College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 4 - Problems - Page 148: 12


(a) The distance traveled during this time is $223.5~m$ (b) The acceleration magnitude is $0.99~m/s^2$

Work Step by Step

(a) $\Delta x = v_{ave}~t$ $\Delta x = \frac{v_f+v_0}{2}~t$ $\Delta x = (\frac{27.3~m/s+17.4~m/s}{2})~(10.0~s)$ $\Delta x = 223.5~m$ The distance traveled during this time is $223.5~m$ (b) We can find the acceleration: $v_f = v_0+at$ $a = \frac{v_f-v_0}{t}$ $a = \frac{27.3~m/s-17.4~m/s}{10.0~s}$ $a = 0.99~m/s^2$ The acceleration magnitude is $0.99~m/s^2$
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.