College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 2 - Kinematics: Description of Motion - Learning Path Questions and Exercises - Exercises - Page 62: 46


(a) The object will travel in the +x-direction and then reverse its direction (b) $t=23\, \mathrm{s}$ (c) $v=-41\,\mathrm{m/s}$

Work Step by Step

(a) The object will travel in the +x-direction and then reverse its direction. When it passes the origin, it has momentum in the +x-direction. Inertia causes the object to continue travelling along the x-direction a bit longer. The negative acceleration causes the object to slow down, momentarily stop and reverse its direction. (b) To calculate the time required for the object to return to the origin, we use $x=0$, $x_0=0$, $v_0=40\,\mathrm{m/s}$ and $a=-3.5\mathrm{m/s^2}$ in the kinematic equation: \begin{align*} x=&x_0+v_0t+{1\over 2}at^2\\ 0=&0+40t+{1\over 2}(-3.5)t^2\\ t(40-{3.5\over 2}t)=&0\\ t=0 \,\,\mathrm{or}\,\,& t={40\times 2\over 3.5}=23\,\mathrm{s} \end{align*} From the moment the object first passes the origin, it takes about 23 s for it to reutrn to the origin. (c) With $v_0=40\,\mathrm{m/s}$, $a=-3.5\,\mathrm{m/s^2}$ and $t=23\,\mathrm{s}$, the kinematic equation yields, \begin{align*} v&=v_0+at\\ &=40+(-3.5)(23)\\ &=-41\,\mathrm{m/s} \end{align*} The velocity of the object when it returns to the origin is $-41\,\mathrm{m/s}$.
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