Factors affecting braking distance 1 - AQA GCSE Physics Exam Questions

Practise 26 exam-style questions on Factors affecting braking distance 1 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.

26 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

An icy road increases the braking distance of a car.

Explain why an icy road does not change the thinking distance.

Model answer

Thinking distance is the distance travelled during the driver's reaction time, and it depends only on the speed and the reaction time. During this time the brakes have not yet been applied, so the friction between the tyres and the road has no effect on it.

Explanation

Road conditions affect braking distance; driver conditions (tiredness, alcohol, distractions) affect thinking distance. Keeping the two parts separate is the key to these questions.

Mark scheme

1 mark
thinking distance depends only on the speed of the car and the driver's reaction time
allow thinking distance = speed × reaction time
1 mark
the brakes have not been applied during the thinking distance, so the friction between the tyres and the road does not matter

Question 2

2 marks

In winter, councils spread salt and grit on roads that are likely to become icy.

Explain how this reduces the braking distances of vehicles.

Model answer

Salt lowers the freezing point of water, so it melts ice or stops it forming, and grit makes the road surface rougher. Both increase the friction between the tyres and the road, so a larger braking force can act without the wheels skidding, the deceleration is greater and the braking distance is shorter.

Explanation

The braking force on a car is limited by the friction between its tyres and the road. Anything that increases that friction, such as removing ice or roughening the surface, allows a bigger deceleration.

Mark scheme

1 mark
salt melts / stops the formation of ice, and grit makes the surface rougher
allow either idea
1 mark
this increases the friction between the tyres and the road, so a bigger braking force / deceleration is possible without skidding, so the braking distance is shorter

Question 3

3 marks

Explain why a car may skid if the driver brakes very hard on an icy road, and why this is dangerous.

Model answer

On ice the maximum friction force between the tyres and the road is very small. If the driver brakes very hard, the braking force needed is bigger than this friction, so the wheels lock and the tyres slide over the ice. The driver then loses control and cannot steer, and because the sliding tyres grip even less the car takes a very long distance to stop.

Explanation

Pressing the brake harder does not always stop a car faster - the deceleration is limited by tyre grip. On ice, gentle braking started earlier is safer than sudden hard braking.

Mark scheme

1 mark
the braking force needed is greater than the maximum friction force between the tyres and the icy road
max 3
1 mark
so the wheels lock / stop turning and the tyres slide over the road
1 mark
the driver loses control / cannot steer
1 mark
sliding tyres grip less, so the deceleration is small and the braking distance is long

Question 4

4 marks

A car of mass 1200 kg is travelling at 15 m/s on an icy road. The maximum braking force that can act before the wheels skid is 1200 N.

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

Calculate the minimum braking distance of the car.

Model answer

F = m a: deceleration = 1200 ÷ 1200 = 1.0 m/s²
v² − u² = 2 a s
0² − 15² = 2 × (−1.0) × s
s = 225 ÷ 2.0 = 112.5 m

Explanation

On ice the friction limits the braking force, so the deceleration is small. Find the deceleration with F = m a, then use v² − u² = 2 a s with a final velocity of zero. Over 100 m to stop from about 34 mph shows how dangerous ice is.

Mark scheme

1 mark
deceleration = 1200 ÷ 1200 = 1.0 (m/s²)
an answer of 112.5 (m) scores 4 marks; allow 110 or 113
1 mark
0 − 15² = 2 × (−1.0) × s
allow ecf from an incorrect deceleration
1 mark
s = 225 ÷ 2.0
1 mark
s = 112.5 (m)
allow 110 or 113

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