Graphical Analysis of Motion

Orientation

Lesson goal: build accurate physics fluency for graphical analysis of motion and use that fluency to support clear HSC-style scientific writing.

This page is materialised into the MentorMind course shell from existing teaching, textbook, and eduKG material. Use it as the main lesson surface; use the tutor for targeted repair, worked examples, and concise writing feedback.

Syllabus inquiry question

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

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.

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.

Content

Displacement-time graphs

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

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

Reading the graph:

Velocity-time graphs

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

Key relationships:

Acceleration-time graphs

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

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

Graph Comparison: Three Types of Motion

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

Worked Examples

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.

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.

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.

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)

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.

Practice Questions

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.

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)

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

Multiple Choice Questions

Test your understanding with these interactive questions:

Summary

• 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.

Self-Assessment

Check your understanding:

After studying this section, you should be able to:

• Calculate velocity from the gradient of an s-t graph
• Calculate acceleration from the gradient of a v-t graph
• Calculate displacement from the area under a v-t graph
• Distinguish between uniform and non-uniform motion on graphs
• Sketch graphs given a description of motion

Scientific Writing And Exam Support

When answering questions from this lesson, separate:

• the physical quantity being discussed,
• the model or law being applied,
• the mathematical relationship, including units,
• the conclusion in words.

For explanation questions, write in the pattern: claim -> physics reason -> consequence. For calculation questions, state the formula, substitute with units, calculate, then interpret the answer.

Maintenance Loop

One-minute retrieval:

1. State the key law, model, or relationship used in this lesson.
2. Identify one common misconception that would lead to a wrong answer.
3. Write one sentence that links the calculation or evidence back to the physical meaning.