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Answer to Question #14616 in Mechanics | Relativity for CJ

Question #14616
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.
Expert's answer
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.

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