Steady speed: a rectangle
6 m/s for 4 s
Distance = 6 × 4 = 24 m
24 squares of 1 m/s × 1 s = 1 m
Velocity–time graphs Why is the area under a velocity–time graph the distance travelled?
Higher · AQA · 45 minutes
Students decide whether a car pulling away from the lights has caught a cyclist where their velocity–time lines cross, then find out why the area under the graph is the distance. Slides to present, a vote that shows who's stuck, practice and homework.
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The lights turn green. A cyclist rides past them at a steady 6 m/s just as a car pulls away from them, speeding up by a m/s every second. Move a. Where the lines cross, has the car caught the cyclist?
They cross at (3, 6)
The idea in one minute
A velocity–time graph shows how fast something is going at each time. Its gradient is the acceleration: the change in velocity ÷ the time, in m/s². The area under it is the distance: at a steady speed, distance = speed × time is a rectangle, and the same holds for triangles and trapezia. A flat line is a steady speed, not stopped; a line going down is slowing down.
The lesson goes picture first. Students decide whether a car has caught a cyclist where their lines cross before anyone explains, then see a steady speed's distance as a rectangle of squares, find acceleration as a gradient and split a journey into a triangle, a rectangle and a trapezium. It is Higher, with the reading and the gradients also within reach at Foundation; the stretch estimates the area under a curve with trapezia.
By the end, students can
- read a velocity–time graph: speeding up, steady and slowing down
- work out acceleration as the gradient, in m/s²
- work out distance as the area under the graph, with triangles and trapezia
- tell a velocity–time graph from a distance–time graph
Key words
- velocity
- acceleration
- deceleration
- gradient
- area under a graph
- trapezium
Before this lesson
Students should find a gradient from two points and the area of a triangle and a trapezium. Distance–time graphs (the lesson before this one) help most.
Taught in Distance–time graphs.
The lesson, step by step
45 minutes in total. The stretch is optional.
5 min 1 · Hook · students think
Has the car caught her?
Students decide whether crossing lines on a velocity–time graph mean the car has caught the cyclist.
Say“The lights turn green. A cyclist rides past at a steady 6 m/s just as a car pulls away from the lights, speeding up. Where the lines cross, has the car caught the cyclist? Yes or no, and why?”
Ask- What is the same for both where the lines cross?
- Before that moment, who was faster?
Watch forSaying yes because crossing lines meant meeting on a distance–time graph: keep it, the next step is about exactly that.
For the teacher
Give a minute of silent thinking, then vote. Many will say yes, carrying over the last lesson's crossing lines.
Don't explain yet. The question is the point: until the crossing, the cyclist was faster every second, so the car must still be behind. How far behind? That needs the area.
Take a vote with Ask the class: has the car caught the cyclist at the crossing? Then move a: the crossing moves, but the answer never changes.
Shorter: Vote once; skip moving a (−2 min).
12 min 2 · Explain · you explain
Area is distance, gradient is acceleration
A rectangle of squares at steady speed; the gradient as acceleration; then triangles and a trapezium.
Say“Read the axes first: up is velocity, how fast, not how far.”
Ask- 6 m/s for 4 seconds: how far? Where is that on the graph?
- If a gets bigger, what happens to the line? To the acceleration?
Watch forLeaving out the ½ for the triangle: it is half of the rectangle around it.
Speeding up: gradient and triangle
From rest to 6 m/s in 3 s
Acceleration = 6 ÷ 3 = 2 m/s²
Distance = ½ × 3 × 6 = 9 m
A whole journey: trapezium
Up to 6 m/s in 2 s, steady to 5 s, down to 0 at 7 s
Parallel sides 3 s and 7 s, height 6 m/s
½ × (3 + 7) × 6 = 30 m
Or: 6 + 18 + 6 = 30 m
For the teacher
First slide: 6 m/s for 4 seconds is 24 m. Under the line is a rectangle of 24 squares, each 1 m/s for 1 s, 1 m. So distance is the area under the graph.
Second slide: y = ax goes up a m/s every second. The gradient is the acceleration, 6 ÷ 3 = 2 m/s² when a = 2. a = 0 is flat: a steady speed, no acceleration.
Third slide: from rest to 6 m/s in 3 s, the area is a triangle, half the 3 × 6 rectangle: 9 m. The average speed is 3 m/s. This answers the hook: at the crossing the car has gone half as far as the cyclist.
Fourth slide: a whole journey is a trapezium: ½ × (3 + 7) × 6 = 30 m, or a triangle, a rectangle and a triangle. Going down at the end is slowing down, −3 m/s², still forwards. Last, the slide that compares it with a distance–time graph.
Vote before moving a (Ask the class): what will the line look like when a is 0?
Shorter: Skip the slide with a to move, and the comparison slide (−4 min).
8 min 3 · Your turn · students write
Worked, then yours
Each worked example beside a twin that students solve themselves.
Say“Watch me do one. Then you do its twin, which is almost the same.”
Ask- Gradient or area? So which units?
Watch forA deceleration written as a positive number: slowing down is a negative acceleration.
Acceleration
From 0 to 12 m/s in 4 s
Gradient = 12 ÷ 4 = 3 m/s²
Then theirs
From 5 m/s to 15 m/s in 5 s
Slowing down
From 20 m/s to 0 in 5 s
Gradient = (0 − 20) ÷ 5 = −4 m/s²
Then theirs
From 18 m/s to 6 m/s in 4 s
Distance: a trapezium
From 4 m/s to 10 m/s in 6 s
½ × (4 + 10) × 6 = 42 m
Then theirs
From 2 m/s to 8 m/s in 5 s
For the teacher
Work the first example aloud, writing each line. Then students solve its twin while you watch who uses the change in velocity, 15 − 5, not 15.
Do the same for each pair. The second pair catches students who drop the minus; the third, those who multiply the final speed by the time.
Open the answers slide once most have finished their twins.
Shorter: Do the first and third pairs only (−3 min).
15 min 4 · Who's stuck? · students vote
Vote, then practise
One question shows who's stuck; then practice in levels, and a stretch.
Say“Choose an answer on your own. Nobody talks until everyone has chosen.”
Ask- What are the axes? So what does a flat line say?
Watch forReading a flat line as stopped, as on a distance–time graph.
Show your working on paper. Velocities are in m/s, times in seconds. Write a negative acceleration with a minus sign. The stretch is optional.
The question that shows who's stuck
- A velocity–time graph is a flat line at 5 m/s from 0 to 10 seconds. What does it show?
In your Classroom, as they answer (an example) - A8
- B11
- C4
- D3
Each wrong answer is a different mistake. While they answer, your Classroom shows who chose which, so you know in seconds whom to help with what.
- A car's velocity rises steadily from 0 to 20 m/s in 4 seconds. What is its acceleration in m/s²?
- A runner keeps a steady 8 m/s for 15 seconds. How many metres does she run?
- From rest, a car speeds up steadily to 10 m/s in 4 seconds. How many metres does it go in those 4 seconds?
- On a velocity–time graph a straight line goes from (0, 12) to (6, 0). What is the acceleration in m/s²?
- A velocity–time line slopes down from 12 m/s to 0. What is the object doing?
- Two cars' velocity–time lines cross at (4, 10). What does that tell you?
- A train speeds up from 0 to 20 m/s in 10 seconds, stays at 20 m/s for 30 seconds, then slows to 0 in 20 seconds. How many metres does it travel?
- A car going at 4 m/s speeds up steadily for 10 seconds and covers 90 m. What is its velocity at the end, in m/s?
- As the lights turn green, a cyclist rides past at a steady 6 m/s and a car pulls away from rest, speeding up by 2 m/s every second. After how many seconds does the car catch the cyclist?
- Stretch: a curved velocity–time graph goes through (0, 0), (2, 3) and (4, 8), and bends upwards. Two trapezia estimate the distance in the first 4 seconds as ___ m. Is that an overestimate or an underestimate? ___
Support · For question 3: draw it. The area is a triangle, base 4 s and height 10 m/s: ½ × base × height. For question 4: the gradient is the change in velocity ÷ the time, and the velocity goes down.
