AQA GCSE Combined Science: Trilogy Physics Paper 2 (Higher), 2024: Question 5

12 marks · Standard Demand difficulty · Extended Answer

Calculate and explain how speed affects stopping and braking distance, including determining initial velocity, identifying the speed-squared relationship, and calculating momentum before braking.

Practise this question

Question

The question page shows Question 5 about a safety test investigating how the speed of a car affects stopping distance. Part 05.1 gives acceleration 5.8 m/s² for 2.5 s and final velocity 20 m/s, asking for the initial velocity in a 4-mark calculation. Part 05.2 asks how measured reaction time can be used to calculate thinking distance, worth 1 mark. A graph labelled Figure 8 plots braking distance in metres on the y-axis against speed in m/s on the x-axis; the curve rises steeply and is non-linear, reaching about 20 m at 14 m/s. Part 05.3 asks which relationship matches the graph, with options braking distance proportional to 1/speed, speed, or speed squared. Part 05.4 gives braking force 6250 N, deceleration 5.0 m/s², and braking distance 14.4 m, then asks for the car’s momentum before braking in a 6-mark calculation using the equation sheet and Figure 8.
Question text

05 A safety test was carried out to determine how the speed of a car affects the

stopping distance of the car.

05.1 At the start of the test the car was moving slowly.

Then the car accelerated at 5.8 m/s2 for 2.5 s.

The final velocity of the car was 20 m/s.

Calculate the initial velocity of the car.

Use the Physics Equations Sheet.

[4 marks]

Initial velocity = m/s

05.2 The reaction time of the driver was measured.

How can the reaction time of the driver be used to calculate the thinking distance?

[1 mark]

The car was driven at a constant speed. The driver applied the maximum braking

force, and the braking distance was measured.

The test was repeated at different speeds.

Figure 8 shows the results.

Figure 8

05.3 Which of the following gives the relationship between the speed and the

braking distance?

[1 mark]

*20* Tick ( ) one box.

braking distance ∝

speed

braking distance ∝ speed

braking distance ∝ speed2

05.4 During one test the brakes were applied with a force of 6250 N.

The deceleration of the car was 5.0 m/s2.

The braking distance of the car was 14.4 m.

Determine the momentum of the car before the brakes were applied.

Use the Physics Equations Sheet.

Use Figure 8.

[6 marks]

Momentum = kg m/s

Mark scheme

Show the mark scheme The mark scheme lists answers for Question 5 in a table with columns for question number, answers, extra information, mark, and AO/specification reference. For 05.1 it awards marks for using acceleration equals change in velocity over time, calculating delta v as 14.5, and finding initial velocity as 5.5 m/s, with follow-through allowed. For 05.2 it accepts multiplying the speed of the car by the reaction time. For 05.3 it gives braking distance proportional to speed squared. For 05.4 it awards marks for using force equals mass times acceleration to find mass 1250 kg, reading speed as about 12 m/s from the graph, then using momentum equals mass times velocity to get 15000 kg m/s, with tolerance on graph reading and follow-through for earlier errors. The total for Question 5 is 12 marks.

AO /

Question Answers Extra information Mark

Spec. Ref.

05.1 Δv 1 AO2

5.8 =

2.5 6.5.4.1.5

Δv = 5.8 × 2.5 1

Δv = 14.5 1

v = (20 – 14.5) = 5.5 (m/s) allow use of an incorrectly 1

calculated Δv if the correct

equation has been used

AO /

Spec. Ref.

05.2 multiply the speed of the car by 1 AO3

the reaction time 6.5.4.3.2

AO /

Spec. Ref.

05.3 braking distance ∝ speed2 1 AO3

– – – 6.5.4.3.4

AO /

Spec. Ref.

05.4 6250 = m × 5.0 1 AO2

6.5.4.2.2

6250 1 6.5.5.1

m =

5.0

m = 1250 (kg) 1

v = 12 (m/s) 1

p = 1250 × 12 allow a substitution using a 1

value of v in the range 11.8 to

12.2

allow a correct substitution using

their incorrectly calculated value

for m using the correct equation

p = 15 000 (kgm/s) allow a calculation using a value 1

16 of v in the range 11.8 to 12.2

allow a correct calculation using

their incorrectly calculated value

for m using the correct equation

– – –

Total Question 5 12

Question 6

How to answer it

Stopping distance, reaction time and momentum

What this question tests

This question checks your ability to use the SUVAT-style equation for acceleration, interpret a graph, identify a proportional relationship, and use F = ma and p = mv . It also tests whether you can show working clearly and link a driver’s reaction time to thinking distance.

