Question

You testify as an "expert witness" in a case involving an accident in which car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. You find that the slope of the hill is θ = 12.0°, that the cars were separated by distance d = 25.0 m when the driver of car A put the car into a slide (it lacked any automatic anti-brake-lock system), and that the speed of car A at the onset of braking was v0 = 18.0 m/s.

(a) With what speed did car A hit car B if the coefficient of kinetic friction was 0.60 (dry road surface)?
____ m/s

(b) What was the speed if the coefficient of kinetic friction was 0.10 (road surface covered with wet leaves)?
____ m/s

Answer 1

distance between cars d = 25 m

initial speed of car A , v₀ = 18 m/s

angle φ = 12⁰

a)

if  coefficient of kinetic friction μ_k = 0.60

apply Newton's second law of motion ,

       mg sinφ - f_k = ma   ............... (1)

here, frictional force f_k = μ_kN

                                   = μ_kmgcosφ   ........... (2)

substitute eq (2) in eq (1) , we get

     mg sinφ - μ_kmgcosφ = ma

       gsinφ - μ_kgcosφ = a   .................... (3)

therefore acceleration a = gsinφ - μ_kgcosφ

                                    = (9.8 m/s²)sin12 - (0.6)(9.8 m/s²)cos12

                                    = 2.03 - 5.7515

                                    ˜ -3.72 m/s²

from kinematic equation ,

                 v² - v_o² = 2ad

                 v = [v_o² + 2ad]^1/2

                   = [(18 m/s)² + 2(-3.72 m/s²)(25 m)]^1/2

                    = 11.75 m/s

b)

coefficient of kinetic friction μ_k = 0.10

from eq (3) ,

  acceleration a = gsinφ - μ_kgcosφ

                        = (9.8 m/s²)sin12 - (0.1)(9.8 m/s²)cos12

                        = 2.03 - 0.9585

                        ˜ -1.07 m/s²

from kinematic equation ,

                 v² - v_o² = 2ad

                 v = [v_o² + 2ad]^1/2

                   = [(18 m/s)² + 2(1.07 m/s²)(25 m)]^1/2

                    = 19.43 m/s