# Graphical representation of motion

> A-Level Physics · CIE 9702
> Source: https://www.owlsprep.com/study/cie-9702-u2-graphical-representation-of-motion/

This sub-topic explains how to interpret and use displacement-time and velocity-time graphs to solve kinematics problems. You will learn to calculate key motion quantities from gradients and areas under graphs for uniform and non-uniform motion.

**Prerequisites:** [Basic kinematic quantities (displacement, velocity, acceleration)](https://www.owlsprep.com/study/cie-9702-u2-basic-kinematic-quantities/); Calculating gradient and area under a curve

## Learning objectives

- Interpret displacement-time and velocity-time graphs for motion
- Calculate gradient to find velocity/acceleration from graphs
- Calculate area under velocity-time graph to find displacement
- Solve uniform and non-uniform motion problems graphically

## Displacement-Time (s-t) Graphs

**Displacement-Time Graph** — A graph with time on the horizontal x-axis and displacement on the vertical y-axis. The gradient at any point equals the instantaneous velocity of the object.

*Example:* A straight line with positive gradient represents constant velocity

For any s-t graph, the gradient between two points gives the average velocity over that time interval. A horizontal line means gradient is zero, so velocity is zero (object is stationary). A straight line means constant gradient, so velocity is constant.

**Worked example:** An object moves along a straight line. Its displacement at 5 time points is: (0s, 0m), (2s, 10m), (4s, 20m), (6s, 20m), (8s, 0m). Draw the s-t graph and find the velocity in each interval.

1. 1. Connect the points with straight line segments for each interval. First interval 0s to 2s: use $v = \frac{\Delta s}{\Delta t}$
2. $$v = \frac{10 - 0}{2 - 0} = 5 \text{ m s}^{-1}$$
3. 2. Interval 2s to 4s:
4. $$v = \frac{20 - 10}{4 - 2} = 5 \text{ m s}^{-1}$$
5. 3. Interval 4s to 6s: displacement is constant
6. $$v = \frac{20 - 20}{6 - 4} = 0 \text{ m s}^{-1}$$
7. 4. Interval 6s to 8s: object returns to origin
8. $$v = \frac{0 - 20}{8 - 6} = -10 \text{ m s}^{-1}$$

> **info**
>
> A negative gradient on an s-t graph always means the object is moving in the negative direction, back towards the starting point.

## Velocity-Time (v-t) Graphs

**Velocity-Time Graph** — A graph with time on the x-axis and velocity on the y-axis. Gradient equals acceleration, and total area between the graph and x-axis equals total displacement.

*Example:* A straight line through the origin represents constant acceleration from rest

Gradient between two points on a v-t graph gives average acceleration over that interval. A horizontal line means gradient is zero, so acceleration is zero (constant velocity). A straight line means constant acceleration.

Areas above the x-axis are positive displacement, areas below are negative. To get total distance travelled, sum the absolute values of all areas.

**Worked example:** A car accelerates from rest at $2 \text{ m s}^{-2}$ for 5s, travels at constant velocity for 10s, then decelerates to rest in 5s. Find total displacement.

1. 1. Calculate maximum velocity after acceleration:
2. $$v = u + at = 0 + (2)(5) = 10 \text{ m s}^{-1}$$
3. 2. The v-t graph forms a trapezium. Area of trapezium gives displacement:
4. $$s = \frac{1}{2}(a + b)h = \frac{1}{2}(20 + 10) \times 10 = 150 \text{ m}$$
5. 3. Alternative split into three shapes confirms the result:
6. $$(\frac{1}{2} \times 5 \times 10) + (10 \times 10) + (\frac{1}{2} \times 5 \times 10) = 150 \text{ m}$$

> **Exam tip:** Always check if the question asks for displacement or distance. For distance, add all areas regardless of sign.

## Graphs for Non-Uniform Motion

When velocity or acceleration is not constant, graphs are curved instead of straight lines, but the same core rules for gradient and area still apply.

For a curved s-t graph, the gradient of the tangent to the curve at a point gives the instantaneous velocity at that point. Increasing gradient means increasing velocity (acceleration), while decreasing gradient means deceleration.

For a curved v-t graph, the gradient of the tangent gives instantaneous acceleration. To find the area under a curved graph, you can estimate it by counting unit squares in an exam.

**Worked example:** A curved s-t graph for an accelerating car has a tangent at $t=3\text{s}$ that passes through $(1\text{s}, 2\text{m})$ and $(5\text{s}, 20\text{m})$. Find instantaneous velocity at $t=3\text{s}$.

1. 1. Instantaneous velocity equals gradient of the tangent. Calculate change in displacement and change in time:
2. $$\Delta s = 20 - 2 = 18 \text{ m}, \Delta t = 5 - 1 = 4 \text{ s}$$
3. 2. Calculate gradient:
4. $$v = \frac{\Delta s}{\Delta t} = \frac{18}{4} = 4.5 \text{ m s}^{-1}$$

> **note**
>
> When counting squares, first calculate the value of one square by multiplying the x-axis unit value by the y-axis unit value, then multiply by the number of squares.

## Common pitfalls

- **Wrong:** Confusing gradient meanings: taking gradient of a v-t graph to get velocity
  - Why it fails: Each graph type has a different physical meaning for gradient, mixing them up leads to wrong answers
  - Correct: Memorize: gradient of s-t = velocity, gradient of v-t = acceleration
- **Wrong:** Treating negative area as positive when calculating total displacement from v-t graphs
  - Why it fails: Negative velocity means motion in the opposite direction, so displacement in that region is negative
  - Correct: Add areas with their sign for displacement, add absolute values for distance
- **Wrong:** Calculating gradient of the whole curved graph instead of tangent gradient for instantaneous values
  - Why it fails: For non-uniform motion, velocity/acceleration changes, so average gradient is not equal to the instantaneous value
  - Correct: Draw a tangent to the curve at the required time point and calculate gradient of the tangent only
- **Wrong:** Ignoring the y-intercept when interpreting motion graphs
  - Why it fails: At $t=0$, the y-value gives the initial displacement (s-t) or initial velocity (v-t), which changes the final result
  - Correct: Always check the y-intercept first to account for initial conditions of motion

## Cheatsheet

| Graph Type | Gradient Meaning | Area Under Graph Meaning |
| --- | --- | --- |
| Displacement-Time (s-t) | Velocity | N/A |
| Velocity-Time (v-t) | Acceleration | Displacement |
| Straight line | Constant gradient quantity | - |
| Horizontal line | Zero gradient quantity | - |
| Curved line | Tangent = instantaneous value | Count squares to estimate area |

## What's next

Graphical representation of motion is the foundation for solving more complex kinematics problems, including projectile motion and connected systems. Understanding how gradients and areas relate to physical quantities also prepares you for topics like Newton's laws of motion, where you will analyse force-time and force-displacement graphs. Mastery of this sub-topic is critical for both multiple choice and structured questions in CIE 9702, as it frequently appears combined with other kinematics or dynamics concepts. Next, you will learn the kinematic equations for constant acceleration, which connect directly to the graphical methods you learned here.

- [Projectile motion](https://www.owlsprep.com/study/cie-9702-u2-projectile-motion/)
- [Uniformly accelerated motion equations](https://www.owlsprep.com/study/cie-9702-u2-uniformly-accelerated-motion-equations/)
- [Free fall](https://www.owlsprep.com/study/cie-9702-u2-free-fall/)

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