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: 40

Answer

$k = 2.0~m/s^3$

Work Step by Step

$x(t) = x_0+\int_{0}^{t}~v(t)~dt$ $x(t) = x_0+\int_{0}^{t}~kt^2~dt$ $x(t) = \frac{k}{3}t^3-9.0$ At t = 3.0 s; $\frac{k}{3}t^3-9.0~m = 9.0~m$ $\frac{k}{3}(3.0~s)^3-9.0~m = 9.0~m$ $(9.0~s^3)~k = 18.0~m$ $k = 2.0~m/s^3$
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