Question

please explain.thank you

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 (see the figure). You find that the slope of the hill is θ = 11.0°, that the cars were separated by distance d = 20.0 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 17.0 m/s. With what speed did car A hit car B if the coefficient of kinetic friction was (a) 0.600 (dry road surface) and (b) 0.150 (road surface covered with wet leaves)?

Answer 1

Given V_0 = 17 m/s theta = 11.0 degree and d = 20 m for mu_k = 0.6 lets focus on car A free body diagram is along the road there are two forces one is complement of mg which is mg sin theta down the road and friction f_k up the road here f_k = mu_k, N = mu_k mg cos theta = mu_k mg cos theta - mg sin theta therefore ma_ret = mu_k mg cos theta - mg sin theta a_net = mu_k g cos theta = g sin theta a_net = 0.6 times 9.8 times cos 11 degree - 9.8 sin 11 degree a_net = 3.9 m/s^2 retanding acceleration up the road \r\n let ar A collides with B with velocity v then form kinematics for constant retardation V^2 = V^2_0 - 2 a_ret d where d = distance covered therefore V^2 = (17)^2 - 2 times 3.9 times 20 = 133 v =