University Physics With Modern Physics Technology Update 13th Edition Young Solutions Manual by Kakalots

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MOTION ALONG A STRAIGHT LINE

2.1.

2.2.

IDENTIFY: x  vav-x t SET UP: We know the average velocity is 6.25 m/s. EXECUTE: x  vav-x t  250 m EVALUATE: In round numbers, 6 m/s  4 s = 24 m ≈ 25 m, so the answer is reasonable. x IDENTIFY: vav-x  t 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 t 1166  106 s (b) For the round trip, x2  x1 and x  0. The average velocity is zero. EVALUATE: The average velocity for the trip from the nest to the release point is positive. IDENTIFY: Target variable is the time  t it takes to make the trip in heavy traffic. Use Eq. (2.2) that relates the average velocity to the displacement and average time. x x SET UP: vav-x  so x  vav-x t and t   t vav-x EXECUTE: Use the information given for normal driving conditions to calculate the distance between the two cities:

EXECUTE: (a) vav-x 

2.3.

x  vav-x t  (105 km/h)(1 h/60 min)(140 min)  245 km

Now use vav-x for heavy traffic to calculate t ;  x is the same as before: x 245 km   350 h  3 h and 30 min. vav-x 70 km/h The trip takes an additional 1 hour and 10 minutes. EVALUATE: The time is inversely proportional to the average speed, so the time in traffic is (105/70)(140 min)  210 min t 

2.4.

x . Use the average speed for each segment to find the time t traveled in that segment. The average speed is the distance traveled 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. 280 m 200 m  700 s.  400 s and the westward run takes EXECUTE: (a) The eastward run takes time 40 m/s 50 m/s 480 m  44 m/s. The average speed for the entire trip is 1100 s x 80 m   073 m/s. The average velocity is directed westward. (b) vav-x  t 1100 s

IDENTIFY: The average velocity is vav-x 

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2-1

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2-2

2.5.

2.6.

Chapter 2 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: Given two displacements, we want the average velocity and the average speed. x SET UP: The average velocity is vav-x  and the average speed is just the total distance walked t divided by the total time to walk this distance. EXECUTE: (a) Let +x be east. x  600 m  400 m  200 m and t  280 s  360 s  640 s. So x 200 m vav-x    0312 m/s. t 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. x . Use x(t ) to find x for each t. IDENTIFY: The average velocity is vav-x  t SET UP: x (0)  0, x (200 s)  560 m, and x (400 s)  208 m 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  760 m/s (c) vav-x  200 s EVALUATE: The average velocity depends on the time interval being considered. (a) IDENTIFY: Calculate the average velocity using Eq. (2.2). x SET UP: vav-x  so use x(t ) to find the displacement  x for this time interval. t EXECUTE: t  0 : x  0 (b) vav-x 

2.7.

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 x 120 m   120 m/s t 100s (b) IDENTIFY: Use Eq. (2.3) to calculate vx (t ) and evaluate this expression at each specified t.

Then vav-x 

dx  2bt  3ct 2  dt EXECUTE: (i) t  0 : vx  0 SET UP: vx 

(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  0

EXECUTE: 2b  3ct so t 

2.8.

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. 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.

Motion Along a Straight Line

SET UP: The instantaneous velocity is vx (t ) 

2-3

dx d (280 m  (124 m/s)t  (00450 m/s3 )t 3 )  . dt dt

dx  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. x . We can find the displacement  t for each IDENTIFY: The average velocity is given by vav-x  t 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), EXECUTE: vx (t ) 

2.9.

vx  2 30 m/s for t  20 s to t  30 s. When the velocity is constant, x  v x t. EXECUTE: (a) For t  0 to t  20 s, x  (20 m/s)(20 s)  40 m. For t  20 s to t  30 s, x  (30 m/s)(10 s)  30 m. For the first 3.0 s, x  40 m  30 m  70 m. The distance traveled is also

x 70 m   233 m/s. The average speed is also 2.33 m/s. t 30 s (b) For t  20 s to 3.0 s, x  ( 30 m/s)(10 s)   3 0 m. For the first 3.0 s, x  40 m  ( 30 m)  10 m. The dog runs 4.0 m in the +x-direction and then 3.0 m in the 7.0 m. The average velocity is vav-x 

x-direction, so the distance traveled is still 7.0 m. vav-x 

2.10.

2.11.

x 10 m   033 m/s. The average speed is t 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. 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. 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

2.12.

EVALUATE: The sign of v x shows the direction the car is moving. v x is constant when x versus t is a straight line. v IDENTIFY: aav-x  x . a x (t ) is the slope of the v x versus t graph. t SET UP: 60 km/h  167 m/s 0  167 m/s 167 m/s  0  17 m/s 2 .  17 m/s 2 . (ii) aav-x  EXECUTE: (a) (i) aav-x  10 s 10 s (iii) vx  0 and aav-x  0. (iv) vx  0 and aav-x  0. (b) At t  20 s, v x is constant and a x  0. At t  35 s, the graph of v x versus t is a straight line and ax  aav-x  17 m/s 2 .

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

2-4

2.13.

Chapter 2

vx . t SET UP: Assume the car is moving in the  x direction. 1 mi/h  0447 m/s, so 60 mi/h  2682 m/s, 200 mi/h  89.40 m/s and 253 mi/h  1131 m/s. EXECUTE: (a) The graph of v x versus t is sketched in Figure 2.13. The graph is not a straight line, so the acceleration is not constant. 8940m/s  2682 m/s 2682 m/s  0  128 m/s 2 (ii) aav-x   350 m/s 2 (b) (i) aav-x  200 s  21 s 21 s 1131m/s  8940 m/s  0718 m/s 2 . The slope of the graph of v x versus t decreases as t (iii) aav-x  53 s  200 s increases. This is consistent with an average acceleration that decreases in magnitude during each successive time interval. EVALUATE: The average acceleration depends on the chosen time interval. For the interval between 0 and 1131m/s  0  213 m/s 2 . 53 s, aav-x  53 s IDENTIFY: The average acceleration for a time interval  t is given by aav-x 

Figure 2.13 2.14.

2.15.

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 16.0 m/s. dv d ((0860 m/s3 )t 2 ) . SET UP: ax (t )  x  dt dt dv EXECUTE: ax (t )  x  (172 m/s3 )t . When vx  160 m/s, t  4313 s. At this time, a x  742 m/s 2 . dt EVALUATE: The acceleration of this car is not constant. dv dx IDENTIFY and SET UP: Use vx  and a x  x to calculate vx (t ) and a x (t ) dt dt dx 2  200 cm/s  (0 125 cm/s )t EXECUTE: vx  dt dv ax  x  0125 cm/s 2 dt (a) At t  0, x  500 cm, vx  200 cm/s, ax  2 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

Motion Along a Straight Line

2-5

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 255 cm/s (e) The graphs are sketched in Figure 2.15.

Figure 2.15

2.16.

EVALUATE: The acceleration is constant and negative. v x 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 IDENTIFY: Use Eq. (2.4), with t  10 s in all cases. SET UP: v x is negative if the motion is to the left. 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

2.17.

(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. v IDENTIFY: The average acceleration is aav-x  x . Use vx (t ) to find v x at each t. The instantaneous t dvx . acceleration is a x  dt SET UP: vx (0)  300 m/s and vx (500 s)  550 m/s. EXECUTE: (a) aav-x 

vx 550 m/s  300 m/s   0500 m/s 2 t 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 2.17. (b) ax 

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 a x (t ) is a linear function of t.

Figure 2.17

2-6

2.18.

Chapter 2 IDENTIFY: vx (t )  SET UP:

dv dx and a x (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/s6 )t 5 and ax (t )  960 m/s 2  (300 m/s 6 )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 2.18. EVALUATE: For the entire time interval from t  0 to t  200 s, the velocity v x is positive and x increases. While a x is also positive the speed increases and while a x is negative the speed decreases.

Figure 2.18 2.19.

IDENTIFY: Use the constant acceleration equations to find v0x and ax  (a) SET UP: The situation is sketched in Figure 2.19. x  x0  700 m

t  7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  50 m/s EXECUTE: Use x  x0   0 x x  t , so v0 x    2 t 700s vx  v0 x 150 m/s  50 m/s   143 m/s2  t 700s EVALUATE: The average velocity is (700 m)/(700s)  100 m/s The final velocity is larger than this, so (b) Use vx  v0 x  a xt , so ax 

2.20.

the antelope must be speeding up during the time interval; v0x  vx and a x  0 IDENTIFY: In (a) find the time to reach the speed of sound with an acceleration of 5g, and in (b) find his speed at the end of 5.0 s if he has an acceleration of 5g. SET UP: Let  x be in his direction of motion and assume constant acceleration of 5g so the standard kinematics equations apply so vx  v0 x  a xt. (a) vx  3(331 m/s)  993 m/s, v0 x  0, and ax  5 g  490 m/s 2  (b) t  50 s

Motion Along a Straight Line EXECUTE: (a) vx  v0 x  a xt and t 

2-7

vx  v0 x 993 m/s  0   203 s. Yes, the time required is larger ax 490 m/s 2

than 5.0 s. (b) vx  v0 x  axt  0  (490 m/s 2 )(50 s)  245 m/s. 2.21.

EVALUATE: In 5 s he can only reach about 2/3 the speed of sound without blacking out. IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) 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/s2 . 2( x  x0 ) 2(150 m)

2( x  x0 ) 2(150 m)  v  vx  (b) x  x0   0 x t gives t    00667 s   2  v0 x  vx 450 m/s 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. IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) 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  a xt gives

EVALUATE: We could also use vx  v0 x  a xt to find t  2.22.

vx  v0 x 7314 m/s  0   2440 m/s 2 . 23 t 300  10 s  v0 x  vx   0  7314 m/s  23 (b) x  x0   t   (300  10 s)  110 m  2   2 ax 

EVALUATE: We could also use x  x0  v0 xt  12 a xt 2 to calculate x  x0 : x  x0  12 (2440 m/s 2 )(300  102 3 s) 2  110 m, which agrees with our previous result. The acceleration

2.23.

of the ball is very large. IDENTIFY: Assume that the acceleration is constant and apply the constant acceleration kinematic equations. Set |ax | equal to its maximum allowed value. SET UP: Let  x be the direction of the initial velocity of the car. a x  2 250 m/s 2 . 105 km/h  2917 m/s. EXECUTE: v0 x  1 2917 m/s. vx  0. vx2  v02x  2ax ( x  x0 ) gives

vx2  v02x 0  (2917 m/s)2   170 m. 2a x 2(250 m/s2 ) 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. IDENTIFY: In (a) we want the time to reach Mach 4 with an acceleration of 4g, and in (b) we want to know how far he can travel if he maintains this acceleration during this time. SET UP: Let  x be the direction the jet travels and take x0  0 With constant acceleration, the equations x  x0 

2.24.

vx  v0 x  a xt and x  x0  v0 xt  12 a xt 2 both apply. ax  4 g  392 m/s 2 , vx  4(331 m/s)  1324 m/s,

and v0 x  0 EXECUTE: (a) Solving vx  v0 x  a xt for t gives t 

vx  v0 x 1324 m/s  0   338 s. ax 392 m/s 2

(b) x  x0  v0 xt  12 a xt 2  12 (392 m/s 2 )(338 s) 2  224  104 m  224 km. EVALUATE: The answer in (a) is about ½ min, so if he wanted to reach Mach 4 any sooner than that, he would be in danger of blacking out.

