Calculating stopping distances

It is important to be able to:

  • estimate how the stopping distance for a vehicle varies with different speeds
  • calculate the work done in bringing a moving vehicle to rest

The diagram shows some typical stopping distances for an average car in normal conditions.

Bar chart showing the thinking and braking distances of a car at different speeds. The greater the speed, the longer the thinking and braking takes.Bar chart showing the thinking and braking distances of a car at different speeds. The greater the speed, the longer the thinking and braking takes.

Some typical stopping distances

Travelling at 20 mph (32 km/h):

  • thinking distance = 6 m
  • braking distance = 6 m
  • total stopping distance = 12 m

Travelling at 40 mph (64 km/h):

  • thinking distance = 12 m
  • braking distance = 24 m
  • total stopping distance = 36 m

Travelling at 70 mph (112 km/h):

  • thinking distance = 21 m
  • braking distance = 75 m
  • total stopping distance = 96 m

It is important to note that the thinking distance is proportional to the starting speed. This means that it increases proportionally as speed increases - ie if speed doubles, thinking distance also doubles.

However, the braking distance increases by a factor of four each time the starting speed doubles.

For example, if a car doubles its speed from 30 mph to 60 mph, the thinking distance will double from 9 m to 18 m and the braking distance will increase by a factor of four from 14 m to 56 m.

Braking forces - Higher

The braking distance increases four times each time the starting speed doubles. This is because the work done in bringing a car to rest means removing all of its kinetic energy.

Work done by brakes = loss of kinetic energy

Work done = braking force × distance

W = F \times d

kinetic\ energy = \frac{1}{2} \times mass \times (velocity^{2})

KE=\frac{1}{2} \times m \times v^{2}

This means that: F \times d  = \frac{1}{2} \times m \times v^{2}

So for a fixed maximum braking force, the braking distance is proportional to the square of the velocity.

Example thinking distance calculation

A car travels at 12 m/s. The driver has a reaction time of 0.5 s and sees a cat run into the road ahead. What is the thinking distance as the driver reacts?

distance = speed × time

d = v \times t

d = 12~m/s \times 0.5~s

\underline{thinking\ distance = 6~m}

Example braking distance calculation - Higher

The car in the previous example has a total mass of 900 kg. With a braking force of 2,000 N, what will the braking distance be?

F \times d  = \frac{1}{2} \times m \times v^{2}

d=\frac{1}{2}\frac{mv^{2}}{F}

d=\frac{1}{2} \times \frac{900 \times 12^{2}}{2,000}

\underline{braking\ distance = 32\ m}

Example stopping distance calculation

What is the stopping distance for the car above?

stopping distance = thinking distance + braking distance

stopping distance = 6 + 32

stopping distance = 38 m

Question

Calculate the stopping distance for the car and driver in the example above when travelling at 24 m/s.

thinking\ distance =  24~m/s \times 0.5~s = 12~m

braking\ distance  = \frac{1}{2} \times \frac{900 \times 24^{2}}{2000}  = 130\ m

\underline{stopping\ distance = 12 + 130 = 142~m}

Estimates example - Higher

Estimate the braking force needed to stop a family car from its top speed on a single carriageway in 100 m.

Using values of ~1,600 kg and ~27 m/s

F \times d = \frac{1}{2} \times m \times v^{2}

F = \frac{1}{2}\frac{mv^{2}}{d}

F = \frac{1}{2}\frac{1600 \times 27^{2}}{100}

\underline{braking\ force\ is ~5,800\ N}

Question

Estimate the force needed to decelerate a lorry from its top speed on a single carriageway in 100 m.

Using values of ~36,000 kg and ~22 m/s

F \times d = \frac{1}{2} \times m \times v^{2}

F = \frac{1}{2}\frac{mv^{2}}{d}

F = \frac{1}{2} \times\frac{36000 \times 22^{2}}{100}

\underline{braking\ force\ is ~87000\ N}