant.

(

2.51.

)

(a) IDENTIFY: Integrate a x (t ) to find vx (t ) and then integrate vx (t ) to find x(t ). t

SET UP: vx = v0 x +  ax dt , a x = At − Bt 2 with A = 1.50 m/s3 and B = 0.120 m/s 4 . 0

EXECUTE: vx = v0 x +  ( At − Bt 2 ) dt = v0 x + 12 At 2 − 13 Bt 3 0

At rest at t = 0 says that v0 x = 0, so vx = 12 At 2 − 13 Bt 3 = 12 (1.50 m/s3 )t 2 − 13 (0.120 m/s 4 )t 3 vx = (0.75 m/s3 )t 2 − (0.040 m/s 4 )t 3 t

SET UP: x − x0 +  vx dt 0

EXECUTE: x = x0 +  ( 12 At 2 − 13 Bt 3 ) dt = x0 + 16 At 3 − 121 Bt 4 0

At the origin at t = 0 says that x0 = 0, so x = 16 At 3 − 121 Bt 4 = 16 (1.50 m/s3 )t 3 − 121 (0.120 m/s 4 )t 4 x = (0.25 m/s3 )t 3 − (0.010 m/s 4 )t 4 dx dv and a x (t ) = x . dt dt dvx dvx (b) IDENTIFY and SET UP: At time t, when vx is a maximum, , the = 0. (Since a x = dt dt maximum velocity is when ax = 0. For earlier times a x is positive so vx is still increasing. For later

EVALUATE: We can check our results by using them to verify that vx (t ) =

times a x is negative and vx is decreasing.) dvx = 0 so At − Bt 2 = 0 dt One root is t = 0, but at this time vx = 0 and not a maximum.

EXECUTE: a x =

The other root is t =

A 1.50 m/s3 = = 12.5 s B 0.120 m/s 4

At this time vx = (0.75 m/s3 )t 2 − (0.040 m/s 4 )t 3 gives vx = (0.75 m/s3 )(12.5 s) 2 − (0.040 m/s 4 )(12.5 s)3 = 117.2 m/s − 78.1 m/s = 39.1 m/s. EVALUATE: For t < 12.5 s, a x > 0 and vx is increasing. For t > 12.5 s, a x < 0 and vx is decreasing.

Motion Along a Straight Line 2.52.

2-25

IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. t

Instead, use vx = v0 x +  ax dt and x = x0 +  vx dt . Use the values of vx and of x at t = 1.0 s to evaluate v0x and x0 . SET UP:

 t dt = n + 1 t n

n +1

, for n ≥ 0. t

EXECUTE: (a) vx = v0 x +  α tdt = v0 x + 12 α t 2 = v0 x + (0.60 m/s3 )t 2 . vx = 5.0 m/s when t = 1.0 s gives 0

v0 x = 4.4 m/s. Then, at t = 2.0 s, vx = 4.4 m/s + (0.60 m/s3 )(2.0 s)2 = 6.8 m/s. t

(b) x = x0 +  (v0 x + 12 α t 2 ) dt = x0 + v0 xt + 16 α t 3. x = 6.0 m at t = 1.0 s gives x0 = 1.4 m. Then, at 0

t = 2.0 s, x = 1.4 m + (4.4 m/s)(2.0 s) + 16 (1.2 m/s3 )(2.0 s)3 = 11.8 m. (c) x (t ) = 1.4 m + (4.4 m/s)t + (0.20 m/s3 )t 3. vx (t ) = 4.4 m/s + (0.60 m/s3 )t 2 . a x (t ) = (1.20 m/s3 )t. The

graphs are sketched in Figure 2.52. EVALUATE: We can verify that a x =

dvx dx and vx = . dt dt

Figure 2.52 2.53.

IDENTIFY: The sprinter’s acceleration is constant for the first 2.0 s but zero after that, so it is not constant over the entire race. We need to break up the race into segments. v +v  SET UP: When the acceleration is constant, the formula x − x0 =  0 x x  t applies. The average  2 

velocity is vav-x =

Δx . Δt

 v + v   0 + 10.0 m/s  EXECUTE: (a) x − x0 =  0 x x  t =   (2.0 s) = 10.0 m. 2  2    (b) (i) 40.0 m at 10.0 m/s so time at constant speed is 4.0 s. The total time is 6.0 s, so Δx 50.0 m vav-x = = = 8.33 m/s. Δt 6.0 s (ii) He runs 90.0 m at 10.0 m/s so the time at constant speed is 9.0 s. The total time is 11.0 s, so 100 m vav-x = = 9.09 m/s. 11.0 s (iii) He runs 190 m at 10.0 m/s so time at constant speed is 19.0 s. His total time is 21.0 s, so 200 m vav-x = = 9.52 m/s. 21.0 s EVALUATE: His average velocity keeps increasing because he is running more and more of the race at his top speed.