2-8 2.25.

Chapter 2 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), a x 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  a xt and x  x0  v0 xt  12 a xt 2 both apply since the acceleration is constant.

EXECUTE: Solving vx  v0 x  a xt for v0x gives v0 x   a xt  Then x  x0  v0 xt  12 a xt 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   a xt   (588 m/s 2 )(36  102 3 s)  21 m/s  47 mph

2.26.

IDENTIFY: In (a) the hip pad must reduce the person’s speed from 2.0 m/s to 1.3 m/s over a distance of 2.0 cm, and we want the acceleration over this distance, assuming constant acceleration. In (b) we want to find out how the acceleration in (a) lasts. SET UP: Let  y be downward. v0 y  20 m/s, v y  13 m/s, and y  y0  0020 m The equations

 v0 y  v y  v y2  v02y  2a y ( y  y0 ) and y  y0    t apply for constant acceleration. 2   EXECUTE: (a) Solving v y2  v02y  2a y ( y  y0 ) for ay gives

ay 

2.27.

v y2  v02y 2( y  y0 )



(13 m/s)2  (20 m/s) 2   58 m/s2  59 g. 2(0020 m)

 v0 y  v y  2( y  y0 ) 2(0020 m) (b) y  y0     12 ms.  t gives t  v  v 2  0 m/s  13 m/s 2 0y y   EVALUATE: The acceleration is very large, but it only lasts for 12 ms so it produces a small velocity change. 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)

vx  v0 x 5.0  103 m/s   1.6 ms. ax 3.1 106 m/s2 EVALUATE: (c) The calculated a is less than 450,000 g so the acceleration required doesn’t rule out this hypothesis. IDENTIFY: Apply constant acceleration equations to the motion of the car. SET UP: Let  x be the direction the car is moving. (b) vx  v0 x  axt gives t 

2.28.

EXECUTE: (a) From Eq. (2.13), with v0 x  0, ax 

2.29.

vx2 (20m/s)2   167m/s2  2( x  x0 ) 2(120m)

(b) Using Eq. (2.14), 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. v IDENTIFY: The average acceleration is aav-x  x . For constant acceleration, Eqs. (2.8), (2.12), (2.13) t and (2.14) apply. SET UP: Assume the shuttle travels in the +x direction. 161 km/h  4472 m/s and 1610 km/h  4472 m/s. 100 min  600 s

Motion Along a Straight Line

2-9

vx 4472 m/s  0   559 m/s 2 t 800 s 4472 m/s  4472 m/s  774 m/s 2 (ii) aav-x  600 s  800 s  v  vx   0  4472 m/s  (b) (i) t  800 s, v0 x  0, and vx  4472 m/s. x  x0   0 x t   (800 s)  179 m.  2   2 EXECUTE: (a) (i) aav-x 

(ii) t  600 s  800 s  520 s, v0 x  4472 m/s, and vx  4472 m/s.

 v  v   4472 m/s  4472 m/s  4 x  x0   0 x x  t    (520 s)  128  10 m.  2   2 EVALUATE: When the acceleration is constant the instantaneous acceleration throughout the time interval equals the average acceleration for that time interval. We could have calculated the distance in part (a) as x  x0  v0 xt  12 axt 2  12 (559 m/s 2 )(800 s) 2  179 m, which agrees with our previous calculation.

2.30.

IDENTIFY: The acceleration a x is the slope of the graph of v x versus t. SET UP: The signs of v x and of a x 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) v x 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 . For t  0 to 4.5 s, x  x0  (80 cm/s)(45 s)  12 (13 cm/s 2 )(45 s) 2  228 cm. For t  0 to 7.5 s, x  x0  (80 cm/s)(75 s)  12 (13 cm/s 2 )(75 s) 2  234 cm

(d) The graphs of a x and x versus t are given in Figure 2.30.

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

Figure 2.30 2.31.

(a) IDENTIFY and SET UP: The acceleration a x at time t is the slope of the tangent to the v x versus t curve at time t. EXECUTE: At t  3 s, the v x versus t curve is a horizontal straight line, with zero slope. Thus a x  0 At t  7 s, the v x 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 

2-10

Chapter 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 a x  112 m/s 2  EVALUATE: a x  0 when v x is constant, a x  0 when v x is positive and the speed is increasing, and a x  0 when v x 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 a x  0 t  5 s x  x0  ? x  x0  v0xt ( a x  0 so no

1 a t2 2 x

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 a xt 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. 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) a x  2 112 m/s 2 t  4 s x0  230 m x  x0  ?

x  x0  v0 xt  12 a xt 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 v x is always positive so the displacement is always positive and displacement and distance traveled are the same. The average velocity for time interval  t is vav-x  x/t  For t  0 to

2.32.

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 Figure 2.37 in the textbook. 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, v x is positive and a x  0. For car A, a x is positive and v x increases with t. EXECUTE: (a) The motion diagrams for the cars are given in Figure 2.32a. (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.

Motion Along a Straight Line

2-11

(c) The graphs of v x versus t for each car are sketched in Figure 2.32b. (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 x B in the x-t graph. This happens at t  3 s. (f) Car B passes car A when x B moves above x A in the x-t graph. This happens at t  1 s. EVALUATE: When a x  0, the graph of v x versus t is a horizontal line. When a x is positive, the graph of v x versus t is a straight line with positive slope.

Figure 2.32a-b 2.33.

IDENTIFY: For constant acceleration, Eqs. (2.8), (2.12), (2.13) and (2.14) apply. SET UP: Take  y to be downward, so the motion is in the  y direction. 19,300 km/h  5361 m/s, 1600 km/h  4444 m/s, and 321 km/h  892 m/s. 40 min  240 s. EXECUTE: (a) Stage A: t  240 s, v0 y  5361 m/s, v y  4444 m/s. v y  v0 y  a yt gives v y  v0 y

4444 m/s  5361 m/s   205 m/s 2 . t 240 s Stage B: t  94 s, v0 y  4444 m/s, v y  892 m/s. v y  v0 y  a yt gives ay 

ay 

v y  v0 y t



892m/s  4444 m/s  38 m/s 2 . 94 s

Stage C: y  y0  75 m, v0 y  892 m/s, v y  0. v 2y  v02y  2a y ( y  y0 ) gives ay 

v 2y  v02y

2( y  y0 ) is upward.



0  (892 m/s)2  530 m/s2 . In each case the negative sign means that the acceleration 2(75 m)

 v0 y  v y   5361 m/s  4444 m/s  (b) Stage A: y  y0   t    (240 s)  697 km. 2 2      4444m/s  892 m/s  Stage B: y  y0    (94 s)  25 km.  2

2.34.

Stage C: The problem states that y  y0  75 m 0075km. The total distance traveled during all three stages is 697 km  25 km  0075 km  722 km. EVALUATE: The upward acceleration produced by friction in stage A is calculated to be greater than the upward acceleration due to the parachute in stage B. The effects of air resistance increase with increasing speed and in reality the acceleration was probably not constant during stages A and B. IDENTIFY: Apply the constant acceleration equations to the motion of each vehicle. The truck passes the car when they are at the same x at the same t  0. SET UP: The truck has a x  0. The car has v0 x  0. Let  x be in the direction of motion of the vehicles. Both vehicles start at x0  0. The car has aC  320 m/s 2 . The truck has vx  200 m/s.

2-12

Chapter 2 EXECUTE: (a) x  x0  v0 xt  12 a xt 2 gives xT  v0Tt and xC  12 aCt 2 . Setting xT  xC gives t  0 and v0T  12 aCt , so t 

2v0T 2(200 m/s)   125 s. At this t, xT  (200 m/s)(12 5 s)  250 m and aC 320 m/s 2

x  12 (320 m/s 2 )(125 s) 2  250 m. The car and truck have each traveled 250 m.

(b) At t  125 s, the car has vx  v0 x  axt  (320 m/s 2 )(125 s)  40 m/s. (c) xT  v0Tt and xC  12 aCt 2 . The x-t graph of the motion for each vehicle is sketched in Figure 2.34a. (d) vT  v0T . vC  aCt. The vx -t graph for each vehicle is sketched in Figure 2.34b. EVALUATE: When the car overtakes the truck its speed is twice that of the truck.

Figure 2.34a-b 2.35.

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  2 980 m/s 2 v0 y  ?

v 2y  v02y  2a y ( y  y0 )

EXECUTE: v0 y  2a y ( y  y0 )  2(980 m/s 2 )(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  1 294 m/s a y  2 980 m/s 2 t  ?

y  y0  v0 yt  12 a yt 2

EXECUTE: With y  y0  0 this gives t  

2v0 y ay



2(294 m/s) 980 m/s2

 0600 s

EVALUATE: We can use v y  v0 y  a yt to show that with v0 y  294 m/s, v y  0 after 0.300 s. 2.36.

IDENTIFY: The rock has a constant downward acceleration of 9.80 m/s 2. We know its initial velocity and position and its final position. SET UP: We can use the kinematics formulas for constant acceleration. EXECUTE: (a) y  y0  30 m, v0 y  180 m/s, a y  98 m/s 2 . The kinematics formulas give v y   v02y  2a y ( y  y0 )   (180 m/s) 2  2(98 m/s 2 )(30 m)  302 m/s, so the speed is 30.2 m/s.

Motion Along a Straight Line

(b) v y  v0 y  a yt and t 

2.37.

2.38.

v y  v0 y ay



303 m/s  180 m/s 98 m/s 2

2-13

 492 s.

EVALUATE: The vertical velocity in part (a) is negative because the rock is moving downward, but the speed is always positive. The 4.92 s is the total time in the air. IDENTIFY: The pin has a constant downward acceleration of 9.80 m/s 2 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 yt  a yt 2 . We know that y  y0  0, so 2 2v0 y 2(820 m/s) t 2 2  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. 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  v02y  2a y ( y  y0 )   (950 m/s)2  2(980 m/s 2 )(360 m)  444 m/s

(b) t 

2.39.

v y  v0 y ay



444 m/s  950 m/s 98 m/s 2

 0517 s

EVALUATE: The putty is stopped by the ceiling, not by gravity. 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 constantacceleration formulas v y  v0 y  a yt and y  y0  v0 yt  12 a yt 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 yt  12 a yt 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 yt gives v0 y   a yt   (371 m/s 2 )(425 s)  158 m/s (c) The graphs are sketched in Figure 2.39.

