A train has a length of 80.1 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 = 13.3 s, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time t = 30.4 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.
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Expert's answer
2012-09-25T11:14:10-0400
Let the indeces "c" and "t" for the car and the train, respectively. t1 = 13.3 s: Vc * t1 - a * t1^2 / 2 = L t2 = 30.4 s, t2' = t2 - t1 = 17.1 s. Vc * t2' - (a * t1^2 + a * t2'^2 / 2) = - L
Vc * 13.3 - a * 13.3^2 / 2 = 80.1 Vc * 17.1 - (a * 13.3^2 + a * 17.1^2 / 2) = - 80.1
Vc = (a * 13.3^2 / 2 + 80.1) / 13.3 Vc * 17.1 - (a * 13.3^2 + a * 17.1^2 / 2) = - 80.1
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