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 (Fig. 6-25). You find that the slope of the hill is u " 12.03, that the cars were separated by distance d " 24.0 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. With what speed did car A hit car B if the coefficient of kinetic friction was (a) 0.60 (dry road surface) and (b) 0.10 (road surface covered with wet leaves)

A) 12.08 m/s

B) 19.39 m/s

Explanation:

A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:

mg(sinθ) – F_k = ma

Where; F_k is frictional force due to kinetic friction given by the formula;

F_k = (μ_k) × F_n

F_n is normal force given by mgcosθ

Thus;

F_k = μ_k(mg cosθ)

We now have;

mg(sinθ) – μ_k(mg cosθ) = ma

Dividing through by m to get;

g(sinθ) – μ_k(g cosθ) = a

a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)

a = -3.71 m/s²

We are told that distance d = 24.0 m and v_o = 18 m/s

Using newton's 3rd equation of motion, we have;

v = √(v_o² + 2ad)

v = √(18² + (2 × -3.71 × 24))

v = 12.08 m/s

B) Now, μ_k = 0.10

Thus;

a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)

a = 1.08 m/s²

Using newton's 3rd equation of motion, we have;

v = √(v_o + 2ad)

v = √(18² + (2 × 1.08 × 24))

v = 19.39 m/s

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