u = a/g
a = u*g = 0.53*(-9.8) = -5.194 m/s^2
V^2 = Vo^2 + 2a*d = 0
Vo^2 = 2*5.194*11 = 114.3
Vo = 10.7 m/s.
A car moving at some speed hits the brakes and skids to a stop after 11 m on a level road. If the coefficient of friction for the road conditions of wet asphalt is 0.53, what was the car's original speed before braking?
I have used multiple modes to find the answer and have solved for velocities ranging from 14-78.
Please help?
1 answer