AQA GCSE Combined Science: Trilogy Physics Paper 2 (Higher), June 2025: Question 6
11 marks · High Demand difficulty · Extended Answer
Analyze motion in a triathlon using a speed-time graph to determine forward force from work done, describe velocity in circular motion, and interpret changing deceleration.
Practise this questionQuestion
Question text
06 A triathlon race consists of swimming, cycling and running.
06.1 One athlete in a triathlon swims at a speed of 1.5 m/s.
How does the athlete’s swimming speed compare with the typical speed of a
person walking?
[1 mark]
Tick ( ) one box.
athlete’s swimming speed > typical walking speed
athlete’s swimming speed = typical walking speed
athlete’s swimming speed < typical walking speed
Figure 8 shows a speed–time graph for the athlete for the whole triathlon.
Figure 8
06.2 The cycling part of the race was on a straight road.
The work done by the forward force on the bicycle was 110 000 J.
Determine the mean forward force on the bicycle.
Use the Physics Equations Sheet.
[6 marks]
Mean forward force = N
Figure 8 is repeated below.
Figure 8
06.3 The first part of the run was on a circular track.
The time taken for the athlete to run round the track was 100 s.
Describe the velocity of the athlete during the first 100 s of the run.
Use Figure 8.
[2 marks]
06.4 After running round the track once, the rest of the run was on a
straight, horizontal road.
Explain how Figure 8 shows that the magnitude of the deceleration of the
athlete increased.
[2 marks]
Mark scheme
Show the mark scheme
Question 6
AO /
Question Answers Extra information Mark
Spec. Ref.
06.1 athlete’s swimming speed = 1 AO1
typical walking speed 6.5.4.1.2
AO /
Spec. Ref.
AO3
06.2 time taken = 1400 subsequent marks may be 1 6.5.2
awarded if time taken is 6.5.4.1.5
determined from the graph
7.5 + 6.2 allow mean speed calculated 1
mean speed = using a maximum speed
between 7.4 m/s and 7.6 m/s
(= 6.85 m/s)
distance (= 6.85 × 1400) allow a distance consistent with 1
= 9590 (m) their calculated mean speed
subsequent marks may only be
awarded if values of speed and
time from the graph have been
used to determine a distance
subsequent marks may be
awarded using their value of
total distance
110 000 = F × 9590 1
110 000
F = 1
9590
F = 11(.470…) (N) – TRILOGY – 8464/P1/2H –
06.2 OR
cont.
time taken = 1400 (1) subsequent marks may be
awarded if time taken is
determined from the graph
area of triangle (1) allow area calculated using a
= 0.5 × 1400 × (7.5 - 6.2) maximum speed between 7.4
(= 910) m/s and 7.6 m/s
and 17
area of rectangle
= 1400 × 6.2 (= 8680)
distance (1) allow a distance consistent with
(= 910 + 8680) = 9590 (m) their calculated areas
subsequent marks may only be
awarded if values of speed and
time from the graph have been
used to determine a distance
subsequent marks may be
awarded using their value of
total distance
110 000 = F × 9590 (1)
110 000 (1)
F =
9590
F = 11(.470…) (N) – (1) – 8464/P/2H –
06.2 OR
cont.
time taken = 1400 (1) subsequent marks may be
awarded if time taken is
determined from the graph
6.2 - 7.5 (1) allow acceleration calculated
acceleration= using a maximum speed
1400
between 7.4 m/s and 7.6 m/s
(= (-)0.00092857…)
subsequent marks may be
awarded if their calculated value
for acceleration is rounded
distance (1) allow a distance consistent with
6.22 − 7.52 their calculated acceleration
(= )
(−)2 × 0.00092857 …
18 = 9590 (m)
subsequent marks may only be
awarded if values of speed and
time from the graph have been
used to determine a distance
subsequent marks may be
awarded using their value of
total distance
110 000 = F × 9590 (1)
110 000 (1)
F =
9590
F = 11(.470…) (N) – (1) TRILOGY – 8464/P/2H –
AO /
Spec. Ref.
06.3 constant magnitude (of velocity) allow speed was constant 1 AO2
allow speed was 4 m/s
allow (magnitude of) velocity
was 4 m/s
do not accept constant velocity
direction (of velocity) kept 1 AO1
changing
if no other mark awarded allow 1 6.5.4.1.3
mark for the velocity is
constantly changing 19
AO /
Spec. Ref.
06.4 the gradient is equal to the 1 AO3
deceleration / acceleration 6.5.4.1.5
(and) the (magnitude of the) 1
gradient increased
Total Question 6 11
How to answer it
Triathlon Physics: Speed-Time Graphs, Work & Vectors
What this question tests
- Typical speeds: Recalling standard approximate speeds for everyday human activities (walking, running, cycling).