Figure 2.39 EVALUATE: The answers can be checked several ways. For example, v y  0, v0 y  158 m/s, and a y   37 m/s 2 in v y2  v02y  2a y ( y  y0 ) gives y  y0 

2.40.

v y2  v02y 2a y



0  (158 m/s) 2 2(371 m/s 2 )

 336 m,

which agrees with the height calculated in (a). IDENTIFY: Apply constant acceleration equations to the motion of the lander. SET UP: Let  y be positive. Since the lander is in free-fall, a y  16 m/s 2 .

2-14

Chapter 2 EXECUTE: v0 y  08 m/s, y  y0  50 m, a y  1 16 m/s2 in v 2y  v02y  2a y ( y  y0 ) gives v y  v02y  2a y ( y  y0 )  (08 m/s) 2  2(16 m/s 2 )(50 m)  41 m/s.

2.41.

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. 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 yt  12 a yt 2 gives d  (490 m/s 2 )t 2 and t  (b) t 

2.42.

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. 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  250 s, a y  980 m/s 2 . y  y0  v0 yt  12 a yt 2  12 (980 m/s 2 )(250 s) 2  306 m. The building is 30.6 m tall. (b) v y  v0 y  a yt  0  (980 m/s 2 )(250 s)  245 m/s (c) The graphs of a y , v y and y versus t are given in Figure 2.42. Take y  0 at the ground.

 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 2.42 2.43.

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  1 225 m/s 2 , v0 y  0. v 2y  v02y  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  v02y  2a y ( y  y0 ) gives

y  y0 

v 2y  v02y 2a y



0  (486 m/s) 2 2(980 m/s 2 )

 121 m and y  121 m  525 m  646 m.

(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  v02y  2a y ( y  y0 ) gives

Motion Along a Straight Line

2-15

v y  2 (486 m/s) 2  2(980 m/s 2 )(525 m)  112 m/s. Then v y  v0 y  a yt gives

t

v y  v0 y ay



112 m/s  486 m/s 980 m/s 2

 164 s.

(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 yt  12 a yt 2 gives t 

2( y  y0 ) 2(525 m)   216 s. The rocket strikes the launch pad ay 225 m/s2

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 yt 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 blast-off phase and with negative slope of 980 m/s 2 after engine failure. During each phase y  y0  v0 yt  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 2.43. EVALUATE: In part (b) we could have found the time from y  y0  v0 yt  12 a yt 2 , finding v y first allows us to avoid solving for t from a quadratic equation.

Figure 2.43 2.44.

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 yt  12 a yt 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 yt  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 yt  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 yt  12 a yt 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

t





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

2-16

Chapter 2 (d) v0 y  500 m/s, a y  980 m/s 2 , v y  0. v 2y  v02y  2a y ( y  y0 ) gives

y  y0 

v 2y  v02y 2a y



0  (500 m/s) 2 2(980 m/s 2 )

 128 m. The maximum height is 41.3 m above the ground.

(e) The graphs of a y , v y , and y versus t are given in Figure 2.44. Take y  0 at the ground. EVALUATE: The sandbag initially travels upward with decreasing velocity and then moves downward with increasing speed.

Figure 2.44 2.45.

IDENTIFY: Use the constant acceleration equations to calculate a x and x  x0  (a) SET UP: vx  224 m/s, v0 x  0, t  0900 s, a x  ? vx  v0 x  a xt

EXECUTE: ax 

vx  v0 x 224 m/s  0   249 m/s2 t 0900 s

(b) ax /g  (249 m/s 2 )/(980 m/s 2 )  254 (c) x  x0  v0 xt  12 axt 2  0  12 (249 m/s 2 )(0900 s) 2  101 m (d) SET UP: Calculate the acceleration, assuming it is constant: t  140 s, v0 x  283 m/s, vx  0 (stops), a x  ? vx  v0 x  a xt

EXECUTE: ax 

vx  v0 x 0  283 m/s   202 m/s2 t 140s

ax /g  (202 m/s 2 )/(980 m/s 2 )  206; ax  2 206 g

2.46.

If the acceleration while the sled is stopping is constant then the magnitude of the acceleration is only 20.6g. But if the acceleration is not constant it is certainly possible that at some point the instantaneous acceleration could be as large as 40g. EVALUATE: It is reasonable that for this motion the acceleration is much larger than g. 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 yt  12 a yt 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  v02y  2a y ( y  y0 ) gives

y  y0 

v 2y  v02y 2a y



0  (185 m/s) 2 2(980 m/s 2 )

 175 m.

Motion Along a Straight Line

2-17

(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 2.46. v y  v0 y 185 m/s EVALUATE: The time for the egg to reach its maximum height is t    189 s. The ay 98 m/s 2 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 2.46 2.47.

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 EXECUTE:

2 y  y0  v0 yt  12 a yt 2 gives h  12 gtE2 and h  12 g EntEn . Combining these two equations gives

2.48.

2 t   175 s  2 2 gtE2  g En tEn and g En  g  E   (980 m/s 2 )    00868 m/s . t 18  6 s    En  EVALUATE: The acceleration due to gravity is inversely proportional to the square of the time of fall. 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 yt gives

t

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 yt 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 2.48. 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. © Copyright 2012 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.

2-18

Chapter 2

Figure 2.48 2.49.

IDENTIFY: Two stones are thrown up with different speeds. (a) Knowing how soon the faster one returns to the ground, how long it will take the slow one to return? (b) Knowing how high the slower stone went, how high did the faster stone go? SET UP: Use subscripts f and s to refer to the faster and slower stones, respectively. Take  y to be upward and y0  0 for both stones. v0f  3v0s  When a stone reaches the ground, y  0 The constantacceleration formulas y  y0  v0 yt  12 a yt 2 and v y2  v02y  2a y ( y  y0 ) both apply. EXECUTE: (a) y  y0  v0 yt  12 a yt 2 gives a y  

v and ts  tf  0s  v0f

  

2v0 y t

 Since both stones have the same a y ,

v0f v0s  tf ts

 13  (10 s)  33 s

(b) Since v y  0 at the maximum height, then v y2  v02y  2a y ( y  y0 ) gives a y  

v02y 2y

 Since both

2.50.

v  v2 v2 have the same a y , 0f  0s and yf  ys  0f   9 H  yf ys  v0s  EVALUATE: The faster stone reaches a greater height so it travels a greater distance than the slower stone and takes more time to return to the ground. IDENTIFY: We start with the more general formulas and use them to derive the formulas for constant acceleration. t

SET UP: The general formulas are vx  v0 x  Ñax dt and x  x0  Ñvx dt. t

EXECUTE: For constant acceleration, these formulas give vx  v0 x  Ñax dt  v0 x  ax Ñdt  v0 x  axt and

2.51.

t t t t 1 x  x0  Ñvx dt  x0  Ñ(v0 x  a xt )dt  x0  v0 x Ñdt  a x Ñtdt  x0  v0 xt  a xt 2 . 0 0 0 0 2 EVALUATE: The general formulas give the expected results for constant acceleration. 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 )  Ña y dt and y  y0  Ñv y dt. t

EXECUTE: (a) v y (t )  Ña y dt  Ñ(280 m/s3 )tdt  (140 m/s3 )t 2 t

y  y0  Ñv y dt  Ñ(140 m/s3 )t 2dt  (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.

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

Motion Along a Straight Line 2.52.

2-19

IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. Instead, use Eqs. (2.17) and (2.18). Use the values of v x and of x at t  10 s to evaluate v0x and x0 . SET UP:

t

dt 

1 n 1 t , for n  0. n 1 t

EXECUTE: (a) vx  v0 x  Ñtdt  v0 x  12 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/s 3 )(20 s) 2  68 m/s. t

(b) x  x0  Ñ(v0 x  12 t 2 )dt  x0  v0 xt  16 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20m/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 ax  At  Bt 2 with A  150 m/s3 and B  0120 m/s 4

2.53.

(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  Ñax dt 0

EXECUTE: vx  v0 x  Ñ( 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  Ñvx dt 0

1 Bt 4 EXECUTE: x  x0  Ñ( 12 At 2  13 Bt 3 )dt  x0  61 At 3  12 0

At the origin at t  0 says that x0  0, so 1 1 x  16 At 3  12 Bt 4  16 (150 m/s3 )t 3  12 (0120 m/s 4 )t 4

x  (025 m/s3 )t 3  (0010 m/s 4 )t 4

dv dx and ax (t )  x  dt dt dvx dvx , the maximum  0 (Since a x  (b) IDENTIFY and SET UP: At time t, when v x is a maximum, dt dt velocity is when a x  0 For earlier times a x is positive so v x is still increasing. For later times a x is EVALUATE: We can check our results by using them to verify that vx (t ) 

negative and v x is decreasing.)

2-20

Chapter 2

dvx  0 so At  Bt 2  0 dt One root is t  0, but at this time vx  0 and not a maximum. EXECUTE: ax 

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

2.54.

EVALUATE: For t  125 s, a x  0 and v x is increasing. For t  125 s, a x  0 and v x is decreasing. IDENTIFY: a (t ) is the slope of the v versus t graph and the distance traveled is the area under the v versus t graph. SET UP: The v versus t graph can be approximated by the graph sketched in Figure 2.54. EXECUTE: (a) Slope  a  0 for t  13 ms. (b) 1 hmax  Area under v-t graph  ATriangle  ARectangle  (13 ms)(133 cm/s)  (25 ms  13 ms)(133cm/s)  025 cm 2 133 cm/s  10  105 cm/s 2 . (c) a  slope of v-t graph. a (05 ms)  a (10 ms)  13 ms a (15ms)  0 because the slope is zero

1 (d) h  area under v-t graph. h(05 ms)  ATriangle  (05 ms)(33 cm/s)  83 103 cm. 2 1 h(10 ms)  ATriangle  (10 ms)(100cm/s)  50  102 cm. 2 1 h(15 ms)  ATriangle  ARectangle  (13 ms)(133 cm/s)  (02 ms)133 cm/s 011 cm 2 EVALUATE: The acceleration is constant until t  13 ms, and then it is zero. g  980 cm/s 2 . The acceleration during the first 1.3 ms is much larger than this and gravity can be neglected for the portion of the jump that we are considering.

Figure 2.54 2.55.

