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 = 14.5°, that the cars were separated by distance d = 22.0 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 = 18.5 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
How can this be solved? The book says , a = g sin angle – coefficient cos angle = a negative numver. But when I try it, the number is positive…
Best Answer
Let the $x$ axis be along the road, downwards and let the $y$ axis be perpendicular to the road upwards. Projection of gravitational acceleration on the x-axis is $g\sin{\theta}$ and $-g\cos{\theta}$ on the y-axis. Friction is along the negative $x$ axis and is $F_{xf}=-\gamma{v_{x}}$. Newton along $x$ is $$\dot{v}_{x}=g\sin{\theta}-\frac{\gamma}{m}{v_{x}}$$ the solution is $${v_{x}}(t)=\frac{mg}{\gamma}\sin{\theta}-\frac{mc}{\gamma}e^{-\frac{\gamma}{m}t}$$ At $t=0$ the speed was $v_{0x}$, so $$c=g\sin{\theta}-\frac{\gamma}{m}v_{0x}$$ And the solution with initial conditions is $${v_{x}}(t)=(1-e^{-\frac{\gamma}{m}t})\frac{mg}{\gamma}\sin{\theta}+v_{0x}e^{-\frac{\gamma}{m}t}$$ Use your numbers and calculate