Question

You have been hired as an expert witness in a court case involving an automobile accident. The accident involved car A of mass 1430 kg which crashed into stationary car B of mass 1160 kg .The driver of car A applied his brakes 15 m before he skidded and crashed into car B. After the collision, car A slid 18 m while car B slid 30 m.The coefficient of kinetic friction between the locked wheels and the road was measured to be 0.60. Show that the driver of car A was exceeding the 55-mi/h (90 km/h) speed limit before applying the brakes. Find this speed

Answer 1

due to friction ,

acceleration of cars , a = -u * g

a = -0.60 * 9.8

a = -5.88 m/s^2

speed of car A after collision, vA = sqrt(2 * a * d)

vA = sqrt(2 * 5.88 * 18)

vA = 14.55 m/s

speed of car B after collision ,

vB = sqrt(2 * 5.88 * 30)

vB = 18.8 m/s

Now, let the initial speed of car A is u

using conservation of momentum

vA * mA + mB * vB = mA * u

1430 * u = 1430 * 14.55 + 1160 * 18.8

u = 29.8 m/s

u = 29.8 * 3600/1000 kmh

u = 107.23 km/h

the speed of car A was 107.23 km/h before collision