A train has a length of 81.1 m and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just reaches the end of the train. The car is moving with a constant velocity. At a time t = 11.6 s, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time t = 36.3 s, the car is again at the rear of the train. Find the magnitudes of (a) the car's velocity and (b) the train's acceleration. g

Answer: a) vcar = 7 m/s; b) a train = 0.65 m/s^2

Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.

We have for the car

distance = v car * t, considering the length of train (81.1 m) travel by the car during the first 11.6 s

the v car = distance/time = 81.1 m/11.6s = 7 m/s

In order to calculate the acceleration we have to use the kinematic equation for the train from the rest

distance train = (a * t^2) / 2

distance train : distance travel by the car at constant speed

so distance train = (vcar*36.35) m=421 m

the a traiin = (2 * 421 m) / (36s) ^2=0.65 m/s^2