Displacement-time graphs — position over time — Physics

Before this

Velocity-time graphs
Distance-time and speed-time graphs · coming soon
Gradient — steepness and direction · coming soon

This leads to

Solving kinematics problems step by step
Projectile motion
Differentiation from first principles · coming soon

The same journey, told a different way

In the last lecture, a car made a 16-second journey in five stages. It sped up, cruised, slowed to a stop, reversed, and stopped again. A velocity-time graph told that story, and the area under it told us where the car ended up: 56 m in front of its start.

But the v-t graph hides one thing in plain view. To find where the car is at any moment, you have to add up areas. Is there a graph that shows the position directly, at a glance?

There is. Plot the car's position against time, and you get a displacement-time graph. Every point on it answers one question: where is the object at this instant? And here is the surprising part. The new graph still contains the velocity. You only have to know where to look. This lecture shows where.

What a point on the graph means

A displacement-time graph, or s-t graph, has time $t$ on the horizontal axis and displacement $s$ on the vertical axis. Displacement is the position measured from a fixed starting point, with a direction. As always, we choose one direction as positive at the start.

So a point at $s = 20$ m means the object is 20 m on the positive side of its start. A point at $s = 0$ means it is exactly at its start. A point below the axis means it is on the negative side. Notice what this does not tell you. A point below the axis says nothing about which way the object is moving at that moment. It only says where the object is.

The gradient is the velocity

Take the simplest motion: an object moving at a steady 4 m/s. Every second, its displacement grows by 4 m. On an s-t graph, that is a straight line that rises 4 m for every 1 s across.

Now look at the gradient of that line. The gradient is the rise divided by the run. On an s-t graph, the rise is a change in displacement, $Δ s$ , and the run is a change in time, $Δ t$ . So the gradient is $Δ s /Δ t$ . That is exactly the definition of velocity.

Reading an s-t graph

The gradient of a displacement-time graph is the velocity: $v = \frac{Δ s}{Δ t}$ .

A steeper line means a faster object. A positive gradient means motion in the positive direction, a negative gradient means motion in the negative direction, and a zero gradient means the object is stationary.

Check the units, as you did for the v-t graph. The gradient is metres divided by seconds, which is m/s. The units confirm that the gradient is a velocity.

Here is a three-stage journey. Each stage is a straight line, so each has one constant velocity.

A three-stage journey. The dashed triangle on the first stage shows the gradient: Δs = 20 m over Δt = 5 s, so v = 4 m/s.

Stage 1 rises 20 m in 5 s, so $v = 20 \div 5 = 4$ m/s. Stage 2 is flat, so $v = 0$ : the object is standing still. Stage 3 rises from 20 m to 35 m in 5 s, so $v = 15 \div 5 = 3$ m/s. The stage 3 line is less steep than stage 1, and the numbers agree: 3 m/s is slower than 4 m/s.

Lines, curves and what they mean

A straight line has the same gradient everywhere, so the velocity is constant. That means the acceleration is zero. So on an s-t graph, a straight line, even a steep one, means no acceleration.

Acceleration shows up as a curve. If the curve gets steeper as time goes on, the velocity is increasing: the object is speeding up. If the curve gets flatter, the velocity is shrinking: the object is slowing down.

Compare this with the last lecture, because this is where students mix the two graphs up:

Shape	On an s-t graph	On a v-t graph
Horizontal line	stationary	constant velocity
Straight sloping line	constant velocity	constant acceleration
Curve	accelerating	changing acceleration

The same shape means different things on the two graphs. The axis label decides, every time.

Gradients on a curve: the tangent

How do you find a gradient when the line is curved? A curve has a different steepness at every point. So we need the gradient at one instant.

Draw a tangent: a straight line that touches the curve at that point and runs in the same direction as the curve there. The gradient of the tangent is the gradient of the curve at that instant. And that is the velocity at that instant.

Here is the car's journey from the last lecture, drawn as an s-t graph. Slide the tangent along it and compare its gradient with the velocity the v-t graph gave.

Explore — the tangent's gradient is the velocity

Move the time. The dashed tangent touches the curve at that instant. Watch its gradient, and where it becomes zero.

Time: 10 s

The car journey as an s-t graph, with the tangent at 10 s, and the gradient at five moments:

Stages A to E of the journey. At 10 s the tangent rises 12 m over 3 s, so the velocity is 4 m/s.

t (s)	s (m)	Gradient = v (m/s)	What the car is doing
2	4	4	speeding up (curve getting steeper)
6	32	8	steady (straight line)
10	60	4	slowing down (curve getting flatter)
12	64	0	stopped for an instant (top of the hill)
15	57	−2	reversing (gradient negative)

Look at what the tangent shows. At 12 s, the curve reaches its highest point, and the tangent there is flat. A flat tangent means zero velocity: the car has stopped for an instant before it starts to reverse. After 12 s, the tangent slopes downwards, and the velocity is negative. Every value matches the v-t graph from the last lecture. The two graphs tell the same story.

That is it. That is the whole idea. Position is the main character on an s-t graph, and velocity is its steepness.

