You are testifying as an expert witness in a case involving an accident in which

Question

You are testifying 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 downhill. You find that the slope of the hill relative to the horizontal is (theta)=12.0 degrees, that the cars were separated by a distance d=24.0m when the driver of car A put the car into a slide, and that the speed of car A at the onset of braking was V(initial)=18.0m/s. With what speed did car A hit car B if the coefficient of kinetic friction was (a) 0.60 for dry road and (b) 0.10 for wet road

Explanation / Answer

The car decelerates the force decelerating the car is the frictional force minus the component of gravity pushing pushing the car downhill. F = ma F fr = mu F normal - mg sin12 ma = mu m g cos 12 - mg sin12 a = mu g cos 12 - g sin 12 this is the a that we will use to find the final speed using the kinematic equation Vf^2 = Vo^2 + 2ad (a) mu = 0.60 a = 0.60*9.8m/s^2*cos12 - 9.8m/s^2*sin12 = -3.71m/s^2 Vf^2 = (18m/s)^2 - 2*3.71m/s^2*24m = 146 m^2/s^2 Vf = 12 m/s (b) mu = 0.10 tan(12) = .21 so we need a coefficient of friction of 0.21 in order for even a parked car not to slip on this hill - the first collision was ugly; this will be a disaster a = 0.1*9.8m/s^2*cos12 - 9.8m/s^2*sin12 = 1.08m/s^2 !! the car is accelerating! Vf^2 = (18m/s)^2 + 2*1.08m/s^2*24m = 146 m^2/s^2 Vf = 19.4 m/s

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