Question

Police investigators, examining the scene of an accident involving two cars, measure 72 m long skid marks on one of the cars, which nearly came to a stop before colliding. The coefficient of kinetic friction between rubber and the pavement is about 0.80. Estimate the initial speed of that car, assuming a level road.

Short answer

The initial speed of the car was $33.60\frac{\mathrm{m}}{\mathrm{s}}$.

Step-by-step solution

Step 1. Given data and assumption

After braking, the car stops after moving a distance. During that time, the friction force act on the car in the opposite direction of the motion.

Given data:

The final speed of the car is $v_{\mathrm{f}}=0$as at the end, it stops.

The coefficient of kinetic friction is ${\mu }_{\mathrm{k}}=0.80$.

The length of the skid marks is $\Delta x=72\mathrm{m}$.

Assumption:

Let the initial velocity of the car be $v_{\mathrm{i}}$.

Let the mass of the car be m.

Let a be the deceleration of the box due to the friction force acting on the box.

Step 2. Calculation of the distance before coming to rest

The normal force acting on the car is $N=mg$.

The kinetic friction force on the car is

$\begin{array}{c}{f}_{\mathrm{k}}={\mu }_{\mathrm{k}}N\\ ={\mu }_{\mathrm{k}}mg\end{array}$

Then, to find the magnitude of the deceleration of the car,

$\begin{array}{c}ma=f_{\mathrm{k}}\\ ma={\mu }_{\mathrm{k}}mg\\ a={\mu }_{\mathrm{k}}g\end{array}$

Now, to find the value of $v_{\mathrm{i}}$,

$\begin{array}{c}{v}_{\mathrm{f}}^2=v_{\mathrm{i}}^2-2a\Delta x\\ v_{\mathrm{f}}^2=v_{\mathrm{i}}^2-2{\mu }_{\mathrm{k}}g\Delta x\\ 0^2=v_{\mathrm{i}}^2-\left(2\times 0.80\times 9.80\frac{\mathrm{m}}{{\mathrm{s}}^2}\times 72\mathrm{m}\right)\\ v_{\mathrm{i}}=33.60\frac{\mathrm{m}}{\mathrm{s}}\end{array}$

Hence, the initial speed of the car was $33.60\frac{\mathrm{m}}{\mathrm{s}}$.