- Graph analysis: Reading values, calculating time intervals, and determining distance travelled from the area under a speed-time graph.
- Forces & Work: Applying the work done equation ( W = F × s ) to calculate mean forward force.
- Scalar vs Vector: Explaining why an object moving in a circle has a changing velocity despite constant speed.
- Non-linear motion: Linking the changing gradient of a curved speed-time graph to increasing deceleration.
Comparing Swimming Speed to Typical Walking Speed
Recall of typical everyday speeds (AO1)
✅ Correct Answer
Tick the second box:
athlete's swimming speed = typical walking speed
💡 Key Knowledge
AQA specification requires recall of these typical values:
- Walking: ~1.5 m/s
- Running: ~3.0 m/s
- Cycling: ~6.0 m/s
Since the swimmer moves at 1.5 m/s , it exactly matches typical walking speed.
Determining Mean Forward Force from the Cycling Section
Multi-step calculation: Area under graph & Work Done formula (AO3 / Spec 6.5.2 & 6.5.4.1.5)
🔧 Step-by-Step Calculation
- Find the time taken for cycling:
Start time = 500 s , End time = 1900 s
Time = 1900 − 500 = 1400 s - Read initial and final cycling speeds:
Initial speed at 500 s = 7.5 m/s (allow 7.4 – 7.6 m/s)
Final speed at 1900 s = 6.2 m/s - Calculate the distance travelled (Area under trapezoid):
Method A (Mean speed):
Mean speed = (7.5 + 6.2) ÷ 2 = 6.85 m/s
Distance (s) = Mean speed × Time = 6.85 × 1400 = 9590 m
Method B (Splitting into rectangle + triangle):
Rectangle area = 1400 × 6.2 = 8680 m
Triangle area = 0.5 × 1400 × (7.5 − 6.2) = 0.5 × 1400 × 1.3 = 910 m
Total distance = 8680 + 910 = 9590 m - State the equation linking Work Done, Force, and Distance:
Work Done = Force × distance ( W = F × s ) - Rearrange to solve for Forward Force ( F ):
110 000 = F × 9590
F = 110 000 ÷ 9590 = 11.470... N - Final Answer:
11 N or 11.5 N
🧠 Exam Technique & Mark Breakdown
- Mark 1: Determining time interval = 1400 s.
- Mark 2: Calculating mean speed (6.85 m/s) OR splitting area correctly.
- Mark 3: Determining total distance = 9590 m.
- Mark 4: Substituting values into W = F × s .
- Mark 5: Correct rearrangement: F = 110 000 ÷ 9590 .
- Mark 6: Correct final answer with appropriate rounding (11 N or 11.5 N).
❌ Common Traps & Lost Marks
- Using total time (1900 s): Students frequently forgot to subtract 500 s and used 1900 s directly.
- Treating the shape as a simple rectangle: Using only 7.5 × 1400 or 6.2 × 1400 completely overlooks the sloping line.
- Misreading the vertical scale: Each small grid square vertically represents 0.1 m/s (10 small squares = 1 m/s). Final speed is 2 small squares above 6.0 = 6.2 m/s.
Velocity on a Circular Track
Scalar speed vs Vector velocity (AO1, AO2 / Spec 6.5.4.1.3)
✅ Model Answer (2 Marks)
- Point 1: The speed was constant (at 4 m/s) / the magnitude of velocity was constant. [1 mark]
- Point 2: The direction was constantly changing, so the velocity was constantly changing. [1 mark]
💡 Key Scientific Concept
Velocity is a vector quantity (it has both magnitude and direction). Speed is a scalar quantity (magnitude only).
Even if an object moves at a constant speed along a circular track, its direction of travel is continuously changing at every single instant. Therefore, its velocity is constantly changing.
❌ Examiner Warning
Do NOT write "velocity was constant". This contradicts physics principles and will immediately lose the mark!
Explaining Increasing Deceleration from a Curved Graph
Interpreting gradients on speed-time graphs (AO3 / Spec 6.5.4.1.5)
✅ Model Answer (2 Marks)
- The gradient of the speed–time graph represents the acceleration / deceleration. [1 mark]
- The gradient becomes steeper (magnitude of the gradient increases) as time goes on. [1 mark]
🧠 Two-Part Explanation Strategy
For questions asking you to "Explain how Figure X shows...":
- State the feature link: State what the graphical feature physically represents (e.g., gradient = acceleration / rate of change of speed).
- Describe the visual change: State what is happening visually to that feature on the graph (e.g., the curve gets steeper / gradient increases).
Topics
Physics · P5: Forces
Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Physics Paper 2 (Higher), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.