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Acceleration and velocity: Car and train problem

Cat29
1. Homework Statement
A train has a length of 118 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 = 14.0 s, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time t = 34.6 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.

2. Homework Equations

x = vt
x = ut + 0.5at^2


3. The Attempt at a Solution

At t = 12s:
distance x = 0.5a14^2 = 98a
x + 118 = v14
14v - 118 = 98a
14v = 98a + 118
v = 7a + 8.43

At t = 34.6s:
distance y = 0.5a34.6^2 = 598.58a
y = v34.6
34.6v = 598.58a
v = 17.3a

Combining the two:
17.3a = 7a + 8.43
10.3a = 8.43

(a) v = 14.2 m/s
(b) a = 0.818 m/s^2


Is this correct? If it isn't, where did I go wrong? Thanks.
 

Answers and Replies

haruspex
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Is this correct?
Pretty much, but you quote too many significant figures in your answer. To justify three, you would need to keep another digit through all the calculations.
 
Cat29
Pretty much, but you quote too many significant figures in your answer. To justify three, you would need to keep another digit through all the calculations.
Okay. Thank you!
 
scottdave
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You can check your correctness, by considering the front of the train, starting at x0 = 118 meters, then you have x(t) = x0 + v0*t + (1/2)*a*t2, with your values of a, v, and t= 14 sec, and see if the constant velocity car and the front of the train reach the same x value at that time. You could do the same with the tail end of the train (starting at x(0) = 0, and see if they get to the same point at 34.6 seconds.
 

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