Need Help with problems involving constant acceleration and constant speed. :/

[irubi005]

Homework Statement

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.60 m/s^2. At the same instant a truck, traveling with a constant speed of 20.5 m/s, overtakes and passes the car.

How far beyond its starting point does the car overtake the truck?

How fast is the car traveling when it overtakes the truck?

Homework Equations

(V_x)²=(V_0x)²+2a_x(x-x₀)

The Attempt at a Solution

First, I realized that time wasn't involved in the equation. That's how I knew how what equation to use. Then i tried converting it into a_x= ((V_x)₂-(V_0x)₂)/2(X-X₀). After plugging in all the numbers: 3.60 m/s^2 being the acceleration and 20.5 i believe is just the absolute value of the velocity so i plugged that in as the second velocity. Through out the problem i was trying to figure out x... which turned out to be x= 420.25/7.20 m= 58.36. But then the question asks how far beyond it's starting point does the car overtake the truck? how do i figure that out after figuring out x, which is the distance at which the truck overpassed the car... can someone help me?

 

[rl.bhat]

The car and the truck start simultaneously at the signal. When they meet again, they must have traveled same distanced in th4e same interval of time.

Hence 1/2*a*t^2 = v*t.

Solve for t and then find x.