# Finding Acceleration

## Homework Statement

A car travelling 70mph can stop in 157ft. What is its rate of acceleration? (assume its constant)

## The Attempt at a Solution

70mph=102.6ft/sec

So if the the acceleration is constant, the graph of velocity will be a line with constant slope. The integral then will be the area of the triangle with heigh 102.6 and width t, the time when velocity=0. We also know that the integral is equal to 157. So, we have 102.6/2*t=157

t=3.05 seconds

Since acceleration is distance/second/second we multiple 102.6*3.05 seconds which is 312.9 and this should be the rate of acceleration. However if I were to use differential equations to create a position function and evaluate it at t=3.05 I don't get 157ft which suggests that I did something wrong. So, what am I doing wrong? Thanks for the help.

This is actually for a calc II class( so maybe I should move it to the calc forum, but I thought it had a distinctly physics flavor to it), so let's say I "wanted" to use integrals, where am I going wrong?

jhae2.718
Gold Member
Since acceleration is distance/second/second we multiple 102.6*3.05 seconds which is 312.9 and this should be the rate of acceleration.

The dimensions of acceleration are length over time squared. So you should divide the acceleration by the time, not multiply. Didn't catch that the first time around. Other than that, you're fine.

Seems like an awful lot of trouble to go through to get the answer.

use simple equations you have initial and final velocity and distance
use v^2 = u^2 + 2as