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: 69


The minimum acceleration is $4.4~m/s^2$

Work Step by Step

At a speed of 30 m/s, the train will reach the crossing in 2 seconds. In the first 0.50 seconds, the car moves 10 meters. Therefore, the car needs to go at least 35 meters in 1.5 seconds. We can find the minimum acceleration as: $y = v_0t+\frac{1}{2}at^2$ $a = \frac{2y - 2v_0t}{t^2}$ $a = \frac{(2)(35~m) - (2)(20~m/s)(1.5~s)}{(1.5~s)^2}$ $a = 4.4~m/s^2$ The minimum acceleration is $4.4~m/s^2$.
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.