Velocity-time graphs — reading motion — Physics

Before this

The SUVAT equations
Distance-time and speed-time graphs · coming soon
Gradient — steepness and direction · coming soon

This leads to

A story told by one line

Here is a graph of a car's velocity during a short journey. Before you read any further, try to tell its story. When is the car speeding up? When is it steady? When is it slowing down? And what is happening after 12 seconds, where the line drops below the axis?

The velocity of a car over 16 s, in five stages, A to E.

Most students can say something about stages A, B and C. Stages D and E usually cause confusion. A velocity below zero seems strange, and it is not obvious how far the car has travelled at all.

By the end of this lecture, you will read this graph like a sentence. You will be able to find the acceleration in every stage without any formula from memory, find how far the car travelled, and find where it ends up. Only two ideas are needed: the gradient of the line, and the area under it.

The axes: what a point on the graph means

A velocity-time graph, or v-t graph, has time on the horizontal axis and velocity on the vertical axis. Each point on the line answers one question: how fast, and in which direction, was the object moving at that instant?

Velocity has a direction, so it can be negative. As in the last lecture, we choose one direction as positive, here "forwards". A point above the time axis means the car is moving forwards at that moment. A point below the axis means it is moving backwards. A point on the axis means it is stopped, if only for an instant.

Idea 1: the gradient is the acceleration

Acceleration is the change in velocity divided by the time taken: $a = \frac{Δ v}{Δ t}$ . Now look at what the gradient of a line means. The gradient is the rise divided by the run. On a v-t graph, the rise is a change in velocity, and the run is a change in time. So the gradient is $\frac{Δ v}{Δ t}$ . That is exactly the acceleration.

Reading a v-t graph

The gradient of a v-t graph is the acceleration. The area between the line and the time axis is the displacement.

This one idea tells you a lot from the shape alone:

Line sloping up: the velocity is increasing, so the acceleration is positive. A steeper line means a bigger acceleration.
Horizontal line: the velocity is not changing. The acceleration is zero, and the object moves at constant velocity. It is not stopped, unless the line is on the axis.
Line sloping down: the acceleration is negative.

Read the units too. The gradient is m/s divided by s, which is m/s². The units confirm that the gradient is an acceleration.

Apply this to the journey. In stage A, the velocity rises from 0 to 8 m/s in 4 s. So $a = \frac{8 - 0}{4} = 2$ m/s². In stage B, the line is flat, so $a = 0$ . In stage C, the velocity falls from 8 m/s to 0 in 4 s, so $a = \frac{0 - 8}{4} = - 2$ m/s².

Idea 2: the area is the displacement

The last lecture promised a link between the SUVAT equations and graphs. Here it is.

Start with the simplest case. In stage B, the car moves at 8 m/s for 4 s. It travels $8 \times 4 = 32$ m. On the graph, stage B is a rectangle 4 s wide and 8 m/s tall. Its area is also $8 \times 4 = 32$ . That is not a coincidence. For constant velocity, displacement is velocity × time, and a rectangle's area is height × width. They are the same calculation.

Now stage A, where the velocity changes. The shape under the line is a triangle. Its area is $\frac{1}{2} \times 4 \times 8 = 16$ . In the last lecture you saw that, for constant acceleration, displacement = average velocity × time, and the average velocity is halfway between the start and end velocities. In stage A that is 4 m/s, so the displacement is $4 \times 4 = 16$ m. The triangle's area gives the same number. In fact, the area of a trapezium under a straight line is $\frac{1}{2} (u + v) t$ , which is exactly the SUVAT equation $s = \frac{1}{2} (u + v) t$ . The graph and the equation are the same idea, drawn and written.

Does the area rule work when the line is curved? Yes. Cut the time into very thin strips. In each thin strip, the velocity hardly changes, so the strip is almost a rectangle, and its area is almost exactly the displacement in that short time. Add up all the strips, and you get the total displacement. The thinner the strips, the better the answer. That "adding up thin strips" is the central idea of a part of mathematics called integration.

Below the axis: going backwards

Now stages D and E. At $t = 12$ s the line crosses the axis. The velocity is zero there: the car has stopped, just for an instant. After that, the velocity is negative. The car is reversing.

The area of a region below the axis counts as negative displacement: movement in the backwards direction. Stage D is a triangle with area $\frac{1}{2} \times 2 \times 4 = 4$ , below the axis, so its displacement is −4 m. Stage E is the same: −4 m.

