Question

A car is traveling at v = 16m/s . The driver applies the brakes and the car decelerates at a= -4.0 m/s^2 . What is the stopping distance?

Answer 1

vi = 16

d = vi t - 2t^2

0 = vi - 4t
t = vi/4 = 4

d = 16*4 - 2(4^2)
= 2*16
=32 meters