Chapter 11, Problem 46A

To determine

The distance travel by the car after the brakes are applied

Answer to Problem 46A

  $66\text{m}$

Explanation of Solution

Given:

The weight of the car, $mg=14700\text{N}$

Initial velocity of the car, $v_i=25\text{}\text{m/s}$

Since finally the car stops, thus the final velocity of the car, $v_f=0\text{m/s}$

The average force between the tires and the road is, $F=7100\text{N}$

Formula used:

The kinetic energy of an object of mass $m$ travelling at a speed of $v$ is given by,

  $KE=\frac{1}{2}m{v}^2$

The work done on a object is given by,

  $W=Fd\cos \theta$

Where $F$ is the applied force, $d$ is the displacement of the object, and $\theta$ is the angle between the force and the displacement.

And, the work done on an object is given by the change in its kinetic energy,

  $W=K{E}_f-K{E}_i$

Where $K{E}_f$ is the final kinetic energy, and $K{E}_i$ is the initial kinetic energy.

Calculation:

Thus, combining above formulas, it can be written,

  $\begin{array}{c}W=K{E}_f-K{E}_i\\ Fd\cos \theta =K{E}_f-K{E}_i\end{array}$

Since, the force is applied in opposite direction of the displacement of the car, thus $\theta =180°$ .

Using all the given values in above, the distance travelled by the car after the brakes are applied will be,

  $\begin{array}{c}W=K{E}_f-K{E}_i\\ \left(7100\right)d\cos 180°=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2\\ -\left(7100\right)d=\frac{1}{2}m\left(0\right)-\frac{1}{2}\left(\frac{14700}{9.8}\right){\left(25\right)}^2\\ d\approx 66\text{m}\end{array}$

Conclusion:

Thus, the distance travelled by the car after the brakes are applied is $66\text{m}$ .