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: 42

Answer

(a) After t = 0, the particle first reaches a turning point at t = 1 s. (b) $a = -6.28~m/s^2$

Work Step by Step

(a) $v(t) = (2.0~m/s)~sin(\pi~t)$ We can find turning points when $v(t) = 0$. Note that $v(t) = 0$ when $t = 0, 1, 2, 3, 4,...$ After t = 0, the particle first reaches a turning point at t = 1 s. (b) $a(t) = \frac{dv}{dt}$ $a(t) = (2.0~m/s)(\pi)~cos(\pi~t)$ We can then find the acceleration at t = 1 s; $a = (2.0~m/s)(\pi)~cos(\pi~t)$ $a = (2.0~m/s)(\pi)~cos(\pi)$ $a = (2.0~m/s)(\pi)(-1)$ $a = -6.28~m/s^2$
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