For the teacher
Everyone chooses before anyone explains. Then ask three questions in turn: what is the up axis? What is the gradient of a flat line? How far does 5 m/s go in 10 seconds?
Questions 1–2 build fluency; 3–4 change the look; 5–6 ask for reasons; 7–8 are exam-style; 9 is the hook's race. The stretch estimates the area under a curve.
Students show their working on paper or the board; the quiz keeps their final answers.
If many chose standing still, put a distance–time graph and a velocity–time graph with the same flat line side by side and ask what each says, before they practise.
Shorter: After the vote, questions 1–4 only (−6 min).
Longer: The stretch, then: are trapezia under a curve too much or too little? (+10 min).
5 min 5 · Exit · students show
Gradient or area?
One acceleration and one distance, each named as gradient or area, show who has both ideas.
Say“Find the acceleration and the distance, and say which part of the graph gives each.”
Watch forThe final speed times the time, 8 × 6, for the distance.
Show your working; one sentence for the last part.
- A cyclist speeds up steadily from 2 m/s to 8 m/s in 6 seconds. Find the acceleration and the distance travelled. Say which part of the velocity–time graph gives each.
For the teacher
A cyclist speeds up steadily from 2 m/s to 8 m/s in 6 seconds. Sketch the velocity–time graph.
Find the acceleration and the distance, and answer the exit question.
The answers show who knows which number comes from the gradient and which from the area: start the next lesson with a graph to label if few do. Set the homework.
Longer: Sketch the distance–time graph of the same ride: why does it curve? (+5 min).
One to one: before each graph, ask the student to say what the axes are; for every number they work out, ask whether it came from a gradient or an area, and what its units are.
Homework: gradients and areas
Sketch a velocity–time graph for each question before working anything out. Say whether each answer comes from a gradient or an area, and give its units.
- A car speeds up steadily from 0 to 24 m/s in 8 seconds. Find its acceleration.
- A motorbike slows down steadily from 15 m/s to a stop in 3 seconds. Find its acceleration.
- A train goes at a steady 12 m/s for 25 seconds. How far does it go?
- A car speeds up from 0 to 10 m/s in 5 seconds, keeps 10 m/s for 20 seconds, then slows to a stop in 10 seconds. Find the total distance.
- A car going at 6 m/s speeds up steadily for 8 seconds and covers 80 m. Find its final velocity and its acceleration.
- A velocity–time curve goes through (0, 0), (2, 2), (4, 6) and (6, 12). Use three trapezia to estimate the distance in the first 6 seconds. Is your estimate too big or too small? Why?
Print the worksheet with working space and homework. Teacher answers come with your Library copy.
Teaching velocity–time graphs: common questions
Why is the area under a velocity–time graph the distance?
At a steady speed, distance = speed × time, and under a flat line that is a rectangle's area: across is time, up is speed. Cut any journey into thin strips and each strip is nearly a rectangle, so adding them gives the whole area. For straight lines that is triangles and trapezia.
How do students tell velocity–time and distance–time graphs apart?
Read the up axis first. On a distance–time graph a flat line is stopped and the gradient is the speed; on a velocity–time graph a flat line is a steady speed, the gradient is the acceleration and the area is the distance. Comparing the same flat line on both is quick and sticks.
Is velocity the same as speed here?
At GCSE these graphs are of motion in one direction, so the velocity is the speed. Velocity can be negative for going backwards, but a line going down towards 0 on a velocity–time graph is only slowing down.
Is velocity–time graphs on the Foundation or Higher tier?
The area under a graph is Higher (AQA A15). Foundation students can still read a velocity–time graph and find an acceleration as a gradient, the first half of this lesson. The stretch, estimating the area under a curve with trapezia, is Higher.
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Curriculum reference: AQA GCSE Mathematics 8300: A15 (gradients and areas under graphs, interpreted on velocity–time graphs; Higher), with A14 (graphs in real contexts) and R14 (the gradient of a straight line as a rate of change).
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