Acceleration and velocity: Car and train problem

[Cat29]

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.

Homework Equations

x = vt
x = ut + 0.5at^2

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[/B]

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

 

[haruspex]

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]

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.