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