Drag care race, finding distance and time

[Prio]

EDIT***: sorry I figured out how to do this problem and don't need help anymore...:\
Disregard and even delete this thread.

Homework Statement

A race-car driver buys a car (CAR A) that can accelerate at +5.9 m/s² . The racer decides to race against another driver with a car (CAR B) that can move with a constant acceleration of +3.6 m/s². Both start from rest, but CAR B leaves 1.0 s before the driver of CAR A.

Find the time it takes CAR A to overtake CAR B

Find the distance the two drivers travel during this time.

Find the velocities of both cars when they are side by side.

Homework Equations

I'm not sure if this is what to use, but
Final velocity = Initial velocity + at
or maybe
d = (initial velocity)(t) + 1/2 (a)(t)²

The Attempt at a Solution

Because car B has a 1 second lead, it will have established a 3.6 m lead on car A and already entered acceleration when car A gets off the line. I would think then that car B has an initial velocity of 3.6 while car A's is 0. I'm trying to find t where d₁=2₂

So d₁ = 0 + 1/2 (5.9)(t)²
d₂ = 3.6 + 1/2 (3.6)(t)²

if these are set equal to equal to each other and the first is solved for t, I get t²=2.95

then I plugged this number into the t² in the other equation and get 3.6 + 1/2(3.6)(2.95)

= 8.91s

Did I do this right?

vvvvv

EDIT***: sorry I figured out how to do this problem and don't need help anymore...:\
Disregard and even delete this thread.

 

[nasu]

You can do it this way if you like it.
The distance traveled by the first car in 1 s is not 3.6 m. Look again at your second formula.
Then, when you write again the distance traveled by the first car, it has an initial velocity of 3.6 m/s. You should have this term too (x=xo+vo*t + 1/2 a*t^2)