2-26

Chapter 2

2.54.

IDENTIFY: We know the vertical position of the lander as a function of time and want to use this to find its velocity initially and just before it hits the lunar surface. dy SET UP: By definition, v y (t ) = , so we can find vy as a function of time and then evaluate it for the dt desired cases. dy EXECUTE: (a) v y (t ) = = −c + 2dt. At t = 0, v y (t ) = −c = −60.0 m/s. The initial velocity is 60.0 m/s dt downward. (b) y (t ) = 0 says b − ct + dt 2 = 0. The quadratic formula says t = 28.57 s ± 7.38 s. It reaches the

surface at t = 21.19 s. At this time, v y = −60.0 m/s + 2(1.05 m/s 2 )(21.19 s) = −15.5 m/s. EVALUATE: The given formula for y(t) is of the form y = y0 + v0yt + 12 at2. For part (a), v0y = −c = −60

m/s. 2.55.

IDENTIFY: In time tS the S-waves travel a distance d = vStS and in time tP the P-waves travel a

distance d = vPtP . SET UP: tS = tP + 33 s EXECUTE:

  d d 1 1 = + 33 s. d  −  = 33 s and d = 250 km. vS vP 3.5 km/s 6.5 km/s  

EVALUATE: The times of travel for each wave are tS = 71 s and tP = 38 s. 2.56.

IDENTIFY: A rock is thrown upward from the edge of a roof and eventually lands on the ground a distance H below the edge. Its acceleration is g downward, so the constant-acceleration equations apply. SET UP: The time to reach the maximum height is Tmax and the time after throwing to reach the ground is 3Tmax. The equations v y = v0 y + a yt and v 2y = v02 y + 2a y ( y − y0 ) apply. Call the y-axis upward with y =

0 at the edge of the roof. At the highest point, vy = 0 and at the ground y = –H. We want the initial speed V0 in terms of H. Making a sketch helps to organize the information, as in Fig. 2.56.

Figure 2.56 EXECUTE: At the highest point: v y = v0 y + a yt gives 0 = V0 – gTmax, so V0 = gTmax.

At the ground: After the rock was thrown, it took time Tmax to reach the highest point and time 3Tmax to reach the ground. Therefore the time to fall from the highest point to the ground was 2Tmax. Apply v y = v0 y + a yt to the interval from the highest point to the ground to find the speed vy when the rock reaches the ground. vy = 0 + (–g)(2Tmax) = –2gTmax. Now look at the entire motion from the instant the rock is thrown up until the instant it reaches the ground a distance H below the edge of the roof. We = –2gTmax at ground level. Therefore v 2y = v02 y + 2a y ( y − y0 ) have already found that V0 = gTmax and vy Browsegrades.net © Copyright 2020 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Motion Along a Straight Line

2-27

  V  gives (–2gTmax)2 = V02 + 2(–g)(–H). Using Tmax = V0/g gives  −2 g  0   = V02 + 2 gH  g  

→

2 gH . 3 EVALUATE: It might seem like a good idea to check the special case if H = 0 (the edge of the roof is at ground level). Our result seems to give V0 = 0 in that case. But if the roof were at ground level, the time for the entire up-and-down trip would take 2Tmax, not 3Tmax, so our result would not apply. Checking units in the final answer would be a good idea. V0 =

2.57.

IDENTIFY: The average velocity is vav-x =

Δx . Δt

SET UP: Let + x be upward. 1000 m − 63 m EXECUTE: (a) vav-x = = 197 m/s 4.75 s 1000 m − 0 (b) vav-x = = 169 m/s 5.90 s 63 m − 0 = 54.8 m/s. When the velocity isn’t 1.15 s constant the average velocity depends on the time interval chosen. In this motion the velocity is increasing.

EVALUATE: For the first 1.15 s of the flight, vav-x =

2.58.

IDENTIFY: The block has a constant westward acceleration, so we can use the constant-acceleration equations. Δx 1 SET UP: The equations x = x0 + v0 xt + axt 2 , vx = v0 x + axt , vx2 = v02x + 2ax ( x − x0 ) , and vav -x = Δt 2 apply. EXECUTE: (a) The target variable is the time for the block to return to x = 0. Using 1 1 x = x0 + v0 xt + axt 2 gives 0 = (12.0 m/s)t + (−2.00 m/s 2 )t 2 . So t = 12.0 s. 2 2 (b) The block instantaneously stops when it reaches it maximum distance east, so vx = v0 x + axt gives 0

= 12.0 m/s + (–2.00 m/s)t, which gives t = 6.00 s. Using vx2 = v02x + 2ax ( x − x0 ) we have 0 = (12.0 m/s)2 + 2(–2.00 m/s 2 )Δx , which gives Δx = 36.0 m. EVALUATE: To check, we can use vav -x =

0 + 12.0 m/s  Δx to find Δx . This gives Δx = vav Δt =   (6.00 s) = Δt 2  

36.0 m, which agrees with our result in (b). 2.59.

