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: 35

Answer

(a) $t = 0$ or $t = 3~s$ (b) At t = 0: $x = 12~m$ $a = -18~m/s^2$ At t = 3 s: $x = -15~m$ $a = 18~m/s^2$

Work Step by Step

(a) $x(t) = 2t^3-9t^2+12$ $v(t) = \frac{dx}{dt} = 6t^2-18t$ We can find $t$ when $v = 0$ $6t^2-18t = 0$ $(6t - 18)~t = 0$ $t = 0$ or $t = 3~s$ (b) $x(t) = 2t^3-9t^2+12$ $a(t) = \frac{dv}{dt} = 12t-18$ At t = 0: $x = 12~m$ $a = -18~m/s^2$ At t = 3 s: $x = 2(3)^3-9(3)^2+12$ $x = -15~m$ $a = (12)(3)-18$ $a = 18~m/s^2$
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