A train has a length of 102 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 = 18.3 s, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time t = 38.0 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.

atrain = 0.5658 m/s^2

vocar = 10.75 m/s

Explanation:

we have the following dа ta:

vcar = ?

acar = 0

votrain = 0

atrain = ?

t1 = 18.3 s

L = 102 m

t2 = 38 s

For time t1, the distance that the car advances is the same as the distance that the train advances, to this we must add the length of the train. For time t2, the distance the car travels is the same as the train travels. Therefore, an equation equals:

x = vo*t + (a*t^2) / 2

for the car, we have:

dcar = vocar*t

for the train, we have:

dtrain = (atrain*t^2) / 2

dcar = dtrain + L = (atrain*t1^2) / 2 + L = vocar*t1 (For the time 1, t1)

For the time 2, t2:

dcar = dtrain = (atrain*t2^2) / 2 = vocar*t2

Clearing vocar:

vocar = (atrain*t2) / 2 (eq. 1)

(atrain*t1^2) / 2 + L = vocar*t1 = (atrain*t2) * t1

Clearing atrain:

(atrain*t2) * t1 - (atrain*t1^2) = L

atrain = (2*L / (t2*t1 - t1^2) = (2*102) / (38*18.3 - 18.3^2) = 0.5658 m/s^2

In eq. 1:

vocar = (0.5658*38) / 2 = 10.75 m/s