AQA GCSE Combined Science: Trilogy Physics Paper 2 (Higher), 2018: Question 6

13 marks · Standard Demand difficulty · Extended Answer

Calculate and explain thinking, braking and stopping distances for vehicles using graphs, energy transfers and motion equations.

Practise this question

Question

The question page contains four linked physics parts labelled 06.1 to 06.4 about vehicle motion and stopping distance. Part 06.1 shows a straight-line distance–time graph for a car travelling at 15 metres per second, with time from 0 to 1.0 seconds on the x-axis and distance from 0 to about 18 metres on the y-axis; students must determine the extra distance travelled when reaction time increases from 0.50 s to 0.82 s, for 2 marks. Part 06.2 asks why the temperature of the brakes increases when the brakes are used, using ideas about energy, for 2 marks; part 06.3 states that a lorry travels 84 m with constant acceleration 2.0 m/s squared to reach 19 m/s and asks for the initial velocity using the Physics Equations Sheet, for 3 marks. Part 06.4 shows a graph of thinking distance, braking distance and stopping distance against speed in km/h, with a key identifying a straight solid line for thinking distance and two upward-curving dashed lines for braking and stopping distance; students must describe the relationships and include factors affecting the gradient of the lines, for 6 marks.
Question text

06.1 Figure 8 shows the distance-time graph for a car travelling at 15 m/s

Figure 8

When the driver is tired, his reaction time increases from 0.50 seconds to

0.82 seconds.

Determine the extra distance the car would travel before the driver starts braking.

[2 marks]

Distance = m

06.2 When the brakes are used, the temperature of the brakes increases.

Explain why. Use ideas about energy in your explanation.

[2 marks]

A lorry travels 84 m with a constant acceleration of 2.0 m/s2 to reach a

06.3

velocity of 19 m/s

Calculate the initial velocity of the lorry.

*17* Use the Physics Equations Sheet.

[3 marks]

Initial velocity = m/s

06.4 Figure 9 shows how the thinking distance, braking distance and stopping distance for

a car vary with the speed of the car.

Figure 9

Describe the relationships shown in Figure 9

You should include factors that would affect the gradient of the lines.

[6 marks]

Mark scheme

Show the mark scheme The mark scheme is a table with rows for questions 06.1 to 06.4, giving answers, extra information, mark allocation and assessment objective references. For 06.1 it awards marks for obtaining distances from the graph or using 15 multiplied by 0.32 seconds to get an extra distance of 4.8 m; for 06.2 it credits a decrease in kinetic energy and an increase in the internal or thermal energy store of the brakes, allowing work done by friction. For 06.3 it uses the equation v squared minus u squared equals 2as, substituted as 19 squared minus u squared equals 2 times 2 times 84, leading to u equals 5 m/s. For 06.4 it gives level-based descriptors and indicative content including that thinking distance is directly proportional to speed, braking distance is proportional to speed squared, stopping distance is the sum of thinking and braking distances, and factors such as drugs, alcohol, tiredness, distractions, poor brakes, poor tyres, wet or icy roads, and mass increase the distances and the gradients.

AO /

Question Answers Extra information Mark

Spec. Ref.

06.1 an answer between 4.7 (m) and AO2

4.9 (m) scores 2 marks 6.5.4.3

either:

7.5 (m) and 12.3 (m) from the allow 7.5 (m) and between 12.2 1

graph (m) and 12.4 (m)

or

15 (m/s) × 0.32 (s) using speed

extra distance = 4.8 (m) 1

06.2 there is a decrease in kinetic allow work is done by friction (on 1 AO1

energy of the car the brakes) 6.1.1.1

6.5.4.3.4

so this (causes) the internal / 1

thermal energy store of the

brakes to increase

06.3 an answer of 5 (m/s) scores 3 AO2

marks 6.5.4.1.5

192 − u2 = 2 × 2 × 84 1

u2 = 192 − (2 × 2 × 84) 1

u = �192- (2 × 2 × 84)

u = 5 (m/s) 1

AO /

Question Answers Mark

Spec. Ref.

06.4 Level 3: Scientifically relevant facts, events or processes are 5–6 AO3

identified and given in detail to form an accurate account.

Level 2: Scientifically relevant facts, events or processes are 3-4 AO1

identified and their relevance is clear. The account is not fully AO3

accurate.

Level 1: Facts, events or processes are identified and simply 1–2 AO1

stated but their relevance is not clear.

No relevant content 0

Indicative content 6.5.4.3.1

• use of drugs, alcohol, tiredness and distractions would increase

the thinking distance

• thinking distance increases with speed

• thinking distance is directly proportional to speed

• use of drugs, alcohol, tiredness and distractions would increase

the gradient of thinking distance

• poor brakes, poor tyres, wet / icy roads and mass would

increase the braking distance

• braking distance increases with speed

• braking distance increases at an increasing (accept greater)

rate (with speed)

• poor brakes, poor tyres, wet/icy roads and mass would increase

the gradient of braking distance

• braking distance is directly proportional to speed squared

• stopping distance = thinking distance + braking distance

• factors that increase thinking and / or braking distance would

increase the gradient of stopping distance

• stopping distance increases at an increasing (accept greater)

rate (with speed)

Total 13

How to answer it

Stopping distance, braking, and motion graphs

What this question tests

You need to read values from a graph, use speed × time to find distance, explain energy transfer during braking, use the SUVAT equation to find an unknown speed, and describe patterns on a graph using scientific language.

Exam focus: 06.1–06.4 mix graph reading, calculations, and extended explanation. Marks are awarded for correct values, correct equations, clear units, and physics vocabulary.

