AQA GCSE Physics Physics Paper 2 (Higher), June 2024: Question 7
18 marks · Standard Demand difficulty · Short Answer
Analyze train motion using a velocity-time graph, calculate distance, determine deceleration, and explain stopping distances including the effect of alcohol.
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Question text
07 Figure 12 shows a velocity–time graph for a train travelling between two stations.
Figure 12
07.1 Determine the distance travelled by the train in the first 600 s of the journey.
[3 marks]
27 Distance = m
07.2 Explain what happens to the braking force as the train decelerates.
Use information from Figure 12.
[3 marks]
07.3 Determine the maximum deceleration of the train.
[3 marks]
28Deceleration = 2
m/s
07.4 Another train travels at a speed of 60 m/s.
A constant braking force of 270 000 N causes the train to decelerate and stop.
mass of train = 240 000 kg
Calculate the distance travelled while the braking force is applied.
Use the Physics Equations Sheet.
[6 marks]
Distance travelled =29 m
07.5 It is illegal for train drivers to drink alcohol before driving a train.
Explain how drinking alcohol would affect the stopping distance of a train.
[3 marks]
Mark scheme
Show the mark scheme
Question 7
AO /
Question Answers Extra information Mark
Spec. Ref.
07.1 1 1 AO2
( 2 × 56 × 220) = 6160
4.5.6.1.5
(56 × 380) = 21280
(6160 + 21280) = 27 440 (m) allow a correctly calculated total 1
distance from an incorrectly
calculated area of the rectangle
and / or the triangle
AO /
Spec. Ref.
07.2 the gradient is less after 720 s allow the gradient is less after 1 AO3
(velocity decreases to) 20 m/s 4.5.6.1.5
4.5.6.2.2
so the deceleration is smaller 1
so the braking force is smaller 1
AO /
Spec. Ref.
07.3 correct section of line identified judge by values used 1 AO2
4.5.6.1.5
attempt to calculate a gradient allow use of correct values 1
using values from the correct obtained from the section of the
section of the graph graph after 720 s
eg
(–)36
gradient =
correct calculation using their allow a correct calculation using 1
correct values correct values obtained from the
eg a = (–)0.3 (m/s2) section of the graph after 720 s
if no other marks scored,
an answer that rounds to
0.16 (m/s2) –scores 1 mark – –
AO /
Spec. Ref.
07.4 (–)270 000 = 240 000 × a 1 AO2
4.5.6.1.5
(–)270 000 1 19
a =
240 000
a = (–) 1.125 (m/s2) the equation F = ma must have 1
been used to score subsequent
marks
0 = 602 + (2 × (–1.125) × s) allow a correct substitution using 1
their value of deceleration
3600 allow a correct re-arrangement 1
s = using their value of deceleration
2.25
s = 1600 (m) allow a correct calculation using 1
their value of deceleration
OR
Ek = 2 × 240 000 × 60 (1)
= 432 000 000 (1)
Δ Ek = work done (1)
the equation Ek = 2 mv must
have been used to score
subsequent marks
432 000 000 = 270 000 × s (1) allow a correct substitution using
their value of Ek
432 000 000 allow a correct re-arrangement
s = (1) using their value of Ek
270 000
s = 1600 (m) (1) allow a correct calculation using
their value of Ek
– – –
07.4 OR
cont.
p = 240 000 × 60 (= 14 400 000)
(1)
14 400 000
270 000 = (1)
t
t = 53.333…(s) (1) allow t = 53 (s)
the equation
change in momentum
F =
time taken
must have been used to score
subsequent marks
mean speed = = 30 (1)
s = 30 × 53.333… (1) allow a correct substitution using
their value of t
s = 1600 (m) (1) allow a correct calculation using
their value of t
AO /
Spec. Ref.
07.5 stopping distance includes both 1 AO1
braking distance and thinking 4.5.6.3.1
distance 4.5.6.3.2
alcohol increases driver’s 1
reaction time
which will increase the thinking 1
distance so stopping distance
increases
Total Question 7 18
How to answer it
Forces and Motion: Velocity-Time Graphs & Stopping Distances
What this question tests
This question assesses your ability to extract and interpret data from velocity-time graphs, calculate acceleration, distance, and work done using standard physics equations, apply Newton's second law (F = ma), and explain factors affecting stopping distances, such as reaction time and braking force.
