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Describing Chapter Motion: Kinematics in One Dimension
73. For the car to pass the train, the car must travel the length of the train AND the distance the train 95 km h t or travels. The distance the car travels can thus be written as either d car vcar t
d car Ltrain vtrain t 1.10 km 75 km h t . To solve for the time, equate these two expressions for the distance the car travels. 1 .1 0 k m 9 5 k m h t 1 .1 0 k m 7 5 k m h t t 0 .0 5 5 h 3 .3 m in 20 km h
The distance the car travels during this time is d
95 km h 0 .0 5 5 h 5 .2 2 5 k m 5 .2 k m .
If the train is traveling the opposite direction from the car, then the car must travel the length of the train MINUS the distance the train travels. Thus the distance the car travels can be written as either d car 95 km h t or d car 1.10 km 75 km h t . To solve for the time, equate these two expressions for the distance the car travels. 1 .1 0 k m 9 5 k m h t 1 .1 0 k m 7 5 k m h t t 6 .4 7 1 0 3 h 2 3 .3 s 170 km h The distance the car travels during this time is d 74. For the baseball, v0 motion ) is v
v
2
9 5 km h
6 .4 7 1 0
3
h
0 .6 1 k m .
0, x
x0
3.5 m , and the final speed of the baseball (during the throwing
v2 2x v02 x0
2
44 m s . The acceleration is found from Eq. 2-11c.
2a x x0 a 44 m s 0 2 3 .5 m 280 m s 2
v
2 0
75. (a) Choose upward to be the positive direction, and y0
2
0 the at ground. The rocket has v0
0,
a 3.2 m s , and y 1200 m when it runs out of fuel. Find the velocity of the rocket when it runs out of fuel from Eq 2-11c, with x replaced by y. 2 v12200 m v0 2 a y y0
v1200 m
2 v0
2a y
y0
0 2 3 .2 m s 2
1200 m
8 7 .6 4 m s
88 m s
The positive root is chosen since the rocket is moving upwards when it runs out of fuel. (b) The time to reach the 1200 m location can be found from equation (2-11a). v1200 m v0 87.64 m s 0 v1200 m v0 at1200 m t1200 m 2 7 .3 9 s 2 7 s a 3 .2 m s 2 (c) For this part of the problem, the rocket will have an initial velocity v0
2
87.64 m s , an
acceleration of a 9.8 m s , and a final velocity of v 0 at its maximum altitude. The altitude reached from the out-of-fuel point can be found from equation (2-11c). v 2 v12200 m 2a y 1200 m
ymax
12 0 0 m
0 v12200 m 2a
1 2 00 m
8 7 .6 4 m s 2 9. 8 m s 2
2
1 2 00 m 3 90 m
15 9 0 m
(d) The time for the "coasting" portion of the flight can be found from Eq. 2-11a. v v0 0 8 7 .6 4 m s v v1200 m atcoast t c oa st 8 .9 4 s a 9 .8 m s 2
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