## Physics (10th Edition)

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