AQA A-Level Physics Paper 1, June 2023: Question 2
10 marks · Medium difficulty · Short Answer
Calculate the time taken for a jet-powered car to complete a run, determine its kinetic energy and acceleration-related power percentages from a speed-distance graph, and deduce whether its deceleration is within safe limits.
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Question text
02 In 2021 the world land speed record was 1230 km h−1.
This was the average speed achieved by a jet-powered car in two runs. Each run was
measured over a distance of 1.61 km.
02.1 The average speed for one of these runs was 343 m s−1.
Calculate, in s, the time taken for the car to complete the other run.
[2 marks]
6 time = s
02.2 Engineers are designing a new jet-powered car to break this record.
Figure 1 shows the variation of speed with distance for the car, as predicted by the
engineers.
Figure 1
The car reaches its maximum acceleration when it is 5600 m from the start.
At this point the mass of the car is 6.50 × 103 kg.
Determine the kinetic energy of the car at its maximum acceleration.
[2 marks]
7 kinetic energy = J
02.3 At any point on the graph in Figure 1, the acceleration is given by:
acceleration = speed × gradient of line
When the car is at its maximum acceleration, the power input to the jet engines
is 640 MW.
*06* Calculate the percentage of the input power used to accelerate the car at its
maximum acceleration.
[4 marks]
percentage of input power = %
02.4 Scientists recommend that the average deceleration of the driver of the car should be
less than 3g.
Deduce whether the average deceleration is less than 3g.
[2 marks]
Mark scheme
Show the mark scheme
Question Answers Additional comments/Guidelines Mark AO
02.1 Conversion of 1230 km h−1 to m s−1 Expect to see 342 m s-1 (341.7) 2 AO1
OR
-1 AO2
Calculates time for 343 m s run Expect to see 4.69 s
OR
Expect to see 9.42 s
Calculates total time (using total distance, 3.22 km, and
speed record)
OR Expect to see 340.3 m s-1
Calculates unknown speed
Answer that rounds to 4.73 (s) Do not accept 2sf for final answer.
02.2 speed from graph: −1 Accept 445 – 455 m s-1 2 AO3
450 m s
AO2
Use of their speed and KE equation to give consistent
answer Expect to see 6.6 × 108 (J)
Expect to see 450 m s−1 for their speed
02.3 MAX three from: 4 AO1
AO2
Evidence for gradient may be on figure
2× AO3
Allow ECF from 02.2
• Use of graph to determine gradient 450
• = 0.080(4)
5600
• Uses (their) speed and (their) gradient to give
• Expect to see 450 × 0.08
acceleration −2
= 36(.2) m s
• Use of F = m × (their a) to give resultant force 5
• Expect to see 2.35 × 10 N
• Use of P = (their F)× (their speed) 5
• Expect to see 450 × 2.35 × 10
= 106 MW
Final answer between 16% and 17%
Reject power that is calculated assuming a
constant speed.
02.4 Identifies distance decelerating allow 7000 m to 7600 m 2 AO3
AND
max velocity = (470 ±5) m s−1 allow answer consistent with their distance
that rounds to 15 or 16
Uses suvat equation(s)
−2 give full credit to calculations that show that
to get a = (−) 15 m s which is less than 3g (so yes).
an acceleration of 3g would stop the car in a
(much) shorter distance, with a statement that
this means that the actual acceleration must
10 be (much) less than 3g.
For MP2 allow calculation of
gradient × average speed to give
a = (−) 15 m s−2 which is less than 3g (so yes)
Total 10
How to answer it
Jet-Powered Car Mechanics Study Guide
What this question tests
This multi-part mechanics question tests your ability to manipulate kinematic equations, extract and interpret data from a speed-distance graph, calculate kinetic energy, link power with force and velocity using instantaneous rates of change, and evaluate safety decelerations using equations of motion or graphical analysis.
Calculating Run Times & Unit Conversions
✅ Correct Answer
4.73 s (Accept 4.7 s or higher)
💡 Key Knowledge
- Convert km h⁻¹ to m s⁻¹ by dividing by 3.6 (or multiplying by 1000 and dividing by 3600).
- Average speed = Total distance / Total time.
🧠 Exam Technique
There are multiple valid routes to the answer (e.g., finding the overall average speed from the 1230 km h⁻¹ record, finding individual run times, or calculating the unknown speed first). Choose the method you find most intuitive to avoid arithmetic slips.