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  vx  SET UP: When the acceleration is constant, the formula x  x0   0 x  t applies. The average  2  velocity is vav-x 

x . t

 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 x 500 m vav-x    833 m/s. t 60 s

Motion Along a Straight Line

2.56.

2.57.

2-21

(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. IDENTIFY: The average speed is the total distance traveled divided by the total time. The elapsed time is the distance traveled divided by the average speed. SET UP: The total distance traveled is 20 mi. With an average speed of 8 mi/h for 10 mi, the time for that 10 mi  125 h. first 10 miles is 8 mi/h 20 mi  50 h. The second 10 mi EXECUTE: (a) An average speed of 4 mi/h for 20 mi gives a total time of 4 mi/h 10 mi  27 mi/h. must be covered in 50 h  125 h  375 h. This corresponds to an average speed of 375 h 20 mi  167 h. The second 10 mi must (b) An average speed of 12 mi/h for 20 mi gives a total time of 12 mi/h 10 mi  24 mi/h. be covered in 167 h  125 h  042 h. This corresponds to an average speed of 042 h 20 mi  125 h. But 1.25 h was already (c) An average speed of 16 mi/h for 20 mi gives a total time of 16 mi/h spent during the first 10 miles and the second 10 miles would have to be covered in zero time. This is not possible and an average speed of 16 mi/h for the 20-mile ride is not possible. EVALUATE: The average speed for the total trip is not the average of the average speeds for each 10-mile segment. The rider spends a different amount of time traveling at each of the two average speeds. dx dv IDENTIFY: vx (t )  and a x  x . dt dt d n n 1 (t )  nt , for n  1. SET UP: dt EXECUTE: (a) vx (t )  (900 m/s3 )t 2  (200 m/s 2 )t  900 m/s. ax (t )  (180 m/s3 )t  200 m/s 2 . The graphs are sketched in Figure 2.57. (b) The particle is instantaneously at rest when v x (t )  0. v0 x  0 and the quadratic formula gives





1 200  (200)2  4(900)(900) s  111 s  048 s. t  0627 s and t  159 s. These results 180 agree with the vx -t graphs in part (a). t

t  0627 s the slope of the vx -t graph is negative and at t  159 s it is positive, so the same answer is deduced from the vx (t ) graph as from the expression for a x (t ). (d) vx (t ) is instantaneously not changing when a x  0. This occurs at t 

200 m/s 2

 111 s. 180 m/s3 (e) When the particle is at its greatest distance from the origin, vx  0 and a x  0 (so the particle is starting to move back toward the origin). This is the case for t  0627 s, which agrees with the x-t graph in part (a). At t  0627 s, x  245 m. (f) The particle’s speed is changing at its greatest rate when a x has its maximum magnitude. The a x -t graph in part (a) shows this occurs at t  0 and at t  200 s. Since v x is always positive in this time interval, the particle is speeding up at its greatest rate when a x is positive, and this is for t  200 s. © Copyright 2012 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.

2-22

Chapter 2 The particle is slowing down at its greatest rate when a x is negative and this is for t  0. EVALUATE: Since a x (t ) is linear in t, vx (t ) is a parabola and is symmetric around the point where |vx (t )| has its minimum value ( t  111 s ). For this reason, the answer to part (d) is midway between the

two times in part (c).

Figure 2.57 2.58.

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  c  2dt. At t  0, v y (t )  c  600 m/s. The initial velocity is 60.0 m/s EXECUTE: (a) v y (t )  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 + 2.59.

1 2

at2. For part (a), v0y = c = 60m/s.

IDENTIFY: In time tS the S-waves travel a distance d  vStS and in time t P 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. 2.60.

IDENTIFY: The average velocity is vav-x 

x . The average speed is the distance traveled divided by the t

elapsed time. SET UP: Let  x be in the direction of the first leg of the race. For the round trip, x  0 and the total distance traveled is 50.0 m. For each leg of the race both the magnitude of the displacement and the distance traveled are 25.0 m. x 250 m   125 m/s. This is the same as the average speed for this leg of the race. EXECUTE: (a) |vav-x | t 200 s (b) |vav-x | 

x 250 m   167 m/s. This is the same as the average speed for this leg of the race. t 150 s

500 m  143 m/s. 350 s EVALUATE: Note that the average speed for the round trip is not equal to the arithmetic average of the average speeds for each leg. (d) The average speed is

Motion Along a Straight Line

2.61.

IDENTIFY: The average velocity is vav-x 

2-23

x . t

SET UP: Let  x be upward. 1000 m  63 m  197 m/s EXECUTE: (a) vav-x  4.75 s 1000 m  0  169 m/s (b) vav-x  5.90 s 63 m  0  54.8 m/s. When the velocity isn’t constant 1.15 s the average velocity depends on the time interval chosen. In this motion the velocity is increasing. (a) IDENTIFY and SET UP: The change in speed is the area under the a x versus t curve between vertical lines at t  2.5 s and t  7.5 s. EXECUTE: This area is 12 (4.00 cm/s 2  8.00 cm/s 2 )(7.5 s  2.5 s)  30.0 cm/s

EVALUATE: For the first 1.15 s of the flight, vav-x  2.62.

This acceleration is positive so the change in velocity is positive. (b) Slope of v x versus t is positive and increasing with t. The graph is sketched in Figure 2.62.

Figure 2.62 EVALUATE: The calculation in part (a) is equivalent to vx  ( aav-x )t. Since a x is linear in t, 2.63.

aav-x  ( a0 x  ax )/2. Thus aav-x  12 (4.00 cm/s 2  8.00 cm/s 2 ) for the time interval t  2.5 s to t  7.5 s. IDENTIFY: Use information about displacement and time to calculate average speed and average velocity. Take the origin to be at Seward and the positive direction to be west. distance traveled (a) SET UP: average speed  time EXECUTE: The distance traveled (different from the net displacement ( x  x0 )) is

76 km  34 km  110 km Find the total elapsed time by using vav-x  Seward to Auora: t 

x x  x0  to find t for each leg of the journey. t t

x  x0 76 km   08636 h vav-x 88 km/h

x  x0 34 km   04722 h vav-x 72 km/h Total t  08636 h  04722 h  1336 h 110 km  82 km/h Then average speed  1336 h x , where  x is the displacement, not the total distance traveled. (b) SET UP: vav-x  t

Auora to York: t 

42 km  31 km/h l336 h EVALUATE: The motion is not uniformly in the same direction so the displacement is less than the distance traveled and the magnitude of the average velocity is less than the average speed. 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. For the whole trip he ends up 76 km  34 km  42 km west of his starting point. vav-x 

2.64.

2-24

Chapter 2 EXECUTE: t  0 to 14.0 s: x  x0  v0 xt  12 axt 2  12 (1.60 m/s 2 )(14.0 s) 2  157 m. At t  14.0 s, the speed is vx  v0 x  axt  (1.60 m/s 2 )(14.0 s)  22.4 m/s. In the next 70.0 s, ax  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, ax  3.50 m/s 2 and vx  0.

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. v v v (a) IDENTIFY: Calculate the average acceleration using aav-x  x  x 0 x . Use the information about t t the time and total distance to find his maximum speed. SET UP: v0 x  0 since the runner starts from rest. vx2  v02x  2ax ( x  x0 ) gives x  x0 

2.65.

t  40 s, but we need to calculate vx , the speed of the runner at the end of the acceleration period.

EXECUTE: For the last 91 s  40 s  51 s the acceleration is zero and the runner travels a distance of d1  (51 s)vx (obtained using x  x0  v0 xt  12 axt 2 ).

During the acceleration phase of 4.0 s, where the velocity goes from 0 to vx , the runner travels a distance v v  v d 2   0 x x  t  x (40 s)  (20 s)vx 2  2 

The total distance traveled is 100 m, so d1  d 2  100 m This gives (51 s)vx  (20 s)vx  100 m

vx 

100 m  1408 m/s 71 s

vx  v0 x 1408 m/s  0   35 m/s 2  t 40 s (b) For this time interval the velocity is constant, so aav-x  0 Now we can calculate aav-x : aav-x 

EVALUATE: Now that we have v x we can calculate d1  (51 s)(1408 m/s)  718 m and d 2  (20 s)(1408 m/s)  282 m So, d1  d 2  100 m, which checks.

vx  v0 x , where now the time interval is the full 9.1 s of the race. t We have calculated the final speed to be 1408 m/s, so 1408 m/s aav-x   15 m/s 2  91 s EVALUATE: The acceleration is zero for the last 5.1 s, so it makes sense for the answer in part (c) to be less than half the answer in part (a). (d) The runner spends different times moving with the average accelerations of parts (a) and (b). IDENTIFY: Apply the constant acceleration equations to the motion of the sled. The average velocity for a x . time interval  t is vav-x  t SET UP: Let  x be parallel to the incline and directed down the incline. The problem doesn’t state how much time it takes the sled to go from the top to 14.4 m from the top. 256 m  144 m  560 m/s. 25.6 to 40.0 m: EXECUTE: (a) 14.4 m to 25.6 m: vav-x  200 s 400 m  256 m 576 m  400 m vav-x   720 m/s. 40.0 m to 57.6 m: vav-x   880 m/s. 200 s 200 s

2.66.

Motion Along a Straight Line

2-25

(b) For each segment we know x  x0 and t but we don’t know v0x or v x . Let x1  144 m and

x x v v  x x x2  256 m. For this interval  1 2   2 1 and at  v2  v1. Solving for v2 gives v2  12 at  2 1 . t t  2  v v  x x Let x2  256 m and x3  400 m. For this second interval,  2 3   3 2 and at  v3  v2 . Solving  2  t

x3  x2 . Setting these two expressions for v2 equal to each other and solving for t 1 1 a gives a  2 [( x3  x2 )  ( x2  x1)]  [(400 m  256 m)  (256 m  144 m)]  080 m/s 2. t (200 s) 2 for v2 gives v2   12 at 

Note that this expression for a says a 

vav-23  vav-12 , where vav-12 and vav-23 are the average speeds for t

successive 2.00 s intervals. (c) For the motion from x  144 m to x  256 m, x  x0  112 m, ax  080 m/s 2 and t  200 s.

x  x0 1 112 m 1  2 a xt   (080 m/s 2 )(200 s)  480 m/s. t 200 s 2 (d) For the motion from x  0 to x  144 m, x  x0  144 m, v0 x  0, and vx  48 m/s. x  x0  v0 xt  12 a xt 2 gives v0 x 

2( x  x0 ) 2(144 m)  v  vx    60 s. x  x0   0 x  t gives t   2  v0 x  vx 48 m/s (e) For this 1.00 s time interval, t  100 s, v0 x  48 m/s, a x  080 m/s 2 . x  x0  v0 xt  12 a xt 2  (48 m/s)(100 s)  12 (080 m/s 2 )(100 s) 2  52 m.

