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: 34

Answer

The space vehicle moved a distance of 460 m between t = 2.0 s and t = 6.0 s.

Work Step by Step

$a = \frac{\Delta v}{\Delta t} = \frac{162 ~m/s -85 ~m/s}{10.0 ~s} = 7.7 ~m/s^2$ Let $x_1$ be the position at $t = 2.0 ~s$. Then, $x_1 = v_0t + \frac{1}{2}at^2$ $x_1 = (85 ~m/s)(2.0 ~s) + \frac{1}{2}(7.7 ~m/s^2)(2.0 ~s)^2 = 190 ~m$ Let $x_2$ be the position at $t = 6.0 ~s$. $x_2 = v_0t + \frac{1}{2}at^2$ $x_2 = (85 ~m/s)(6.0 ~s) + \frac{1}{2}(7.7 ~m/s^2)(6.0 ~s)^2 = 650 ~m$ $\Delta x = x_2 - x_1 = 650 ~m - 190 ~m = 460 ~m$ The space vehicle moved a distance of 460 m between t = 2.0 s and t = 6.0 s.
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