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 - Page 62: 64


Bob is in jail.

Work Step by Step

We can find the distance the car moves in 0.60 seconds; $d = v~t = (50~m/s)(0.60~s)$ $d = 30~m$ We can find the distance $x$ the car travels during the braking period; $x = \frac{v^2-v_0^2}{2a}$ $x = \frac{0-(50~m/s)^2}{(2)(-10~m/s^2)}$ $x = 125~m$ The car travels a total distance of 125 m + 30 m which is 155 meters. Since the car needed to stop within 150 meters, Bob is in jail.
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.