Question

By what percent does the braking distance of a car decrease, when the speed of the car is reduced by 10.3 percent? Braking distance is the distance a car travels from the point when the brakes are applied to when the car comes to a complete stop.

Answer 1

The braking distance is the distance traveled by a car experiencing a braking force until it comes to rest.

Our initial energy is solely kinetic:
$E_i = \frac{1}{2}mv^2$

And, since the car goes to rest, it is no longer in motion. It will have no kinetic energy.
$E_f = 0$

Therefore, there was work done by the braking force.

$W_B = E_f - E_i = -\frac{1}{2}mv^2$

Recall the definition of work:
$W = F\cdot \Delta x$

Or in this case, since the displacement and breaking force are antiparallel:
$W = -F_B\Delta x$

This is equivalent to the dissipation of kinetic energy:
$W = -F_B\Delta x = -\frac{1}{2}mv^2$

Now, to visualize this, let's rearrange the equation to solve for displacement.

$\Delta x =\frac{mv^2}{2F_B}$

There is a direct, SQUARE relationship between necessary braking distance speed.

If the speed was reduced by 10.3 percent, its new speed is only 89.7% percent of the original, so:
$\Delta x' =\frac{m(0.897v)^2}{2F_B}$

$\Delta x' = 0.8046\Delta x$

The reduction by a percentage is:
$1 - 0.8046 = 0.1954 \\\\\boxed{= 19.54\%}$