Part (a) — Calculate the initial velocity

Question 05.1

💡 Key knowledge

  • Use v = u + at .
  • v = final velocity, u = initial velocity.
  • a = acceleration, t = time.
  • Rearrange to find u : u = v - at .

📐 Calculations

  1. Write the equation: v = u + at
  2. Substitute the values: 20 = u + (5.8 × 2.5)
  3. Calculate the change in velocity: 5.8 × 2.5 = 14.5
  4. Rearrange: u = 20 - 14.5
  5. Answer: u = 5.5 m/s

✅ Correct answer

Initial velocity = 5.5 m/s

Marking note: full marks were awarded for correct equation, substitution, calculation, and final answer with units.

🧠 Exam technique

  • Always use the equation sheet if one is provided.
  • Show the substitution first — this can earn method marks even if a later arithmetic slip happens.
  • Include units in your final answer.

❌ Common errors

  • Using a = v/t instead of v = u + at .
  • Forgetting to rearrange for the initial velocity.
  • Writing 14.5 m/s² instead of 14.5 m/s for the change in velocity.
  • Leaving the answer as 5.5 with no unit.

Part (b) — Use reaction time to find thinking distance

Question 05.2

💡 Key knowledge

Thinking distance is how far the car travels during the driver’s reaction time, before the brakes are applied.

The relationship is:

thinking distance = speed × reaction time

✅ Correct answer

Multiply the speed of the car by the reaction time.

This is the exact idea needed for the 1 mark.

🧠 Exam technique

  • Use the words speed and reaction time.
  • If you know the actual speed and reaction time, you can calculate thinking distance directly.

❌ Common errors

  • Confusing thinking distance with braking distance.
  • Using braking force or deceleration instead of speed.
  • Writing a vague statement like “use it in a formula” without naming the formula idea.

Part (c) — Relationship between speed and braking distance

Question 05.3

💡 Key knowledge

The graph curves upwards, so braking distance increases more quickly as speed increases.

For a car braking at a fixed force, braking distance is proportional to the square of speed.

✅ Correct answer

braking distance ∝ speed²

🧠 Exam technique

  • Read the pattern from the curve: it is not a straight line.
  • The correct proportional statement must include the square.

❌ Common errors

  • Saying braking distance ∝ speed because the graph rises with speed.
  • Saying braking distance ∝ 1/speed which would decrease as speed increases.

Part (d) — Determine the momentum before braking

Question 05.4

💡 Key knowledge

  • Use F = ma to find the mass.
  • Then use p = mv for momentum.
  • Use the graph to find the speed for a braking distance of 14.4 m .
  • Momentum unit is kg m/s .

📐 Calculations

  1. Find the mass using F = ma
    6250 = m × 5.0
  2. Rearrange:
    m = 6250 / 5.0 = 1250 kg
  3. Use Figure 8
    A braking distance of 14.4 m corresponds to a speed of about 12 m/s .
  4. Find momentum using p = mv
    p = 1250 × 12
  5. Answer:
    p = 15 000 kg m/s

✅ Correct answer

Momentum = 15 000 kg m/s

The mark scheme allowed a speed in the range about 11.8 to 12.2 m/s from the graph.

🧠 Exam technique

  • Use the graph carefully: read the speed from the braking distance, not the other way round.
  • Show each rearrangement step clearly.
  • If your graph reading is slightly off but sensible, you can still gain method marks if the rest is correct.
  • Always include units in both mass and momentum.

❌ Common errors

  • Using the braking distance value as if it were speed.
  • Forgetting to convert the force and acceleration into mass first.
  • Writing momentum with the wrong unit, such as N or kg/s .
  • Rounding too early. Keep 12 m/s until the final step.

🧠 How the marks were awarded

  • 1 mark for using F = ma correctly.
  • 1 mark for rearranging to find mass.
  • 1 mark for the correct mass value.
  • 1 mark for reading the graph speed correctly.
  • 1 mark for using p = mv .
  • 1 mark for the correct final momentum with units.

Quick examiner-style recap

💡 Key facts to remember

  • v = u + at
  • thinking distance = speed × reaction time
  • braking distance ∝ speed²
  • F = ma
  • p = mv

❌ Biggest traps

  • Mixing up braking distance and thinking distance.
  • Not using the square relationship in part (c).
  • Dropping units in calculation questions.
  • Not showing the rearrangement step, which can lose method marks.

✅ Full-mark answers summary

  • (a) 5.5 m/s
  • (b) Multiply the speed by the reaction time
  • (c) braking distance ∝ speed²
  • (d) 15 000 kg m/s

Topics

Physics · P5: Forces

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 2 (Higher), 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.