Below the axis, and moving backwards: not the same thing

This is the hard part of the lecture. Spend time here. Two different facts are easy to confuse:

Where the line is (above or below the axis) tells you which side of the start the object is on.
Which way the line slopes (up or down) tells you which way the object is moving.

Look at the car between 12 s and 16 s. The line slopes down, so the car is moving in the negative direction. But the line is still far above the axis. The car is still more than 50 m in front of its start. It is reversing towards the start, not behind it. A negative gradient does not mean "behind the start". It means "moving in the negative direction".

Stop and think

In the last lecture, two cars' lines crossed on a v-t graph, and the cars were not side by side. Now two runners' lines cross on an s-t graph at $t = 6$ s. What does the crossing mean this time?

Show the answer

It means the runners are at the same position at 6 s: they meet, or one overtakes the other. On an s-t graph, the height of a line is the position, so equal heights mean equal positions. On a v-t graph, equal heights mean equal velocities, which is why the cars were not side by side. The same crossing tells two different stories, depending on the axis.

Displacement and distance

Displacement is the change in position, with a sign. Distance is the total length of the path, always positive. They are equal only when the object never turns back.

On an s-t graph, the displacement at any time is simply the height of the line. To find the distance, follow the line and add up every rise and every fall as positive amounts. For the car: up 64 m to the top of the hill, then down 8 m. The distance is $64 + 8 = 72$ m, but the displacement at 16 s is only 56 m. These are the same two numbers the areas gave in the last lecture.

Worked examples

Worked Example 1 — A three-stage journey

For the three-stage graph above, find the velocity in each stage, the final displacement and the total distance.

Stage 1 (0 to 5 s): $v = \frac{20 - 0}{5 - 0} = 4$ m/s. Stage 2 (5 to 8 s): flat, so $v = 0$ . Stage 3 (8 to 13 s): $v = \frac{35 - 20}{13 - 8} = 3$ m/s.

Final displacement: the height at 13 s, which is 35 m. Distance: $20 + 0 + 15 = 35$ m.

Check: the distance equals the displacement here because the line never goes down. The object never turned back.

Worked Example 2 — Turning back past the start

An object's s-t graph rises in a straight line from $s = 0$ to $s = 30$ m at $t = 6$ s, then falls in a straight line to $s = - 10$ m at $t = 11$ s. Take the first direction of motion as positive. Find (a) the velocity from 6 s to 11 s, (b) the distance travelled, and (c) the average velocity and the average speed over the 11 s.

(a) $v = \frac{Δ s}{Δ t} = \frac{- 10 - 30}{11 - 6} = \frac{- 40}{5} = - 8$ m/s. The minus sign means the object moves in the negative direction, at 8 m/s.

(b) The object goes 30 m out, then 40 m back (from 30 m to −10 m). Distance $= 30 + 40 = 70$ m. The displacement is only −10 m.

(c) Average velocity uses displacement: $- 10 \div 11 \approx - 0.91$ m/s. Average speed uses distance: $70 \div 11 \approx 6.36$ m/s. They differ a lot, because the object turned back.

Check: from 0 to 6 s, $v = 30 \div 6 = 5$ m/s. Distance from speeds: $5 \times 6 + 8 \times 5 = 30 + 40 = 70$ m. The two methods agree.

Worked Example 3 — A gradient from a tangent

On the car's s-t graph, find the velocity at $t = 2$ s by drawing a tangent.

At $t = 2$ s, the car is at $s = 4$ m. A tangent drawn carefully at that point passes through $(1, 0)$ and $(3, 8)$ . Choose two points on the tangent that are far apart, because a big triangle gives a more accurate gradient.

v = \frac{8 - 0}{3 - 1} = \frac{8}{2} = 4 m/s

Check with the v-t graph: in stage A the acceleration is 2 m/s² from rest, so at 2 s, $v = 2 \times 2 = 4$ m/s. The tangent and the v-t graph agree. In an exam, a hand-drawn tangent gives an answer close to this, not exactly. Examiners accept a small range.

One error appears again and again when students find a distance from an s-t graph. Find it in the solution below.

Spot the error — how far did it travel?

An object's s-t graph rises from 0 to 24 m in 4 s, then falls to −6 m at 10 s. A student finds the distance travelled. One line is wrong. Tap it.

The student's solution, with the error marked:

1. Stage 1: Δs = 24 − 0 = 24 m. Correct.
2. Stage 2: Δs = −6 − 24 = −30 m. Correct: this is the change in displacement.
3. Velocity in stage 2 = −30 ÷ 6 = −5 m/s. Correct.
4. Distance = 24 + (−30) = −6 m. Wrong. Adding the signed changes gives the displacement, not the distance. Distance counts every movement as positive: 24 + 30 = 54 m.
5. So the object travelled 6 m. Wrong, because it follows from line 4. A distance can never be negative, and the object clearly moved more than 24 m.

Three classic mistakes

Trap 1 — Reading an s-t graph as if it were a v-t graph

On an s-t graph, a flat line means stopped. On a v-t graph, a flat line means moving at constant velocity. The last lecture warned that this would come up. Read the vertical axis label before you read the shape.

