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: 41

Answer

(a) The velocity is zero again at $t = 20~s$ (b) $x = 667~m$

Work Step by Step

(a) $v(t) = v_0+\int_{0}^{t}~a(t)~dt$ $v(t) = 0+\int_{0}^{t}~(10-t)~dt$ $v(t) = 10t-\frac{1}{2}t^2$ We can then find $t$ when $v(t) = 0$; $10t-\frac{1}{2}t^2 = 0$ $(10-\frac{1}{2}t)~t = 0$ $t = 0$ or $t = 20$ Therefore, velocity is zero again at $t = 20~s$. (b) $x(t) = x_0+\int_{0}^{t}~v(t)~dt$ $x(t) = 0+\int_{0}^{t}~10t-\frac{1}{2}t^2~dt$ $x(t) = 5t^2-\frac{1}{6}t^3$ We can then find $x(t)$ at $t = 20 ~s$; $x = 5t^2-\frac{1}{6}t^3$ $x = (5)(20)^2-\frac{1}{6}(20)^3$ $x = 667~m$
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