Asked by Anonymous
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?
Answers
Answered by
Damon
vi = 16
d = vi t - 2t^2
0 = vi - 4t
t = vi/4 = 4
d = 16*4 - 2(4^2)
= 2*16
=32 meters
d = vi t - 2t^2
0 = vi - 4t
t = vi/4 = 4
d = 16*4 - 2(4^2)
= 2*16
=32 meters
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