Car collision

1. Sep 7, 2009

Bryon

1. The problem statement, all variables and given/known data

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

2. Relevant 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

3. 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!

2. Sep 7, 2009

kuruman

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

3. Sep 7, 2009

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?

4. Sep 7, 2009

Jebus_Chris

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