Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 48

Answer

The gazelle is able to escape from the cheetah with 7.5 meters to spare.

Work Step by Step

We can find the gazelle's constant speed. $v = a~t = (4.6~m/s^2)(5.0~s)$ $v = 23~m/s$ Note that the cheetah gains on the gazelle the whole time they are both running. We can find the distance the cheetah can run before it must stop. $d = v~t = (30~m/s)(15~s) = 450~m$ To escape, the gazelle has 15 seconds to travel $450~m-170~m$ which is 280 meters. We can find the distance $x_1$ the gazelle travels during the 5.0 second acceleration period. $x_1 = \frac{1}{2}at^2$ $x_1 = \frac{1}{2}(4.6~m/s^2)(5.0~s)^2$ $x_1 = 57.5~m$ We can find the distance $x_2$ the gazelle could run in the next 10 seconds. $x_2 = v~t = (23~m/s)(10~s)$ $x_2 = 230~m$ The total distance the gazelle can travel in 15 seconds is 230 m + 57.5 m which is 287.5 meters. Since the gazelle only needed to run 280 meters to escape, the gazelle is able to escape from the cheetah with 7.5 meters to spare.
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