This gives two different totals, and they answer two different questions:

Displacement (where did the car end up?): add the areas with their signs. $16 + 32 + 16 - 4 - 4 = 56$ m. The car finishes 56 m in front of its starting point.
Distance travelled (how far did the wheels roll?): add the sizes of the areas, ignoring signs. $16 + 32 + 16 + 4 + 4 = 72$ m.

Watch the car move along the road as you move through the journey below. The shaded area always matches the car's displacement.

Graph — the area becomes the position

Move the time slider. Watch the shaded area grow, and the car move along the road above the graph.

Time: 6 s

Each stage of the journey, read from the graph:

Stage	Time (s)	Gradient: $a$ (m/s²)	Area: displacement (m)	Displacement so far (m)
A	0–4	+2	+16	16
B	4–8	0	+32	48
C	8–12	−2	+16	64
D	12–14	−2	−4	60
E	14–16	+2	−4	56

The car is furthest forward, 64 m, at 12 s. It then reverses 8 m and finishes 56 m from the start, after travelling 72 m in total.

Two objects on one graph

A v-t graph can show two objects at once. The higher line at any moment belongs to the faster object at that moment. The steeper line belongs to the object with the bigger acceleration. And where the lines cross, the two objects have the same velocity at that instant. That last sentence is where a very common mistake begins. Test yourself.

Predict, then reveal — where the lines cross

Two cars, P and Q, start side by side at $t = 0$ . P moves at a steady 6 m/s. Q starts from rest and accelerates at 1.5 m/s². Their v-t lines cross at $t = 4$ s. At that moment, where are the cars? Choose one, then press "Show me".

The question: P moves at a steady 6 m/s. Q starts from rest beside it and accelerates at 1.5 m/s². Their v-t lines cross at 4 s. Where are the cars then?

Up to 4 s, the area under P is 24 m and the area under Q is 12 m.

Answer: P is 12 m ahead. Crossing lines mean equal velocities, not equal positions. Position comes from the area: P has covered 24 m and Q only 12 m. Q catches P at $t = 8$ s, when both areas equal 48 m.

Stop and think

A ball is thrown straight up. At the very top of its flight, its velocity is zero. Is its acceleration zero there too?

Show the answer

No. Gravity does not switch off at the top. The acceleration is −9.8 m/s² (taking up as positive) for the whole flight, on the way up, at the top, and on the way down. On a v-t graph, the flight is one straight line with the same gradient everywhere. The line passes through zero at the top, but its gradient, the acceleration, never changes. Zero velocity and zero acceleration are completely different things.

Worked examples

Worked Example 1 — A cyclist

A cyclist starts from rest and accelerates steadily to 8 m/s in 4 s. She then rides at constant speed for 6 s. Find her acceleration in the first stage and her total displacement.

The v-t graph rises in a straight line from (0, 0) to (4, 8), then runs flat to (10, 8).

Acceleration = gradient of the first part: $a = \frac{8 - 0}{4 - 0} = 2$ m/s².

Displacement = area = triangle + rectangle:

s = \frac{1}{2} \times 4 \times 8 + 6 \times 8 = 16 + 48 = 64 m

Check with SUVAT for the first stage: $s = \frac{1}{2} (0 + 8) \times 4 = 16$ m. The graph and the equation agree.

Worked Example 2 — A ball thrown upwards

A ball is thrown straight up at 15 m/s. Take up as positive and $g = 9.8$ m/s². Describe its v-t graph. Find when it reaches the top, how high it goes, and what happens next.

The acceleration is −9.8 m/s² throughout, so the graph is one straight line starting at +15 m/s with gradient −9.8.

At the top, $v = 0$ : $0 = 15 - 9.8 t$ , so $t = \frac{15}{9.8} \approx 1.53$ s.

Height = area of the triangle above the axis: $\frac{1}{2} \times 1.53 \times 15 \approx 11.5$ m.

After 1.53 s, the line continues straight on, below the axis: the ball is falling, so its velocity is negative. At about 3.06 s, the triangle below the axis has the same area as the one above. The total displacement is zero: the ball is back at the height it was thrown from, moving down at 15 m/s. The distance travelled is about $2 \times 11.5 = 23$ m.

Worked Example 3 — The opening journey, in full

For the car at the start of this lecture, find the acceleration in each stage, the final displacement, the total distance and the average velocity over the 16 s.

Gradients: A +2 m/s², B 0, C −2 m/s², D −2 m/s², E +2 m/s².

Notice D and E. In D, the acceleration is negative while the velocity is also negative. The car is speeding up backwards. In E, the acceleration is positive while the velocity is negative. The car is slowing down while still reversing. The sign of the acceleration alone does not tell you "speeding up" or "slowing down". Compare it with the sign of the velocity.

