# Particles under constant acceleration

1. Jun 3, 2007

### Aerosion

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

Okay, wordy. Read through though, plz

The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with an acceleration of -5.60 m/s^2 for 4.20s, making straight skid marks 62.4m long ending at the tree. At what speed does it hit the tree?

2. Relevant equations

Vf=Vi+at

3. The attempt at a solution

So I saw I had a=-5.60m/s^2, t=4.20s, xf=62.4m, and Vf=0 (since after he hit the tree, he stopped).

So I plugged them into Vf=Vi+at (or rather Vi=Vf-at), or 0-(-5.60)*4.20, to give 23.52 m/s. It seems wrong, though. Suggestions or advice?

2. Jun 3, 2007

### Pythagorean

you jumped ahead by setting Vf to 0, you simply want to solve for Vf to find out what speed it is going when it hits the tree (as the question asks).

they also give you a time and distance, so you have an average velocity to help figure out the initial velocity in terms of the final velocity (or vice versa)

After it hits the tree, it experiences another negative acceleration as it crumples against the tree, each piece of the car at different rates; but that would be a different problem all together.

Last edited: Jun 3, 2007
3. Jun 3, 2007

### Aerosion

Oh...so I'm to find the average velocity, right? Umm...that's x/t, which is 62.4/4.20, which gives 14.9.

Then, I would have to find the initial velocity, uh huh? xf=xi+Vxit+1/2at^2...Vi=26.6.

Then Vavg=Vi+Vf/2 would come to play...14.9=26.6+x/2, gives a different answer, Vf=3.10m/s. It looks a little bit more correct, so thanks.