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: 26

Answer

(a) The length of the incline is 63 meters. (b) It takes 7.1 seconds to reach the bottom of the incline.

Work Step by Step

(a) The acceleration down the incline is $g~sin(\theta)$. Therefore; $d = \frac{v^2-v_0^2}{2a}$ $d = \frac{(15~m/s)^2-(3.0~m/s)^2}{(2)(9.80~m/s^2)~sin(10^{\circ})}$ $d = 63~m$ The length of the incline is thus 63 meters. (b) $t = \frac{v-v_0}{a}$ $t = \frac{(15~m/s)-(3.0~m/s)}{9.8~m/s^2~sin(10^{\circ})}$ $t = 7.1~s$ It takes 7.1 seconds to reach the bottom of the incline.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.