# Acceleration/Speed Problem

1. Dec 30, 2012

### theultimate6

1. The problem statement, all variables and given/known data
A 108 m long train starts from rest (at t = 0) and accelerates uniformly. At the same time (at t = 0), a car moving with constant speed in the same direction reaches the back end of the train. At t = 12 s the car reaches the front of the train. However, the train continues to speed up and pulls ahead of the car. At t = 32 s, the car is left behind the train. Determine,
a. the car’s speed
b. the trains acceleration

3. The attempt at a solution
When the car reached the front of the train it moved a distance of $$x+108$$ where x is the distance covered by the train

from $$x=Vot+0.5at^2$$

$$x=0.5at^2 =0.5*a*12^2 =72a$$

the distance covered by the car in the first 12 seconds is $$72a+108$$

that means its speed is
$$v=x/t = (72a+108)/12 =6a+9$$

by 32 seconds the car has moved a distance of

$$x=vt$$
$$=(6a+9) *(32-12) =20(6a+9)=120a+180$$

the train covered this distance + 108 meters:
distance covered by the train : $$120a+180+108 =120a+288$$

the distance covered by the train is also

$$V+0.5a*20^2+108$$

Where $$V= 72a$$

Therefore : $$72a+0.5a*20^2+108=120a+288$$

Solving for $$a$$ we get

$$a=0.5263 m/s^2$$

therefore $$V_c =12.1578 m/s$$

2. Dec 30, 2012

### SammyS

Staff Emeritus
What you have below occurs during the time from the 12 second mark to the 32 second mark.
You should not have added the 108 in the above equation.

What you have for the following velocity is actually the distance the train traveled during the first 12 seconds.

The train's velocity at the 12 second mark is the same as the car's velocity (Do you know why?) and is given by $\ \ V=6a+9\ .$
Instead of considering the situation from the 12 second mark to the 32 second mark, you could have ised the fact that from the 0 second mark to the 32 second mark, both the car and train travel the same distance.