Answer to Question #14616 in Mechanics | Relativity for CJ

Question

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.

Answer 1

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

Vc = (a * 13.3^2 / 2 + 80.1) / 13.3
((a * 88.5 + 80.1) / 13.3) *
17.1 - (a * 176.9 + a * 146.2) = - 80.1

Vc = (a * 13.3^2 / 2 + 80.1) /
13.3
((a * 88.5 + 80.1) / 13.3) * 17.1 - (a * 176.9 + a * 146.2) = -
80.1
113.7 * a + 103.0 - 323.1 * a = - 80.1
209.4 * a = 183.1
a = 0.874
m/s^2.

Vc = (0.874 * 13.3^2 / 2 + 80.1) / 13.3 = 11.835
m/s.
Answer:
(a) the car's velocity is 11.835 m/s.
(b) the train's
acceleration is 0.874 m/s^2.