AQA AS Level Mathematics Paper 1, June 2025: Question 16

2 marks · Easy difficulty · Short Answer

Calculate the speed from a linear displacement-time graph and sketch a more realistic displacement-time graph for a car starting from rest.

Practise this question

Question

Figure 1 shows a displacement-time graph with a straight line from the origin (0, 0) to (5, 12). Part (a) asks to find the speed of the car. Part (b) provides Figure 2, an empty set of axes with displacement from 0 to 12 metres and time from 0 to 5 seconds, asking to draw a more realistic displacement-time graph for the car starting from rest.
Question text

16 The displacement–time graph, Figure 1, shows the first 5 seconds of the motion of a

car which starts from rest and travels 12 metres.

Figure 1

Displacement (m)

O 5 Time (s)

16 (a) Find the speed of the car.

[1 mark]

16 (b) On Figure 2, draw a more realistic displacement–time graph to show the first

5 seconds of motion of the car.

[1 mark]

Figure 2

Displacement (m)

O 5 Time (s)

Mark scheme

Show the mark scheme Mark scheme for Question 16: Part (a) gives 1 mark (B1) for 2.4 m s^-1, condoning missing units. Part (b) gives 1 mark (B1) for drawing a curve that is convex downwards at the origin (gradient starting near zero) and reaches or passes through approximately (5, 12).

Q Marking instructions AO Marks Typical solution

16(a) Obtains 2.4 m s–1 1.1b B1 2.4 m s–1

ACF

Condone missing units

Subtotal 1

16(b) Draws a graph which is convex 3.5c B1

at the origin and reaches or

passes through approximately

(5,12)

Subtotal 1

Question 16 Total 2

How to answer it

Interpreting & Refining Displacement–Time Graphs

📋 What this question tests

This question assesses your foundational understanding of kinematics graphs in Mechanics:

  • Interpreting the gradient of a displacement–time (s–t) graph as velocity or speed.
  • Critiquing mathematical models: recognising why a constant non-zero gradient contradicts a starting state of rest.
  • Sketching realistic curved displacement–time graphs that represent acceleration from rest (convex shape at the origin).
Part 16 (a)

Calculating Speed from a Linear Model

1 Mark • Assessment Objective 1.1b

📐 Step-by-Step Calculation

  1. Identify gradient formula:
    Speed = Gradient of displacement–time graph = (Change in displacement) / (Change in time)
  2. Substitute graph coordinates:
    Line passes through (0, 0) and (5, 12):
    Speed = (12 − 0) / (5 − 0) = 12 / 5
  3. Evaluate final value:
    Speed = 2.4 m s⁻¹

✅ Correct Answer & Mark Scheme

Answer: 2.4 m s⁻¹ or 12/5 m s⁻¹

[B1] Awarded for obtaining 2.4 m s⁻¹ (or any equivalent fraction / decimal form). Missing units are condoned.

💡 Key Knowledge

  • On a displacement–time graph, Gradient = Velocity . Since motion is unidirectional here, this equals speed.
  • A straight line represents constant velocity (zero acceleration).

❌ Common Errors

  • Inverted ratio: Calculating time / distance = 5 / 12 ≈ 0.417.
  • Finding area under the curve: Attempting (1/2) × 5 × 12 = 30. (Area under a displacement–time graph has no standard physical meaning).
Part 16 (b)

Refining the Model to Reflect Starting from Rest

1 Mark • Assessment Objective 3.5c

✅ Correct Curve Specifications

A hand-drawn curve that satisfies both of the following criteria:

  1. Convex at the origin: Starts at (0, 0) with a horizontal tangent (gradient = 0) and curves upwards with an increasing gradient.
  2. Correct endpoint: Passes through or reaches approximately the point (5, 12) .
[B1] Awarded for a curve that is convex at the origin and reaches or passes through approximately (5, 12).

🧠 Exam Technique & Examiner Insights

  • Spot the clue in the question stem: The car "starts from rest". This means at t = 0, initial speed u = 0.
  • Since velocity is the gradient, the graph must start flat (gradient = 0) at the origin.
  • Figure 1 was unrealistic because a straight line through the origin implies an instantaneous jump from 0 m s⁻¹ to 2.4 m s⁻¹ at t = 0 (infinite acceleration).

📐 How to Draw the Graph on Figure 2

Place your pencil at the origin (0, 0):

  • Start with a flat, horizontal slope along the time axis at t = 0.
  • Smoothly curve upwards so that the steepness increases as time increases (representing positive acceleration).
  • End precisely at or very close to the coordinate point aligned with 5 seconds on the horizontal axis and 12 metres on the vertical axis.

❌ Common Errors

  • Concave curve (decelerating): Drawing a curve that starts steep at (0,0) and flattens out towards (5, 12). This would mean the car started at high speed and slowed down.
  • Missing the endpoint: Drawing a parabolic curve that ends noticeably above or below the level of 12 m at 5 s.
  • Drawing another straight line: A straight line cannot represent a car starting from rest and accelerating.

💡 Physical Reality Check

Assuming constant acceleration from rest: s = (1/2)at² . This is a quadratic curve of the form s ∝ t², which has a vertex (gradient = 0) at the origin and curves steeply upwards—precisely confirming the required convex shape!

Topics

Mechanics · Q: Kinematics

Question and mark scheme from the AQA AS Level Mathematics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.