Braking distance depends on speed

1. Stopping distance has two parts. $stopping\; distance=thinking\; distance+braking\; distance$.
2. Thinking distance is travelled before braking begins. For a fixed reaction time, $d=vt$, so thinking distance is directly proportional to initial speed.
3. Braking distance is travelled while the brakes slow the vehicle. It depends on the vehicle's initial kinetic energy and the work done by the braking force.

Work done removes kinetic energy

1. A moving vehicle initially stores kinetic energy. $E_{\mathrm{k}}=\frac{1}{2}m{v}^2$.
2. The braking force does work against the motion. For a constant braking force, $W=Fd$, where $d$ is the braking distance.
3. Bringing the vehicle to rest transfers its initial kinetic energy. Therefore $Fd=\frac{1}{2}m{v}^2$.
4. Rearranging gives the braking-distance model. $d=\frac{m{v}^2}{2F}$.
5. For constant mass and braking force, $d\propto v^2$. Doubling speed makes braking distance four times larger, while tripling speed makes it nine times larger.
6. A larger mass increases braking distance for the same braking force. More kinetic energy must be transferred because $E_{\mathrm{k}}$ is directly proportional to mass.
7. A smaller braking force increases braking distance. Worn brakes, wet or icy roads and reduced tyre grip limit the force available, so a greater distance is required to transfer the same energy.

Estimating stopping distance

1. Convert speed to metres per second before calculating. A useful estimate is $30mph\approx 13{m\; s}^{-1}$ and $60mph\approx 27{m\; s}^{-1}$.
2. Calculate thinking distance with $d=vt$. A typical alert reaction time may be estimated as about $0.7s$ when no value is provided.
3. Calculate braking distance using energy or kinematics. Use $Fd=\frac{1}{2}m{v}^2$ when force is known, or use a valid constant-acceleration equation when deceleration is known.
4. Add the two distances only at the end. Thinking distance and braking distance respond differently to increasing speed.