What is the shortest distance he can break?

Question

Rafael is driving his car at 32 m/s. What is the shortest distance in which he can brake and stop if the coefficient of static friction between the tires and the road is 0.61?

Answer 1

Most physics problems can be solved using the Conservation of Energy principle.

Using the GUESS Method:

Givens:

v = 32 m/s

µ = 0.61 (kinetic)

Assumptions:

g = 9.81 m/s^2

Unknowns:

x = distance in m

Equations:

N = F_g = mg

F_µ = µN = µF_g = µmg

KE = (1/2) mv^2

W = Fx (Work is force times distance)

Solve:

Conservation of energy gives

KE = W

Substituting gives

(1/2) mv^2 = F_µ x

(1/2) mv^2 = µmgx

mv^2 = 2µmgx

Solving for distance (x) gives

x = mv^2 / 2µmg

Simplifying by cancelling out the mass to gives

x = v^2 / 2µg

Substitute:

x = v^2 / 2µg

x = 32^2 / 2(0.61)(9.81)

ANSWER:

x = 85.56 m

Therefore, the minimum braking distance is 85.56 meters.

Answer 2

F_w = -μmgd = -1/2mv₀²

μ = 0.61

g = 9.81 m/s²

^v₀ = 32 m/s

d = v₀²/2μg = (32)²/(2*0.61*9.81) = 85.56 m