# The train eventually passes the car

## Homework Statement

The train (92m long) starts from rest, constant accel. at t=0s
The car at t=0s just reaches the end of the train. The car is moving at a constant velocity.
At time t=14s the car just reaches the front of the train.
The train eventually pulls ahead of the car, and at t=28s the car is again at the rear of the train. Find the magnitudes of a) the cars velocity b) the trains acceleration

v=d/t
vf=vi + at
d=vit +1/2at^2

## The Attempt at a Solution

I calculated the car travelled a distance of 92m in 14s from the rear to the front of the train at constant velocity using v=d/t=92/14=6.57m/s. In another 14s, the car travelling at 6.57m/s will have travelled an additional distance of d=vt=(6.57)(14)=92m for a total distance of 184m in 28s. So the answer i got for a) The cars velocity was 6.57m/s. I know this is wrong, but i do not understand why? Please help?
The trains acceleration i do not know how to figure that out at all.

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The car didnt travel 92 meters in 14 seconds though. The train is moving too, so that means the car traveled much farther.

I only used one of your equations to solve this problem: d= d_0 + vit +1/2at^2

I think the easiest way to solve this problem is to set up the coordinate system so that it is attached to the back of the train (the train will have zero velocity, acceleration, etc). Try to set up two equations with two unknowns for the car remembering that each variable has to be the relative to the coordinate system you are using.

I hope this helps a little, this problem was actually quite hard in my opinion.

no, that doesn't help me at all actually...thanks for trying though

collinsmark
Homework Helper
Gold Member
Hello 1irishman,

BishopUser has given you some good insight.

The problem statement specifies that the car reaches the front of the train at t = 14 sec. But that position is not where the front of the train used to be (at t = 0 sec), but rather it is where the front of the train is at time t = 14 sec. It is a larger distance than 92 m because the train is moving too.

The first step in solving this problem is to formulate equations for dt(t) and dc(t), where:
dt(t) is the position of the train as a function of time, and
dc(t) is the position of the car as a function of time.

As BishopUser has pointed out, it is probably easiest to define dt(t) as the position of the back of the train (such that dt(0) = 0), but this decision is really up to you. You could define dt(t) as the position of the front of the train if you choose to (such that dt(0) = 92 m), or some other part of the train.

Whichever way you choose to define them, create expressions for dt(t) and dc(t).

Once you have these equations, you can move on to the second step. You know the relationship between dt(t) and dc(t) at two additional times, t = 14 s and t = 28 s. Use these relationships to create two simultaneous equations (which can be expressed solely in terms of the car's initial velocity and the train's acceleration). You'll end up with two equations and two unknowns. You can then solve for the unknowns.

Thank you collinsmark...i will see if i can figure it out. I appreciate your additional help with this hard question. The prof hasn't even taught us how to approach these types of questions, so i was at a complete loss....