Time for Car 1 and Car 2 Collision: Solving the Puzzle

[Bryon]

Homework Statement

I am trying to find the time a collision occurs of car 1 that is traveling 31m/s and can accelerate at -1.8m/s and car 2 that is traveling at a constant velocity of 6m/s.

Homework Equations

v(final)^2=v(initial)+ 2a(x(final) - x(initial))
v(final) = v(initial) + at
x(final) = x(initial) + v(initial)t + .2at^2

The Attempt at a Solution

I found change in velocity of car 1 over the 30 meter distance.

v(final)^2 = 31^2 - 2(-1.8)(-30) = 28,837
28.837 = 31 + (-1.8)t ...t = 0.996

the distance car 2 traveled over the 0.996s is 5.976m

so adding the distance car 2 traveled plus the distance car 1 is initially from car 2...

v(final)^2 = 31^2 - 2(-1.8)(-35.976) = 28.835
28.835 = 31 + (-1.8)t.....t =1.204s

Which 1.204 seconds turned out to be the wrong answer. Would I have to find the relative velocity between the cars over the 30 meters? Would the relative velocity be the average over the 30m? I am not sure what else to look at.

Thanks for the help!

 

[kuruman]

Can you state the problem exactly as it is given? Specifically, how far apart are the cars initially?

 

[Bryon]

Here is the problem: A certain automobile can decelerate at 1.8 m/s^2. Traveling at a constant car 1 = 31m/s, this car comes up behind a car traveling at a constant car 2 = 6m/s. The driver of car 1 applies the brakes until it is just 30m behind the slower car. Call the instant which the brakes are applied t = 0. At what time does the inevitable collision occur?

 

[Jebus_Chris]

A collision is when their positions are the same. So create two equations, 1 for each car, that model each cars position.
$$x=x_o+v_ot-\frac{1}{2}gt^2$$