Determine the distance travelled by the train in the first 600 s.
✅ Correct Answer
Distance = 27440 m
💡 Key Knowledge
- The distance travelled is equal to the area under a velocity-time graph.
- Split the shape up to 600 s into standard geometric shapes: a triangle and a rectangle.
- Triangle area = ½ × base × height = ½ × 220 × 56 = 6160 m .
- Rectangle area = base × height = (600 − 220) × 56 = 380 × 56 = 21280 m .
📐 Calculation Steps
- Step 1: Calculate triangle area = 0.5 × 220 × 56 = 6160
- Step 2: Calculate rectangle width = 600 − 220 = 380
- Step 3: Calculate rectangle area = 380 × 56 = 21280
- Step 4: Add areas together = 6160 + 21280 = 27440 m
❌ Common Errors
- Treating the entire shape up to 600 s as a single massive triangle and using base = 600.
- Misreading scale values on the time or velocity axes.
Explain what happens to the braking force as the train decelerates. Use information from Figure 12.
✅ Correct Answer
The gradient of the graph is less steep after 720 s, meaning the deceleration is smaller, therefore the braking force is smaller.
💡 Key Knowledge
- The gradient of a velocity-time graph represents acceleration (or deceleration).
- Newton's Second Law: Force equals mass times acceleration ( F = ma ). Since mass is constant, a smaller deceleration requires a smaller force.
🧠 Exam Technique
This is a multi-step causal explanation. Ensure you explicitly quote or reference graph features (gradient getting shallower after 720 s) before deducing the physics consequence (force).
Determine the maximum deceleration of the train.
✅ Correct Answer
Deceleration = 0.3 m/s² (allow 0.28 to 0.33)
📐 Calculation Steps
- Step 1: Identify the steepest downward slope (between 720 s and 1000 s, velocity drops from 20 to 0 m/s).
- Step 2: Change in velocity = 0 − 20 = −20 m/s (or just use magnitude 20).
- Step 3: Time taken = 1000 − 720 = 280 s. Alternative section: between 600s and 720s: Δv = 20 − 56 = −36, Δt = 120s.
- Step 4: Gradient calculation = 36 / 120 = 0.3 m/s² .
❌ Common Errors
- Using the wrong time interval or mixing up coordinate points from different sections of the graph.
Calculate the distance travelled while the braking force is applied. (Mass = 240,000 kg, Braking Force = 270,000 N, Initial speed = 60 m/s).
✅ Correct Answer
Distance travelled = 1600 m
📐 Method 1: Using Work Done & Kinetic Energy
- Step 1: Calculate initial kinetic energy (Ek = ½mv²) = ½ × 240,000 × 60² = 432,000,000 J
- Step 2: State that Work Done = Initial Kinetic Energy (all energy is dissipated by brakes).
- Step 3: Rearrange Work Done = Force × distance (ΔW = F × s) → s = Work / Force
- Step 4: s = 432,000,000 ÷ 270,000 = 1600 m
🧠 Alternative Method (F = ma & suvat)
- Calculate acceleration: a = F / m = 270,000 / 240,000 = 1.125 m/s²
- Use equation: v² = u² + 2as
- 0 = 60² + (2 × −1.125 × s)
- s = 3600 / 2.25 = 1600 m
Explain how drinking alcohol would affect the stopping distance of a train.
✅ Correct Answer
Stopping distance increases because alcohol increases the driver's reaction time, which increases the thinking distance.
💡 Key Knowledge
- Stopping Distance = Thinking Distance + Braking Distance.
- Thinking distance is affected by tiredness, drugs, alcohol, and distractions.
- Braking distance is affected by road/track conditions, tyre/brake condition, and vehicle mass. Alcohol does not change the mechanical braking distance.
❌ Common Errors
- Vaguely stating "stopping distance increases" without explaining *why* (failing to mention reaction time or thinking distance).
- Incorrectly claiming alcohol increases the mechanical braking distance of the train.
Topics
Physics · P5: Forces
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Higher), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.