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
Physics (10th Edition)
by Young, David; Stadler, Shane
Published by
Wiley
ISBN 10:
1118486897
ISBN 13:
978-1-11848-689-4
Chapter 2 - Kinematics in One Dimension - Problems - Page 49: 31
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
