Calculating Speed of Car A Hitting Car B in Different Friction Conditions

You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B

Car A slid into the rear of car B, which was stopped at a red light along a road headed down a hill. The slope of the hill is θ= 12.0°, the cars were separated by distance d = 24.0 m when the driver of car A put the car into a slide, and the speed of car A at the onset of braking was Vo= 18.0 m/s. The coefficient of kinetic friction was 0.60 for a dry road surface and 0.10 for a road surface covered with wet leaves.

With what speed did car A hit car B?

(a) Coefficient of kinetic friction = 0.60 (dry road surface)

Explanation: Downward acceleration of car A along the slope= g sinθ - μ g 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. Therefore, v = 12.08 m/s.

(b) Coefficient of kinetic friction = 0.10 (road surface covered with wet leaves)

Explanation: In this 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 x 1.079 x 24 = 324 + 52 = 376. Thus, v = 19.4 m/s.

With what speed did car A hit car B if the coefficient of kinetic friction was (a) 0.60 and (b) 0.10?

Explanation: (a) Speed of car A hitting car B with a coefficient of kinetic friction of 0.60 was 12.08 m/s. (b) Speed of car A hitting car B with a coefficient of kinetic friction of 0.10 was 19.4 m/s.