Acceleration
From 4 m/s to 13 m/s in 6 s
a = Δv ÷ t = (13 − 4) ÷ 6 = 1.5 m/s²
Speed, velocity and acceleration How quickly does a car get up to speed, and how far does it take to stop?
Foundation and Higher · AQA · 45 minutes
Students find out, before anyone explains, that a car going twice as fast needs four times as far to brake, then learn what physics adds to the maths of motion graphs: vectors, units, typical speeds, a = Δv ÷ t, the equation sheet's v² − u² = 2as and free fall. Slides, a vote that shows who's stuck, virtual velocity–time scenes, practice and homework.
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A car brakes, slowing by d m/s every second: about 7 on a dry road, less in the wet. Across: its speed as it brakes (m/s); up: how far it goes before it stops (m). A child steps out 30 m ahead. Move d: from what speeds can it stop in time? Double the speed: does the distance double?
They cross at (20.49, 30)
The idea in one minute
Speed is how fast something goes; velocity is its speed in a direction, so it is a vector, as displacement is. Acceleration is how quickly the velocity changes: a = Δv ÷ t, the change in velocity over the time, in m/s². When there is no time to use, v² − u² = 2as links the speeds before and after with the distance. Anything falling freely near the Earth speeds up by about 9.8 m/s every second, until air resistance grows and it reaches a steady terminal velocity.
The lesson builds on the maths lessons on distance–time and velocity–time graphs: gradients and areas are not taught again. It adds the physics: which quantities have a direction, what typical speeds are, units, the equations as the equation sheet gives them, and free fall. Students then work out with the equations what a car or a cyclist needs, set it in a virtual velocity–time scene, and see whether it does what they predicted.
By the end, students can
- tell speed from velocity, and distance from displacement (scalars and vectors)
- give typical speeds for walking, running, cycling, cars and sound
- work out acceleration with a = Δv ÷ t, a deceleration as negative, in m/s²
- use v² − u² = 2as to find a speed, a distance or an acceleration
- describe free fall at 9.8 m/s² and how a falling object reaches terminal velocity
Key words
- scalar
- vector
- displacement
- velocity
- acceleration
- deceleration
- free fall
- terminal velocity
Before this lesson
Builds on the maths lessons on distance–time and velocity–time graphs (gradient and area). Students should square, square-root and rearrange a formula.
Next: Newton's laws of motion.
The lesson, step by step
30 minutes in total. Each step says what changes for this length.
3 min 1 · Hook · students think
How far to stop?
Students change how hard a car brakes on a graph of braking distance against speed, and find which speeds stop in time.
For this length: One value of d only, then move on
Say“A car brakes hard. Across is its speed when the brakes go on, up is how far it travels before it stops. A child steps out 30 metres ahead. From what speeds can the car stop in time?”
Ask- What does d stand for, in words? What would make it smaller?
- Guess first: twice the speed, twice the distance?
- Why does the curve bend upwards, not go up in a straight line?
Watch forReading the curve as a journey, distance against time, after the maths lessons: across is the speed here. Ask what the axes are first.
For the teacher
Give a minute to think, then read where the curve crosses the flat line: at d = 7 (a dry road) the car stops within 30 m from up to about 20 m/s; at d = 3 (a wet road) only from about 13 m/s, 30 mph.
Then the guess: at d = 7, from 10 m/s the car needs about 7 m, from 20 m/s about 29 m, four times as far. Don't name the equation yet: v² − u² = 2as explains the four times later in the lesson.
Take a show of hands on the guess before anyone moves anything: most say twice the speed, twice the distance. Leave d at 7 (a dry road) and read the curve at 10 m/s and at 20 m/s together.
Shorter: One value of d only, then move on (−2 min).
8 min 2 · Explain · you explain
What physics adds
Scalars and vectors, typical speeds, a = Δv ÷ t with its units, v² − u² = 2as, free fall and terminal velocity.
For this length: Skip the graphs slide; say terminal velocity in a sentence
Say“You know the graphs from maths. Physics adds three things: direction, units and a sense of size.”
Ask- You walk 300 m to a shop and 300 m back. How far have you gone? Where are you now?
- Is 150 m/s a sensible speed for a cyclist? What is?
- Which letter is missing from v² − u² = 2as, compared with a = Δv ÷ t?
Watch forDividing the final velocity, not the change, by the time: 13 ÷ 6 instead of (13 − 4) ÷ 6.
v² − u² = 2as: a final speed
From rest, a = 3 m/s² over 24 m
v² − 0² = 2 × 3 × 24 = 144
v = √144 = 12 m/s
Free fall
Dropped from rest, falling for 2 s
v = 9.8 × 2 = 19.6 m/s
For the teacher
Speed or velocity: distance and speed have a size only (scalars); displacement and velocity have a size and a direction (vectors). Walk 300 m to a shop and back: distance 600 m, displacement 0. Acceleration and force are vectors too; mass and time are scalars.
Typical speeds: walking about 1.5 m/s, running 3 m/s, cycling 6 m/s, a car in town 13 m/s (30 mph), on a motorway 31 m/s (70 mph), sound in air 330 m/s. Use them to check answers: a person walking at 15 m/s is a mistake.
Acceleration: a = Δv ÷ t, the change in velocity ÷ the time. Its unit is m/s per second, m/s². Slowing down gives a negative acceleration, a deceleration. On a velocity–time graph it is the gradient, as in the maths lesson.
v² − u² = 2as: u is the velocity at the start, v at the end, s the distance. Use it when no time is given. Rearrange it in steps: v² = u² + 2as, or s = (v² − u²) ÷ 2a.
Back to the hook: braking from 20 m/s at −7 m/s², 0² − 20² = 2 × (−7) × s, so s = 400 ÷ 14 ≈ 29 m. The distance goes with v², not v: twice the speed makes v² four times as big, and the braking distance four times as long.
Free fall: near the Earth anything falling with no air resistance speeds up at about 9.8 m/s². With air resistance, a skydiver's drag grows with speed until it equals their weight: no resultant force, so no more acceleration, a steady terminal velocity. Their velocity–time graph curves and flattens.
Graphs in physics: the gradient of a distance–time graph is a speed in m/s, of a velocity–time graph an acceleration in m/s²; the area under a velocity–time graph is the distance in m (Higher). For a curved distance–time graph, the speed at one moment is the gradient of a tangent (Higher).
Students write each equation in their book with its units under each letter. The equation sheet gives v² − u² = 2as; a = Δv ÷ t and s = v t are to be learnt.
Shorter: Skip the graphs slide; say terminal velocity in a sentence (−4 min).
5 min 3 · Your turn · students write
Worked, then yours
Each worked example beside a twin that students solve themselves.
For this length: Do the first and third pairs only
Say“Write u, v, a, s and t first, with what you know. Then choose the equation.”
Ask- Is there a time in the question? So which equation?
Watch forForgetting the square root at the end of v² − u² = 2as, giving v² as the speed.
Acceleration
A train: 10 m/s to 25 m/s in 30 s
a = (25 − 10) ÷ 30 = 0.5 m/s²
Then theirs
A bus: 3 m/s to 12 m/s in 6 s
Slowing down
A cyclist: 9 m/s to rest in 3 s
a = (0 − 9) ÷ 3 = −3 m/s²
Then theirs
A car: 24 m/s to rest in 6 s
v² − u² = 2as: a speed
From rest, a = 2 m/s² over 16 m
v² = 0 + 2 × 2 × 16 = 64
v = √64 = 8 m/s
Then theirs
From 3 m/s, a = 2 m/s² over 4 m
v² − u² = 2as: a braking distance
15 m/s to rest at a = −3 m/s²
0² − 15² = 2 × (−3) × s
s = −225 ÷ −6 = 37.5 m
Then theirs
16 m/s to rest at a = −4 m/s²
For the teacher
Work the first example aloud, writing u, v, a, s, t down the side. Students solve its twin while you watch for 12 ÷ 6 instead of (12 − 3) ÷ 6.
The second pair catches a deceleration written as positive; the third, v² given as the speed and 3 not squared; the fourth, the signs: −225 ÷ −6 is positive, a distance.
Open the answers slide once most have finished their twins.
Shorter: Do the first and third pairs only (−3 min).
9 min 4 · Who's stuck? · students vote
Vote, then practise
One question shows who's stuck; then practice in levels with a virtual velocity–time scene.
For this length: After the vote, questions 1–7 only
Say“Choose an answer on your own. Nobody talks until everyone has chosen.”
Ask- Is there a time in the question? So which equation?
Watch forChoosing an answer without writing anything down: ask each student for u, v, a and s with their numbers before they defend their choice.
Show your working on paper. Velocities are in m/s, times in seconds, distances in metres. Write a deceleration with a minus sign. The scenes are virtual experiments: work out what you expect first, then move only the setting the question names. The stretch is optional.
The question that shows who's stuck
- A car speeds up steadily from 4 m/s to 12 m/s over a distance of 32 m. What is its acceleration?
In your Classroom, as they answer (an example) - A5
- B8
- C11
- D4
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 bus speeds up from 2 m/s to 8 m/s in 4 seconds. What is its acceleration, in m/s²?
- Which is closest to a typical speed for a person walking?
- A runner does one 400 m lap of a track in 80 seconds and finishes where she started. Which is right?
- Sort each quantity: a scalar (a size only) or a vector (a size and a direction).They sort 8 items into Scalar and Vector.
- Virtual experiment. A car starts at 4 m/s and speeds up for 6 seconds. Set its acceleration a so that it reaches 13 m/s at the end. Work out what a should be first, then set it.They change an interactive scene (velocity–time) until it does it.
- Virtual experiment. A cyclist starts from rest and speeds up at 0.5 m/s². How long does she take to travel 25 m? Work it out first: her speed after 25 m with v² − u² = 2as, then the time with a = Δv ÷ t. Then set the time t and check the distance under the line.They change an interactive scene (velocity–time) until it does it.
- Virtual experiment. A car going at 12 m/s brakes steadily for 3 seconds. Set its acceleration a so that it has just stopped at the end of the 3 seconds. Work out what a should be first, then set it.They change an interactive scene (velocity–time) until it does it.
- The braking car in the last scene went from 12 m/s to rest at −4 m/s². Use v² − u² = 2as to work out its braking distance, in metres. Does it agree with the distance under the line in the scene?
- From rest, a car accelerates at 2 m/s² over 25 m. Use v² − u² = 2as to find its final speed, in m/s.
- A train slows down steadily from 30 m/s to 10 m/s over 400 m. Find its acceleration, in m/s².
- A skydiver falls at a steady 55 m/s, her terminal velocity. Which is true?
- Stretch: a car is going at 108 km/h. That is ___ m/s. It brakes at −6 m/s². Its braking distance is ___ m.
Support · For the scenes: work out what you expect first with the equations, then change only the setting the question lets you move and watch the velocity–time line and its readings. Does it do what you predicted?
For the teacher
Everyone chooses before anyone explains. Ask a student for each answer to say how they got it, then ask the room which line of working they disagree with.
Questions 1–4 build fluency with a = Δv ÷ t, typical speeds, and speed against velocity; 5–7 are the scenes (work it out, then set it and check); 8–10 use v² − u² = 2as and check a scene with it; 11 is terminal velocity; the stretch starts in km/h.
Students show their working on paper or the board; the quiz keeps their final answers.
If the vote splits, ask one student for each answer to write their working on the board, line by line, and let the class find the line where each went a different way. Then the practice. The scenes are a virtual experiment, a model of motion at constant acceleration, not a substitute for measuring real motion.
Shorter: After the vote, questions 1–7 only (−6 min).
Longer: The stretch, then: why is a car's braking distance four times as long when its speed doubles? (+10 min).
5 min 5 · Exit · students show
Stopping a car
One acceleration and one braking distance show who can choose and use each equation.
Say“A car brakes steadily from 20 m/s to rest in 4 seconds. Find its acceleration, then the distance it needs to stop.”
