Question

Because of your expert physics skills, you have been hired as an analyst in a court case involving an automobile accident. The accident involved a 2,402 kg car (car A) which approached a stationary car of mass 1,506 kg (car B). The driver of car A slammed on his brakes(locking his wheels) 18 m before he/she crashed into car B. After the collision, car A slid 19 m forward while car B slid 22 m forward (Note: the cars did not stick together upon colliding). The coefficient of kinetic friction between the locked wheels and the road was measured to be 0.78. Calculate the speed of car A, in mph, at the moment he began braking. (1 mph = 0.447 m/s) fyi: The speed limit was 55 mph.

Answer 1

Step-by-step explanation:

We shall calculate the negative work done by friction on both the cars which exhausted the kinetic energy of car A

work done by friction on car A

μ mg x d

μ is coefficient of friction mg is weight of the car and d is distance of slippage

= .78x 2402 x 9.8 x (18 +19)

= 679352.85 J

Work done by friction on car B

μ is coefficient of friction mg is weight of the car and d is distance of slippage

= .78x 1506 x 9.8 x 22

= 253261 J

Total = 932613.85 J

Now if v be the velocity of car A

1/2 m v ² = 932613.85

v² = (932613.85 x2) / 2402

= 776.53

v = 27.87 m /s

= 27.87 / .447

= 62.34 mph