3.3 Safety in public transport — Learn

Prerequisites: Section 3.1, especially speed, acceleration and $v^2 - u^2 = 2as$ ; Section 3.2, especially $F = ma$ and Newton's first law; Section 1.1, especially the kinetic energy store.

By the end of this section you can: work out a thinking distance, a braking distance and a stopping distance; explain the factors that affect each one; explain why large decelerations are dangerous; and, at Higher tier, estimate the force in a large deceleration.

1. Stopping distance

The stopping distance of a vehicle is the total distance it travels from the moment the driver sees a hazard to the moment the vehicle stops. It has two parts.

The thinking distance is the distance travelled while the driver reacts.
The braking distance is the distance travelled while the brakes act.

\text{stopping distance} = \text{thinking distance} + \text{braking distance}

2. Thinking distance and reaction time

The reaction time is the time between seeing a hazard and starting to brake. It varies from person to person. A typical value is between 0.2 s and 0.9 s.

During the reaction time the vehicle keeps moving at its original speed, so

\text{thinking distance} = \text{speed} \times \text{reaction time}

Worked example 1. A car travels at 15 m/s. The driver has a reaction time of 0.80 s. Calculate the thinking distance.

\begin{aligned} s &= v t \\ &= 15 \times 0.80 \\ &= 12\,\text{m} \end{aligned}

The reaction time is longer when a driver is tired, or has taken drugs or alcohol. Distractions inside or outside the vehicle also make it longer. A longer reaction time means a longer thinking distance, and so a longer stopping distance.

You can measure a reaction time with a simple ruler test. A partner holds a ruler vertically above your open hand. Your partner drops it without warning, and you catch it as quickly as you can. The distance the ruler falls before you catch it measures your reaction time: a longer distance means a slower reaction. Repeating the test and taking an average gives a more reliable result.

3. Braking distance

The braking distance depends on the speed of the vehicle, the braking force, and the conditions of the road, the brakes and the tyres.

The greater the speed, the greater the braking distance. The braking distance increases with the square of the speed, so doubling the speed makes the braking distance four times as large.
A wet or icy road gives less grip, so the same braking force produces a smaller deceleration and a longer braking distance.
Worn brakes or worn tyres reduce the braking force, which also makes the braking distance longer.

When the driver brakes, the friction force between the brakes and the wheels does work. This work reduces the kinetic energy of the vehicle, and the temperature of the brakes rises. The kinetic energy is transferred to the thermal store of the brakes and the surroundings.

Worked example 2. A car of mass 1200 kg travels at 15 m/s. Its braking distance is 25 m. Calculate the deceleration.

\begin{aligned} v^2 - u^2 &= 2 a s \\ 0^2 - 15^2 &= 2 \times a \times 25 \\ -225 &= 50 a \\ a &= \frac{-225}{50} \\ &= -4.5\,\text{m/s}^2 \end{aligned}

The deceleration is $4.5\,\text{m/s}^2$ .

The stopping distance of this car is $12 + 25 = 37\,\text{m}$ : a thinking distance of 12 m from Worked example 1, plus a braking distance of 25 m.

The greater the speed, the greater the braking force needed to stop the vehicle in a certain distance. A greater braking force gives a greater deceleration.

Watch out. Thinking distance and braking distance do not grow in the same way. Thinking distance is proportional to speed, so doubling the speed doubles it. Braking distance depends on the square of the speed, so doubling the speed makes it four times as large. If a question doubles the speed, do not double the whole stopping distance. Work out the two parts separately.

4. Estimating the force in a large deceleration

You can use $v^2 - u^2 = 2as$ to find the deceleration, and then $F = ma$ to find the force.

Worked example 3. Calculate the braking force on the car in Worked example 2.

\begin{aligned} F &= m a \\ &= 1200 \times 4.5 \\ &= 5400\,\text{N} \\ &= 5.4 \times 10^3\,\text{N} \end{aligned}

Worked example 4. A car of mass 1500 kg brakes from 24 m/s to rest in a distance of 36 m. Calculate the deceleration and the braking force.

\begin{aligned} v^2 - u^2 &= 2 a s \\ 0^2 - 24^2 &= 2 \times a \times 36 \\ -576 &= 72 a \\ a &= -8.0\,\text{m/s}^2 \end{aligned}

The deceleration is $8.0\,\text{m/s}^2$ , which is close to the acceleration of a falling object. The braking force is

\begin{aligned} F &= m a \\ &= 1500 \times 8.0 \\ &= 12000\,\text{N} \end{aligned}

A deceleration this large is dangerous. If a vehicle stops suddenly, the people inside continue to move forwards until something stops them. Without a seat belt, that something is the dashboard or the windscreen.

5. The dangers of large decelerations

A large deceleration can be dangerous in two ways.

The brakes may overheat. The kinetic energy of the vehicle becomes thermal energy in the brakes. A long, hard braking effort can make the brakes too hot, so they work less well.
The vehicle may lose control. If the braking force is too large for the grip of the tyres, the wheels lock and the vehicle skids. A skidding vehicle cannot be steered.

Both effects mean that a driver should brake firmly enough to stop, and gently enough to keep control.

Try it: speed and stopping distance

The figure works out the thinking distance and the braking distance for a range of speeds. You can change the speed and the braking deceleration. The reaction time is fixed at 0.60 s.

Predict first. Do not move the sliders yet. If the speed doubles, what happens to the thinking distance? What happens to the braking distance? Write down both answers.

Test. Set the speed to 10 m/s and read the two distances. Now set the speed to 20 m/s and read them again. Then lower the braking deceleration, as if the road were wet, and watch the braking distance.

Explain. Why does doubling the speed double the thinking distance but quadruple the braking distance? Write your answer in two or three sentences. Use the equations $\text{thinking distance} = vt$ and $v^2 - u^2 = 2as$ .

Progress check

Return to the chapter problem. On a dry road the braking distance from 12 m/s is 9.0 m. On a wet road it is 14 m. The reaction time is 0.60 s, and the hazard is 20 m ahead. Work out the stopping distance on each road, and decide whether the bus stops in time on each one.

You should now be able to

work out a stopping distance from a thinking distance and a braking distance;
use $\text{thinking distance} = \text{speed} \times \text{reaction time}$ ;
describe how tiredness, alcohol, drugs and distractions affect reaction time;
describe a simple test to measure reaction time;
explain how speed, road conditions and the condition of the vehicle affect the braking distance;
explain how friction transfers the kinetic energy of a vehicle to the thermal store of the brakes;
explain why large decelerations are dangerous;
estimate the force in a large deceleration, using $v^2 - u^2 = 2as$ and $F = ma$ (Higher tier).