AQA GCSE Combined Science: Trilogy Physics Paper 2 (Foundation), June 2025: Question 3

11 marks · Low Demand difficulty · Short Answer

Analyze a three-day cycling journey by determining distance from a scale map, calculating weight and distance from work done, interpreting motion graphs and bar charts, and explaining scalars and displacement.

Practise this question

Question

Question 3 presents several parts regarding a cycling journey: 03.1 shows a mobile phone screen displaying a map with a grid where a scale bar indicates 1 cm : 20 m, asking for distance between start and junction; 03.2 asks to calculate weight using weight = mass × gravitational field strength with mass 15 kg and g = 9.8 N/kg; 03.3 shows a straight upward-sloping velocity-time graph and asks how it shows constant acceleration; 03.4 provides forward force 250 N and work done 4500 J to find distance using distance = work done / force; 03.5 shows a winding route from start to end and asks to compare distance with displacement; 03.6 is a multiple-choice question on why distance is a scalar quantity; 03.7 displays a bar chart of distance travelled on Monday, Tuesday, and Wednesday, asking to calculate the total distance.
Question text

03 A student went on a cycling journey that lasted for three days.

The student recorded information about the journey using a cycling app on a

mobile phone.

03.1 At the start of the journey, the student travelled along a straight road to a

road junction.

A map on the cycling app showed the road.

Figure 5 shows the map.

The scale of the map is 1 cm : 20 m

Figure 5

Calculate the distance travelled by the student between the start and the

road junction.

Use Figure 5.

[2 marks]

Distance travelled = m

03.2 The bicycle had a mass of 15 kg.

gravitational field strength = 9.8 N/kg

Calculate the weight of the bicycle.

Use the equation:

weight = mass × gravitational field strength

[2 marks]

12 Weight = N

03.3 Figure 6 shows a velocity–time graph for a downhill part of the journey.

Figure 6

The gradient of the graph represents acceleration.

How does Figure 6 show that the acceleration was constant?

[1 mark]

03.4 As the student accelerated downhill, the forward force on the student was 250 N.

The work done by the forward force was 4500 J.

Calculate the distance travelled as the student accelerated downhill.

Use the equation:

work done

distance travelled =

force

[2 marks]

Distance travelled =13 m

03.5 Most of the roads on the journey were not straight.

Figure 7 shows a map of the route taken for the whole journey.

Figure 7

The line in Figure 7 represents the total distance travelled when the student reached

the end of the journey.

How does the total distance travelled compare with the magnitude of the

displacement of the student from the start?

[1 mark]

Tick ( ) one box.

Total distance travelled is less than the displacement.

Total distance travelled is equal to the displacement.

Total distance travelled is greater than the displacement.14

03.6 Why is distance a scalar quantity?

[1 mark]

Tick ( ) one box.

*13*Distance has direction only.

Distance has magnitude only.

Distance has direction and magnitude.15

03.7 Figure 8 shows the distance travelled by the student each day.

Figure 8

Calculate the total distance travelled by the student.

Use Figure 8.

[2 marks]

Total distance travelled = km

Mark scheme

Show the mark scheme Mark scheme for Question 3: 03.1 awards 1 mark for 3.0 × 20 and 1 mark for 60 m (allow length 2.9 to 3.1 cm); 03.2 awards 1 mark for 15 × 9.8 and 1 mark for 147 N (allow 150 N); 03.3 awards 1 mark for line has a constant gradient or the line is straight; 03.4 awards 1 mark for 4500 / 250 and 1 mark for 18 m; 03.5 awards 1 mark for total distance travelled is greater than the displacement; 03.6 awards 1 mark for distance has magnitude only; 03.7 awards 2 marks for 158 km (1 mark for reading 56, 40, and 62 from graph). Total marks: 11.

Question 3

AO /

Question Answers Extra information Mark

Spec. Ref.

03.1 distance travelled = 3.0 × 20 both marks may be awarded if a 1 AO3

length in the range 2.9 to 3.1 cm 6.5.4.1.1

is used

distance travelled = 60 (m) 1

AO /

Spec. Ref.

03.2 weight = 15 × 9.8 1 AO2

6.5.1.3

weight = 147 (N) allow weight = 150 (N) 1

AO /

Spec. Ref.

03.3 the line has a constant gradient allow the line is straight 1 AO2

6.5.4.1.5

do not accept directly

proportional

AO /

Spec. Ref.

03.4 4500 1 AO2

distance travelled = 6.5.2

distance travelled = 18 (m) 1

AO /

Spec. Ref.

03.5 total distance travelled is greater 1 AO1

than the displacement 6.5.4.1.1

AO /

Spec. Ref.

03.6 distance has magnitude only 1 AO1

6.5.1.1

– – 8464/P/2F – 6.5.4.1.1

AO /

Spec. Ref.

