Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 39

Answer

(a) $t = 2$ or $t=5$ (b) At t = 2 s: $a = -3~m/s^2$ At t = 5 s: $a = 3~m/s^2$

Work Step by Step

(a) $v(t) = t^2-7t+10$ We can find the turning points when $v(t) = 0$; $t^2-7t+10 = 0$ $(t-5)(t-2) = 0$ $t = 2$ or $t=5$ (b) $a(t) = \frac{dv}{dt} = 2t-7$ At t = 2 s: $a = (2)(2)-7$ $a = -3~m/s^2$ At t = 5 s: $a = (2)(5)-7$ $a = 3~m/s^2$
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