EVALUATE: With x  0 at the top of the hill, x(t )  v0 xt  12 axt 2  (040 m/s 2 )t 2 . We can verify that 2.67.

t  60 s gives x  144 m, t  80 s gives 25.6 m, t  100 s gives 40.0 m, and t  120 s gives 57.6 m. IDENTIFY: When the graph of v x 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 v x 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 v x 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 m. 2   (b) x  x0  800 m  120 m  920 m (c) For t  0 to 10.0 s, ax 

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 2.67. 200 s EVALUATE: When v x and a x are both positive, the speed increases. When v x is positive and a x is negative, the speed decreases. ax 

2-26

Chapter 2

Figure 2.67 2.68.

IDENTIFY: When the graph of v x 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. For t  0 to 5.0 s, v x is positive and the ball moves in the  x direction. For t  50 s to 20.0 s, v x is negative and the ball moves in the x direction. The acceleration a x is the slope of the v x versus

t graph. SET UP: For the t  0 to t  50 s segment, v0 x  0 and vx  300 m/s. For the t  50 s to t  200 s segment, v0 x  2 200 m/s and vx  0.

 v  v   0  300 m/s  EXECUTE: (a) For t  0 to 5.0 s, x  x0   0 x x  t    (50 m/s)  750 m. The ball 2   2    200 m/s  0  travels a distance of 75.0 m. For t  50 s to 20.0 s, x  x0    (150 m/s)  1500 m. The 2   total distance traveled is 750 m  1500 m  2250 m. (b) The total displacement is x  x0  750 m  (1500 m)  750 m. The ball ends up 75.0 m in the negative x-direction from where it started. 300 m/s  0  600 m/s2 . For t  50 s to 20.0 s, (c) For t  0 to 5.0 s, ax  50 s 0  (200 m/s) ax   133 m/s 2 . The graph of a x versus t is given in Figure 2.68. 150 s (d) The ball is in contact with the floor for a small but nonzero period of time and the direction of the velocity doesn’t change instantaneously. So, no, the actual graph of vx (t ) is not really vertical at 5.00 s. EVALUATE: For t  0 to 5.0 s, both v x and a x are positive and the speed increases. For t  50 s to 20.0 s, v x is negative and a x is positive and the speed decreases. Since the direction of motion is not the same throughout, the displacement is not equal to the distance traveled.

Figure 2.68

Motion Along a Straight Line 2.69.

2-27

IDENTIFY and SET UP: Apply constant acceleration 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  a xt 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: v0 x  a x (50 s), t  50 s, x  x0  150 m

x  x0  v0 xt  12 a xt 2 gives 150 m  (25 s 2 )ax  (125 s 2 )ax and ax  40 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 (40 m/s 2 )(50 s) 2  500 m

EVALUATE: The ball is speeding up so it travels farther in the second 5.0 s interval than in the first. In fact, x  x0 is proportional to t 2 since it starts from rest. If it goes 50.0 m in 5.0 s, in twice the time (10.0 s) it should go four times as far. In 10.0 s we calculated it went 50 m  150 m  200 m, which is four times 50 m. 2.70.

IDENTIFY: Apply x  x0  v0 xt  12 a xt 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 a xt 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 quadratic





1 100  (100)2  4(00500)(200) s  (100  775) s. The collision occurs at 0100 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. formula gives t 

(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 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 2.70. 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 2.70 2.71.

IDENTIFY: Apply constant acceleration equations to the motion of the two objects, you and the cockroach. You catch up with the roach when both objects are at the same place at the same time. Let T be the time when you catch up with the cockroach. SET UP: Take x  0 to be at the t  0 location of the roach and positive x to be in the direction of motion of the two objects. roach: v0 x  150 m/s, a x  0, x0  0, x  120 m, t  T

2-28

Chapter 2 you: v0 x  080 m/s, x0  2 090 m, x  120 m, t  T , a x  ? Apply x  x0  v0 xt  12 a xt 2 to both objects: EXECUTE: roach: 120 m  (150 m/s)T , so T  0800 s you: 120 m  (090 m)  (080 m/s)T  12 axT 2 210 m  (080 m/s)(0800 s)  12 ax (0800 s) 2

210 m  064 m  (0320 s 2 ) ax

ax  46 m/s 2 

 v  vx  EVALUATE: Your final velocity is vx  v0 x  a xt  448 m/s Then x  x0   0 x  t  210 m, which  2 

2.72.

2.73.

checks. You have to accelerate to a speed greater than that of the roach so you will travel the extra 0.90 m you are initially behind. IDENTIFY: The insect has constant speed 15 m/s during the time it takes the cars to come together. SET UP: Each car has moved 100 m when they hit. 100 m  10 s. During this time the grasshopper travels a distance EXECUTE: The time until the cars hit is 10 m/s of (15 m/s)(10 s)  150 m. EVALUATE: The grasshopper ends up 100 m from where it started, so the magnitude of his final displacement is 100 m. This is less than the total distance he travels since he spends part of the time moving in the opposite direction. 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 2.73a. Let d be the distance that the auto initially is behind the truck, so x0 (auto)  d and x0 (truck)  0 Let T be the time it takes the auto to catch the truck. Thus at time T the truck has undergone a displacement x  x0  400 m, so is at x  x0  400 m  400 m The auto has caught the truck so at time T is also at x  400 m

Figure 2.73a (a) SET UP: Use the motion of the truck to calculate T: x  x0  400 m, v0 x  0 (starts from rest), a x  210 m/s 2 , t  T x  x0  v0 xt  12 a xt 2

Since v0 x  0, this gives t  EXECUTE: T 

2( x  x0 ) ax

2(400 m)

 617 s 210 m/s2 (b) SET UP: Use the motion of the auto to calculate d: x  x0  400 m  d , v0 x  0, a x  340 m/s 2 , t  617 s x  x0  v0 xt  12 a xt 2

Motion Along a Straight Line

2-29

EXECUTE: d  400 m  12 (340 m/s 2 )(617 s) 2

d  648 m  400 m  248 m (c) auto: vx  v0 x  axt  0  (340 m/s 2 )(617 s)  210 m/s truck: vx  v0 x  axt  0  (210 m/s 2 )(617 s)  130 m/s (d) The graph is sketched in Figure 2.73b.

Figure 2.73b

2.74.

EVALUATE: In part (c) we found that the auto was traveling faster than the truck when they came abreast. The graph in part (d) agrees with this: at the intersection of the two curves the slope of the x-t curve for the auto is greater than that of the truck. The auto must have an average velocity greater than that of the truck since it must travel farther in the same time interval. IDENTIFY: Apply the constant acceleration equations to the motion of each car. The collision occurs when the cars are at the same place at the same time. SET UP: Let  x be to the right. Let x  0 at the initial location of car 1, so x01  0 and x02  D. The cars collide when x1  x2 . v0 x1  0, a x1  a x , v0 x 2  2 v0 and a x 2  0. EXECUTE: (a) x  x0  v0 xt  12 a xt 2 gives x1  12 a xt 2 and x2  D  v0t. x1  x2 gives 1 a t2 2 x

 v0t  D  0. The quadratic formula gives t 

physical, so t 









1 a t2 2 x

 D  v0t.

1 v0  v02  2ax D . Only the positive root is ax

1 v0  v02  2ax D . ax

(b) v1  axt  v02  2ax D  v0 (c) The x-t and vx -t graphs for the two cars are sketched in Figure 2.74. EVALUATE: In the limit that a x  0, D  v0t  0 and t  D/v0 , the time it takes car 2 to travel distance D. In the limit that v0  0, t 

2D , the time it takes car 1 to travel distance D. ax

Figure 2.74

2-30

2.75.

Chapter 2 IDENTIFY: The average speed is the distance traveled divided by the time. The average velocity is vav-x 

x . t

SET UP: The distance the ball travels is half the circumference of a circle of diameter 50.0 cm so is 1  d  1  (500 cm)  785 cm. Let  x be horizontally from the starting point toward the ending point, so 2 2  x equals the diameter of the bowl.

2.76.

1d 2

785 cm  785 cm/s. t 100 s x 500 cm   500 cm/s. (b) The average velocity is vav-x  t 100 s EVALUATE: The average speed is greater than the magnitude of the average velocity, since the distance traveled is greater than the magnitude of the displacement. IDENTIFY: The acceleration is not constant so the constant acceleration equations cannot be used. Instead, t dv use a x (t )  x and x  x0  Ñvx (t )dt. 0 dt 1 n 1 t SET UP:  t n dt  for n  0. n 1 EXECUTE: (a) The average speed is



EXECUTE: (a) x(t )  x0  Ñ[   t 2 ]dt  x0   t  13  t 3. x  0 at t  0 gives x0  0 and 0

dvx  2 t  (400 m/s3 )t. dt (b) The maximum positive x is when vx  0 and a x  0. vx  0 gives    t 2  0 and x(t )   t  13  t 3  (400 m/s)t  (0667 m/s3 )t 3. ax (t ) 

t

 400 m/s   141 s. At this t, a x is negative. For t  141 s,  200 m/s3

x  (400 m/s)(141 s)  (0667 m/s3 )(141 s)3  377 m.

2.77.

EVALUATE: After t  141 s the object starts to move in the  x direction and goes to x   as t  . IDENTIFY: Apply constant acceleration equations to each vehicle. SET UP: (a) It is very convenient to work in coordinates attached to the truck. Note that these coordinates move at constant velocity relative to the earth. In these coordinates the truck is at rest, and the initial velocity of the car is v0 x  0 Also, the car’s acceleration in these coordinates is the same as in coordinates fixed to the earth. EXECUTE: First, let’s calculate how far the car must travel relative to the truck: The situation is sketched in Figure 2.77.

Figure 2.77 The car goes from x0  2 240 m to x  515 m So x  x0  755 m for the car. Calculate the time it takes the car to travel this distance: a x  0600 m/s 2 , v0 x  0, x  x0  755 m, t  ? x  x0  v0 xt  12 a xt 2

2( x  x0 ) 2(755 m)   1586 s ax 0600 m/s 2 It takes the car 15.9 s to pass the truck. t

Motion Along a Straight Line

2-31

(b) Need how far the car travels relative to the earth, so go now to coordinates fixed to the earth. In these coordinates v0 x  200 m/s for the car. Take the origin to be at the initial position of the car. v0 x  200 m/s, a x  0600 m/s 2 , t  1586 s, x  x0  ?

x  x0  v0 xt  12 axt 2  (200 m/s)(1586 s)  12 (0600 m/s 2 )(1586 s)2 x  x0  3172 m  755 m  393 m

2.78.

