# Speed, Velocity and Motion Graphs

> CIE IGCSE Physics · 0625 (2026-2028)
> Source: https://www.owlsprep.com/study/cie-0625-u1-speed-velocity-and-motion-graphs/

This guide covers core definitions of speed and velocity, plus interpretation of distance-time and velocity-time graphs aligned to CIE IGCSE Physics 0625 syllabus, including extended-level calculation skills.

**Prerequisites:** [Scalar and vector quantities](https://www.owlsprep.com/study/cie-0625-u1-scalars-vectors/); [Basic linear graph interpretation](https://www.owlsprep.com/study/cie-0625-maths-graph-basics/)

## Learning objectives

- Define and distinguish between scalar speed and vector velocity
- Calculate average speed using total distance divided by total time
- Interpret distance-time and velocity-time graphs for Core syllabus
- Calculate acceleration and distance from velocity-time graphs for Extended syllabus

## Core: Speed and Velocity Definitions

All motion quantities are classified as either scalar (magnitude only) or vector (magnitude + direction). Speed and velocity are two foundational motion quantities that follow this classification.

**Speed** — Scalar quantity measuring the rate of change of distance travelled, standard units m/s or km/h

*Example:* A car travelling at 25 m/s moves 25 metres every second, regardless of its direction of travel.

**Velocity** — Vector quantity measuring the rate of change of displacement (straight-line distance from start to end in a specific direction), standard units m/s or km/h with direction

*Example:* A car travelling north at 25 m/s has a velocity of 25 m/s north; if it turns south at the same speed, its velocity becomes 25 m/s south.

**Worked example:** A runner completes a 400 m circular lap in 80 seconds, ending exactly where they started. Calculate their average speed and average velocity.

1. Step 1: Calculate average speed using total distance divided by total time

   $$average\ speed = \frac{total\ distance}{total\ time} = \frac{400\ m}{80\ s} = 5\ m/s$$
2. Step 2: Calculate average velocity using total displacement divided by total time. Displacement is 0 m as the runner starts and ends at the same point

   $$average\ velocity = \frac{total\ displacement}{total\ time} = \frac{0\ m}{80\ s} = 0\ m/s$$

> **tip**
>
> If an object returns to its starting position, its average velocity will always be 0, no matter how fast it travelled.

> **Exam tip:** Examiners frequently ask to compare average speed and average velocity for objects that return to their start point, so always check if displacement is zero first.

## Core: Distance-Time Graphs

Distance-time graphs plot total distance travelled (y-axis) against time taken (x-axis). The gradient of any section of the graph equals the speed of the object over that time interval.

- Horizontal line = object is stationary (speed = 0 m/s)
- Straight sloped line = object is moving at constant speed
- Steeper gradient = faster constant speed
- Curved line = object is accelerating or decelerating (changing speed)

**Worked example:** A distance-time graph for a cyclist has a straight line from (0 s, 0 m) to (10 s, 50 m), then a horizontal line from (10 s, 50 m) to (20 s, 50 m). Calculate the cyclist's speed in the first 10 seconds, and state their speed between 10 and 20 seconds.

1. Step 1: Calculate gradient of the first section to find speed

   $$speed = \frac{change\ in\ distance}{change\ in\ time} = \frac{50\ m - 0\ m}{10\ s - 0\ s} = 5\ m/s$$
2. Step 2: The line between 10 and 20 s is horizontal, so gradient is 0, meaning speed = 0 m/s (the cyclist is stationary)

> **Exam tip:** Never calculate speed from the y-value of a distance-time graph directly; always use the gradient of the line segment.

## Core: Velocity-Time Graph Basics

Velocity-time graphs plot velocity (y-axis) against time (x-axis). For Core tier candidates, you only need to interpret these graphs qualitatively, not perform calculations.

- Positive (upward) gradient = object is accelerating (speeding up)
- Horizontal line = object is moving at constant velocity (no acceleration)
- Negative (downward) gradient = object is decelerating (slowing down)
- Line on x-axis = object is stationary (velocity = 0 m/s)

**Worked example:** A velocity-time graph has a positive gradient from 0 s to 5 s, a horizontal line from 5 s to 15 s, and a negative gradient from 15 s to 20 s where it meets the x-axis. Describe the object's motion over each interval.

