Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 2 - Exercises and Problems - Page 31: 49


Please see the work below.

Work Step by Step

We know that the distance traveled in the first fifteen minutes can be determined as $d_1=v_1t_1$ We plug in the known values to obtain: $d_1=(20\frac{mi}{h})(\frac{15}{60})=5mi$ The required distance to be covered is 20 mi in 25 min so $v_2=\frac{20}{\frac{25}{60}}$ $v_2=48\frac{mi}{h}$
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.