How far behind is the automobile when it reaches a speed of 38.9 m/s?

[bigzee20]

Homework Statement

An automobile and train move together along
parallel paths at 38.9 m/s.
The automobile then undergoes a uniform
acceleration of −4 m/s2 because of a red light
and comes to rest. It remains at rest for 71.7 s,
then accelerates back to a speed of 38.9 m/s
at a rate of 3.52 m/s2.
How far behind the train is the automobile
when it reaches the speed of 38.9 m/s, as-
suming that the train speed has remained at
38.9 m/s? Answer in units of m.

Ok i know that the total distance that the train covered in those 71.7s is 2789.13m now what else do i need to do to find out how far behind is the automobile? How can i tell the total distance traveled by the car? is it d = vi(t)+1/2a(t)^2?

 

[Redbelly98]

You need to consider the car's motion as 3 separate calculations:

1. Deceleration at -4 m/s^2
2. At rest
3. Acceleration at 3.52 m/s^2

How much time does each of those take?

 

[bigzee20]

Ok i got the times
1. 9.725 sec
2. 71.7s
3. 11.05s

 

[Redbelly98]

Okay, all that's left is to figure out how far the car and the train move during those times.