Question:Traffic Court.You are called as an expert witness in a trial for a traffic violation. The facts are these: A driver slammed on his brakes and came to a stop with constant acceleration. Measurements of his tires and the skid marks on the pavement indicate that he locked his car’s wheels, the car traveled before stopping, and the coefficient of kinetic friction between the road and his tires was 0.750 . He was charged with speeding in a 45- mi/h zone but pleads innocent. What is your conclusion: guilty or innocent? How fast was he going when he hit his brakes?

The given data can be listed below as:

• The distance traveled by car is d =192 ft .
• The kinetic frictional coefficient is${\mu }_k=0.750$ .
• The velocity of the driver is v =45-mi/h .

The velocity of an object is directly proportional to the distance traveled by that object in unit time. The change in velocity in unit time gives acceleration.

The equation of the net force acting on the car is expressed as:

F = ma …(i)

Here,F is the net force, m is the mass of the driver and a is the acceleration of the driver.

From the question, it has been identified that the net force acting on the car is the frictional force. The frictional force acting on the car is given as:

$f={\mu }_{k}mg$ …(ii)

Here, f is the frictional force, ${\mu }_k$ is the kinetic frictional coefficient and g is the acceleration due to gravity.

Equaling the equation (i) and (ii)

$$\begin{gathered} ma={\mu }_{k}mg \\ a={\mu }_{k}g \end{gathered}$$

Substitute the values in the above equation.

$$\begin{gathered} a=\left(0.750\right)\left(32.2{\text{ft/s}}^{\text{2}}\right) \\ =24.2{\text{ft/s}}^{\text{2}} \end{gathered}$$

The equation of the speed of the driver is expressed as:

$v^2=u^2+2ad$

Here, is the final and is the initial speed, is the distance travelled by the car.

As the driver was at rest initially, then the initial velocity of the driver is zero.

Substitute the values in the above equation.

$$\begin{gathered} v^2={\left(0\right)}^2+2\left(24.2{\text{ft/s}}^{\text{2}}\right)\left(192\text{ft}\right) \\ =\left(48.4{\text{ft/s}}^{\text{2}}\right)\left(192\text{ft}\right) \\ =\left(9292.8{\text{ft}}^2{\text{/s}}^{\text{2}}\right) \\ v=96.4\text{ft/s} \end{gathered}$$

The final velocity of the driver is given as:

$$\begin{gathered} v=96.4ft/s\times 0.305m/ft\times {10}^{3}km/m\times 0.621mi/km\times 3600s/h \\ =65.7mi/h \end{gathered}$$

Thus, the driver is guilty.

He was going at 65.7 mi/h when he hit the brakes.