Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 67: 67


The cart rolled 6.9 meters.

Work Step by Step

The cart's acceleration is $g~sin(3^{\circ})$. We can compare the distance covered by the cart and Julie to find the time $t$. $\frac{1}{2}at^2 = \frac{1}{2}[g~sin(3^{\circ})]~t^2+20~m$ $\frac{1}{2}at^2 - \frac{1}{2}[g~sin(3^{\circ})]~t^2=20~m$ $t^2 = \frac{40~m}{a-g~sin(3^{\circ})}$ $t = \sqrt{\frac{40~m}{2.0~m/s^2-(9.80~m/s^2)~sin(3^{\circ})}}$ $t = 5.186~s$ We can find the distance the cart has rolled. $x = \frac{1}{2}(g~sin(3^{\circ}))~t^2$ $x = \frac{1}{2}(9.80~m/s^2)~sin(3^{\circ})~(5.186~s)^2$ $x = 6.9~m$ The cart rolled 6.9 meters.
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