Chapter 6, Problem 22P

To determine

The factor at which minimum braking distance increases if the speed of a car is increased by 50%.

Answer to Problem 22P

Solution: Braking distance is increased by 2.25 times the initial distance if the speed is increased by 50%.

Explanation of Solution

Given: Speed is increased by $50\%\text{ i}\text{.e}\text{.}v=150\%=1.5v$

Formula Used:

By conservation of energy, total work done by the force is equal to kinetic energy.

  $\begin{array}{l}W=K.E\end{array}$

 $\begin{array}{c}F\times s=\frac{1}{2}m{v}^2\end{array}$

Where, F is the force exerted, s is the braking distance, mis the mass of the car; v is the speed of the car.

Calculations:

Find out the distance s from the above formula

  $s_{}={\frac{m{v}_{}}{2F}}^2$ .... (1)

This is when speed is v

If speed is increased by $50\%$ , substitute vas 1.5 v

  $\begin{array}{l}s={\frac{m(1.5v)}{2F}}^2\end{array}$

  $\begin{array}{c}s={\frac{2.25\times mv}{2F}}^2\mathrm{......}(2)\end{array}$

Comparing equations 1 and 2, the distance is increased by 2.25 times the initial distance, if the speed is increased by $50\%$ .

Conclusion: Braking distance is increased by 2.25 times the initial distance if the speed of the car is increased by 50%.