Reading a velocity–time graph

8 minAS Level4.2.2

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

By the end of this lesson

By the end you can use a velocity–time graph to find acceleration from the gradient and displacement from the area, including when part of the graph lies below the time axis.

Key to the symbols

Symbol	Meaning
$v$	velocity, in $\text{m s}^{-1}$
$t$	time, in s
$a$	acceleration, in $\text{m s}^{-2}$
$g$	acceleration due to gravity; take $g = 10\ \text{m s}^{-2}$
$\Delta$	"the change in", so $\Delta v$ means the change in $v$

Uniformly means at a constant rate. Decelerates means slows down: the acceleration points opposite to the velocity.

One graph, two questions

A velocity–time graph answers two different questions, and examiners test both.

How fast is the velocity changing? That is the gradient: acceleration.
How far has the particle moved? That is the area between the graph and the time axis: displacement.

Why area? At a constant velocity, displacement = velocity × time. On the graph, that's the area of a rectangle: height $v$ times width $t$ .

When the graph slopes or curves, cut the area into many thin rectangles and add them up. That's exactly what integration does.

Definition

Velocity–time graph facts

Gradient $= \dfrac{\Delta v}{\Delta t} =$ acceleration.
Area between the graph and the time axis $=$ displacement.
Area below the axis counts as negative displacement, because the velocity is negative there.
Total distance is the sum of the sizes of all the areas, ignoring signs.

What the shapes mean

Feature of the graph	What it means
Horizontal line	constant velocity, so $a = 0$
Straight line sloping up	constant positive acceleration
Straight line sloping down	constant negative acceleration
Crossing the time axis	the particle reverses direction
Touching the time axis	the particle is momentarily at rest
Curve	changing acceleration

Do not confuse this with the displacement–time graph. There, a horizontal line meant "at rest"; here it means "moving at a steady speed".

Worked example: Accelerate, cruise, brake

A train accelerates uniformly from rest to $12\ \text{m s}^{-1}$ in $3\ \text{s}$ , travels at a constant $12\ \text{m s}^{-1}$ for $5\ \text{s}$ , then decelerates uniformly to rest in $4\ \text{s}$ .

Find (a) the acceleration in each stage, (b) the total displacement, (c) the total distance.

Have a go first?

When the graph dips below the axis

Suppose a ball is thrown straight up at $20\ \text{m s}^{-1}$ , taking $g = 10\ \text{m s}^{-2}$ . Take upwards as positive. Gravity reduces the upward velocity by $10\ \text{m s}^{-1}$ every second, so the acceleration is $-10\ \text{m s}^{-2}$ and the velocity after $t$ seconds is $v = 20 - 10t$ , so the graph is a straight line through $v = 0$ at $t = 2\ \text{s}$ and $v = -20$ at $t = 4\ \text{s}$ .

From $t = 0$ to $2$ : area $= \tfrac{1}{2}(2)(20) = +20\ \text{m}$ (going up).
From $t = 2$ to $4$ : area $= \tfrac{1}{2}(2)(-20) = -20\ \text{m}$ (coming back down).

Displacement at $t = 4$ is $20 - 20 = 0$ : the ball is back at the start.

Distance travelled is $20 + 20 = 40\ \text{m}$ : it really did go up and come back.

This is the classic trap. A negative area is not a mistake in the arithmetic; it is the particle moving in the negative direction.

Common mistakes

The area under a velocity–time graph is the distance travelled.

It is the displacement, signed. The distance is the sum of the sizes of the areas, so the two agree only if the velocity never changes sign.

A horizontal velocity–time line means the particle is stationary.

It means constant velocity. The particle is stationary only where the graph is on the axis.

Where the graph crosses the time axis, the particle stops.

It is momentarily at rest at that instant, then continues in the opposite direction.

Negative acceleration means the particle is moving backwards.

In the exam

Always say which question you are answering: gradient for acceleration, area for displacement.

Split the area into triangles, rectangles and trapezia and show the arithmetic. For a trapezium, $\tfrac{1}{2}(\text{sum of the parallel sides})\times\text{height}$ is quicker and safer than three separate pieces.

If part of the graph is below the axis, state clearly that this area is negative when finding displacement, and that you are taking the size of it when finding distance.

Check yourself

CalculateQuestion 1 · Recall

CalculateQuestion 2 · Apply

Answer the 2 questions above to continue.