EVALUATE: In 15.86 s the truck travels x  x0  (200 m/s)(1586 s)  3172 m The car travels 3927 m  3172 m  75 m farther than the truck, which checks with part (a). In coordinates attached to v v  the truck, for the car v0 x  0, vx  95 m/s and in 15.86 s the car travels x  x0   0 x x  t  75 m, 2   which checks with part (a). IDENTIFY: Use a velocity-time graph to find the acceleration of a stone. Then use that information to find how long it takes the stone to fall through a known distance and how fast you would have to throw it upward to reach a given height and the time to reach that height. SET UP: Take  y to be downward. The acceleration is the slope of the v y versus t graph. EXECUTE: (a) Since v y is downward, it is positive and equal to the speed v The v versus t graph has

300 m/s  15 m/s 2  The formulas y  y0  v0 yt  12 a yt 2 , v y2  v02y  2a y ( y  y0 ), and 20 s v y  v y  a yt apply.

slope a y 

(b) v0 y  0 and let y0  0 y  y0  v0 yt  12 a yt 2 gives t 

2y  ay

2(35 m) 15 m/s 2

 068 s.

v y  v0 y  a yt  (15 m/s 2 )(068 s)  102 m/s.

(c) At the maximum height, v y  0 Let y0  0 v y2  v02y  2a y ( y  y0 ) gives v0 y  2a y ( y  y0 )  2(15 m/s 2 )(180 m)  23 m/s v y  v y  a yt gives

t

2.79.

v y  v0 y ay



0  23 m/s 15 m/s 2

 15 s.

EVALUATE: The acceleration is 980 m/s 2 , downward, throughout the motion. The velocity initially is upward, decreases to zero because of the downward acceleration and then is downward and increasing in magnitude because of the downward acceleration. a (t )     t , with   200 m/s 2 and   300 m/s3 (a) IDENTIFY and SET UP: Integrage a x (t ) to find vx (t ) and then integrate vx (t ) to find x(t ) t

EXECUTE: vx  v0 x  Ñax dt  v0 x  Ñ(  t )dt  v0 x  t  12 t 2 t

x  x0  Ñvx dt  x0  Ñ(v0 x   t  12  t 2 )dt  x0  v0 xt  12  t 2  16  t 3

At t  0, x  x0  To have x  x0 at t1  400 s requires that v0 xt1  12  t12  16  t13  0 Thus v0 x   16  t12  12  t1   16 (300 m/s3 )(400 s) 2  12 (200 m/s 2 )(400 s)  400 m/s (b) With v0x as calculated in part (a) and t  400 s, vx  v0 x  t  12 t 2  400 s  (200 m/s 2 )(400 s)  12 (300 m/s3 )(400 s) 2  120 m/s

2-32

2.80.

Chapter 2 EVALUATE: a x  0 at t  067 s For t  067 s, a x  0 At t  0, the particle is moving in the x-direction and is speeding up. After t  067 s, when the acceleration is positive, the object slows down and then starts to move in the  x-direction with increasing speed. IDENTIFY: Find the distance the professor walks during the time t it takes the egg to fall to the height of his head. SET UP: Let  y be downward. The egg has v0 y  0 and a y  980 m/s 2 . At the height of the professor’s head, the egg has y  y0  442 m. EXECUTE:

y  y0  v0 yt  12 a yt 2 gives t 

2( y  y0 ) 2(442 m)   300 s. The professor walks a ay 980 m/s 2

distance x  x0  v0 xt  (120 m/s)(300 s)  360 m. Release the egg when your professor is 3.60 m from the point directly below you. EVALUATE: Just before the egg lands its speed is (980 m/s 2 )(300s)  294 m/s. It is traveling much 2.81.

faster than the professor. IDENTIFY: Use the constant acceleration equations to establish a relationship between maximum height and acceleration due to gravity and between time in the air and acceleration due to gravity. SET UP: Let  y be upward. At the maximum height, v y  0. When the rock returns to the surface, y  y0  0.

EXECUTE: (a) v 2y  v02y  2a y ( y  y0 ) gives a y H   12 v02y , which is constant, so aE H E  aM H M .

 980 m/s 2  a  HM  HE  E   H   264 H .  371 m/s 2   aM    (b) y  y0  v0 yt  12 a yt 2 with y  y0  0 gives a yt  2v0 y , which is constant, so aETE  aMTM .

2.82.

a  TM  TE  E   264T .  aM  EVALUATE: On Mars, where the acceleration due to gravity is smaller, the rocks reach a greater height and are in the air for a longer time. IDENTIFY: Calculate the time it takes her to run to the table and return. This is the time in the air for the thrown ball. The thrown ball is in free-fall after it is thrown. Assume air resistance can be neglected. SET UP: For the thrown ball, let  y be upward. a y  980 m/s 2 . y  y0  0 when the ball returns to its original position. EXECUTE: (a) It takes her

550 m  220 s to reach the table and an equal time to return. For the ball, 250 m/s

y  y0  0, t  440 s and a y  980 m/s 2 . y  y0  v0 yt  12 a yt 2 gives

v0 y   12 a yt   12 (980 m/s 2 )(440 s)  216 m/s.

(b) Find y  y0 when t  220 s. y  y0  v0 yt  12 a yt 2  (216 m/s)(220 s)  12 (980 m/s 2 )(220 s) 2  238 m

2.83.

EVALUATE: It takes the ball the same amount of time to reach its maximum height as to return from its maximum height, so when she is at the table the ball is at its maximum height. Note that this large maximum height requires that the act either be done outdoors, or in a building with a very high ceiling. (a) IDENTIFY: Use constant acceleration equations, with a y  g , downward, to calculate the speed of the diver when she reaches the water. SET UP: Take the origin of coordinates to be at the platform, and take the  y -direction to be downward. y  y0  213 m, a y  980 m/s 2 , v0 y  0 (since diver just steps off), v y  ?

v 2y  v02y  2a y ( y  y0 )

Motion Along a Straight Line

2-33

EXECUTE: v y   2a y ( y  y0 )   2(980 m/s2 )(21 3 m)   20 4 m/s We know that v y is positive because the diver is traveling downward when she reaches the water. The announcer has exaggerated the speed of the diver. EVALUATE: We could also use y  y0  v0 yt  12 a yt 2 to find t  2085 s The diver gains 9.80 m/s of speed each second, so has v y  (980 m/s 2 )(2085 s)  204 m/s when she reaches the water, which checks. (b) IDENTIFY: Calculate the initial upward velocity needed to give the diver a speed of 25.0 m/s when she reaches the water. Use the same coordinates as in part (a). SET UP: v0 y  ?, v y  250 m/s, a y  980 m/s2 , y  y0  213 m v 2y  v02y  2a y ( y  y0 )

EXECUTE: v0 y  2 v 2y  2a y ( y  y0 )   (250 m/s) 2  2(980 m/s 2 )(213 m)  144 m/s (v0 y is negative since the direction of the initial velocity is upward.)

EVALUATE: One way to decide if this speed is reasonable is to calculate the maximum height above the platform it would produce: v0 y  144 m/s, v y  0 (at maximum height), a y  980 m/s 2 , y  y0  ?

v 2y  v02y  2a y ( y  y0 )

y  y0 

2.84.

v 2y  v02y 2a y



0  (144 s) 2  106 m 2(980 m/s)

This is not physically attainable; a vertical leap of 10.6 m upward is not possible. IDENTIFY: The flowerpot is in free-fall. Apply the constant acceleration equations. Use the motion past the window to find the speed of the flowerpot as it reaches the top of the window. Then consider the motion from the windowsill to the top of the window. SET UP: Let  y be downward. Throughout the motion a y  980 m/s 2 . EXECUTE: Motion past the window: y  y0  190 m, t  0420 s, a y  980 m/s 2 . y  y0  v0 yt  12 a yt 2 y  y0 1 190 m 1  2 a yt   (980 m/s 2 )(0420 s)  2466 m/s. This is the velocity of the t 0420 s 2 flowerpot when it is at the top of the window.

gives v0 y 

Motion from the windowsill to the top of the window: v0 y  0, v y  2466 m/s, a y  980 m/s 2 . v 2y  v02y  2a y ( y  y0 ) gives y  y0 

v 2y  v02y 2a y



(2466 m/s) 2  0 2(980 m/s 2 )

 0310 m. The top of the window is

0.310 m below the windowsill. EVALUATE: It takes the flowerpot t 

v y  v0 y ay



2466 m/s 980 m/s 2

 0252 s to fall from the sill to the top of the

window. Our result says that from the windowsill the pot falls 0310 m  190 m  221 m in

0252 s  0420 s  0672 s. y  y0  v0 yt  12 a yt 2  12 (980 m/s 2 )(0672 s) 2  221 m, which checks. 2.85.

(a) IDENTIFY: Consider the motion from when he applies the acceleration to when the shot leaves his hand. SET UP: Take positive y to be upward. v0 y  0, v y  ?, a y  350 m/s 2 , y  y0  0640 m, v 2y  v02y  2a y ( y  y0 )

EXECUTE: v y  2a y ( y  y0 )  2(350 m/s2 )(0 640 m)  6 69 m/s (b) IDENTIFY: Consider the motion of the shot from the point where he releases it to its maximum height, where v  0. Take y  0 at the ground.

2-34

Chapter 2 y0  220 m, y  ?, a y  2 980 m/s2 (free fall), v0 y  669 m/s (from part (a), v y  0 at

SET UP:

maximum height), v 2y  v02y  2a y ( y  y0 ) EXECUTE:

y  y0 

v 2y  v02y 2a y



0  (669 m/s) 2 2(980 m/s 2 )

 229 m, y  220 m  229 m  449 m

(c) IDENTIFY: Consider the motion of the shot from the point where he releases it to when it returns to the height of his head. Take y  0 at the ground. y0  220 m, y  183 m, a y  2 980 m/s2 , v0 y  669 m/s, t  ? y  y0  v0 yt  12 a yt 2

SET UP:

EXECUTE: 183 m  220 m  (669 m/s)t  12 ( 980 m/s 2 )t 2  (669 m/s)t  (490 m/s 2 )t 2 , 490t 2  669t  037  0, with t in seconds. Use the quadratic formula to solve for t: 1 t 669  (669)2  4(490)(037)  06830  07362. Since t must be positive, 980 t  06830 s  07362 s  142 s. EVALUATE: Calculate the time to the maximum height: v y  v0 y  a yt, so t  (v y  v0 y )/a y 





(669 m/s)/(  980 m/s 2 )  068 s It also takes 0.68 s to return to 2.2 m above the ground, for a total time

2.86.

of 1.36 s. His head is a little lower than 2.20 m, so it is reasonable for the shot to reach the level of his head a little later than 1.36 s after being thrown; the answer of 1.42 s in part (c) makes sense. IDENTIFY: The motion of the rocket can be broken into 3 stages, each of which has constant acceleration, so in each stage we can use the standard kinematics formulas for constant acceleration. But the acceleration is not the same throughout all 3 stages.  v0 y  v y  1 2 SET UP: The formulas y  y0    t , y  y0  v0 yt  a yt , and v y  v0 y  a yt apply. 2 2   EXECUTE: (a) Let +y be upward. At t  250 s, y  y0  1094 m and v y  875 m/s. During the next 10.0 s the

 v0 y  v y   875 m/s  1325 m/s  rocket travels upward an additional distance y  y0   t    (100 s)  1100 m. 2 2     The height above the launch pad when the second stage quits therefore is 1094 m  1100 m  2194 m. For the free-fall motion after the second stage quits: y  y0 

v 2y  v02y 2a y



0  (1325 m/s)2 2(98 m/s 2 )

 896 m.