Displacement = $16 + 32 + 16 - 4 - 4 = 56$ m. Distance = 72 m.

Average velocity = displacement ÷ time = $56 \div 16 = 3.5$ m/s.

Three classic mistakes

Trap 1 — Reading a flat line as "stopped"

A horizontal line on a v-t graph means constant velocity. The object is stopped only if the line lies on the time axis itself. On a displacement-time graph, which you will meet next, a flat line does mean stopped, so check the axis label every time.

Trap 2 — Treating crossing lines as a meeting

Where two lines cross on a v-t graph, the velocities are equal. Positions are found from areas, not from where the lines cross.

Trap 3 — Adding areas below the axis as positive

For displacement, area below the axis is negative. For distance, add every area as positive. Read the question carefully, and say which one you are finding.

Try it yourself

Sketch the graph first, every time. Open a hint only when you are stuck.

A train accelerates steadily from rest to 20 m/s in 10 s, travels at 20 m/s for 30 s, then slows steadily to rest in 20 s. Find the acceleration in the first and last stages, the total distance, and the average speed.

Hint 1
The graph is a trapezium: a rise, a flat top and a fall.

Hint 2
Split the area into a triangle, a rectangle and a triangle.

Full answer
First stage: $a = \frac{20}{10} = 2$ m/s². Last stage: $a = \frac{0 - 20}{20} = - 1$ m/s². Distance = $\frac{1}{2} \times 10 \times 20 + 30 \times 20 + \frac{1}{2} \times 20 \times 20 = 100 + 600 + 200 = 900$ m. Average speed = $900 \div 60 = 15$ m/s.
A v-t graph is a straight line from (0 s, 12 m/s) to (6 s, 0 m/s). Describe the motion. Find the acceleration and the displacement.

Hint 1
Is the line going up or down? Where does it end?

Hint 2
The shape under the line is a triangle.

Full answer
The object starts at 12 m/s and slows steadily, stopping at 6 s. $a = \frac{0 - 12}{6} = - 2$ m/s². Displacement = $\frac{1}{2} \times 6 \times 12 = 36$ m.
A toy car's velocity changes steadily from +6 m/s to −6 m/s in 6 s. Find its acceleration, its displacement after 6 s, and the distance it travels.

Hint 1
When is the velocity zero? Split the graph there.

Hint 2
You get two equal triangles, one above the axis and one below.

Full answer
$a = \frac{- 6 - 6}{6} = - 2$ m/s². The velocity is zero at 3 s. Each triangle has area $\frac{1}{2} \times 3 \times 6 = 9$ . Displacement = $9 - 9 = 0$ : the car ends where it started. Distance = $9 + 9 = 18$ m.
Runner A runs at a steady 5 m/s. Runner B starts beside A at the same moment, from rest, accelerates at 1 m/s² until she reaches 6 m/s, then runs at 6 m/s. At what time do their v-t lines cross? When and where does B catch A?

Hint 1
The lines cross when B's velocity is 5 m/s. B reaches 6 m/s at 6 s.

Hint 2
B catches A when their areas are equal. Write each area as an expression in $t$ for $t > 6$ .

Full answer
The lines cross at $t = 5$ s, when B's velocity reaches 5 m/s. That is not where B catches A. Up to 6 s, B covers $\frac{1}{2} \times 6 \times 6 = 18$ m, while A covers 30 m. After that, B's area is $18 + 6 (t - 6)$ and A's is $5 t$ . Equal when $18 + 6 t - 36 = 5 t$ , so $t = 18$ s. Both have run $5 \times 18 = 90$ m.

The bigger idea

You have just used the two central operations of calculus without the names. Finding the gradient of a graph is called differentiation. Finding the area under a graph is called integration. Velocity comes from displacement by the first, and displacement comes back from velocity by the second.

Your phone does this every day. Its accelerometer measures acceleration many times a second. Adding up acceleration over time gives velocity, and adding up velocity gives position. This is one way a phone can keep track of movement in a tunnel, where satellite signals cannot reach it. It is the area under a graph, done by a machine.

Key concepts in this lesson

On a v-t graph, the gradient is the acceleration, because gradient is $\frac{Δ v}{Δ t}$ .
The area between the line and the time axis is the displacement, and for a straight line it matches $s = \frac{1}{2} (u + v) t$ .
Velocity below the time axis means motion in the negative direction, and area below the axis counts as negative displacement.
Displacement adds areas with their signs, while distance adds their sizes.
A horizontal line means constant velocity, not rest, and crossing lines mean equal velocities, not equal positions.
Speeding up or slowing down depends on whether the acceleration and the velocity have the same sign.