Trap 2 — "Going down" means "behind the start"

A line sloping down means motion in the negative direction. It does not mean the object is behind its start. Only a line below the axis means that. The car reversing from 64 m to 56 m slopes down but stays far in front of the start.

Trap 3 — Using the final displacement as the distance

The height of the line at the end is the displacement. If the line ever goes down and up again, the distance is bigger. Add every rise and every fall as positive amounts.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

An s-t graph goes in straight lines from $(0, 0)$ to $(4, 12)$ , then to $(6, 12)$ , then to $(10, 0)$ , with $t$ in s and $s$ in m. Find the velocity in each stage, the final displacement and the distance travelled.

Hint 1
Each stage is a straight line, so find each gradient as rise ÷ run.

Hint 2
The last stage falls from 12 m to 0 m in 4 s, so its rise is −12.

Full answer
Stage 1: $12 \div 4 = 3$ m/s. Stage 2: flat, so 0 m/s (stopped). Stage 3: $(0 - 12) \div (10 - 6) = - 3$ m/s. Final displacement: 0 m, because the object is back at its start. Distance: $12 + 0 + 12 = 24$ m. Check: the object went out and came back at the same speed, so the two moving stages cover the same 12 m.
A toy car starts at $s = 10$ m and moves at a constant $- 2.5$ m/s. Describe its s-t graph. Where is it at $t = 6$ s, and when does it pass its starting point $s = 0$ ?

Hint 1
Constant velocity means a straight line. Its gradient is −2.5 and it starts at a height of 10.

Hint 2
Each second, $s$ falls by 2.5 m. So $s = 10 - 2.5 t$ .

Full answer
The graph is a straight line starting at 10 m and falling with gradient −2.5. At $t = 6$ s: $s = 10 - 2.5 \times 6 = - 5$ m, so the car is 5 m on the negative side of $s = 0$ . It reaches $s = 0$ when $10 - 2.5 t = 0$ , so $t = 4$ s. Notice that the gradient is negative the whole time, but the car is on the positive side until 4 s.
An object starts from rest and accelerates steadily at 2 m/s², so its displacement is $s = t^{2}$ . A tangent to its s-t graph at $t = 3$ s passes through $(1.5, 0)$ and $(5, 21)$ . Find the velocity at 3 s, and check it with a SUVAT equation.

Hint 1
The velocity at 3 s is the gradient of the tangent, not of the curve between two far-apart points.

Hint 2
Gradient $= (21 - 0) \div (5 - 1.5)$ . For the check, use $v = u + a t$ with $u = 0$ .

Full answer
Gradient $= 21 \div 3.5 = 6$ , so $v = 6$ m/s. Check: $v = u + a t = 0 + 2 \times 3 = 6$ m/s. They agree. Also check that the tangent touches the curve: at $t = 3$ , $s = 3^{2} = 9$ m, and the tangent line $s = 6 t - 9$ gives $6 \times 3 - 9 = 9$ m at $t = 3$ .
Runner A starts at $s = 0$ and runs at a steady 4 m/s. At the same moment, runner B starts 30 m ahead and walks at a steady 1 m/s in the same direction. Sketch both s-t lines. When and where does A catch B? What would their v-t graphs look like, and why do those lines never cross?

Hint 1
Write each position as a formula: $s_{A} = 4 t$ and $s_{B} = 30 + t$ .

Hint 2
A catches B when the two positions are equal: $4 t = 30 + t$ .

Full answer
$4 t = 30 + t$ gives $3 t = 30$ , so $t = 10$ s. Then $s = 4 \times 10 = 40$ m. On the s-t graph, A's steeper line starts at 0, B's flatter line starts at 30, and they cross at $(10, 40)$ . On a v-t graph, A is a flat line at 4 m/s and B is a flat line at 1 m/s. Those lines are parallel and never cross, because the runners never have the same velocity. The catch-up shows only on the s-t graph, where height means position. Check: B has walked $10 \times 1 = 10$ m from 30 m, so B is at 40 m too.

The bigger idea

You now have a loop. The gradient of the s-t graph gives the v-t graph. The gradient of the v-t graph gives the acceleration. Going the other way, the area under the v-t graph gives back the displacement. Gradient takes you down the chain, and area brings you back up.

Those two moves have names in mathematics: differentiation finds a gradient at every instant, and integration adds up areas. They are the two halves of calculus. A phone's navigation app uses the first one constantly: it measures your position again and again, and turns each change in position into a speed. You have been doing calculus with your eyes.

Key concepts in this lesson

The gradient of a displacement-time graph is the velocity, because the gradient is $Δ s /Δ t$ .
On an s-t graph, a horizontal line means stationary, a straight sloping line means constant velocity, and a curve means the object is accelerating.
On a curved s-t graph, the velocity at an instant is the gradient of the tangent at that point.
A line below the axis means the object is on the negative side of its start, while a line sloping down means it is moving in the negative direction.
Displacement is the height of the line at the end, while distance adds every rise and fall as a positive amount.