IDENTIFY: A block is moving with constant acceleration on an incline, so the constant-acceleration equations apply. SET UP: All quantities are down the surface of the incline, so choose the x-axis along this surface and pointing downward. We first find the acceleration of the block and then use that to find its speed after sliding 16.0 m starting from rest and how long it takes to slide that distance. EXECUTE: (a) First use vx2 = v02x + 2ax ( x − x0 ) to find ax. (3.00 m/s)2 = 0 + 2ax(8.00 m). ax = 0.5625

m/s2. Now use the same equation to find the speed when the block has moved 16.0 m. vx2 = 0 + 2(0.5625 m/s2)(16.0 m) → vx = 4.24 m/s. 1 (b) Use x = x0 + v0 xt + axt 2 to find the time to travel 16.0 m. 2 1 16.0 m = 0 + 0 + (0.5625 m/s2)t2 → t = 7.54 s. 2 Browsegrades.net © Copyright 2020 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2-28

Chapter 2 EVALUATE:

From vx2 = v02x + 2ax ( x − x0 ) , with v0x = 0, we see that vx2 ∝ ( x − x0 ) , so vx ∝ x − x0 . So

if (x – x0) increases by a factor of 2, vx should increase by a factor of

2 . At the end of 8.00 m, the

block was moving at 3.00 m/s, so at the end of 16.0 m, it should be moving at (3.00 m/s) m/s, which agrees with our result in (a). 2.60.

2 = 4.24

IDENTIFY: Use constant acceleration equations to find x − x0 for each segment of the motion. SET UP: Let + x be the direction the train is traveling. EXECUTE: t = 0 to 14.0 s: x − x0 = v0 xt + 12 a xt 2 = 12 (1.60 m/s 2 )(14.0 s) 2 = 157 m.

At t = 14.0 s, the speed is vx = v0 x + a xt = (1.60 m/s 2 )(14.0 s) = 22.4 m/s. In the next 70.0 s, a x = 0 and x − x0 = v0 xt = (22.4 m/s)(70.0 s) = 1568 m. For the interval during which the train is slowing down, v0 x = 22.4 m/s, a x = −3.50 m/s 2 and vx = 0. vx2 = v02x + 2ax ( x − x0 ) gives x − x0 =

vx2 − v02x 0 − (22.4 m/s) 2 = = 72 m. 2a x 2(−3.50 m/s 2 )

The total distance traveled is 157 m + 1568 m + 72 m = 1800 m. EVALUATE: The acceleration is not constant for the entire motion, but it does consist of constant acceleration segments, and we can use constant acceleration equations for each segment. 2.61.

IDENTIFY: When the graph of vx versus t is a straight line the acceleration is constant, so this motion

consists of two constant acceleration segments and the constant acceleration equations can be used for each segment. Since vx is always positive the motion is always in the + x direction and the total distance moved equals the magnitude of the displacement. The acceleration a x is the slope of the vx versus t graph. SET UP: For the t = 0 to t = 10.0 s segment, v0 x = 4.00 m/s and vx = 12.0 m/s. For the t = 10.0 s to 12.0 s segment, v0 x = 12.0 m/s and vx = 0.  v + v   4.00 m/s + 12.0 m/s  EXECUTE: (a) For t = 0 to t = 10.0 s, x − x0 =  0 x x  t =   (10.0 s) = 80.0 m. 2  2     12.0 m/s + 0  For t = 10.0 s to t = 12.0 s, x − x0 =   (2.00 s) = 12.0 m. The total distance traveled is 92.0 2   m. (b) x − x0 = 80.0 m + 12.0 m = 92.0 m (c) For t = 0 to 10.0 s, a x =

12.0 m/s − 4.00 m/s = 0.800 m/s 2 . For t = 10.0 s to 12.0 s, 10.0 s

0 − 12.0 m/s = −6.00 m/s 2 . The graph of a x versus t is given in Figure 2.61. 2.00 s EVALUATE: When vx and a x are both positive, the speed increases. When vx is positive and a x is ax =

negative, the speed decreases.

Motion Along a Straight Line

2-29

Figure 2.61 2.62.

IDENTIFY: Apply x − x0 = v0 xt + 12 axt 2 to the motion of each train. A collision means the front of the

passenger train is at the same location as the caboose of the freight train at some common time. SET UP: Let P be the passenger train and F be the freight train. For the front of the passenger train x0 = 0 and for the caboose of the freight train x0 = 200 m. For the freight train vF = 15.0 m/s and aF = 0. For the passenger train vP = 25.0 m/s and aP = −0.100 m/s 2 . EXECUTE: (a) x − x0 = v0 xt + 12 axt 2 for each object gives xP = vPt + 12 aPt 2 and xF = 200 m + vFt.

