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: 44

Answer

(a) The rate of deceleration is $\frac{v_0^2}{2~(d - v_0\cdot t_R)}$ (b) The car will be able to stop in time.

Work Step by Step

(a) In a time of $t_R$, the car travels a distance of $v_0\cdot t_R$. The remaining distance is $d - v_0\cdot t_R$. We can find the required acceleration. $a = \frac{0-v_0^2}{2~(d - v_0\cdot t_R)}$ The rate of deceleration is $\frac{v_0^2}{2~(d - v_0\cdot t_R)}$ (b) We can use part (a) to find the minimum rate of deceleration required to stop in time. $a = \frac{v_0^2}{2~(d - v_0\cdot t_R)}$ $a = \frac{(21~m/s)^2}{2~[50~m - (21~m/s)\cdot (0.50~s)]}$ $a = 5.6~m/s^2$ Since the car's maximum deceleration of $6.0~m/s^2$ is greater than the minimum required rate of deceleration ($5.6~m/s^2$), the car will be able to stop in time.
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.