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(b) The red graph of Figure 2.14c best illustrates the speed (distance traveled per time
interval) of the puck as a function of time. It shows the puck gaining speed for
approximately three time intervals, moving at constant speed for about four time
intervals, then slowing to rest during the last two intervals.
(c) The green graph of Figure 2.14d best shows the puck’s acceleration as a function of
time. The puck gains velocity (positive acceleration) for approximately three time
intervals, moves at constant velocity (zero acceleration) for about four time intervals,
and then loses velocity (negative acceleration) for roughly the last two time intervals.
19 CHAPTER 6. 2 (c). The acceleration of the ball remains constant while it is in the air. The magnitude of its
acceleration is the free-fall acceleration, g = 9.80 m/s2. 7. (c). As it travels upward, its speed decreases by 9.80 m/s during each second of its motion.
When it reaches the peak of its motion, its speed becomes zero. As the ball moves
downward, its speed increases by 9.80 m/s each second. 8. (a) and (f). The first jumper will always be moving with a higher velocity than the second.
Thus, in a given time interval, the first jumper covers more distance than the second.
Thus, the separation distance between them increases. At any given instant of time, the
velocities of the jumpers are definitely different, because one had a head start. In a time
interval after this instant, however, each jumper increases his or her velocity by the same
amount, because they have the same acceleration. Thus, the difference in velocities stays
the same. 20 CHAPTER 2 Problem Solutions
2.1 Distances traveled are
∆x 1 = v 1 ( ∆t1 ) = ( 80.0 k m h ) ( 0.500 h ) = 40.0 k m
∆x 2 = v 2 ( ∆t 2 ) = (100 k m h ) ( 0.200 h ) = 20.0 km ∆x 3 = v 3 ( ∆t 3 ) = ( 40.0 k m h ) ( 0.750 h ) = 30.0 k m Thus, the total distance traveled is ∆x = ( 40.0 + 20.0 + 30.0 ) k m = 90.0 k m , and the elapsed
time is ∆t = 0.500 h + 0....
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