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.

Source Lesson Material

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.

Tutor Context

Use this lesson context when the student asks about graphical analysis of motion, related calculations, representations, or scientific writing. Prefer a short diagnostic before re-teaching. Check whether the student is confusing closely related categories such as force, velocity, acceleration, field, energy, momentum, model evidence, or mathematical representation.

Useful tutor diagnostic:

Which quantity is changing here, what causes that change, and what unit should the final answer use?

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.

Source Trace

This content record was materialised from:

• edu/physics-prep/hsc/physics/textbook/module-1/sections/m1-2-graphical-analysis.qmd
• edu/physics-prep/hsc/physics/eduKG/lessons/program/year-11/module-1/T1W1L2-graphical-analysis.qmd
• edu/physics-prep/hsc/physics/eduKG/textbook/_custom-year11/chapters/m1-topic-1-2.qmd