AQA A-Level Mathematics Paper 2, June 2025: Question 12

1 mark · Easy difficulty · Short Answer

Find the displacement of a particle after 8 seconds from its velocity-time graph.

Practise this question

Question

A velocity-time graph showing the velocity v in metres per second of a particle against time t in seconds from t = 0 to t = 8. The line starts at (0, -2) and remains horizontal at v = -2 until t = 2, then rises linearly through (4, 0) to reach (6, 2), and remains horizontal at v = 2 until t = 8. Below the graph are four multiple-choice options to circle: -12, 0, 6, and 12.
Question text

12 A particle moves in a straight line for 8 seconds.

The velocity v, in m s–1, of the particle at time t seconds is shown in the diagram.

v

24 6 8 t

–2

Find the displacement, in metres, of the particle after 8 seconds.

Circle your answer.

[1 mark]

–12 0 6 12

Mark scheme

Show the mark scheme Mark scheme table for Question 12 indicating 1 mark (B1) for AO 2.2a by circling the second answer, with a typical solution of 0.

Q Marking instructions AO Marks Typical solution

12 Circles the second answer 2.2a B1 0

Question 12 Total 1

How to answer it

Displacement from a Velocity-Time Graph

📌 What this question tests

Understanding the graphical interpretation of linear motion: specifically, distinguishing between displacement (signed area between curve and time-axis) and total distance (total scalar area), as well as exploiting rotational symmetry in piecewise linear velocity-time graphs.

Question 12 Breakdown

Multiple Choice: Determining displacement after 8 seconds

✅ Correct Answer

0

Circle the second option: 0

Mark Scheme: B1 (AO 2.2a) for identifying and circling the correct value of 0.

💡 Key Knowledge

  • Displacement (s): Given by the net signed area under the velocity-time graph:
    s = ∫ v dt
  • Areas below the t-axis represent motion in the negative direction and carry a negative sign.
  • Areas above the t-axis represent motion in the positive direction and carry a positive sign.
  • Total distance would sum the absolute magnitudes of the areas, regardless of direction.

📐 Step-by-Step Calculation

  1. Calculate the area below the axis (from t = 0 to t = 4 s):
    The region is a trapezium with parallel vertical/horizontal lengths:
    • Top base along time-axis = 4 - 0 = 4 s
    • Bottom base at v = -2 m s⁻¹ = 2 - 0 = 2 s
    • Height = 2 m s⁻¹
    Displacement₁ = -½ × (4 + 2) × 2 = -6 m
  2. Calculate the area above the axis (from t = 4 to t = 8 s):
    The region is an identical trapezium reflected above the axis:
    • Base along time-axis = 8 - 4 = 4 s
    • Top parallel side at v = 2 m s⁻¹ = 8 - 6 = 2 s
    • Height = 2 m s⁻¹
    Displacement₂ = +½ × (4 + 2) × 2 = +6 m
  3. Sum the signed areas to find total displacement:
    Total Displacement = (-6) + (+6) = 0 m

🧠 Exam Technique & Symmetry Shortcut

Notice the point symmetry about the point (4, 0) :

  • The shape from t = 0 to t = 4 is identical in magnitude and shape to the section from t = 4 to t = 8 , but inverted.
  • Because one lies entirely below the axis and the other entirely above, their signed areas cancel out exactly.
  • Recognising this symmetry allows you to circle 0 in seconds without doing any formal arithmetic!

❌ Common Traps & Misconceptions

  • Circling 12: Confusing displacement with distance travelled. Distance is scalar: |-6| + |6| = 12 m .
  • Circling -12: Doubling the negative region without considering that the particle reverses direction and travels in the positive direction for the last 4 seconds.
  • Circling 6: Only calculating one trapezium's area and forgetting to combine both stages of motion.

Topics

Mechanics · Q: Kinematics

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.