AshPhys

Deducing acceleration

AshPhys · Revision cheat sheet
AS Level9702 AS Ch.2Chapter 2: Accelerated motionLesson 2.4Sections 2.4

Goal: Find acceleration from the gradient of a velocity–time graph.

01The big picture

On a velocity–time graph the gradient is Δv/Δt\Delta v/\Delta t, which is the acceleration. A steeper line means a larger acceleration; a horizontal line means constant velocity (zero acceleration); a line sloping down means the object is slowing.

A straight line means uniform acceleration. For a curved line, draw a tangent at the moment you want and find its gradient with a large triangle.

Gradient [paraphrase]
Change in velocity ÷ change in time, read from a large triangle on the line.
Fig. 1 · A velocity–time graph
A velocity–time graphVelocity against time. A straight line rises from 0 at 0 s to 12 m/s at 4 s, stays flat at 12 m/s until 10 s, then falls in a straight line to 0 at 14 s.v / m s⁻¹t / s12041014A→BB→CC→D
Steeper line = larger acceleration. Flat = constant velocity. Downward slope = slowing down.

02Key equations

Acceleration = gradient of v–t
a=ΔvΔta = \frac{\Delta v}{\Delta t}
Symbols and units
  • vvvelocitym s−1\mathrm{m\,s^{-1}}
  • tttimes\mathrm{s}
  • aaaccelerationm s−2\mathrm{m\,s^{-2}}

03Sample problems with solutions

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

  1. A to B:

    a=12−04−0=3.0 m s−2a = \frac{12 - 0}{4 - 0} = 3.0\ \mathrm{m\,s^{-2}}

  2. B to C: the line is horizontal, so the velocity is constant: a=0a = 0.

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

    a=0−1214−10=−3.0 m s−2a = \frac{0 - 12}{14 - 10} = -3.0\ \mathrm{m\,s^{-2}}
Answer: +3.0 m s⁻², then 0, then −3.0 m s⁻²
Unit check: m s−1s=m s−2 ✓\frac{\mathrm{m\,s^{-1}}}{\mathrm{s}} = \mathrm{m\,s^{-2}}\ \checkmark

04Practice problems

  1. P1
    What does a horizontal line on a velocity–time graph show?
    Final answerConstant velocity (zero acceleration)
    Answer: Constant velocity (zero acceleration)
  2. P2
    A straight line goes from (0 s, 4.0 m s⁻¹) to (5.0 s, 14 m s⁻¹). Find the acceleration.
    Final answer2.0 m s⁻²
    Answer: 2.0 m s⁻²
  3. P3
    A tangent to a curved v–t graph passes through (0 s, 2.0 m s⁻¹) and (4.0 s, 10 m s⁻¹). Find the acceleration there.
    Final answer2.0 m s⁻²
    Answer: 2.0 m s⁻²
  4. P4
    A v–t line falls from 20 m s⁻¹ to 0 in 8.0 s. Find the acceleration.
    Final answer−2.5 m s⁻²
    Answer: −2.5 m s⁻²

05Other useful information

Common mistakes

  • ✗ Wrong: Read a horizontal v–t line as “at rest”.

    ✓ Right: That is an s–t graph. On a v–t graph it means constant velocity.

Exam tips

  • Check the axes first: s–t gradient = velocity; v–t gradient = acceleration.

What the syllabus asks

  • 9702 topic 2: acceleration from the gradient of a v–t graph. Coursebook 2.4.

Explore: Practice questions for this lesson