A table of numbers tells you what happened; a graph tells you the story at a glance, speeding up, slowing down, stopped, reversing. This lesson shows how to take raw motion data and plot it as displacement–time, velocity–time and acceleration–time graphs, so the shape itself reveals how an object moved.
On a displacement–time graph, one object's line is flat (horizontal) for ten seconds.
Before reading: what is that object doing during those ten seconds? Write your answer and one sentence explaining how the flat line tells you.
Plot position up the page against time across, the shape tells the story
A displacement–time graph puts time on the horizontal axis and position on the vertical axis, so the steepness of the line at any moment shows how fast the object is moving.
To build one from data, plot each (time, displacement) pair and join the points. A straight diagonal line means the object covers equal displacement each second, constant velocity. A line that curves and gets steeper means it covers more displacement each second, it is speeding up. A flat line means the position is not changing, the object is at rest.
On a displacement–time graph (position vs time), a straight diagonal line means constant velocity, an upward-curving line means acceleration, and a flat (horizontal) line means the object is at rest.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? On a displacement–time graph, what does a straight diagonal line represent?
Now velocity goes up the page, the slope becomes the acceleration
We just saw how position plots against time. That raises a question: what if we plot velocity instead of position? This card answers it → a velocity–time graph, where a straight sloped line means constant acceleration.
A velocity–time graph plots velocity against time, so a rising straight line means the velocity climbs steadily, constant acceleration, and a flat line means the velocity is unchanging.
Plot each (time, velocity) pair from the data and join the points. On a v–t graph a flat horizontal line means constant velocity (zero acceleration), a straight rising line means constant positive acceleration, and a straight falling line means constant deceleration.
A car's velocity is recorded as: $0\ \text{s} \to 0\ \text{m s}^{-1}$, $2\ \text{s} \to 4\ \text{m s}^{-1}$, $4\ \text{s} \to 8\ \text{m s}^{-1}$, $6\ \text{s} \to 12\ \text{m s}^{-1}$. Describe the v–t graph and state the acceleration.
On a velocity–time graph (velocity vs time): a flat line means constant velocity (zero acceleration), a straight rising line means constant acceleration, and a falling line means deceleration. The gradient equals the acceleration.
Add the highlighted worked method to your notes before the check below.
Quick check: True or false: a flat horizontal line on a velocity–time graph means the object is at rest.
Pattern A, Construct from data
A trolley moves with these readings: at $0,1,2,3\ \text{s}$ its displacement is $0,2,8,18\ \text{m}$ and its velocity is $0,4,8,12\ \text{m s}^{-1}$. In your book: (1) plot the d–t graph (note its curve); (2) plot the v–t graph; (3) state the acceleration and sketch the a–t graph.
Two truths, one lie: which statement about constructing motion graphs is the lie?
Dot point: Use data to construct displacement–time graphs, velocity–time graphs and acceleration–time graphs
The simplest graph of the three, for uniform motion
We just learned to build velocity–time graphs. That raises a question: how do we show acceleration itself over time? This card answers it → an acceleration–time graph, which for uniform motion is just a flat line.
An acceleration–time graph plots acceleration against time; when the acceleration is constant, the graph is simply a horizontal line at that value.
For the uniformly accelerating car above, the acceleration is a steady $2.0\ \text{m s}^{-2}$, so the a–t graph is a flat line at $2.0$. If the object moved at constant velocity, the acceleration would be zero and the a–t graph would lie on the time axis. Reading across the three graphs: a curving d–t graph corresponds to a sloped v–t graph and a flat, non-zero a–t graph.
An acceleration–time graph plots acceleration vs time. For uniformly accelerated motion it is a horizontal line at the constant acceleration value; for constant velocity it lies on the time axis (zero acceleration).
Pause, copy the highlighted statement into your book before the check below.
Last check: Decide if the statement is true or false.
For an object with constant acceleration, the acceleration–time graph is a horizontal line above the time axis.
Odd one out: three of these describe an object at rest. Which one does not?
Dot point: Use data to construct displacement–time graphs, velocity–time graphs and acceleration–time graphs
The gradient of one graph is the value of the next; the area under v–t is the displacement
Use the simulation. Press Thrown up the track ($u = +5.0\ \text{m s}^{-1}$, $a = -1.0\ \text{m s}^{-2}$) and drag time until the tangent on the x–t graph is flat. What is happening at that instant?
Use the simulation. Keep the same preset and drag time to $t = 10\ \text{s}$. True or false: the displacement and the distance travelled are both zero, because the object is back where it started.
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
UnderstandBand 4(4 marks) 1. A cyclist starts from rest and speeds up steadily. Describe the shape of (a) the displacement–time graph and (b) the velocity–time graph for this motion, and explain what the shape of each tells you.
1 mark: d–t graph curves upward (gets steeper) · 1 mark: explanation that steepening curve = increasing velocity · 1 mark: v–t graph is a straight line rising from the origin · 1 mark: explanation that constant slope = constant acceleration
ApplyBand 5(4 marks) 2. A trolley starts from rest and accelerates uniformly to $6.0\ \text{m s}^{-1}$ in $3.0\ \text{s}$, then travels at that constant velocity for a further $2.0\ \text{s}$. Describe the shape of the displacement-time, velocity-time and acceleration-time graphs for this motion.
1 mark: d-t curves upward then becomes a straight slope · 1 mark: v-t rises linearly from 0 to 6.0 then is horizontal · 1 mark: a-t is constant at $2.0\ \text{m s}^{-2}$ then drops to zero · 1 mark: all three described over the correct time intervals
Dot point: Use data to construct displacement–time graphs, velocity–time graphs and acceleration–time graphs
UnderstandBand 4(3 marks) 3. A student is given a table of position and time readings. Outline the steps required to construct a displacement-time graph from the data, and state one reason the graph is more useful than the table.
1 mark: time on the horizontal axis, displacement on the vertical, both labelled with units · 1 mark: choose a scale and plot the points · 1 mark: a reason, such as the gradient giving velocity directly
Dot point: Use data to construct displacement–time graphs, velocity–time graphs and acceleration–time graphs
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): (a) The displacement–time graph is a curve that starts shallow and becomes progressively steeper. Because the gradient of a d–t graph is the velocity, a steepening curve shows the velocity is increasing, the cyclist is accelerating. (b) The velocity–time graph is a straight line rising from the origin. Because the gradient of a v–t graph is the acceleration, a constant slope shows the acceleration is constant. Together the curving d–t graph and the straight, sloped v–t graph describe steady (uniform) acceleration from rest.
Q2 (4 marks): For the first $3.0\ \text{s}$ the velocity increases uniformly, so the velocity-time graph is a straight line rising from the origin to $6.0\ \text{m s}^{-1}$, and the acceleration-time graph is a horizontal line at $a = 6.0/3.0 = 2.0\ \text{m s}^{-2}$. Because the velocity is increasing, the displacement-time graph curves upward with a steadily steepening gradient. From $3.0\ \text{s}$ to $5.0\ \text{s}$ the velocity is constant, so the velocity-time graph is horizontal at $6.0\ \text{m s}^{-1}$, the acceleration-time graph drops to zero, and the displacement-time graph becomes a straight line of constant gradient $6.0\ \text{m s}^{-1}$.
Q3 (3 marks): Plot the independent variable, time, on the horizontal axis and displacement on the vertical axis, labelling both with their quantity and unit. Choose scales that spread the data across most of the available grid, plot each pair as a point, and draw a smooth line or line of best fit through them rather than joining the points dot to dot. The graph is more useful than the table because relationships become visible directly: the gradient at any point gives the velocity at that instant, a straight section immediately signals constant velocity, and any anomalous reading stands out from the trend in a way it never does in a column of numbers.
You can now construct displacement–time, velocity–time and acceleration–time graphs from data and read their shapes. Next lesson extracts the numbers, gradients and areas, to determine displacement, velocity and acceleration from the graphs.
↩ Back to focus-area hubAt the start you saw a flat line on a displacement–time graph. That object is at rest: its position does not change with time, so it covers no displacement. (A flat line on a velocity–time graph would instead mean constant velocity, still moving.) Revisit the misconceptions card if you mixed these up.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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