Setting xP = xF gives vPt + 12 aPt 2 = 200 m + vFt. (0.0500 m/s 2 )t 2 − (10.0 m/s)t + 200 m = 0. The

(

)

1 +10.0 ± (10.0) 2 − 4(0.0500)(200) s = (100 ± 77.5) s. The collision 0.100 occurs at t = 100 s − 77.5 s = 22.5 s. The equations that specify a collision have a physical solution (real, positive t), so a collision does occur. (b) xP = (25.0 m/s)(22.5 s) + 12 (−0.100 m/s 2 )(22.5 s) 2 = 537 m. The passenger train moves 537 m

quadratic formula gives t =

before the collision. The freight train moves (15.0 m/s)(22.5 s) = 337 m. (c) The graphs of xF and xP versus t are sketched in Figure 2.62. EVALUATE: The second root for the equation for t, t = 177.5 s is the time the trains would meet again if they were on parallel tracks and continued their motion after the first meeting.

Figure 2.62 2.63.

IDENTIFY and SET UP: Apply constant acceleration kinematics equations. Find the velocity at the start of the second 5.0 s; this is the velocity at the end of the first 5.0 s. Then find x − x0 for the first 5.0 s. EXECUTE: For the first 5.0 s of the motion, v0 x = 0, t = 5.0 s.

vx = v0 x + axt gives vx = a x (5.0 s). This is the initial speed for the second 5.0 s of the motion. For the second 5.0 s:

2-30

Chapter 2

v0 x = ax (5.0 s), t = 5.0 s, x − x0 = 200 m. x − x0 = v0 xt + 12 axt 2 gives 200 m = (25 s 2 )ax + (12.5 s 2 ) ax so a x = 5.333 m/s 2 . Use this a x and consider the first 5.0 s of the motion: x − x0 = v0 xt + 12 a xt 2 = 0 + 12 (5.333 m/s 2 )(5.0 s) 2 = 67 m. EVALUATE: The ball is speeding up so it travels farther in the second 5.0 s interval than in the first. 2.64.

IDENTIFY: The rock has a constant acceleration, so we can apply the constant-acceleration equations. 1 SET UP: We use x = x0 + v0 xt + axt 2 and vx = v0 x + axt . The force is applied starting at t = 0 and 2 maintains a constant acceleration of 4.00 m/s2 in the –x direction. Our target variables are the three times when the rock is 24.0 m from where it was when the force began to be applied. We also want its velocity at those instants. 1 EXECUTE: When the rock is 24.0 m on the +x side of the origin, x = x0 + v0 xt + axt 2 gives 2 1 24.0 m = (16.0 m/s)t + (–4.00 m/s 2 )t 2 . Using the quadratic formula we get t = 2.00 s and t = 6.00 s. 2

At these times vx = v0 x + axt gives vx = 16.0 m/s + (–4.00 m/s 2 )(2.00 s) = 8.00 m/s and vx = 16.0 m/s + (–4.00 m/s 2 )(6.00 s) = –8.00 m/s. But the rock is also 24.0 m from the origin when it is 1 (–4.00 m/s2 )t 2 . This quadratic 2 equation has two roots, one of which is negative. We discard that root, leaving only t = 9.29 s. At this

24.0 m on the –x side. In this case we get −24.0 m = (16.0 m/s)t +

time, the velocity is vx = 16.0 m/s + (–4.00 m/s 2 )(9.29 s) = –21.2 m/s. EVALUATE: To find the third time when the rock was 24.0 m from the origin, we had to think about the rock’s behavior. It was not sufficient just to plug into an equation and get all three answers. 2.65.

IDENTIFY: Apply constant acceleration equations to each object. Take the origin of coordinates to be at the initial position of the truck, as shown in Figure 2.65a. Let d be the distance that the car initially is behind the truck, so x0 (car) = −d and x0 (truck) = 0. Let

T be the time it takes the car to catch the truck. Thus at time T the truck has undergone a displacement x − x0 = 60.0 m, so is at x = x0 + 60.0 m = 60.0 m. The car has caught the truck so at time T is also at x = 60.0 m.

Figure 2.65a (a) SET UP: Use the motion of the truck to calculate T: x − x0 = 60.0 m, v0 x = 0 (starts from rest), a x = 2.10 m/s 2 , t = T

x − x0 = v0 xt + 12 axt 2 Since v0 x = 0, this gives t = EXECUTE: T =

2( x − x0 ) ax

2(60.0 m) = 7.56 s 2.10 m/s 2

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