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

## Homework Statement

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.

## Homework Equations

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

## 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.

TSny
Homework Helper
Gold Member
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.)

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

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