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: 36

Answer

(a) The particle reaches minimum velocity at t = 1 s The minimum velocity is -6 m/s (b) The acceleration is zero at t = 1 s

Work Step by Step

(a) $x(t) = 2t^3-6t^2+12$ $v(t) = \frac{dx}{dt} = 6t^2-12t$ $v(t) = 6(t-1)^2-6$ The velocity versus time graph is a parabola with a vertex at (1,-6). This means that the particle reaches minimum velocity at t = 1 s and that the minimum velocity is -6 m/s. (b) $a(t) = \frac{dv}{dt} = 12t-12$ $12t-12 = 0$ $t = 1~s$ The acceleration is zero at t = 1 s.
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