03.7 total distance = 158 (km) 2 AO3

6.5.4.1.4

allow 1 mark for 56, 40 and 62

read from graph

allow 1 mark for a total distance

calculated using two correct

values and one incorrect value

from the bar chart

Total Question 3 11

How to answer it

Cycling Journey: Motion, Forces & Graphs

What this question tests

  • Map scales: Converting measured diagram lengths into real-world distances.
  • Calculations using given formulae: Weight ( W = m × g ) and distance from work done ( s = W / F ).
  • Graph interpretation: Identifying constant acceleration from a velocity–time graph and accurately reading discrete values from a bar chart.
  • Vectors vs Scalars: Defining scalar quantities and comparing scalar distance with vector displacement.
Question 03.1 • 2 Marks

Distance from Scale Map

Calculating real-world distance between the start and the road junction

📐 Step-by-Step Calculation

  1. 1 Measure: On the printed exam paper, measure the straight road from 'Start' to 'Road junction' with a ruler = 3.0 cm (range: 2.9 to 3.1 cm).
  2. 2 Scale conversion: Multiply by 20 m (since 1 cm = 20 m):
    3.0 × 20 = 60 m
Mark scheme: 1 mark for correct measurement & scale multiplication; 1 mark for final distance of 60 m.

❌ Common Errors

  • Guessing the length: Trying to estimate by counting screen grid squares instead of measuring with a ruler.
  • Dividing instead of multiplying: Calculating 20 / 3.0 instead of scaling up.
Question 03.2 • 2 Marks

Calculating Weight

Using mass and gravitational field strength

📐 Step-by-Step Calculation

  1. 1 Identify values: Mass = 15 kg, g = 9.8 N/kg.
  2. 2 Substitute:
    Weight = 15 × 9.8
  3. 3 Final answer: 147 N (allow 150 N if using 10 N/kg).
Mark scheme: 1 mark for substitution ( 15 × 9.8 ); 1 mark for 147 N.

💡 Key Knowledge

  • Weight is the force acting on an object due to gravity (unit: Newtons, N).
  • Mass is the amount of matter in an object (unit: kilograms, kg).
  • Always check units: mass must be in kg to yield weight in N.
Question 03.3 • 1 Mark

Velocity–Time Graph Gradient

Explaining how the graph indicates constant acceleration

✅ Correct Answers (Choose one)

  • "The line has a constant gradient"
  • "The line is straight"
Mark scheme: 1 mark for stating constant gradient or straight line.

❌ Examiner Warning: Direct Proportion Trap

Do NOT write "directly proportional".

For two variables to be directly proportional, the graph must be a straight line passing through the origin (0,0). In Figure 6, the cyclist has an initial velocity greater than zero, so it is not directly proportional.

Question 03.4 • 2 Marks

Work Done & Distance

Rearranging the work done formula

📐 Step-by-Step Calculation

  1. 1 Identify values: Work done = 4500 J, Forward force = 250 N.
  2. 2 Substitute into formula:
    distance = 4500 / 250
  3. 3 Calculate: 18 m
Mark scheme: 1 mark for substitution; 1 mark for correct answer of 18 m.

🧠 Exam Technique

When the equation is provided in the question stem, always write out the substitution stage clearly. Even if you press the wrong button on your calculator, you will still secure the first method mark.

Questions 03.5 & 03.6 • 2 Marks

Distance vs Displacement & Scalars vs Vectors

Understanding the physical nature of motion quantities

✅ Correct Choices

  • 03.5: [✓] Total distance travelled is greater than the displacement.
  • 03.6: [✓] Distance has magnitude only.

💡 Distance vs Displacement

  • Distance (Scalar): The entire length of the path taken. It only has size (magnitude) and no direction.
  • Displacement (Vector): The shortest straight-line distance from start to finish, including a direction.
  • Because the route twists and turns, the path length (distance) must be greater than the straight line connecting start to finish.
Question 03.7 • 2 Marks

Reading Bar Charts

Total distance across three days

📐 Step-by-Step Calculation

  1. 1 Determine grid scale: Each major division = 10 km (5 small squares), so each small grid square = 2 km.
  2. 2 Read daily bars accurately:
    • Monday: 3 squares above 50 = 56 km
    • Tuesday: Exactly on 40 km
    • Wednesday: 1 square above 60 = 62 km
  3. 3 Sum values:
    56 + 40 + 62 = 158 km
Mark scheme: 1 mark for reading 56, 40, and 62 (or carrying out a correct sum with 2 correct readings); 1 mark for final answer 158 km.

❌ Common Reading Traps

  • Misreading small squares: Assuming each small square is 1 km rather than 2 km (e.g. reading Monday as 53 instead of 56).
  • Always check the difference between major gridlines before reading values off a graph!

Topics

Physics · P5: Forces

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 2 (Foundation), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.