vi = 16
d = vi t - 2t^2
0 = vi - 4t
t = vi/4 = 4
d = 16*4 - 2(4^2)
= 2*16
=32 meters
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?
1 answer