The car is moving with a speed v0 = 48 mi/hr up the 4-percent grade, and the driver applies the brakes at point A, causing all wheels to skid. The coefficient of kinetic friction for the rain-slicked road is μk = 0.61. Determine the stopping distance sAB. Repeat your calculations for the case when the car is moving downhill from B to A.

s = 0.2203 miles

Step-by-step explanation:

Given:

- The initial speed vo = 48 mi/hr

- The mass of the car m

- The coefficient of kinetic friction uk = 0.61

- The slope of the road 4-percent grade

Find:

Determine the stopping distance sAB. Repeat your calculations for the case when the car is moving downhill from B to A.

Solution:

- Apply the energy principle with work done against friction:

                             K.E_1 + P.E_1 = Wf + P_E_2 + K.E_2

- The final velocity of car is zero, K.E_2 = 0

  The initial potential energy is set as zero, P.E_1 = 0

                             K.E_1 = Wf + P_E_2

                             0.5*m*vo^2 = Ff*s + m*g*h

Where, Ff is the frictional force:

                             Ff = uk*N

Where, N is the normal contact force between car and road. By equilibrium equation we have:

                             m*g*cos(θ) - N = 0

                             N = m*g*cos(θ)

Hence,

                             Ff = uk*m*g*cos(θ)

- The vertical distance travelled h is:

                             h = s*sin(θ)

- The energy equation is:

                             0.5*m*vo^2 = uk*m*g*cos(θ)*s + m*g*s*sin(θ)

                             0.5*vo^2 = uk*g*cos(θ)*s + g*s*sin(θ)

                             s*(uk*g*cos(θ) + g*sin(θ) ) = 0.5*vo^2

                             s = [0.5*vo^2 / (uk*g*cos(θ) + g*sin(θ) ) ]

- The slope = 4 / 100,

                      s = [0.5*48^2 / (0.61*8052.97*cos(2.29) + 8052.97*sin(2.29) ) ]

                      s = 0.2203 miles