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Displacement–time graphs

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.1Chapter 1: KinematicsLesson 1.4Sections 1.4

Goal: Read velocity from the gradient of a displacement–time graph, including from a tangent to a curve.

01The big picture

The gradient of a displacement–time graph is the velocity. A straight line means constant velocity; a horizontal line means the object is at rest; a negative gradient means it is moving back towards the start.

A curved line means the velocity is changing. The instantaneous velocity is the gradient of the tangent drawn at that time.

Gradient [paraphrase]
Change in displacement ÷ change in time, read from a large triangle on the line.
Fig. 1 · A displacement–time graph
A displacement–time graphDisplacement against time. A straight line rises from 0 m at 0 s to 30 m at 6 s, stays flat at 30 m until 10 s, then falls in a straight line to 0 m at 16 s. A dashed triangle under the first section shows a rise of 30 m over 6 s.s / mt / s30610160A→BB→C at restC→D
Steeper line = faster. Flat = at rest. Downward slope = moving back.

02Key equations

Velocity = gradient
v=ΔsΔtv = \frac{\Delta s}{\Delta t}
Symbols and units
  • ssdisplacementm\mathrm{m}
  • tttimes\mathrm{s}
  • vvvelocitym s−1\mathrm{m\,s^{-1}}

03Sample problems with solutions

Sample 1 · Reading the graph
Use the graph in Fig. 1 to find the velocity during each of the three stages.

  1. A to B:

    v=30−06−0=5.0 m s−1v = \frac{30 - 0}{6 - 0} = 5.0\ \mathrm{m\,s^{-1}}

  2. B to C: the line is horizontal, so the object is at rest: v=0v = 0.

  3. C to D: the gradient is negative, so it is moving back:

    v=0−3016−10=−5.0 m s−1v = \frac{0 - 30}{16 - 10} = -5.0\ \mathrm{m\,s^{-1}}
Answer: +5.0 m s⁻¹, then 0, then −5.0 m s⁻¹ (back towards the start)
Unit check: ms=m s−1 ✓\frac{\mathrm{m}}{\mathrm{s}} = \mathrm{m\,s^{-1}}\ \checkmark

04Practice problems

  1. P1
    What does the gradient of a displacement–time graph represent?
    Final answerVelocity
    Answer: Velocity
  2. P2
    A straight line goes from (0 s, 0 m) to (8.0 s, 40 m). Calculate the velocity.
    Final answer5.0 m s⁻¹
    Answer: 5.0 m s⁻¹
  3. P3
    What does a horizontal line on a displacement–time graph show?
    Final answerThe object is at rest
    Answer: The object is at rest
  4. P4
    A tangent to a curved s–t graph passes through (1.0 s, 0 m) and (5.0 s, 24 m). Find the velocity there.
    Final answer6.0 m s⁻¹
    Answer: 6.0 m s⁻¹

05Other useful information

Common mistakes

  • ✗ Wrong: Read the velocity off the vertical axis.

    ✓ Right: The vertical axis gives displacement; velocity is the gradient.

Exam tips

  • Use a large gradient triangle; a negative gradient means moving back.

What the syllabus asks

  • 9702 topic 2: determine velocity using the gradient of a displacement–time graph. Coursebook 1.4.

Explore: Distance-Time Graph Explorer (simulation) · Practice questions for this lesson