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1.3.2Dynamics

Determining Acceleration from V-T Graphs — SQA National 5 Physics Revision Notes & Topical Questions

5 min read 5 parts Lesson 9 of 22

📚 SQA National 5 Physics Hub / Topical Revision
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Part 1 of 5Gradient Equals Acceleration

1.3.2 Determining Acceleration from V-T Graphs

The same equation a = (v − u) ÷ t can be applied directly to any straight section of a velocity–time graph.

Key idea
  • The gradient of a velocity–time graph is the change in velocity divided by the change in time, which is the acceleration.
Positive slope
  • A line rising to the right has a positive gradient, so the object is speeding up with positive acceleration.
Negative slope
  • A line falling to the right has a negative gradient, so the object is decelerating.
Zero slope
  • A horizontal line has zero gradient, so the acceleration is zero whether the object is stopped or moving steadily.

Part 2 of 5Method

1
Pick a straight section of the graph and read the velocity at the start and at the end of it.
2
Read the times at the start and end and subtract to find the time taken for that section.
3
Substitute v, u and t into a = (v − u) ÷ t, keeping the signs.
4
State the acceleration with unit m/s², and say whether it is acceleration or deceleration.
5
Repeat separately for each section, because the acceleration may differ from one section to the next.

Part 3 of 5Comparing Sections

Comparing Sections
Steeper line
  • A steeper section has a larger gradient and therefore a larger magnitude of acceleration.
Equal gradients
  • Parallel lines show equal accelerations, even if they start at different velocities.
Curved line
  • On a curve the gradient changes, so the acceleration is not constant at National 5 level discussion.
Link to area
  • Gradient gives acceleration while area gives displacement, so the two skills answer different questions.
Feature of graphTells you
GradientAcceleration
Area under lineDisplacement
HeightVelocity at that time

Part 4 of 5Worked Example

Question

  • A graph shows velocity rising from 0 to 12 m/s in the first 4.0 s, constant for 6.0 s, then falling to 0 between 10 s and 12 s.
  • Find the acceleration in the first and last sections.

  1. First sectiona = (12 − 0) ÷ 4.0 = 3.0 m/s².
  2. Middle sectionthe line is horizontal, so a = 0.
  3. Last sectiona = (0 − 12) ÷ (12 − 10) = −12 ÷ 2.0 = −6.0 m/s².

Answer 3.0 m/s², then 0, then −6.0 m/s² (a deceleration of 6.0 m/s²).

Part 5 of 5Exam Checklist & Tips

1.3.2 – Key Points for the Exam

  • Gradient Change in the vertical value divided by change in the horizontal value.
  • Constant acceleration An acceleration that stays the same size and direction.
  • Deceleration Negative acceleration, shown by a downward-sloping line.
  • Straight section A part of the graph where the gradient does not change.

Common Mistakes

  • Using the area under the graph instead of the gradient to find acceleration.
  • Using the total time for the whole graph instead of the time for just one section.
  • Dropping the minus sign for deceleration or quoting the wrong unit.
  • Using a point where the line changes slope when reading a value for a single section.

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