Question

 A police investigator measures straight skid marks 27 m long in an accident investigation. Assuming a friction force is 8.00 x10^3 and a car mass is 1.00 x10^3 kg, what was the minimum speed of the car when the brakes locked?

Answer 1

The energy-work theorem says

$$\frac{mv^2}{2}=Fd$$

Thus

$$v=\sqrt{2Fd/m}\\ =\sqrt{2*8*10^3*27/1.0*10^3}=20.8\:\rm m/s$$