Factors affecting braking distance 2 - AQA GCSE Physics Exam Questions

Practise 29 exam-style questions on Factors affecting braking distance 2 for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

29 questions
1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

Describe how applying the brakes slows down a moving car. Refer to work done and energy in your answer.

Model answer

When the brakes are applied, the brake pads press on the discs and the friction force between them does work. This work reduces the kinetic energy of the car, so the car slows down, and the energy is transferred to thermal energy, so the temperature of the brakes increases.

Explanation

Braking is an example of work done against friction. The work done by the friction force equals the kinetic energy removed from the car.

Mark scheme

1 mark
the friction force between the brake pads and the brake discs / wheels does work
1 mark
this reduces the kinetic energy of the car and increases the temperature of the brakes
allow kinetic energy is transferred to thermal energy in the brakes

Question 2

2 marks

A car is carrying a heavy load.

Explain why a greater braking force is needed to stop the car in a certain distance than when it is empty, if it is travelling at the same speed.

Model answer

The loaded car has a greater mass, so at the same speed it has more kinetic energy (Eₖ = ½ m v²). The brakes must do more work to remove this energy, and since work done = force × distance, stopping in the same distance needs a greater braking force.

Explanation

Kinetic energy is proportional to mass, so a heavier vehicle needs proportionally more braking force to stop in the same distance. Without it, the braking distance would be longer.

Mark scheme

1 mark
the loaded car has a greater mass, so it has more kinetic energy at the same speed
allow for the same deceleration, a greater mass needs a greater force (F = m a)
1 mark
more work must be done by the brakes, and work done = force × distance, so for the same distance the force must be greater

Question 3

3 marks

A car of mass 900 kg is travelling at 20 m/s.

Show that the braking force needed to stop the car in a distance of 25 m is about 7000 N.

Model answer

Eₖ = ½ m v² = 0.5 × 900 × 20² = 180 000 J
Work done by brakes = kinetic energy lost: F × 25 = 180 000
F = 180 000 ÷ 25 = 7200 N, which is about 7000 N

Explanation

In a 'show that' question you must reach a more precise value than the one given. Here 7200 N rounds to 'about 7000 N'.

Mark scheme

1 mark
Eₖ = 0.5 × 900 × 20² = 180 000 (J)
allow a = 20² ÷ (2 × 25) = 8.0 m/s²
1 mark
180 000 = F × 25
allow F = 900 × 8.0
1 mark
F = 7200 (N)
7200 must be seen

Question 4

4 marks

A car is travelling at 30 m/s when the driver sees a hazard. After the thinking distance, there are only 40 m left in which to stop. The maximum deceleration of the car on this road without skidding is 8.0 m/s².

Use the equation:
(final velocity)² − (initial velocity)² = 2 × acceleration × distance

Calculate the deceleration needed to stop in 40 m, and explain what is likely to happen.

Model answer

v² − u² = 2 a s
0² − 30² = 2 × a × 40
a = −900 ÷ 80 = −11.25 m/s², so a deceleration of 11 m/s² is needed.
This is greater than the maximum deceleration of 8.0 m/s² that the tyres can provide without skidding. So the car cannot stop within 40 m: if the driver brakes harder the wheels will lock and the car will skid out of control, and it will hit the hazard.

Explanation

There is a limit to how hard a car can brake, set by the grip of the tyres. If the deceleration needed is bigger than that limit, the only safe solution is to have been travelling more slowly in the first place.

Mark scheme

1 mark
0 − 30² = 2 × a × 40
allow 900 = 80 × a
1 mark
deceleration = 11 (m/s²)
allow 11.25 or −11.25
1 mark
this is greater than the maximum deceleration of 8.0 m/s², so the car cannot stop in 40 m
allow ecf from their deceleration
1 mark
if the driver brakes harder the wheels lock and the car skids / the driver loses control, and the car hits the hazard

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