4  Graphical Analysis

4.1 Syllabus inquiry question

  • How is the motion of an object moving in a straight line described and predicted?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 9:

Graphs are compact summaries of motion. A single line can show both the story of an object and the mathematics used to predict it.

4.2 Learning Objectives

  • Interpret displacement-time, velocity-time, and acceleration-time graphs.
  • Relate gradients and areas to physical quantities.
  • Recognise uniform and non-uniform motion from graph shapes.
  • Use graph features to describe motion in words.

4.3 Content

4.3.1 Displacement-time graphs

The gradient gives velocity. A straight line means constant velocity. A curve means the velocity changes.

Interactive Exploration

Hover over the graph below to see how the gradient at each point relates to instantaneous velocity.

Reading the graph:

Feature Physical Meaning
Gradient at a point Instantaneous velocity
Positive gradient Moving in positive direction
Negative gradient Moving in negative direction
Zero gradient Momentarily at rest
Straight line Constant velocity
Curve Changing velocity

4.3.2 Velocity-time graphs

The gradient gives acceleration. The area under the curve gives displacement.

Key relationships:

Feature Physical Meaning
Gradient Acceleration
Area above time axis Positive displacement
Area below time axis Negative displacement
Total area (signed) Net displacement
Horizontal line Constant velocity (zero acceleration)

4.3.3 Acceleration-time graphs

The area under the curve gives the change in velocity. A horizontal line indicates constant acceleration.

Note

For constant acceleration, the a-t graph is a horizontal line, and the area equals \(\Delta v = a \times t\).

4.3.4 Graph Comparison: Three Types of Motion

The table below summarizes what different graph shapes mean for each type of motion:

Motion Type s-t Graph v-t Graph a-t Graph
At rest Horizontal line Line at v = 0 Line at a = 0
Constant velocity Straight line (slope ≠ 0) Horizontal line Line at a = 0
Constant acceleration Parabola Straight line Horizontal line
Changing acceleration Complex curve Curve Varying line

4.4 Worked Examples

4.4.1 Example 1: Velocity from an s-t graph

A displacement-time graph is a straight line from \(s = 0\) m at \(t = 0\) s to \(s = 30\) m at \(t = 6\) s.

  1. Gradient: \(v = \frac{\Delta s}{\Delta t} = \frac{30 - 0}{6 - 0}\)
  2. \(v = 5.0\) m/s
  3. Motion is uniform in the positive direction.

4.4.2 Example 2: Acceleration from a v-t graph

A velocity-time graph rises linearly from 2 m/s to 14 m/s over 4.0 s.

  1. \(\Delta v = 14 - 2 = 12\) m/s
  2. \(a = \frac{12}{4.0} = 3.0\) m/s\(^2\)
  3. Acceleration is constant and positive.

4.4.3 Example 3: Displacement from a v-t graph

Velocity increases uniformly from 4 m/s to 10 m/s over 5.0 s.

  1. Area is a trapezium: \(s = \frac{(v_1 + v_2)}{2} \times t\)
  2. \(s = \frac{(4 + 10)}{2} \times 5.0 = 35\) m
  3. Displacement equals the area under the curve.

4.4.4 Interactive: Compare Motion Diagrams and Graphs

Below is a motion diagram showing an object decelerating. Compare it with the v-t graph above to see the relationship.

Connection to the v-t graph:

  • The velocity arrows in the motion diagram correspond to the v-t graph values
  • As dots get closer together, velocity decreases (shown in v-t graph)
  • The object reverses direction when v = 0 (at t = 4 s)

4.5 Common Misconceptions

Common Misconceptions

  • Misconception: The height of a v-t graph gives displacement. Correction: Displacement is the area under the curve.

  • Misconception: A flat s-t graph means acceleration is zero. Correction: It means velocity is zero.

  • Misconception: A curved v-t graph always means non-uniform acceleration. Correction: Only a changing gradient indicates changing acceleration.

4.6 Practice Questions

4.6.1 Easy (2 marks)

A straight-line s-t graph has gradient 3.0 m/s. State the velocity and describe the motion.

  • Correct velocity (1)
  • Motion description (1)

Answer: Velocity = 3.0 m/s. The object moves at constant velocity in the positive direction.

4.6.2 Medium (4 marks)

A v-t graph is a straight line from 0 m/s at 0 s to 12 m/s at 6 s. Calculate acceleration and displacement.

  • Acceleration from gradient (2)
  • Displacement from area (2)

Answer: - \(a = 12/6 = 2.0\) m/s\(^2\) - \(s = \frac{1}{2} \times 6 \times 12 = 36\) m (area of triangle)

4.6.3 Hard (5 marks)

An a-t graph shows 2.0 m/s\(^2\) for 3.0 s, then -1.0 m/s\(^2\) for 2.0 s. The object starts at 5.0 m/s. Find final velocity and total displacement.

  • Velocity change from area under a-t (2)
  • Final velocity (1)
  • Displacement from v-t construction or average velocity (2)

Solution:

Phase 1 (0-3 s): - \(\Delta v_1 = 2.0 \times 3.0 = 6.0\) m/s - \(v\) at end of phase 1: \(5.0 + 6.0 = 11.0\) m/s - \(s_1 = 5.0(3.0) + \frac{1}{2}(2.0)(3.0)^2 = 15 + 9 = 24\) m

Phase 2 (3-5 s): - \(\Delta v_2 = -1.0 \times 2.0 = -2.0\) m/s - Final velocity: \(11.0 - 2.0 = 9.0\) m/s - \(s_2 = 11.0(2.0) + \frac{1}{2}(-1.0)(2.0)^2 = 22 - 2 = 20\) m

Answers: Final velocity = 9.0 m/s, Total displacement = 44 m

4.7 Multiple Choice Questions

Test your understanding with these interactive questions:

4.8 Summary

Key Takeaways
  • Graphs encode motion in gradients and areas.
  • Straight lines indicate constant rates of change.
  • Areas under v-t and a-t graphs produce displacement and velocity changes.
  • Graph interpretation links visual data to equations.

4.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: