# Distance and Acceleration for One Car to Catch Another

1. Oct 15, 2014

### hitemup

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

You are traveling at a constant speed vM, and there is a car in front of you traveling with a speed vA. You notice that vM>vA, so you start slowing down with a constant acceleration a when the distance between you and the other car is x. What relationship between a and x determines whether or not you run into the car in front of you?

2. Relevant equations

x = v0*t - 1/2*a*t^2

3. The attempt at a solution

After a time t, the distance between the cars must be something larger than x if they don't want to crash, so:

vM*t-1/2*a*t^2 - vA*t > x

t(vM-1/2*a*t-vA)>x
(vM-vA-1/2*a*t)>x/t

What should be the next step after this?

2. Oct 15, 2014

### NTW

I believe you are complicating things too much. It's simple... A uniform deceleration has to cancel the relative velocity within a given distance...

And I would choose another equation...

3. Oct 15, 2014

### hafiz ns

we can easily find by using...
s=u+at

4. Oct 15, 2014

### NTW

Remember: you know two things, that are enough: 1) the difference of velocity with respect to the other car Vm - Va. Let's call it U. 2) you also know the distance x to the other car when you start braking with a uniform negative acceleration a.

Thus, you should use a equation that gives a as a function of U and x... That is, acceleration as a function of velocity and space. You probably know the equation as giving velocity as a function of space and acceleration.

Just solve for acceleration...

5. Oct 15, 2014

### hitemup

http://www.sketchtoy.com/63345550

I've done this. But what can be said for a and x just looking at this equation?
(vM- vA)^2 /(2a) must be less than x for cars not to crash?

6. Oct 15, 2014

### NTW

The negative acceleration to be calculated by a = U2/2*x is exactly the necessary to avoid contact... More deceleration will equalize the speeds earlier, keeping the distance, and less deceleration would result in a crash...

Last edited: Oct 15, 2014