Overall difficulty: Standard Demand

This is a typical GCSE Physics question: one graph calculation, one short explanation, one SUVAT calculation, and one longer description using data from a graph.

06.1 Extra distance before braking

Figure 8: reaction time increases from 0.50 s to 0.82 s

✅ Correct answer

Extra distance = 4.8 m

The mark scheme allows answers between 4.7 m and 4.9 m.

💡 Key knowledge

  • Distance travelled = speed × time
  • The car travels at 15 m/s
  • Extra reaction time = 0.82 s − 0.50 s = 0.32 s

🧠 Exam technique

Use the graph to read the distance at both reaction times, then subtract. You can also calculate directly using the extra time.

Top answers showed either:

  • 7.5 m and 12.3 m from the graph
  • or 15 × 0.32 = 4.8 m

❌ Common errors

  • Using the total reaction time instead of the extra time
  • Forgetting units: the answer must be in m
  • Reading values inaccurately from the graph

📐 Calculations: step by step

  1. Find the extra reaction time: 0.82 s − 0.50 s = 0.32 s
  2. Use distance = speed × time
  3. distance = 15 m/s × 0.32 s = 4.8 m

Why this gets full marks: one mark for the correct method/value from the graph or calculation, and one mark for the final correct extra distance.

06.2 Why brake temperature increases

✅ Correct answer

When the brakes are used, the car’s kinetic energy decreases. This energy is transferred by friction to the brakes, increasing their internal / thermal energy.

💡 Key knowledge

  • Brakes use friction
  • Friction does work on the brakes
  • Energy is transferred from kinetic to thermal/internal energy

🧠 Exam technique

To score both marks, link the two ideas:

1) kinetic energy decreases
2) thermal/internal energy of the brakes increases

The examiner accepted “work is done by friction” as a valid alternative link.

❌ Common errors

  • Saying only “the brakes get hot” without explaining energy transfer
  • Talking about force without mentioning kinetic energy
  • Not using the term internal energy or thermal energy

06.3 Initial velocity of the lorry

A lorry travels 84 m with constant acceleration of 2.0 m/s² to reach 19 m/s

✅ Correct answer

Initial velocity = 5 m/s

The mark scheme says an answer of 5 m/s scores full marks.

💡 Key knowledge

  • Use the SUVAT equation: v² = u² + 2as
  • v = final velocity = 19 m/s
  • a = 2.0 m/s²
  • s = 84 m
  • u = initial velocity

🧠 Exam technique

Write the equation first, then rearrange carefully. Full marks are for the correct substitution, rearrangement, and answer.

Keep your units throughout: speed in m/s, distance in m, acceleration in m/s².

❌ Common errors

  • Using the wrong SUVAT equation
  • Forgetting to square the speed
  • Putting 84 into the equation as a speed instead of a distance
  • Leaving the answer as a negative or failing to take the square root

📐 Calculations: step by step

  1. Use v² = u² + 2as
  2. Substitute: 19² = u² + 2 × 2.0 × 84
  3. Rearrange: u² = 19² − (2 × 2 × 84)
  4. Calculate: u² = 361 − 336 = 25
  5. Square root: u = 5 m/s

Mark breakdown: the mark scheme awards 1 mark for the equation/substitution, 1 mark for the rearrangement, and 1 mark for the final answer.

06.4 Describe the relationships in Figure 9

Thinking distance, braking distance and stopping distance against speed

✅ What a top-level answer says

  • Thinking distance increases directly with speed
  • Braking distance increases with speed and at an increasing rate
  • Stopping distance = thinking distance + braking distance
  • Therefore stopping distance also increases at an increasing rate

💡 Key knowledge

  • Thinking distance depends on reaction time and speed
  • Higher speed means the car travels further in the same reaction time
  • Braking distance increases more steeply because higher speed means more kinetic energy to remove
  • Bad road conditions or poor brakes make the gradient steeper

🧠 Exam technique

The question asks for relationships and also factors that affect the gradient.

Strong answers mentioned both the graph pattern and causes, for example:

  • drugs, alcohol, tiredness, distraction → increase thinking distance gradient
  • poor brakes, poor tyres, wet/icy roads, greater mass → increase braking distance gradient

Use comparative language: increases more rapidly, steeper gradient, directly proportional.

❌ Common errors

  • Only describing one line and ignoring the others
  • Saying “the graph goes up” without explaining the relationship
  • Mixing up thinking distance and braking distance
  • Not mentioning anything about gradient

📐 How to describe the graph clearly

  1. Thinking distance: straight line, so it increases in direct proportion to speed.
  2. Braking distance: curve gets steeper, so it increases by a greater amount at higher speeds.
  3. Stopping distance: also curves upwards because it is the sum of the other two distances.

❌ Examiner commentary: where marks were lost

  • Students often wrote only “higher speed means longer stopping distance” — this is true but not enough for full marks.
  • Top responses explained why the lines have different gradients.
  • To reach the highest level, students needed several relevant points written clearly and scientifically.

✅ Model 6-mark answer

Thinking distance increases as speed increases and it is directly proportional to speed, so the graph is a straight line. Braking distance also increases as speed increases, but it increases at a greater rate, so the curve becomes steeper. Stopping distance is the sum of thinking distance and braking distance, so it also increases with speed and gets steeper at higher speeds. Factors that increase the gradient include alcohol, drugs, tiredness or distraction for thinking distance, and poor brakes, poor tyres, wet or icy roads, or greater mass for braking distance.

Topics

Physics · P1: Energy · P5: Forces

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 2 (Higher), 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.