Graphical representation of motion

A-Level Physics· Unit 2: Kinematics, Topic 2· 15 min read

1. Displacement-Time (s-t) Graphs★★☆☆☆⏱ 5 min

📘 Definition

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 $v = \frac{Δ s}{Δ t}$
2
$v = \frac{10 - 0}{2 - 0} = 5 \text{ m s}^{-1}$
3
1. Interval 2s to 4s:
4
$v = \frac{20 - 10}{4 - 2} = 5 \text{ m s}^{-1}$
5
1. Interval 4s to 6s: displacement is constant
6
$v = \frac{20 - 20}{6 - 4} = 0 \text{ m s}^{-1}$
7
1. Interval 6s to 8s: object returns to origin
8
$v = \frac{0 - 20}{8 - 6} = -10 \text{ m s}^{-1}$

2. Velocity-Time (v-t) Graphs★★★☆☆⏱ 6 min

📘 Definition

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 $2 m s^{- 2}$ 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
1. 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
1. 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.

3. Graphs for Non-Uniform Motion★★★★☆⏱ 4 min

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 $t = 3 s$ that passes through $(1 s, 2 m)$ and $(5 s, 20 m)$ . Find instantaneous velocity at $t = 3 s$ .

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
1. Calculate gradient:
4
$v = \frac{\Delta s}{\Delta t} = \frac{18}{4} = 4.5 \text{ m s}^{-1}$

4. Common Pitfalls

Wrong move:

Confusing gradient meanings: taking gradient of a v-t graph to get velocity

Why:

Each graph type has a different physical meaning for gradient, mixing them up leads to wrong answers

Correct move:

Memorize: gradient of s-t = velocity, gradient of v-t = acceleration

Wrong move:

Treating negative area as positive when calculating total displacement from v-t graphs

Why:

Negative velocity means motion in the opposite direction, so displacement in that region is negative

Correct move:

Add areas with their sign for displacement, add absolute values for distance

Wrong move:

Calculating gradient of the whole curved graph instead of tangent gradient for instantaneous values

Why:

For non-uniform motion, velocity/acceleration changes, so average gradient is not equal to the instantaneous value

Correct move:

Draw a tangent to the curve at the required time point and calculate gradient of the tangent only

Wrong move:

Ignoring the y-intercept when interpreting motion graphs

Why:

At $t = 0$ , the y-value gives the initial displacement (s-t) or initial velocity (v-t), which changes the final result

Correct move:

Always check the y-intercept first to account for initial conditions of motion

5. Quick Reference 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

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Interpret v-t graph for accelerating object
2021 · 2
Plot s-t graph from experimental data
2023 · 1
Calculate displacement from v-t graph

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.