AQA GCSE Physics Physics Paper 2 (Foundation), November 2020: Question 7
15 marks · Standard Demand difficulty · Short Answer
Calculate speed, time, acceleration, and force for an aircraft using graphs, equations of motion, and work done.
Practise this questionQuestion
Question text
07.1 An aircraft travels at a constant velocity.
How is the velocity of the aircraft different to the speed of the aircraft?
[1 mark]
07.2 Figure 11 shows one of the engines on the aircraft.
Figure 11
Air is taken into the front of the engine and pushed out of the back of the engine.
Explain the effect this has on the engine.
[2 marks]
07.3 Figure 12 shows a distance-time graph for the aircraft.
Figure 12
Determine the speed of the aircraft.
[3 marks]
Speed = m/s
07.4 Write down the equation that links acceleration (a), change in velocity (Δv) and time
taken (t).
[1 mark]
07.5 At a different stage of the flight, the aircraft was travelling at a velocity of 250 m/s.
The aircraft then decelerated at 0.14 m/s2.
Calculate the time taken for the aircraft to decelerate from 250 m/s to 68 m/s.
[4 marks]
Time = s
07.6 Write down the equation that links distance (s), force (F) and work done (W).
[1 mark]
07.7 When the aircraft landed, it travelled 2000 m before stopping.
The work done to stop the aircraft was 140 000 000 J.
Calculate the mean force used to stop the aircraft.
[3 marks]
Mean force = N
Mark scheme
Show the mark scheme
Question 7
AO /
Question Answers Extra information Mark
Spec. Ref.
07.1 velocity includes direction allow velocity is a vector 1 AO1
(quantity) and speed is a scalar 4.5.6.1.3
(quantity)
07.2 (an equal) force from the air 1 AO2
pushes on the engine/aircraft 4.5.6.2.3
in the opposite direction only scores if first marking point 1
scored
accept to the left or forwards
if no other marks scored, allow
1 mark for pushes the engine
forwards
07.3 correct value for distance and 1 AO2
corresponding time 4.5.6.1.4
(eg 12 000 m and 50 s)
their change in distance this mark may be awarded if 1
v =
their change in time distance and/or time are
incorrectly read from the graph
speed = 240 (m/s) allow a correctly calculated 1
answer using their values of
distance and time from the
graph
07.4 acceleration = 1 AO1
change in velocity 4.5.6.1.5
time taken
or
Δv
a =
t
07.5 250 – 68 = 182 1 AO2
4.5.6.1.5
182 this mark may be awarded if the 1
0.14 =
t change in velocity is
14 incorrectly/not calculated
182 this mark may be awarded if the 1
t =
0.14 change in velocity is
incorrectly/not calculated
t = 1300 (seconds) allow a correctly calculated 1
answer using a change in
velocity incorrectly/not
calculated
07.6 work done = force × distance 1 AO1
4.5.2
or
W = F s
07.7 140 000 000 = force × 2000 1 AO2
4.5.2
140 000 000 1
force =
2000
force = 70 000 (newtons) 1
Total 15
How to answer it
Aircraft Motion and Forces Study Guide
This GCSE Physics question assesses your understanding of motion, vectors vs. scalars, Newton's third law in jet engines, interpreting distance-time graphs, rearranging and applying equations for acceleration, and calculating work done and forces.
Speed vs. Velocity
✅ Correct Answer
Velocity includes a direction, whereas speed does not (or velocity is a vector quantity and speed is a scalar quantity).
💡 Key Knowledge
- Scalar: Has magnitude (size) only. Example: speed (e.g., 200 m/s).
- Vector: Has both magnitude and a specific direction. Example: velocity (e.g., 200 m/s due North).
Jet Engine Forces
✅ Correct Answer
An equal force from the air pushes on the engine/aircraft in the opposite direction (forwards).
💡 Key Knowledge
This is a direct application of Newton's Third Law of Motion: whenever two objects interact, the forces they exert on each other are equal and opposite.
❌ Common Errors
Students often forget to mention the direction of the resulting force or incorrectly state that the engine pushes air backwards without linking it to the reaction force pushing the aircraft forwards.
Distance-Time Graph Calculation
📐 Step-by-Step Calculation
- Choose points from the graph: Use the end point where distance = 12,000 m and time = 50 s (or any clear point on the straight line).
- State the formula: speed = distance / time
- Substitute values: 12000 / 50
- Calculate final answer: 240 m/s
🧠 Exam Technique
Since the graph is a straight line, the speed is constant. You can pick any convenient point along the line to calculate the gradient, but choosing the maximum values makes the math straightforward.
Equation for Acceleration
✅ Correct Answer
acceleration = change in velocity / time taken (or a = Δv / t )
🧠 Exam Technique
Learn standard formula symbols from the equation sheet. Words or standard symbols are both accepted by examiners, but symbols must be standard (use a , v or Δv , and t ).
Calculating Deceleration Time
📐 Step-by-Step Calculation
- Find change in velocity (Δv): 250 - 68 = 182 m/s
- Rearrange acceleration formula: t = change in velocity / acceleration
- Substitute values: t = 182 / 0.14
- Calculate final time: 1300 s
❌ Common Errors
Watch out for subtraction errors when finding the change in velocity. Ensure you divide the change in velocity by acceleration, rather than multiplying them together.
Equation Linking Work Done and Distance
✅ Correct Answer
work done = force × distance (or W = F s )
💡 Key Knowledge
Work done is equal to the energy transferred when a force moves an object through a distance. Units: Work done in Joules (J), Force in Newtons (N), Distance in metres (m).
Calculating Mean Stopping Force
📐 Step-by-Step Calculation
- State the rearranged formula: force = work done / distance
- Substitute values: 140,000,000 / 2000
- Calculate final force: 70,000 N
🧠 Exam Technique
Check all large numbers with many zeros carefully before punching them into your calculator to avoid missing a zero. Always include correct units ( N ) in your final answer line.
Topics
Physics · P5: Forces · P1: Energy
Question and mark scheme from the AQA GCSE Physics examination, Physics Paper 2 (Foundation), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.