Last lesson you built the graphs; now you mine them. The gradient of a displacement–time graph is the velocity. The gradient of a velocity–time graph is the acceleration. And the area under a velocity–time graph is the displacement. Master these three reading rules and a single graph hands you every motion quantity without a stopwatch.
A displacement–time graph is a straight line that rises $20\ \text{m}$ over $4\ \text{s}$.
Before reading: how could you use the slope of this line to find the object's velocity? Write your method and the value you expect.
Slope measures how fast position changes, which is velocity
The slope of a displacement–time line is "how much position changes per second", and that is precisely the definition of velocity.
For a straight d–t line, $\text{velocity} = \dfrac{\text{rise}}{\text{run}} = \dfrac{\Delta s}{\Delta t}$. A steeper line means a higher velocity. For a curved d–t graph, the instantaneous velocity at a point is the gradient of the tangent drawn there.
The gradient of a displacement–time graph equals the velocity: $v = \dfrac{\Delta s}{\Delta t}$ for a straight line, or the gradient of the tangent at a point on a curve. A steeper line means a higher velocity.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? A displacement–time line rises $30\ \text{m}$ over $5\ \text{s}$. What is the velocity?
The same slope idea, one level up
We just saw that the slope of a d–t graph is velocity. That raises a question: what does the slope of a velocity–time graph tell us? This card answers it → it is the acceleration, the rate at which velocity changes.
The slope of a velocity–time line is "how much velocity changes per second", which is exactly acceleration.
So $a = \dfrac{\Delta v}{\Delta t} = \dfrac{v - u}{t}$, where $u$ is the initial velocity and $v$ the final velocity read off the graph.
A velocity–time graph is a straight line rising from $4\ \text{m s}^{-1}$ at $t = 0$ to $20\ \text{m s}^{-1}$ at $t = 8\ \text{s}$. Find the acceleration.
The gradient of a velocity–time graph equals the acceleration: $a = \dfrac{\Delta v}{\Delta t} = \dfrac{v - u}{t}$. Read the initial velocity $u$ and final velocity $v$ off the graph and divide the change by the time.
Add the highlighted worked method to your notes before the check below.
Quick check: A v–t line falls from $18\ \text{m s}^{-1}$ to $6\ \text{m s}^{-1}$ over $4\ \text{s}$. What is the acceleration?
Pattern A, Extract multiple quantities
A velocity–time graph is a straight line from $0\ \text{m s}^{-1}$ at $t=0$ to $12\ \text{m s}^{-1}$ at $t=6\ \text{s}$. In your book: (1) find the acceleration from the gradient; (2) find the displacement from the area under the line (a triangle); (3) state the initial and final velocities.
Fill the gaps. On a displacement-time graph the gradient gives the . On a velocity-time graph the gradient gives the and the area under the line gives the .
Dot point: Use data to analyse motion graphs to describe the displacement, velocity and acceleration of an object
Why an area on the graph equals a distance on the road
We just learned that gradients give rates of change. That raises a question: can a graph also give an accumulated quantity, like total displacement? This card answers it → yes, the area under a velocity–time graph is the displacement.
Velocity multiplied by time gives displacement, and on a velocity–time graph "velocity times time" is exactly the area between the line and the time axis.
For a rectangle (constant velocity), area $= v \times t$. For a triangle (uniform acceleration from rest), area $= \tfrac{1}{2}\, t \times v$. Splitting a graph into rectangles and triangles lets you find displacement for any straight-line v–t graph. This area-as-displacement idea links all five quantities, displacement $s$, initial velocity $u$, final velocity $v$, acceleration $a$ and time $t$, which is exactly what the equations of motion capture next lesson.
The area under a velocity–time graph equals the displacement. For constant velocity it is a rectangle ($s = vt$); for uniform acceleration it is a triangle or trapezium. Gradients give rates; areas give accumulated displacement.
Pause, copy the highlighted statement into your book before the check below.
Last check: Decide if the statement is true or false.
The displacement of an object is given by the area under its displacement–time graph.
Match each graph feature with what it tells you:
Dot point: Analyse motion graphs to determine the relationships between displacement, initial velocity, final velocity, acceleration and time
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.
