Question

A policeman investigating an accident measures the skid marks left by a car on the horizontal road. He determines that the distance between the point that the driver slammed on the brakes (thereby locking the wheels) and the point where the car came to a stop was 28.0 m. From a reference manual he determines that the coefficient of kinetic friction between the tires and the road under the prevailing conditions was 0.300. How fast was the car going when the driver applied the brakes

Answer 1

Answer:

The car was 12.8m/s fast when the driver applied the brakes.

Explanation:

The equations of motion of the car in the horizontal and vertical axes are:

$x: f_k=ma\\\\y: N-mg=0$

Since the kinetic friction is defined as $f_k=\mu_kN$ and $N=mg$ we have:

$\mu_kmg=ma\\\\a=\mu_kg$

Next, from the kinematics equation of speed in terms of distance, we have:

$v^2=v_0^2-2ax\\\\v^2=v_0^2-2\mu_kgx$

Since the car came to a stop, the final velocity $v$ is zero, and we get:

$0=v_0^2-2\mu_kgx\\\\v_0=\sqrt{2\mu_kgx}$

Finally, plugging in the known values, we obtain:

$v_0=\sqrt{2(0.300)(9.81m/s^2)(28.0m)}\\\\v_0=12.8m/s$

It means that the car was 12.8m/s fast when the driver applied the brakes.