Question

If the coefficient of kinetic friction between tires and dry pavement is 0.75, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 25.5 m/s ?

Answer 1

Let us assume the initial direction of the velocity as the positive direction.

Gravitational acceleration = g = 9.81 m/s²

Mass of the automobile = m

Normal force on the automobile from the ground = N

N = mg

Coefficient of kinetic friction between tires and dry pavement = = 0.75

Friction force on the automobile = f

f = -N (Negative as it directed opposite to the initial velocity of the automobile)

f = -mg

Acceleration of the automobile = a

ma = f

ma = -mg

a = -g

a = -(0.75)(9.81)

a = -7.36 m/s²

Initial velocity of the automobile = V₁ = 25.5 m/s

Final velocity of the automobile = V₂ = 0 m/s (Comes to a stop)

Distance the automobile travels = D

V₂² = V₁² + 2aD

(0)² = (25.5)² + 2(-7.36)D

D = 44.2 m

Shortest distance in which the automobile can be stopped by locking the brakes when travelling at 25.5 m/s = 44.2 m