# Determine shortest distance a car could stop without causing serious injury

1. Dec 15, 2012

### cdornz

1. The problem statement, all variables and given/known data
It may be assumed that the human body can withstand an acceleration of 3g's without sustaining serious injury. A person is driving a car at 60 miles per hour. Determine the shortest distance such that the car could be brought to a stop (at constant acceleration) without the driver sustaining serious injury.

2. Relevant equations
Ag's = a/g
d = (Vi + Vf)/2 * t (d stands for displacement)

3. The attempt at a solution
I converted the 3g's into an acceleration so I at least have a comparable number when I try and get the answer. 3*32ft/s2 = 96ft/s2

I haven't done this type of problem in quite a few months, so I forget where to start per say. I did convert the 60 miles per hour to feet per second, 88ft/s. I don't remember if the equation I listed above is the right one to use in this situation, especially since I don't know time or the displacement.

2. Dec 15, 2012

### TSny

If you are initially traveling at 88 ft/s, how much time will it take to come to rest if you are decelerating at 96 ft/s per second?

Then you can use your distance formula.

(There are also other formulas for constant acceleration. There's a formula that will give you the distance directly from the initial and final velocities and the acceleration without needing the time.)

3. Dec 15, 2012

### ap123

You need to use an equation that doesn't involve time.
You can use
v^2 = u^2 + 2as

4. Dec 15, 2012

### cdornz

thank you! I figured I had the information I needed, I just wasn't sure about the equation necessary to find the solution.