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: 66


The cat can escape out the window.

Work Step by Step

We can find the time at which the two animals have the same speed. $(0.85~m/s^2)~t = (1.5~m/s)-(0.10~m/s^2)~t$ $t = \frac{1.5~m/s}{0.95~m/s^2}$ $t = 1.58~s$ We can find the distance the cat moves in this time. $x_c = \frac{1}{2}at^2$ $x_c = \frac{1}{2}(0.85~m/s^2)(1.58~s)^2$ $x_c = 1.06~m$ We can find the distance the dog moves in this time. $x_d = v_0t+\frac{1}{2}at^2$ $x_d = (1.5~m/s)(1.58~s)-\frac{1}{2}(0.10~m/s^2)(1.58~s)^2$ $x_d = 2.25~m$ At t = 1.58 s, the cat is 0.44 meters from the window and the dog is 0.75 meters from the window. After this time, the cat is faster than the dog so the cat can escape out the window.
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.