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: 34

Answer

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

Work Step by Step

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