An accident involves a car of mass 1964 kg which approachs a stationary car of mass 1020 kg. The driver of car A applies his brakes 12.3 meters before he crashes into car B. After the collision, car A slides 12.3 meters while car B slides 31.2 meters. The coefficient of kinetic friction between the locked wheels and the road for both cars is 0.630.
How fast was the driver of car A going before he hit the brakes?
I know this is a conservation of energy problem but it has friction involved so I dont know how to appoach the problem
2007-12-04 14:56:02 · 1 answers · asked by nameless 1 in Science & Mathematics ➔ Physics
a = - (0.630)(9.80665) = - 6.1781895 m/s^2
Va2 = √((2)(6.1781895)(12.3) = 12.32816 m/s
Vb2 = √((2)(6.1781895)(31.2) = 19.63464 m/s
Va1 = (1964*12.32816 + 1020*19.63464)/1964
Va1 = 22.52538 m/s
Va0 = √(22.52538^2 + 2*6.1781895*12.3)
Va0 = 25.67832 m/s ≈ 25.7 m/s ≈ 92.4 kph
2007-12-04 15:49:23 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