Answer:
μk = (Vf - Vc)/(T×g)
Explanation:
Given
Vi = initial velocity of the car
Vf = final velocity of the car
T = Time of application of brakes
g = acceleration due to gravity (known constant)
Let the mass of the car be Mc
Assuming only kinetic frictional force acts on the car as the driver applies the brakes,
The n from Newtown's second law of motion.
Fk = Mc×a
Fk = μk×Mc×g
a = (Vf - Vc)/T
Equating both preceding equation.
μk×Mc×g = Mc × (Vf - Vc)/T
Mc cancels out.
μk = (Vf - Vc)/(T×g)
