## College Physics (4th Edition)

We can find the rate of deceleration: $F = ma$ $a = \frac{F}{m}$ $a = \frac{84,000~N}{55,200~kg}$ $a = 1.52~m/s^2$ We can find the required distance for the train to stop: $v_f^2 = v_0^2+2a\Delta x$ $\Delta x = \frac{v_f^2-v_0^2}{2a}$ $\Delta x = \frac{0-(26.8~m/s)^2}{(2)(-1.52~m/s^2)}$ $\Delta x = 236~m$ Since the train requires a distance of 236 meters to stop, and the truck is only 184 meters ahead, the train can not be stopped in time.