Question

You have been hired as an expert witness in a court case involving an automobile accident. The accident involved a car of mass 1910 kg (car A) which approached a stationary car of mass 1000 kg (car B). The driver of car A applied his brakes 13 m before he crashed into car B. After the collision, car A slid 17 m while car B slid 30 m. The coefficient of kinetic friction between the locked wheels and the road was measured to be 0.80. Reconstruct the accident by calculating all the relevant velocities.

The speed of car A just before the breaks were applied.

km/h

The speed of car A just before the collision.

km/h

The speed of car A just after the collision.

km/h

The speed of car B just after the collision.

km/h

Answer #1

Let's compute speeds after collision using energy

.5*mA*vA^2=mA*g*µk*dA

simplify

vA=sqrt(2*g*µk*dA)

vA=16.33 m/s = 16.33 * 3.6 km/h =
**58.79 km/h (part c)**

similar for B

vB=21.69 m/s = **78.08 km/h
(part d)**

the speed at impact was

1910*vAi=1910*16.33+1000*21.69

solve for vAi

vAi=27.69 m/s = **99.68 km/h
(part b)**

The KE before the brakes were applied less the

work done by the brakes equals the KE at the moment of impact.

Using µk assumes skidding which is in the driver's favor since µs is higher. Let's work with µk since there are apparently 13 m of skid marks before the collision

.5*mA*vA^2=mA*g*µk*13+.5*mA*vAi^2

solve for vA

vA=sqrt(2*g*µk*13+vAi^2)

plug in the numbers

31.15 m/s = **112.14 km/h
(part a)**

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