If a driver brakes from 15 m/s and then brakes again from 30 m/s with the same braking force, what happens to the braking distance and why?
With the same braking force, the braking distance increases by a factor of 4 when speed doubles from 15 m/s to 30 m/s. This is because the work done by the brakes is $W = Fd$ and it must remove the carโs kinetic energy $E_k = \tfrac12 mv^2$, so $d = \tfrac{\tfrac12 mv^2}{F} \propto v^2$. Doubling $v$ makes $v^2$ four times larger, so the stopping distance is four times longer.
What the question is really asking
You are comparing two stops where only the initial speed changes. Since the braking force stays the same, the only way to remove more energy is to brake over a longer distance.
Link braking distance to energy
The car starts with kinetic energy $$E_k = \tfrac12 mv^2.$$ The brakes do work against the motion: $$W = Fd.$$ To stop, the work done by the braking force must equal the kinetic energy that needs to be removed: $$Fd = \tfrac12 mv^2.$$
How braking distance depends on speed
Rearrange for distance: $$d = \frac{\tfrac12 mv^2}{F}.$$ If $m$ and $F$ are constant, then $$d \propto v^2.$$ So when the speed doubles from $15\,\text{m/s}$ to $30\,\text{m/s}$, $$\frac{d_2}{d_1} = \left(\frac{30}{15}\right)^2 = 2^2 = 4.$$
What this means in words
Going twice as fast gives the car four times the kinetic energy, and with the same braking force it takes four times the distance to do four times the stopping work.
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