Question

A car traveling at 37 m/s breaks to a stop in 6.0 m and 3.5 s. What was the car’s acceleration?

Answer 1

For decelerated motion

s= vt-at^2/2.

a=2(vt-s)/t^2=2(37•3.5-6)/3.5^2=20.16 m/s^2