Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 2 - Kinematics in One Dimension - Exercises and Problems: 24

Answer

(a) $x = \frac{5}{3}~m$ (b) v = 2 m/s (c) $a = 4~m/s^2$

Work Step by Step

(a) $v(t) = 2t^2$ $x(t) = x_0+\int_{0}^{t}~v_x(t)~dt$ $x(t) = x_0+\int_{0}^{t}~2t^2~dt$ $x(t) = \frac{2}{3}t^3+1~m$ At t = 1 s: $x = \frac{2}{3}(1~s)^3+1~m$ $x = \frac{5}{3}~m$ (b) $v(t) = 2t^2$ At t = 1 s: $v = (2)(1~s)^2$ $v = 2~m/s$ (c) $a(t) = \frac{dv}{dt}$ $a(t) = 4t$ At t = 1 s: $a = (4)(1~s)$ $a = 4~m/s^2$
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