AQA GCSE Combined Science: Trilogy Physics Paper 2 (Higher), 2022: Question 6

9 marks · Standard Demand difficulty · Extended Answer

Use distance–time and velocity–time graphs to determine reaction time and braking distance, and explain why the braking force was not constant.

Practise this question

Question

The page shows Question 06 with three parts. In 06.1, Figure 9 is a graph of thinking distance in metres against speed of car in metres per second; the line is straight and passes through the origin, with an instruction to determine the driver’s reaction time using the Physics Equations Sheet. In 06.2, Figure 10 is a graph of velocity in metres per second against time in seconds for a car braking from 30 m/s down to 0 m/s by about 5 seconds, and the question asks for the braking distance. In 06.3, there is a written question asking the student to explain how the gradient of the line on Figure 10 shows that the resultant force on the car was not constant.
Question text

06 The distance a car travels during the driver’s reaction time is called the

thinking distance.

06.1 Figure 9 shows how thinking distance depends on speed for a car.

Figure 9

Determine the driver’s reaction time.

Use the Physics Equations Sheet.

[3 marks]

20 Reaction time = s

06.2 Figure 10 shows how the velocity of a car changes during braking.

Figure 10

Determine the braking distance of the car.

[3 marks]

Braking distance = m

06.3 Explain how the gradient of the line on Figure 10 shows that the resultant force on the

car was not constant.

[3 marks]

Mark scheme

Show the mark scheme The mark scheme gives 06.1 as a three-mark calculation: use a pair of values from the graph such as 21 = 30 × t, then rearrange to t = 21/30, giving reaction time 0.7 s, with 0.70 s allowed. For 06.2, it awards marks for using the graph scale to see one square represents 5 m, counting about 9 squares, and calculating braking distance as 9 × 5 = 45 m, with a range of acceptable answers around 42.5 m to 50 m. For 06.3, the mark scheme states that the gradient is equal to acceleration, the gradient/acceleration is not constant, and therefore the resultant force is not constant because force is directly proportional to acceleration for constant mass, with the note that F = ma is allowed.

Question 6

AO /

Question Answers Extra information Mark

Spec. Ref.

06.1 21 = 30 × t allow use of any correct pair of 1 AO2

values from the graph 6.5.4.1.2

21 1

t =

t = 0.7 (s) allow 0.70 (s) 1

AO /

Spec. Ref.

06.2 one square represents 5 (m) 1 AO2

6.5.4.1.2

number of squares = 9 allow number of squares in the 1

range 8.5 to 10

braking distance = 9 × 5 = 45 allow an answer in the range 1

(m) 42.5 (m) to 50 (m)

AO /

Spec. Ref.

06.3 gradient is equal to acceleration 1 AO1

gradient / acceleration is not 1 AO3

constant

so resultant force is not constant allow resultant force is not 1 AO3

because resultant force is constant because F = ma

directly proportional to 6.5.4.2.2

acceleration (for constant mass)

Total Question 6 9

How to answer it

Standard Demand

Stopping Distance Graph Skills

What this question tests

Reading values from graphs, using the equation distance = speed × time, finding distance from the area under a velocity–time graph, and linking gradient to acceleration and then to force using F = ma.

Question 06.1

Part (a): Determine the driver’s reaction time

Use Figure 9 and the Physics Equations Sheet

✅ Correct answer

Reaction time = 0.7 s

3 marks total:
• 1 mark for using a correct pair of values from the graph
• 1 mark for rearranging / substituting correctly
• 1 mark for the final answer 0.7 s

💡 Key knowledge

  • Thinking distance = speed × reaction time
  • So reaction time = thinking distance ÷ speed
  • You can use any correct pair of values from the straight-line graph

📐 Calculation

  1. Choose a point from the graph, for example 30 m/s and 21 m .
  2. Use the equation:
    thinking distance = speed × time
  3. Substitute the values:
    21 = 30 × t
  4. Rearrange:
    t = 21 ÷ 30
  5. Answer:
    t = 0.7 s

🧠 Exam technique

  • Write the equation first so the examiner can award method marks.
  • Read graph values carefully from the line, not just from the axes labels.
  • Include the unit s at the end.
  • Because the graph is a straight line through the origin, different correct pairs should give the same answer.

