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.
It is important to note that the thinking distance is proportional to the starting speed. This is because the reaction time is taken as a constant, and distance = speed × time.
Braking forces
However, 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 = 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
\( d = v \times t\)
\( d = {12} \ m/s \times {0.5} \ s\)
\(thinking \ distance = 6 \ m\)
Example braking distance calculation
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?
\( F \times d = \frac{1}{2} \times m \times v^{2}\)
\(d = \frac{1}{2} \times \frac {{m} \times {v}^{2}}{F} \)
\(d = \frac{1}{2} \times \frac{900 \times 12^{2}}{2,000}\)
\(braking \ distance = 32 \ m\)
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.
\(thinking \ distance = 24 \ m/s \times 0.5 \ s = 12 \ m\)
\(braking \ distance = \frac{1}{2} \times \frac {{900} \times {24}^{2}}{2000} = 130~m\)
\(stopping \ distance = 12 + 130 = 142 \ m\)
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
\( F \times d = \frac{1}{2} \times m \times v^{2}\)
\(F = \frac{1}{2} \times \frac {{m} \times {v}^{2}}{d} \)
\(F = \frac{1}{2} \times \frac {{1,600} \times {27}^{2}}{100} \)
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
\( F \times d = \frac{1}{2} \times m \times v^{2}\)
\(F = \frac{1}{2} \times \frac {{m} \times {v}^{2}}{d} \)
\(F = \frac{1}{2} \times \frac {{36,000} \times {22}^{2}}{100} \)
braking force is ~87,000 N