The maximum height above the launch pad that the rocket reaches is 2194 m  896 m  3090 m. 1 (b) y  y0  v0 yt  a yt 2 gives 2194 m  (1325 m/s)t  (49 m/s 2 )t 2 . From the quadratic formula the 2 positive root is t  386 s. (c) v y  v0 y  a yt  1325 m/s  (98 m/s 2 )(386 s)  246 m/s. The rocket’s speed will be 246 m/s just

2.87.

before it hits the ground. EVALUATE: We cannot solve this problem in a single step because the acceleration, while constant in each stage, is not constant over the entire motion. The standard kinematics equations apply to each stage but not to the motion as a whole. IDENTIFY and SET UP: Let  y be upward. Each ball moves with constant acceleration a y  980 m/s 2  In parts (c) and (d) require that the two balls be at the same height at the same time. EXECUTE: (a) At ceiling, v y  0, y  y0  30 m, a y  980 m/s 2  Solve for v0 y  v 2y  v02y  2a y ( y  y0 ) gives v0 y  77 m/s

(b) v y  v0 y  a yt with the information from part (a) gives t  078 s (c) Let the first ball travel downward a distance d in time t. It starts from its maximum height, so v0 y  0 y  y0  v0 y t  12 a yt 2 gives d  (49 m/s 2 )t 2 © Copyright 2012 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

2-35

The second ball has v0 y  23 (77 m/s)  51 m/s In time t it must travel upward 30 m  d to be at the same place as the first ball. y  y0  v0 yt  12 a yt 2 gives 30 m  d  (51 m/s)t  (49 m/s 2 )t 2 

We have two equations in two unknowns, d and t. Solving gives t  059 s and d  17 m (d) 30 m  d  13 m EVALUATE: In 0.59 s the first ball falls d  (49 m/s 2 )(059 s) 2  17 m, so is at the same height as the second ball. 2.88.

IDENTIFY: The teacher is in free-fall and falls with constant acceleration 980 m/s 2 , downward. The sound from her shout travels at constant speed. The sound travels from the top of the cliff, reflects from the ground and then travels upward to her present location. If the height of the cliff is h and she falls a distance y in 3.0 s, the sound must travel a distance h  ( h  y ) in 3.0 s. SET UP: Let  y be downward, so for the teacher a y  980 m/s 2 and v0 y  0. Let y  0 at the top of the cliff. EXECUTE: (a) For the teacher, y  12 (980 m/s 2 )(30 s) 2  441 m. For the sound, h  (h  y )  vst. h  12 (vst  y ) 

1 2

340 m/s30 s  441 m   532 m,

which rounds to 530 m.

(b) v 2y  v02y  2a y ( y  y0 ) gives v y  2a y ( y  y0 )  2(980 m/s 2 )(532 m)  102 m/s. EVALUATE: She is in the air for t  2.89.

v y  v0 y ay



102 m/s 980 m/s2

 104 s and strikes the ground at high speed.

IDENTIFY: The helicopter has two segments of motion with constant acceleration: upward acceleration for 10.0 s and then free-fall until it returns to the ground. Powers has three segments of motion with constant acceleration: upward acceleration for 10.0 s, free-fall for 7.0 s and then downward acceleration of 20 m/s 2 . SET UP: Let  y be upward. Let y  0 at the ground. EXECUTE: (a) When the engine shuts off both objects have upward velocity v y  v0 y  a yt  (50 m/s 2 )(100 s)  500 m/s and are at y  v0 yt  12 a yt 2  12 (50 m/s 2 )(100 s) 2  250 m.

For the helicopter, v y  0 (at the maximum height), v0 y  500 m/s, y0  250 m, and a y  980 m/s 2 . v 2y  v02y  2a y ( y  y0 ) gives y 

v 2y  v02y 2a y

 y0 

0  (500 m/s) 2 2(980 m/s 2 )

 250 m  378 m, which rounds to 380 m.

(b) The time for the helicopter to crash from the height of 250 m where the engines shut off can be found using v0 y  500 m/s, a y  980 m/s 2 , and y  y0  250 m. y  y0  v0 yt  12 a yt 2 gives 250 m  (500 m/s)t  (490 m/s 2 )t 2 . (490 m/s 2 )t 2  (500 m/s)t  250 m  0. The quadratic formula gives

t





1 500  (500) 2  4(490)(250) s. Only the positive solution is physical, so t  139 s. Powers 980

therefore has free-fall for 7.0 s and then downward acceleration of 20 m/s 2 for 139 s  70 s  69 s. After 7.0 s of free-fall he is at y  y0  v0 yt  12 a yt 2  250 m  (500 m/s)(70 s)  12 ( 980 m/s 2 )(70 s) 2  360 m and has velocity vx  v0 x  axt  500 m/s  (980 m/s 2 )(70 s)  186 m/s. After the next 6.9 s he is at y  y0  v0 yt  12 a yt 2  360 m  (186 m/s)(69 s)  12 ( 200 m/s 2 )(69 s) 2  184 m. Powers is 184 m

above the ground when the helicopter crashes. EVALUATE: When Powers steps out of the helicopter he retains the initial velocity he had in the helicopter but his acceleration changes abruptly from 50 m/s 2 upward to 980 m/s 2 downward. Without the jet pack he would have crashed into the ground at the same time as the helicopter. The jet pack slows his descent so he is above the ground when the helicopter crashes.

2-36 2.90.

Chapter 2 IDENTIFY: Apply constant acceleration equations to the motion of the rock. Sound travels at constant speed. SET UP: Let tfall be the time for the rock to fall to the ground and let ts be the time it takes the sound to travel from the impact point back to you. tfall  ts  100 s. Both the rock and sound travel a distance d that is equal to the height of the cliff. Take  y downward for the motion of the rock. The rock has v0 y  0 and a y  980 m/s 2 .

EXECUTE: (a) For the rock, y  y0  v0 yt  12 a yt 2 gives tfall  For the sound, ts 

2d 980 m/s 2

d . Let  2  d . 000303 2  04518  100  0.   196 and d  384 m. 330 m/s

(b) You would have calculated d  12 (980 m/s 2 )(100 s) 2  490 m. You would have overestimated the

2.91.

height of the cliff. It actually takes the rock less time than 10.0 s to fall to the ground. EVALUATE: Once we know d we can calculate that tfall  88 s and ts  12 s. The time for the sound of impact to travel back to you is 12% of the total time and cannot be neglected. The rock has speed 86 m/s just before it strikes the ground. (a) IDENTIFY: Let  y be upward. The can has constant acceleration a y   g  The initial upward velocity of the can equals the upward velocity of the scaffolding; first find this speed. SET UP: y  y0  150 m, t  325 s, a y  2 980 m/s2 , v0 y  ? EXECUTE:

y  y0  v0 yt  12 a yt 2 gives v0 y  1131 m/s

Use this v0 y in v y  v0 y  a yt to solve for v y : v y  205 m/s (b) IDENTIFY: Find the maximum height of the can, above the point where it falls from the scaffolding: SET UP: v y  0, v0 y  1131 m/s, a y  980 m/s2 , y  y0  ? EXECUTE: v 2y  v02y  2a y ( y  y0 ) gives y  y0  653 m

2.92.

The can will pass the location of the other painter. Yes, he gets a chance. EVALUATE: Relative to the ground the can is initially traveling upward, so it moves upward before stopping momentarily and starting to fall back down. IDENTIFY: Both objects are in free-fall. Apply the constant acceleration equations to the motion of each person. SET UP: Let  y be downward, so a y  980 m/s 2 for each object. EXECUTE: (a) Find the time it takes the student to reach the ground: y  y0  180 m, v0 y  0, a y  980 m/s 2 . y  y0  v0 yt  12 a yt 2 gives t 

2( y  y0 ) 2(180 m)   606 s. Superman must reach ay 980 m/s2

the ground in 606 s  500 s  106 s: t  106 s, y  y0  180 m, a y  980 m/s 2 . y  y0  v0 yt  12 a yt 2

y  y0 1 180 m 1  2 a yt   (980 m/s 2 )(106 s)  165 m/s. Superman must have initial speed t 106 s 2 v0  165 m/s.

gives v0 y 

(b) The graphs of y-t for Superman and for the student are sketched in Figure 2.92. (c) The minimum height of the building is the height for which the student reaches the ground in 5.00 s, before Superman jumps. y  y0  v0 yt  12 a yt 2  12 (980 m/s 2 )(500 s) 2  122 m. The skyscraper must be at least 122 m high. EVALUATE: 165 m/s  369 mi/h, so only Superman could jump downward with this initial speed.

Motion Along a Straight Line

2-37

Figure 2.92 2.93.

IDENTIFY: Apply constant acceleration equations to the motion of the rocket and to the motion of the canister after it is released. Find the time it takes the canister to reach the ground after it is released and find the height of the rocket after this time has elapsed. The canister travels up to its maximum height and then returns to the ground. SET UP: Let  y be upward. At the instant that the canister is released, it has the same velocity as the rocket. After it is released, the canister has a y  980 m/s 2 . At its maximum height the canister has v y  0.

EXECUTE: (a) Find the speed of the rocket when the canister is released: v0 y  0, a y  330 m/s 2 , y  y0  235 m. v 2y  v02y  2a y ( y  y0 ) gives v y  2a y ( y  y0 )  2(330 m/s 2 )(235 m)  394 m/s.

For the motion of the canister after it is released, v0 y  394 m/s, a y  980 m/s 2 , y  y0  235 m. y  y0  v0 yt  12 a yt 2 gives 235 m  (394 m/s)t  (490 m/s 2 )t 2 . The quadratic formula gives t  120 s

as the positive solution. Then for the motion of the rocket during this 12.0 s, y  y0  v0 yt  12 a yt 2  235 m  (394 m/s)(120 s)  12 (330 m/s 2 )(120 s) 2  945 m.