❌ Common Errors
- Forgetting to convert units from km h⁻¹ to m s⁻¹.
- Rounding down to 2 significant figures prematurely (e.g., writing 4.7 s as a final answer when 3 sig figs are safer).
📐 Step-by-Step Calculation
- Convert overall record speed: 1230 / 3.6 = 341.67 m s⁻¹
- Total distance for 2 runs: 1.61 km × 2 = 3.22 km = 3220 m
- Total time taken: 3220 / 341.67 = 9.424 s
- Time for the first run ( 343 m s⁻¹ ): 1610 / 343 = 4.694 s
- Time for the other run: 9.424 - 4.694 = 4.73 s
Kinetic Energy at Maximum Acceleration
✅ Correct Answer
6.6 × 10⁸ J (Accept range based on speed values between 445 and 455 m s⁻¹)
💡 Key Knowledge
- Kinetic Energy formula: Eₖ = 0.5 × m × v²
- Reading values accurately from printed grid lines.
🧠 Exam Technique
Locate the distance 5600 m on the x-axis, follow it up to the curve on Figure 1, and read across to the y-axis to find the velocity corresponding to maximum acceleration.
❌ Component Traps
Misreading the scale on the speed axis. Each small grid square represents 10 m s⁻¹ .
📐 Step-by-Step Calculation
- Read speed from graph at 5600 m : v = 450 m s⁻¹
- Identify mass: m = 6.50 × 10³ kg
- Substitute into Eₖ formula: Eₖ = 0.5 × (6.50 × 10³) × (450)²
- Evaluate: Eₖ = 6.59 × 10⁸ J (rounds to 6.6 × 10⁸ J to 2 s.f.)
Power Input and Acceleration Percentage
✅ Correct Answer
Between 16% and 17%
💡 Key Knowledge
- Acceleration from a distance-speed graph gradient: a = v × gradient
- Newton's Second Law: F = m × a
- Mechanical power: P = F × v
🧠 Exam Technique
Draw a clear tangent at d = 5600 m if the curve isn't linear there, or calculate the chord/gradient carefully using large, easily readable coordinate points on the straight-line section.
❌ Common Errors
Assuming acceleration is constant or attempting to use kinematic equations meant for uniform acceleration where a varies.
📐 Step-by-Step Calculation
- Determine gradient at 5600 m : gradient = 450 / 5600 = 0.0804 s⁻¹
- Calculate acceleration: a = 450 × 0.0804 = 36.2 m s⁻²
- Calculate force: F = (6.50 × 10³) × 36.2 = 2.35 × 10⁵ N
- Calculate useful accelerating power: P = (2.35 × 10⁵) × 450 = 1.06 × 10⁸ W = 106 MW
- Calculate percentage: (106 MW / 640 MW) × 100% = 16.5%
Evaluating Deceleration Against Safety Limits
✅ Correct Answer
Yes, deceleration is less than 3g (Calculated acceleration is approximately 15 m s⁻² , whereas 3g = 29.4 m s⁻² )
💡 Key Knowledge
- Standard gravity g = 9.81 m s⁻² (so 3g ≈ 29.4 m s⁻² ).
- Kinematic equations of motion ( suvat ) or work-energy principles can evaluate stopping performance.
🧠 Exam Technique
Identify both the maximum velocity before braking and the stopping distance available from the graph (distance from peak speed to 0 m s⁻¹, approx. 14500 - 7200 = 7300 m ). State your values clearly before performing the suvat calculation.
❌ Common Errors
Using the total distance from the start instead of the actual braking distance interval.
📐 Step-by-Step Calculation
- Initial velocity for braking: u = 470 m s⁻¹ , Final velocity v = 0 m s⁻¹
- Braking distance s ≈ 15000 - 7500 = 7500 m (allow 7000 to 7600 m )
- Use v² = u² + 2as : 0 = 470² + 2a(7500)
- Solve for a : a = -220900 / 15000 = -14.7 m s⁻²
- Compare with 3g : 3 × 9.81 = 29.4 m s⁻² . Since 14.7 < 29.4 , the statement holds true.
Topics
Physics · Practical skills · 3.4 Mechanics and materials · Data analysis
Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.