Question

A car traveling 132 feet per second decelerates at a constant rate to 60 feet per second in three seconds.
A. what is the negative acceleration?
B. How long after the brakes are first applied will it take to come to a complete stop?
C. How far does it travel before coming to a stop from when the brakes are first applied?

Answer 1

change in velocity/change in time = (60-132)/3 = -24 ft/s^2

change in velocity = a t
0 - 132 = -24 t
t = 5.5 seconds to stop

d = Vi t + (1/2) a t^2
= 132(5.5) -.5(24)(5.5)^2
= 426 - 363 = 363 feet