1. 0-5 s: The object is speeding up at constant acceleration
2. 5-15 s: The object is moving at constant velocity with no acceleration
3. 15-20 s: The object is slowing down at constant deceleration, until it stops at 20 s

## Extended Only: Velocity-Time Graph Calculations

For Extended tier candidates, you will be required to perform two key calculations from velocity-time graphs: acceleration from the gradient, and total distance travelled from the area under the graph.

**Acceleration** — Rate of change of velocity, standard units m/s², calculated as $a = \frac{v - u}{t}$ where $v$ = final velocity, $u$ = initial velocity, $t$ = time taken

*Notation:* $a$

**Worked example:** A velocity-time graph has a straight line from (0 s, 0 m/s) to (4 s, 12 m/s), a horizontal line from (4 s, 12 m/s) to (10 s, 12 m/s), and a straight line from (10 s, 12 m/s) to (16 s, 0 m/s). Calculate the acceleration in the first 4 s, and total distance travelled over 16 s.

1. Step 1: Calculate acceleration from gradient of first section

   $$a = \frac{12 - 0}{4 - 0} = 3\ m/s^2$$
2. Step 2: Calculate total area under the graph to find total distance, split into 3 shapes: triangle, rectangle, triangle
3. Area 1 (0-4 s triangle):

   $$0.5 \times 4 \times 12 = 24\ m$$
4. Area 2 (4-10 s rectangle):

   $$6 \times 12 = 72\ m$$
5. Area 3 (10-16 s triangle):

   $$0.5 \times 6 \times 12 = 36\ m$$
6. Step 3: Sum areas for total distance

   $$24 + 72 + 36 = 132\ m$$

> **Exam tip:** You will never be asked to calculate the area under a curved velocity-time graph for 0625, so all graphs will be made of straight line segments that form simple shapes for area calculation.

## Common pitfalls

- **Wrong:** Using total distance to calculate average velocity
  - Why it fails: Velocity is a vector that depends on displacement (start to end straight line distance), not total distance travelled
  - Correct: Calculate average velocity using total displacement divided by total time; if the object returns to its start, average velocity is 0
- **Wrong:** Calculating speed from the gradient of a velocity-time graph
  - Why it fails: The gradient of a velocity-time graph gives acceleration, not speed
  - Correct: Use the gradient of a distance-time graph for speed, and gradient of a velocity-time graph for acceleration
- **Wrong:** Calculating distance from the gradient of a velocity-time graph
  - Why it fails: Gradient gives acceleration, while total distance travelled is equal to the area under the velocity-time graph
  - Correct: Sum the area of all simple shapes under the velocity-time graph to find total distance
- **Wrong:** Ignoring the direction of velocity on graphs that go below the x-axis
  - Why it fails: Velocity is a vector, so negative values mean the object is moving in the opposite direction to the chosen positive direction
  - Correct: Count area below the x-axis as distance travelled in the reverse direction when calculating total distance
- **Wrong:** Calculating average speed as $(initial + final)/2$ for non-uniform acceleration
  - Why it fails: This formula only works for constant acceleration, which is the only case tested, but using total distance over total time works for all scenarios
  - Correct: Always use $average\ speed = total\ distance / total\ time$ to avoid errors

## Cheatsheet

| Graph Type | Gradient Represents | Area Under Graph | Key Interpretation |
| --- | --- | --- | --- |
| Distance-Time | Speed (m/s) | No physical meaning | Horizontal line = stationary, sloped line = constant speed |
| Velocity-Time (Core) | Acceleration (qualitative) | Not assessed | Positive gradient = accelerating, horizontal = constant velocity |
| Velocity-Time (Extended) | Acceleration (m/s², quantitative) | Total distance (m) | Sum area of triangles/rectangles under graph for total distance |

## What's next

Now that you have mastered speed, velocity and motion graphs for CIE IGCSE Physics 0625, you are ready to progress to the next topics in the Motion, Forces and Energy unit. The next core topic is acceleration and Newton's laws of motion, where you will connect the motion graphs you just studied to the forces that cause changes in velocity. Extended candidates can proceed to learn about momentum, which builds directly on velocity and acceleration concepts. Be sure to practice tiered past paper questions on motion graphs, as they appear in almost every exam series, and use the cheatsheet above for quick last-minute revision before your test.

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