A car and a train move together along straight, parallel paths with the same constant cruising speed v_0. At t = 0 the car driver notices a red light ahead and slows down with constant acceleration - a_0. Just as the car comes to a full stop, the light immediately turns green, and the car then accelerates back to its original speed v_0 with constant acceleration a_0. During the same time interval, the train continues to travel at the constant speed v_0.

the question is:

The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v_0 again?

What I figured out so far is that the time the car takes to decelerate is t = -v / -a and the time it takes to reach back to its original velocity is t = v / a.

Well, the distance travelled during constant acceleration is given by:

s = \frac{u+v}{2} = \frac{v_0}{2}\frac{v_0}{a_0}

Now, acceleration is symmetrical, so when he speeds back up, he travels the same distance as slowing down.

So you need to multiply it by 2 giving \frac{v_0^2}{a_0}

You'll need to check my work because it's early in the morning :)

I'm a little confused on what you did. Here is what I'm doing. Since it takes the car to decelerate t = -v / -a and the time it takes to reach back to its original velocity is t = v / a.
now the car goes the same distance from the time it decelerated as it did to the time it accelerated so you can just multiply v/a * 2. i don't know what to do after this step

OK the part i don't get about your explanation is from where you got this

s = \frac{u+v}{2} = \frac{v_0}{2}\frac{v_0}{a_0}

Also after this do i need to add the 2 distances to get total distance travelled by the car so it will look something like

s = \frac{u+v}{2} = \frac{v_0}{2}\frac{v_0}{a_0} + \frac{v_0^2}{a_0}

Now i need to somehow get total distance travveled by the train in that time which i don't know how to get? And once i get that i need to subtract the total distance from the train - the total distance by the car to get how far the car is behind the train.

A little further explanation is needed please.

Thank you