Watch forA positive acceleration for slowing down, or a negative distance.
Write u, v, a, s and t first. Show your working on paper.
- A car brakes steadily from 20 m/s to rest in 4 seconds. What is its acceleration, in m/s²?
- Use v² − u² = 2as to find how far the same car travels while it brakes, in metres.
For the teacher
Students write u, v, a, s, t first, then work out a with a = Δv ÷ t.
Then the distance with v² − u² = 2as, using the a they found.
The answers show who can use both equations with signs. Start the next lesson, Newton's laws, by asking what force slowed the car down. Set the homework.
Longer: Check the distance with the area under the velocity–time graph (+5 min).
One to one: before each calculation ask the student to write u, v, a, s, t with the numbers they know and a question mark for the one they want; then ask which equation has exactly those letters. After each answer, ask if the size makes sense against the typical speeds.
Homework: Speed, velocity and acceleration
For each calculation write u, v, a, s and t first, then the equation, then the numbers. Give units. Take g = 9.8 m/s² and ignore air resistance unless the question says.
- Someone walks at a typical 1.5 m/s for 20 minutes. How far do they walk, in metres?
- A car speeds up from rest to 27 m/s in 9 seconds. Find its acceleration.
- A cyclist speeds up from 5 m/s to 20 m/s at 2.5 m/s². How long does it take?
- A stone is dropped from rest and falls 5 m. Use v² − u² = 2as to find its speed as it lands, to 2 significant figures.
- A dog runs 30 m out to a ball and 30 m back to its owner in 20 seconds. Give its distance, its displacement and its average speed.
- Describe how a skydiver's velocity changes from jumping until reaching terminal velocity, and explain it with the forces on them. Sketch the velocity–time graph.
Print the worksheet with working space and homework. Teacher answers come with your Library copy.
Teaching speed, velocity and acceleration: common questions
How does this lesson fit with the maths lessons on distance–time and velocity–time graphs?
It builds on them. Those lessons teach reading the graphs, gradients and areas; this one doesn't repeat that. It adds what physics needs on top: speed and velocity as scalar and vector, typical speeds, units such as m/s², the equations as the equation sheet gives them, free fall and terminal velocity. Teach the maths lessons first if the graphs are new.
Which equations do students need to learn, and which are given?
For AQA GCSE Physics, students learn s = v t and a = Δv ÷ t. The equation for uniform acceleration, v² − u² = 2as, is on the equation sheet, so the skill is choosing it and rearranging it: use it when no time is given.
What are the velocity–time scenes in the practice?
Virtual experiments: students change one setting, the acceleration or the time, and watch a velocity–time line and its readings change until the car or cyclist does what is asked; the answer is checked from their settings. They are a model of motion at constant acceleration, a quick way to test a prediction, not a substitute for measuring real motion with light gates or a ticker timer.
Is this lesson for Foundation or Higher tier?
Both. Speed, velocity, acceleration, typical speeds and v² − u² = 2as are on both tiers. The area under a velocity–time graph and a tangent's gradient on a distance–time graph are Higher only, and say so on the slides. The stretch question, in km/h, suits Higher.
Why is a deceleration written as a negative acceleration?
Acceleration is the change in velocity ÷ the time, and when an object slows the change is negative: 20 m/s to rest is a change of −20 m/s. Writing it as −5 m/s² keeps v² − u² = 2as giving the right braking distance; answers saying "a deceleration of 5 m/s²" are fine too.
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Yes. Read the lesson, print the worksheet and open its slides without signing in; nothing you draw there is kept. A free teacher account keeps the whole lesson in your Library to change and teach, with the quizzes, the virtual scenes, the answers and the answer slides.
Curriculum reference: AQA GCSE Physics 8463: 4.5.1.1 (scalar and vector quantities), 4.5.6.1.1 to 4.5.6.1.3 (distance and displacement, speed, velocity), 4.5.6.1.4 (the distance–time relationship) and 4.5.6.1.5 (acceleration, v² − u² = 2as, free fall, terminal velocity, velocity–time graphs).
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