University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 63: 2.68

Answer

(a) $x(t) = 4.00t - \frac{2.00}{3}t^3$ $a_x(t) = -4.00t$ (b) The maximum positive displacement is 3.77 meters.

Work Step by Step

(a) $v_x(t) = 4.00 - 2.00t^2$ $a_x(t) = \frac{dv}{dt} = -4.00t$ $x(t) = \int v_x(t) = \int 4.00 - 2.00t^2$ $x(t) = 4.00t - \frac{2.00}{3}t^3 + c$ Since $x(0) = 0$, then $c=0$. $x(t) = 4.00t - \frac{2.00}{3}t^3$ (b) The maximum positive displacement occurs when $v_x(t) = 0$ as this is the point where the velocity changes from positive to negative. $v_x(t) = 0 = 4.00 - 2.00t^2$ $t^2 = 2.00$ $t = \sqrt{2.00}$ We can use this value of $t$ in $x(t)$ to find the maximum positive displacement. $x(t) = 4.00t - \frac{2.00}{3}t^3$ $x_{max} = (4.00)(\sqrt{2.00}) - \frac{2.00}{3}(\sqrt{2.00})^3$ $x_{max} = 3.77 ~m$ The maximum positive displacement is 3.77 meters.
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