Chapter 9, Problem 9A

Suppose you know the amount of work the brakes of a car must do to stop a car at a given speed. How much work must they do to stop a car that is moving four times as fast? How will the stopping distances compare?

To determine

The change in the work done to stop a car if now the car is moving four times as fast, and to compare the stopping distances.

Answer to Problem 9A

The work done and the stopping distance both will be $16$ times of the initial values.

Explanation of Solution

Introduction:

Work Energy theorem states that the work done on an object is equal to the change in kinetic energy of the object,

  $\begin{array}{c}W=\Delta KE\\ =\Delta \left(\frac{1}{2}m{v}^2\right)\end{array}$

Where,

  $m=$ mass

  $v=$ velocity

Also, the change in kinetic energy of the object can be written in terms of applied force $\left(F\right)$ and distance covered $\left(d\right)$ as,

  $\begin{array}{c}F\times d=\Delta KE\\ =\Delta \left(\frac{1}{2}m{v}^2\right)\end{array}$

Let the initial work done in stopping be $W$ , then it can be written in terms of initial velocity $\left(v\right)$ as,

  $W=0-\frac{1}{2}m{v}^2$

Negative sign denotes that it decelerates, or stops the motion of the car.

Now, if the car was moving four times of its initial velocity, then the work done to stop the car will be,

  $\begin{array}{c}W\text{'}=0-\frac{1}{2}m{\left(4v\right)}^2\\ =16W\end{array}$

Hence, the new work done to stop the car will be $16$ times of the initial required work done.

Similarly, using the mentioned formula, the initial stopping distance can be written as,

  $d\propto -v^2$

Then, if the car was moving four times of its initial velocity, then the new stopping distance of the car will be,

  $\begin{array}{c}d\text{'}\propto -{\left(4v\right)}^2\\ =16d\end{array}$

Conclusion:

Hence, both the work done as well as the stopping distance will be $16$ times of the initial values.