## Essential University Physics: Volume 1 (3rd Edition)

We know that the acceleration of the train is given as $a=\mu g$ We plug in the known values to obtain: $a=(0.58)(9.8)=5.684\frac{m}{s^2}$ The required distance is given as $d=\frac{v^2}{2a}$ $d=\frac{(\frac{140}{3.6})^2}{(2)(5.684)}=130m$ Hence, the train will stop in time.