AnalyseBand 5(4 marks) 1. A car's velocity–time graph is a straight line from $5\ \text{m s}^{-1}$ at $t=0$ to $25\ \text{m s}^{-1}$ at $t=10\ \text{s}$. Determine (a) the acceleration and (b) the displacement over this interval. Show your working.
1 mark: $a = (25-5)/10$ · 1 mark: $a = 2.0\ \text{m s}^{-2}$ · 1 mark: area = trapezium $\tfrac{1}{2}(5+25)\times 10$ · 1 mark: displacement $= 150\ \text{m}$
ApplyBand 5(4 marks) 2. A velocity-time graph shows a car accelerating uniformly from $5.0\ \text{m s}^{-1}$ to $25\ \text{m s}^{-1}$ over $8.0\ \text{s}$. Determine the acceleration and the displacement, stating which graph feature you used for each.
1 mark: acceleration from the gradient · 1 mark: $a = 2.5\ \text{m s}^{-2}$ · 1 mark: displacement from the area under the line · 1 mark: $s = 120\ \text{m}$
Dot point: Analyse motion graphs to determine the relationships between displacement, initial velocity, final velocity, acceleration and time
AnalyseBand 6(3 marks) 3. A displacement-time graph is a curve whose gradient decreases steadily to zero and then becomes negative. Describe the motion this represents.
1 mark: object slows while still moving forward · 1 mark: instantaneously at rest where the gradient is zero · 1 mark: then moves back toward the origin
Dot point: Use data to analyse motion graphs to describe the displacement, velocity and acceleration of an object
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 acceleration is the gradient of the velocity–time line: $a = \dfrac{v - u}{t} = \dfrac{25 - 5}{10} = 2.0\ \text{m s}^{-2}$. (b) The displacement is the area under the line, which is a trapezium with parallel sides $u = 5\ \text{m s}^{-1}$ and $v = 25\ \text{m s}^{-1}$ and width $t = 10\ \text{s}$: $s = \tfrac{1}{2}(u + v)\,t = \tfrac{1}{2}(5 + 25)(10) = \tfrac{1}{2}(30)(10) = 150\ \text{m}$. So the car accelerates at $2.0\ \text{m s}^{-2}$ and travels $150\ \text{m}$ in the $10\ \text{s}$.
Q2 (4 marks): The acceleration is the gradient of the velocity-time graph: $$a = \frac{\Delta v}{\Delta t} = \frac{25 - 5.0}{8.0} = 2.5\ \text{m s}^{-2}$$ The displacement is the area under the line. That area is a trapezium with parallel sides $5.0$ and $25$ and width $8.0$: $$s = \frac{1}{2}(5.0 + 25)(8.0) = 120\ \text{m}$$ Equivalently it can be split into a rectangle ($5.0 \times 8.0 = 40\ \text{m}$) and a triangle ($\tfrac{1}{2} \times 20 \times 8.0 = 80\ \text{m}$), which sum to the same $120\ \text{m}$.
Q3 (3 marks): Gradient on a displacement-time graph is velocity. A gradient that decreases steadily while remaining positive means the object is still moving in the positive direction but slowing down, so it has a negative acceleration. Where the gradient reaches zero the velocity is momentarily zero: the object is instantaneously at rest and has reached its maximum displacement. Beyond that point the gradient is negative, so the velocity is negative and the object is moving back toward, and then past, its starting position. The acceleration is negative throughout, which is why the velocity passes smoothly from positive through zero to negative.
You can now read velocity from a d–t gradient, acceleration from a v–t gradient, and displacement from the area under a v–t graph. Next lesson packages these relationships into the algebraic equations of uniformly accelerated motion.
↩ Back to focus-area hubAt the start you had a d–t line rising $20\ \text{m}$ over $4\ \text{s}$. The velocity is the gradient: $\dfrac{20}{4} = 5.0\ \text{m s}^{-1}$. The key insight is that gradient gives a rate (velocity from d–t, acceleration from v–t) while area gives an accumulated quantity (displacement from v–t). Revisit Card 3 if the area idea is still fuzzy.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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