3.3 Safety in public transport — Practise

Work through these on your own. Keep your working neat: the method matters as much as the answer.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A car travels at 18 m/s. The driver has a reaction time of 0.50 s. Calculate the thinking distance.
A car travels at 22 m/s. The reaction time is 0.50 s and the braking distance is 40 m. Calculate the thinking distance and the stopping distance.
A van travels at 16 m/s and stops in a distance of 32 m. Calculate the deceleration.
H A car of mass 1200 kg travels at 20 m/s and stops in a distance of 25 m. Calculate the deceleration and the braking force.
H A van of mass 1300 kg travels at 20 m/s and stops in a distance of 40 m. Calculate the braking force.
Give two factors that can increase the braking distance of a car, and explain how one of them has its effect.
State two factors that can increase the reaction time of a driver, and say what a longer reaction time does to the stopping distance.
Describe how to measure a reaction time with a ruler.
H A car of mass 1500 kg travels at 24 m/s and stops in a distance of 36 m. Calculate the deceleration and the braking force.
A van travels at 16 m/s. The reaction time is 0.50 s and the braking distance is 24 m. Calculate the stopping distance.
A graph of stopping distance against speed curves upwards. Explain why the thinking distance is a straight line on this graph, and why the braking distance curves.
Explain why the brakes of a vehicle become hot when the vehicle stops.

Answers and concise solutions

1. Thinking distance

Answer: $s = vt = 18 \times 0.50 = 9.0\,\text{m}$ . The car keeps moving at 18 m/s while the driver reacts.

2. Thinking distance and stopping distance

Answer: Thinking distance $= 22 \times 0.50 = 11\,\text{m}$ . Stopping distance $= 11 + 40 = 51\,\text{m}$ .

3. Deceleration of the van

Answer: $v^2 - u^2 = 2as$ with $v = 0$ , so $a = \dfrac{u^2}{2s} = \dfrac{16^2}{2 \times 32} = 4.0\,\text{m/s}^2$ . The deceleration is $4.0\,\text{m/s}^2$ .

4. Deceleration and braking force

Answer: The deceleration is $\dfrac{20^2}{2 \times 25} = 8.0\,\text{m/s}^2$ . The braking force is $F = ma = 1200 \times 8.0 = 9600\,\text{N} = 9.6 \times 10^3\,\text{N}$ .

5. Braking force on the van

Answer: The deceleration is $\dfrac{20^2}{2 \times 40} = 5.0\,\text{m/s}^2$ , so $F = ma = 1300 \times 5.0 = 6500\,\text{N} = 6.5 \times 10^3\,\text{N}$ .

6. Two factors that increase braking distance

A wet or icy road increases the braking distance, because there is less grip between the tyres and the road.
Worn brakes or worn tyres increase the braking distance, because the vehicle can transmit a smaller braking force.
A smaller braking force gives a smaller deceleration, so the vehicle takes longer to stop and travels further while it stops.

7. Two factors that increase reaction time

Tiredness increases the reaction time.
Alcohol or drugs increase the reaction time, and so do distractions.
A longer reaction time means a longer thinking distance, so the stopping distance is longer.

8. Measuring a reaction time with a ruler

A partner holds a ruler vertically above the open hand and lets it fall without warning.
Catch the ruler as quickly as possible and measure how far it fell. A longer distance means a slower reaction.
Repeat the test and take an average, so the result is more reliable.

9. Deceleration and braking force

Answer: The deceleration is $\dfrac{24^2}{2 \times 36} = 8.0\,\text{m/s}^2$ . The braking force is $F = ma = 1500 \times 8.0 = 12000\,\text{N} = 1.2 \times 10^4\,\text{N}$ . A deceleration of $8.0\,\text{m/s}^2$ is large and dangerous.

10. Stopping distance of the van

Answer: Thinking distance $= 16 \times 0.50 = 8.0\,\text{m}$ . Stopping distance $= 8.0 + 24 = 32\,\text{m}$ .

11. Why the graph curves upwards

Thinking distance is speed multiplied by the reaction time, so it is proportional to speed; its graph is a straight line through the origin.
Braking distance depends on the square of the speed, so doubling the speed makes it four times as large. Its graph curves upwards.

12. Why the brakes become hot

The moving vehicle has energy in its kinetic store.
The friction force between the brakes and the wheels does work on the brakes as the vehicle slows.
This work transfers energy to the thermal store of the brakes, so their temperature rises.