Physics (10th Edition)

Published by Wiley

Chapter 2 - Kinematics in One Dimension - Problems - Page 49: 31

Answer

$1.0\; m/s^{2}$ left

Work Step by Step

Right is the positive direction: $v^{2}=u^{2}+2ax$ $a=\frac{v^{2}-u^{2}}{2x}=\frac{0^{2}-5.0^{2}}{2(12.5)}=-1.0\; m/s^{2}$ Hence, acceleration is $1.0\; m/s^{2}$ left

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.