AQA AS Level Physics Paper 2, June 2024: Question 19

1 mark · Medium difficulty · Multiple Choice

Determine the speed of an object at t = 5.0 s given its initial speed and an acceleration-time graph.

Practise this question

Question

A multiple-choice question showing a graph of acceleration against time for an object moving in a straight line. The graph is a straight line starting at an acceleration of 2.0 m s^-2 at t = 0 s and rising to 6.0 m s^-2 at t = 5.0 s. Below the graph, text states that the initial speed at t = 0 is 4.0 m s^-1 and asks for the speed when t = 5.0 s, with four multiple-choice options A (10 m s^-1), B (14 m s^-1), C (20 m s^-1), and D (24 m s^-1).
Question text

19 The graph shows the variation with time t of the acceleration of an object moving in a

straight line.

When t = 0 the speed of the object is 4.0 m s−1.

What is the speed of the object when t = 5.0 s?

[1 mark]

A 10 m s−1

B 14 m s−1

C 20 m s−1

D 24 m s−1

Mark scheme

Show the mark scheme The mark scheme table shows question number 19 with the correct answer D and the value 24 m s^-1.

19 D 24 m s−1

How to answer it

Finding Final Speed from an Acceleration-Time Graph

AQA AS Level Physics • Mechanics • Multiple Choice

What this question tests

This question assesses your understanding of graphical analysis in kinematics—specifically, the physical significance of the area under an acceleration-time graph. It tests your ability to calculate change in velocity from graphical data and combine it with initial conditions to find the final speed.

Question 19 Analysis

Determining final speed from an acceleration-time graph

✅ Correct Answer

Option D: 24 m s⁻¹

Mark Allocation: [1 mark] for selecting D.

💡 Key Knowledge

  • The gradient of a velocity-time graph gives acceleration.
  • The area under an acceleration-time graph represents the change in velocity (delta v).
  • Final velocity = Initial velocity + Change in velocity.

🧠 Exam Technique

Instead of struggling to find the equation of the line, treat the region under the graph as standard geometrical shapes (a rectangle topped by a triangle) to quickly calculate the total area.

❌ Common Errors

  • Forgetting initial speed: Stopping at the area calculation (20 m s⁻¹) and selecting option C.
  • Misreading axes: Taking the initial acceleration as 0 instead of 2.0 m s⁻².

📐 Step-by-Step Calculation

  1. Split the area under the graph into two parts between t = 0 s and t = 5.0 s:
    • A bottom rectangular strip from a = 0 to a = 2.0 m s⁻¹:
      Area 1 = base × height = 5.0 s × 2.0 m s⁻² = 10.0 m s⁻¹
    • A top triangular section from a = 2.0 to a = 6.0 m s⁻¹:
      Area 2 = 0.5 × base × height = 0.5 × 5.0 s × (6.0 - 2.0) m s⁻² = 0.5 × 5.0 × 4.0 = 10.0 m s⁻¹
  2. Calculate total change in velocity (delta v):

    delta v = Area 1 + Area 2 = 10.0 + 10.0 = 20.0 m s⁻¹

  3. Add change in velocity to the initial speed:

    v_final = v_initial + delta v = 4.0 m s⁻¹ + 20.0 m s⁻¹ = 24.0 m s⁻¹

Topics

Physics · Practical skills · 3.4 Mechanics and materials · Data analysis

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.