Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 2 - Kinematics in One Dimension - Problems: 33

Answer

39.2 m

Work Step by Step

Distance travelled before car starts decelerating: $x=ut=(20.0)(0.530)=10.6\; m$ Distance travelled while decelerating: $v^{2}=u^{2}+2ax$ $x=\frac{v^{2}-u^{2}}{2a}=\frac{0^{2}-20.0^{2}}{2(-7.00)}\approx28.57\; m$ Stopping distance = 10.6 + 28.57 $\approx$ 39.2 m
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.