Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - Problems: 36

Answer

(a) The time required to catch the box car is 26 s. (b) The distance traveled to reach the box car is 143 m.

Work Step by Step

$t_1 = \frac{v-v_0}{a} = \frac{6.0 ~m/s}{1.4 ~m/s^2} = 4.3 ~s$ It takes 4.3 s to accelerate up to a speed of 6.0 m/s. $x = v\cdot t_1 = (5.0 ~m/s)(4.3 ~s) = 22 ~m$ In 4.3 s, the box car moves a distance of 22 m. The fugitive is moving at 6.0 m/s and the box car is moving at a speed of 5.0 m/s. Therefore, the fugitive will catch the box car at a rate of 1.0 m/s. $t_2 = \frac{x}{v} = \frac{22 ~m}{1.0 ~m/s} = 22 ~s$ (a) The total time to catch the box car is $t_1 + t_2$. $t_1+ t_2 = 4.3 ~s + 22 ~s = 26 ~s$ The time required to catch the box car is 26 s. (b) Let $x_1$ be the distance traveled during the acceleration period. $x_1 = \frac{1}{2}at^2 = \frac{1}{2}(1.4 ~m/s^2)(4.3 ~s)^2 = 13 ~m$ Let $x_2$ be the distance traveled while moving at a constant speed of 6.0 m/s. $x_2 = vt = (6.0 ~m/s)(22 ~s) = 130 ~m$ The total distance is $x_1 + x_2$. $x_1 + x_2 = 13 ~m + 130 ~m = 143 ~m$ The distance traveled to reach the box car is 143 m.
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