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 = 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 Vo = 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) ?

Question

Physics

Mugsy

4 November, 02:31

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 = 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 Vo = 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) ?

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Answers (1)

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Gilbert · Answer 1

a Downward acceleration of car A along the slope

= g sinθ - μ g cosθ

= g (sinθ - μ cosθ)

= 9.8 (sin 12 -.6 x cos 12)

= 9.8 x (.2079 -.5869)

= - 3.714 m / s²

So there will be deceleration

v² = u² - 2 a s

= 18² - 2 x 3.714 x 24

= 324 - 178

= 146

v = 12.08 m / s

b)

In the second case, kinetic friction changes

downward acceleration

= g (sinθ - μ cosθ)

= 9.8 (sin12 -.1 x cos 12)

9.8 (.2079 -.0978)

= 1.079 m / s

there will be reduced acceleration

v² = u² - 2 a s

= 18² + 2 x1.079 x 24

= 324 + 52

= 376

v = 19.4 m / s