Collision of Two Cars: Time, Distance & Speed

[plaguemelove]

1.Two stunt drivers drive directly toward each other. At time, t=0, the two cars are a distance 85.0 m apart, car 1 is at rest, and car 2 is driving towards car 1 at speed 13.8 m/s. Car 1 begins to move towards car 2 at t=0, speeding-up with constant acceleration 3.52 m/s^2. Car 2 continues to move at constant velocity.

a. At what time do the two cars collide?
b.How far did car 1 move during that time?
c.Find the speed of the car just before the collision.

2. No relevant Equations were given.

3.a. At what time do the two cars collide?
Here is what I did:
x(t)=car 1
y(t)=car two
x(t)=x0+v0*t+0.5*a*t^2
y(t)=x0+v0*t+0.5*a*t^2
x(t)=y(t)
0.5*a*t^2=v0*t
t=2*v0/a
t=2(13.8m/s)/3.52m/s^2
t=7.84 s

b.How far did car 1 move during that time?
Here is what I did:
x(t)-x0=v0*t+0.5*a*t^2
x(t)-x0=0.5*3.52m/s^2 * (7.84s)^2
x=108 m

c.Find the speed of the car just before the collision.
Here is what I did.
v(t)=v0+a*t
v(t)-v0=at
v(t)-v0=(3.52 m/s^2)(7.84s)
v=27.60 m/sI am not sure if I did this problem completely correctly because we do not have these sort of problems in our books or are taught in class, because they want us to learn the application of these equations. Therefore I tried. Please tell me if this is correct and if not, lead me into the direction.
Thank you so much.

 

[BruceW]

x(t)=car 1
y(t)=car two
x(t)=x0+v0*t+0.5*a*t^2
y(t)=x0+v0*t+0.5*a*t^2
This is right. (As long as you remember that the constants in each equation are different).
0.5*a*t^2=v0*t
But this is not right, because car 2 starts at a different position than car 1, but in this equation, you are effectively saying that they start in the same place. Also, the velocity of car 2 should be opposite to that of the acceleration of car 1, so you need to remember that v0 = -13.8 m/s (if we take car 1 to be traveling in positive direction, and car 2 to travel in negative direction).