Reading a displacement–time graph

7 minAS Level4.2.2

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can read a displacement–time graph, use its gradient to find velocity, and sketch a graph from a description of a journey.

Key to the symbols

Symbol	Meaning
$s$	displacement (position measured from a fixed starting point), in m
$t$	time, in s
$v$	velocity, in $\text{m s}^{-1}$
$\Delta$	"the change in", so $\Delta s$ means the change in $s$

A graph that answers "where?"

A displacement–time graph plots displacement on the vertical axis and time on the horizontal axis. Each point answers the question "where was the particle at this moment?"

Recall that the gradient of a straight line is "rise over run": the change in the vertical quantity divided by the change in the horizontal quantity.

Here that is

$\text{gradient} = \frac{\Delta s}{\Delta t} = \text{velocity}$

So the geometry of the graph is the motion.

Definition

Gradient of a displacement–time graph

The gradient of a displacement–time graph at a point equals the velocity there. On a straight section the gradient is constant, so the velocity is constant.

What the shapes mean

Feature of the graph	What it means
Straight line sloping up	constant positive velocity
Horizontal line	at rest ( $v = 0$ )
Straight line sloping down	constant negative velocity (moving back)
Steeper line	larger speed
Curve	changing velocity (speeding up or slowing down)
Two graphs crossing	the particles are at the same displacement at that time

The sign of the gradient gives the direction; the size of the gradient gives the speed.

A tangent is a straight line that just touches the curve at one point, going the same way as the curve there. A chord is a straight line joining two points on the curve.

For a curved graph, the gradient at a single instant is the gradient of the tangent at that point. That is the instantaneous velocity. The gradient of a chord between two points gives the average velocity over that interval.

Worked example: Three sections, three velocities

A particle has displacement $0\ \text{m}$ at $t = 0$ , $12\ \text{m}$ at $t = 4\ \text{s}$ , $12\ \text{m}$ at $t = 10\ \text{s}$ and $2\ \text{m}$ at $t = 14\ \text{s}$ . Each section is a straight line.

Find the velocity in each section, and the average velocity over the whole $14\ \text{s}$ .

Have a go first?

Sketching from a story

To sketch a displacement–time graph:

Mark the starting displacement at $t = 0$ .
Break the journey into stages where the velocity is constant.
For each stage, draw a straight line whose gradient is that velocity.
Where the particle stops, draw a horizontal line.
Where it turns round, the gradient changes sign.

A common task is a cyclist who rides away, waits, then returns past the start. The graph rises, flattens, then falls, and the falling line may cross the axis if the cyclist goes beyond the starting point.

Common mistakes

A horizontal section means the particle is moving at a constant speed.

A horizontal section means the displacement is not changing, so the particle is at rest: velocity zero. (A constant non-zero speed gives a sloping straight line.)

The graph shows the path the particle took through space.

It shows displacement against time. A particle moving in a straight line can produce a graph that goes up and down.

A steeper line means a larger displacement.

Steepness measures how fast the displacement changes: it is the velocity, not the size of the displacement.

The average velocity is the total distance divided by the time.

In the exam

When asked for a velocity from a graph, show the division: $\dfrac{\Delta s}{\Delta t}$ with the two readings. Method marks are for using the gradient.

For an instantaneous velocity on a curve, draw the tangent and use two points on it. For an average velocity, use the two endpoints of the interval, not the tangent.

If the graph crosses the time axis, the particle has passed the starting point; the velocity does not suddenly become zero there.

Check yourself

CalculateQuestion 1 · Recall

CalculateQuestion 2 · Apply

Answer the 2 questions above to continue.