❌ Common errors

  • Using the equation the wrong way round, for example doing speed ÷ distance.
  • Forgetting that the question asks for reaction time, not thinking distance.
  • Reading a value that is not actually on the line.
  • Missing the unit, which can make the answer look incomplete.

Examiner insight: this is a straightforward graph-and-equation question. Students usually lost marks by not showing the substitution clearly or by reading the graph inaccurately. Full-mark answers showed one clear pair of values and correct rearrangement.

Question 06.2

Part (b): Determine the braking distance of the car

Use Figure 10

✅ Correct answer

Braking distance = 45 m

The mark scheme allows answers in the range 42.5 m to 50 m.

3 marks total:
• 1 mark for recognising one square = 5 m
• 1 mark for counting about 9 squares under the graph
• 1 mark for calculating about 45 m

💡 Key knowledge

  • On a velocity–time graph, distance travelled = area under the graph.
  • This is not found by using the gradient.
  • You estimate the area by counting squares under the curve.

📐 Calculation

  1. Remember the rule:
    distance = area under a velocity–time graph
  2. Work out what one large square represents.
    • Time axis: 1 s per square
    • Velocity axis: 5 m/s per square
  3. So area of one square =
    1 s × 5 m/s = 5 m
  4. Estimate the number of squares under the curve:
    about 9 squares
  5. Multiply:
    braking distance = 9 × 5 = 45 m

🧠 Exam technique

  • State clearly that you are finding the area under the graph.
  • Show what one square is worth before counting.
  • Because the line is curved, an estimate is expected.
  • If your square count is sensible, you can still gain marks even if it is not exactly 9.

❌ Common errors

  • Using the gradient instead of the area.
  • Thinking each square is 1 m instead of 5 m.
  • Counting squares only under part of the curve.
  • Giving no working, so if the estimate is slightly off there is no method to credit.

Examiner insight: the main discriminator here was whether students knew that the area under a velocity–time graph gives distance. Strong answers explicitly said this, then counted squares systematically. Many weaker answers used the slope, which did not answer the question.

Question 06.3

Part (c): Explain how the gradient shows the resultant force was not constant

Use Figure 10 and Newton’s second law

✅ Correct answer

A full 3-mark answer could be:

The gradient of a velocity–time graph is the acceleration. The gradient is changing, so the acceleration is not constant. Therefore the resultant force is not constant because F = ma and the mass of the car is constant.

3 marks total:
• 1 mark: gradient = acceleration
• 1 mark: acceleration is not constant
• 1 mark: therefore force is not constant because F is proportional to a for constant mass

💡 Key knowledge

  • Gradient of a velocity–time graph = acceleration
  • A changing gradient means changing acceleration
  • For constant mass, F = ma , so force changes when acceleration changes

🧠 Exam technique

  • This is a chain-of-reasoning question: gradient → acceleration → force.
  • Use the word constant because it appears in the question and mark scheme.
  • Mention that the mass is constant to justify using F = ma properly.
  • Do not just say “the line is curved” on its own; explain what that means physically.

❌ Common errors

  • Saying the gradient is speed instead of acceleration.
  • Not linking the changing gradient to changing acceleration.
  • Forgetting the final step to resultant force.
  • Writing that the force is zero because the car stops at 5 s. The question is about whether the force was constant during braking.

Examiner insight: top-level answers made the full link in order. Many students only gained 1 or 2 marks by saying the gradient changes but not stating that gradient means acceleration, or by missing the final use of F = ma .

Quick full-mark summary

✅ Final answers

  • 06.1: reaction time = 0.7 s
  • 06.2: braking distance = 45 m
  • 06.3: gradient = acceleration; gradient changes, so acceleration changes; therefore resultant force is not constant because F = ma for constant mass

🧠 Best exam approach

  • Read graphs carefully and choose clear values.
  • Show every step of your working for calculation marks.
  • For velocity–time graphs: area = distance, gradient = acceleration.
  • In explanation questions, make the physics link explicit.

Topics

Physics · P5: Forces

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 2 (Higher), 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.