(II) If the speed of a car is increased by 50%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver's reaction time.

The speed of the car is increased by 50%.

Let the initialspeed of the car be\({v_1}\), and the final speed of the car be\({v_2}\).

Let the stopping distance of the car with\({v_1}\)speed be\({d_1}\)and that with\({v_2}\)speed be\({d_2}\)

Consider a constant friction force F acting on the car.

Since the change in kinetic energy is equal to work done and kinetic energy is proportional to the square of the speed,the braking distance is also proportional to the square of the speed.

The relation between the speeds as per the question can be given as follows:

\(\begin{array}{c}{v_2} = {v_1} + \frac{{50}}{{100}}{v_1}\\{v_2} = 1.5{v_1}\\\frac{{{v_2}}}{{{v_1}}} = 1.5\end{array}\)

The final kinetic energy will be zero for both speeds of the car to stop. Therefore, the car's change in kinetic energy is \(\frac{1}{2}mv_1^2\) for \({v_1}\) speed, and \(\frac{1}{2}mv_2^2\) for \({v_2}\) speed.

The car lost its kinetic energy by working against the friction force.

Therefore, from the application of the work-energy theorem for both speeds,

\(\frac{1}{2}mv_1^2 = F{d_1}\)… (i)

\(\frac{1}{2}mv_2^2 = F{d_2}\). … (ii)

Now, divide equation (ii) by equation (i) to obtain the relation between the braking distances for both speeds.

\(\begin{array}{l}\frac{{\frac{1}{2}mv_2^2}}{{\frac{1}{2}mv_1^2}} = \frac{{F{d_2}}}{{F{d_1}}}\\{d_2} = {\left( {\frac{{{v_2}}}{{{v_1}}}} \right)^2}{d_1}\\{d_2} = {\left( {1.5} \right)^2}{d_1}\\{d_2} = 2.25{d_1}\end{array}\)

Hence, the braking distance will increase by a factor of 2.25.