(b) Find the maximum height of the canister above its release point: v0 y  394 m/s, v y  0, a y  980 m/s 2 . v 2y  v02y  2a y ( y  y0 ) gives y  y0 

v 2y  v02y 2a y



0  (394 m/s) 2 2(980 m/s 2 )

 792 m. After its

release the canister travels upward 79.2 m to its maximum height and then back down 792 m  235 m to the ground. The total distance it travels is 393 m. EVALUATE: The speed of the rocket at the instant that the canister returns to the launch pad is v y  v0 y  a yt  394 m/s  (330 m/s 2 )(120 s)  790 m/s. We can calculate its height at this instant by

v 2y  v02y  2a y ( y  y0 ) with v0 y  0 and v y  790 m/s. y  y0 

v 2y  v02y 2a y



(790 m/s) 2 2(330 m/s 2 )

 946 m, which

agrees with our previous calculation. 2.94.

IDENTIFY: Both objects are in free-fall and move with constant acceleration 980 m/s 2 , downward. The two balls collide when they are at the same height at the same time. SET UP: Let  y be upward, so a y  2 980 m/s 2 for each ball. Let y  0 at the ground. Let ball A be the one thrown straight up and ball B be the one dropped from rest at height H. y0 A  0, y0 B  H . EXECUTE: (a) y  y0  v0 yt  12 a yt 2 applied to each ball gives y A  v0t  12 gt 2 and yB  H  12 gt 2 . y A  y B gives v0t  12 gt 2  H  12 gt 2 and t 

H . v0

2-38

Chapter 2 (b) For ball A at its highest point, v yA  0 and v y  v0 y  a yt gives t  part (a) gives

v0 . Setting this equal to the time in g

H v0 v2 and H  0 .  v0 g g

EVALUATE: In part (a), using t  H must be less than

 H gH  in the expressions for y A and y B gives y A  yB  H 1  2  .  2v  v0 0  

2v02 in order for the balls to collide before ball A returns to the ground. This is g

because it takes ball A time t 

2v0 to return to the ground and ball B falls a distance g

1 2 gt 2



2v02 during g

2v02 the two balls collide just as ball A reaches the ground and for H greater than this g ball A reaches the ground before they collide. IDENTIFY and SET UP: Use v x  dx/dt and a x  dv x /dt to calculate vx (t ) and a x (t ) for each car. Use this time. When H 

2.95.

these equations to answer the questions about the motion. dx dv EXECUTE: x A   t   t 2 , v Ax  A    2 t , a Ax  Ax  2 dt dt dxB dvBx 2 3 2 xB   t   t , vBx   2 t  3 t , aBx   2  6 t dt dt (a) IDENTIFY and SET UP: The car that initially moves ahead is the one that has the larger v0x  EXECUTE: At t  0, v Ax   and vBx  0 So initially car A moves ahead. (b) IDENTIFY and SET UP: Cars at the same point implies x A  xB 

t   t 2   t 2   t3 EXECUTE: One solution is t  0, which says that they start from the same point. To find the other solutions, divide by t:    t   t   t 2

 t 2  (    )t    0





1 1 (   )  (   )2  4  160  (160)2  4(020)(260)  400 s  173 s 2 040 So x A  xB for t  0, t  227 s and t  573 s t

EVALUATE: Car A has constant, positive ax  Its v x is positive and increasing. Car B has v0 x  0 and a x that is initially positive but then becomes negative. Car B initially moves in the  x-direction but then slows down and finally reverses direction. At t  227 s car B has overtaken car A and then passes it. At t  573 s, car B is moving in the  x-direction as it passes car A again. d ( xB  x A )  If (c) IDENTIFY: The distance from A to B is xB  x A  The rate of change of this distance is dt d ( xB  x A )  0 But this says vBx  v Ax  0 (The distance between A and B this distance is not changing, dt is neither decreasing nor increasing at the instant when they have the same velocity.) SET UP: v Ax  vBx requires   2 t  2 t  3 t 2 EXECUTE: 3 t 2  2(    )t    0







1 1 2(   )  4(   )2  12  320  4(160) 2  12(020)(260) 6 120 t  2667 s  1667 s, so v Ax  vBx for t  100 s and t  433 s t



Motion Along a Straight Line

2-39

EVALUATE: At t  100 s, v Ax  vBx  500 m/s At t  433 s, v Ax  vBx  130 m/s Now car B is slowing down while A continues to speed up, so their velocities aren’t ever equal again. (d) IDENTIFY and SET UP: a Ax  aBx requires 2   2  6 t EXECUTE: t 

   280 m/s 2  120 m/s 2   267 s 3 3(020 m/s3 )

EVALUATE: At t  0, aBx  a Ax , but aBx is decreasing while a Ax is constant. They are equal at

t  267 s but for all times after that aBx  a Ax  2.96.

IDENTIFY: Apply y  y0  v0 yt  12 a yt 2 to the motion from the maximum height, where v0 y  0. The time spent above ymax /2 on the way down equals the time spent above ymax /2 on the way up. SET UP: Let  y be downward. a y  g . y  y0  ymax /2 when he is a distance ymax /2 above the floor. EXECUTE: The time from the maximum height to ymax /2 above the floor is given by ymax /2  12 gt12 . The 2 time from the maximum height to the floor is given by ymax  12 gttot and the time from a height of

ymax /2 to the floor is t2  ttot  t1.

2t1  t2

ymax /2 ymax  ymax /2



1  4.8. 2 1

EVALUATE: The person spends over twice as long above ymax /2 as below ymax /2. His average speed is less above ymax /2 than it is when he is below this height. 2.97.

IDENTIFY: Apply constant acceleration equations to the motion of the two objects, the student and the bus. SET UP: For convenience, let the student’s (constant) speed be v0 and the bus’s initial position be x0  Note that these quantities are for separate objects, the student and the bus. The initial position of the student is taken to be zero, and the initial velocity of the bus is taken to be zero. The positions of the student x1 and the bus x2 as functions of time are then x1  v0t and x2  x0  (1/2) at 2  EXECUTE: (a) Setting x1  x2 and solving for the times t gives t  t

(50m/s)  (0170m/s ) 1







1 v0  v02  2ax0 . a

(50m/s)2  2(0170m/s 2 )(400 m)  955 s and 493 s.

The student will be likely to hop on the bus the first time she passes it (see part (d) for a discussion of the later time). During this time, the student has run a distance v0t  (5m/s)(955 s)  478 m (b) The speed of the bus is (0170 m/s 2 )(955 s)  162 m/s. (c) The results can be verified by noting that the x lines for the student and the bus intersect at two points, as shown in Figure 2.97a. (d) At the later time, the student has passed the bus, maintaining her constant speed, but the accelerating bus then catches up to her. At this later time the bus’s velocity is (0170m/s 2 )(493 s)  838m/s (e) No; v02  2ax0 , and the roots of the quadratic are imaginary. When the student runs at 35m/s, Figure 2.97b shows that the two lines do not intersect: (f) For the student to catch the bus, v02  2ax0  And so the minimum speed is 2(0170m/s 2 )(40m/s)  3688m/s She would be running for a time

369m/s 0170 m/s 2

 217 s, and covers a

distance (3688m/s)(217 s)  800 m However, when the student runs at 3688m/s, the lines intersect at one point, at x  80 m, as shown in Figure 2.97c.

2-40

Chapter 2 EVALUATE: The graph in part (c) shows that the student is traveling faster than the bus the first time they meet but at the second time they meet the bus is traveling faster. t2  t tot  t1

Figure 2.97 2.98.

IDENTIFY: Apply constant acceleration equations to the motion of the boulder. SET UP: Let  y be downward, so a y   g . EXECUTE: (a) Let the height be h and denote the 1.30-s interval as t ; the simultaneous equations h  12 gt 2 , 32 h  12 g (t  t ) 2 can be solved for t. Eliminating h and taking the square root,

t 3  , and t  t 2

t , and substitution into h  12 gt 2 gives h  246 m 1  2/3 (b) The above method assumed that t  0 when the square root was taken. The negative root (with t  0) gives an answer of 2.51 m, clearly not a “cliff.” This would correspond to an object that was initially near the bottom of this “cliff ” being thrown upward and taking 1.30 s to rise to the top and fall to the bottom. Although physically possible, the conditions of the problem preclude this answer. t 

EVALUATE: For the first two-thirds of the distance, y  y0  164 m, v0 y  0, and a y  980 m/s 2 . v y  2a y ( y  y0 )  567 m/s. Then for the last third of the distance, y  y0  820 m, v0 y  567 m/s and

a y  980 m/s 2 . y  y0  v0 yt  12 a yt 2 gives (490 m/s 2 )t 2  (567 m/s)t  820 m  0.

2.99.





1 567  (567)2  4(49)(820) s  130 s, as required. 98 IDENTIFY: Apply constant acceleration equations to both objects. SET UP: Let  y be upward, so each ball has a y   g . For the purpose of doing all four parts with the t

least repetition of algebra, quantities will be denoted symbolically. That is, let y1  h  v0t 

1 2 gt , 2

1 y2  h  g (t  t0 )2 . In this case, t0  100 s. 2 EXECUTE: (a) Setting y1  y2  0, expanding the binomial (t  t0 ) 2 and eliminating the common term 1 2

gt 2 yields v0t  gt0t  12 gt02 . Solving for t: t 

1 gt 2 2 0

gt0  v0



 t0  1  . 2  1  v0 /(gt0 ) 

Substitution of this into the expression for y1 and setting y1  0 and solving for h as a function of v0 yields, after some





1 gt  v 2 0 2 2 0 1 algebra, h  2 gt0  ( gt0  v0 )2 2  49m/s  v0 

Using the given value t0  100 s and g  980m/s 2 ,

h  200 m  (49 m)     98m/s  v0 

Motion Along a Straight Line

2-41

This has two solutions, one of which is unphysical (the first ball is still going up when the second is released; see part (c)). The physical solution involves taking the negative square root before solving for v0 , and yields 82m/s The graph of y versus t for each ball is given in Figure 2.99. (b) The above expression gives for (i), 0.411 m and for (ii) 1.15 km. (c) As v0 approaches 98m/s, the height h becomes infinite, corresponding to a relative velocity at the time the second ball is thrown that approaches zero. If v0  98m/s, the first ball can never catch the second ball. (d) As v0 approaches 4.9 m/s, the height approaches zero. This corresponds to the first ball being closer and closer (on its way down) to the top of the roof when the second ball is released. If v0  49m/s, the first ball will already have passed the roof on the way down before the second ball is released, and the second ball can never catch up. EVALUATE: Note that the values of v0 in parts (a) and (b) are all greater than vmin and less than vmax .

Figure 2.99

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