The shape of the lesson
By the end of the lesson, learners can
- Where the equations come from: In the lab, release the ball from a chosen height and step through the strobe images.
- Motion under gravity: A particle moving freely up and down has acceleration \(g = 10\ \text{m s}^{-2}\) downwards, whether it is rising, at the top, or falling.
- Two stages and two particles: When the acceleration changes, treat each stage separately: the final velocity of one stage is the initial velocity of the next.
- Positive direction: choose it and give every vector a sign.
- List \(s, u, v, a, t\): which three do you know, which one do you want?
- Choose the equation that leaves out the fifth.
- Gravity: \(a = -10\) with up positive, all the way up and down.
- Stages: the end of one stage is the start of the next.
One picture every idea returns to
For constant acceleration in a straight line: \(v = u + at\), \(s = \tfrac{1}{2}(u + v)t\), \(s = ut + \tfrac{1}{2}at^2\), \(s = vt - \tfrac{1}{2}at^2\), \(v^2 = u^2 + 2as\). Each one leaves out one of the five quantities, so pick the equation without the one you neither know nor want.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Five letters, five equations.. When the acceleration is constant, five quantities describe the motion: \(s, u, v, a, t\). Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Where the equations come from | In the lab, release the ball from a chosen height and step through the strobe images. Project tab 2 of the lab (Kinematics lab, equations of motion) and let a student drive it while the class predicts. Model the worked example Choosing the equation on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Motion under gravity | A particle moving freely up and down has acceleration \(g = 10\ \text{m s}^{-2}\) downwards, whether it is rising, at the top, or falling. Model the worked example Thrown up from a height on the board, one line at a time. Your turn: A stone is dropped from rest off a cliff and hits the sea 3.2 s later. Find the height of the cliff and the speed of impact. | Pairs |
| 21 to 30 min | Teach and check: Two stages and two particles | When the acceleration changes, treat each stage separately: the final velocity of one stage is the initial velocity of the next. Model the worked example Meeting in mid-air on the board, one line at a time. | Connect-Extend-Challenge |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 27 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Two stages and two particles, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The suvat equations need CONSTANT acceleration. | If a question gives \(a\) or \(v\) as a function of \(t\), use calculus instead (next lesson). |
| Using \(s = +2\) for the drop to the ground. | Displacement is measured from the point of projection, so landing below it means a NEGATIVE displacement when up is positive. |
| Restart the clock carefully. | If B is released 1 s after A, then B has been moving for \((t - 1)\) s when A has been moving for \(t\) s. |
Constant acceleration: the suvat equations: worksheet
6 questions, 27 marks. Written in the style of the papers; no past paper question is reproduced.
A car travelling at \(8\ \text{m s}^{-1}\) accelerates uniformly at \(1.5\ \text{m s}^{-2}\) for 6 s. Find its final speed and the distance travelled.
A train decelerates uniformly from \(30\ \text{m s}^{-1}\) to \(12\ \text{m s}^{-1}\) over a distance of 420 m. Find the deceleration and the time taken.
A stone is projected vertically upwards at \(18\ \text{m s}^{-1}\) from ground level. Find its greatest height and the length of time for which it is more than 10 m above the ground.
A particle moves with constant acceleration. It travels 20 m in the 2nd second of its motion and 32 m in the 4th second. Find its initial velocity and its acceleration.
A cyclist starts from rest and accelerates uniformly at \(0.5\ \text{m s}^{-2}\) until her speed is \(6\ \text{m s}^{-1}\). She then continues at this speed. Find the time she takes to travel 300 m from the start.
A ball is dropped from rest from a window 20 m above the ground. One second later a second ball is thrown vertically downwards from the same window with speed \(V\ \text{m s}^{-1}\). Both balls reach the ground at the same instant. Find \(V\).
Answer key and marking notes
- \(v = 8 + 1.5 \times 6 = 17\ \text{m s}^{-1}\).
- \(s = 8 \times 6 + \tfrac{1}{2}(1.5)(36) = 48 + 27 = 75\) m.
- \(v^2 = u^2 + 2as\): \(144 = 900 + 840a\), so \(a = -0.9\). Deceleration \(0.9\ \text{m s}^{-2}\).
- \(s = \tfrac{1}{2}(u + v)t\): \(420 = 21t\), so \(t = 20\) s.
- Greatest height: \(0 = 324 - 20s\), so \(s = 16.2\) m.
- At 10 m: \(10 = 18t - 5t^2\), so \(5t^2 - 18t + 10 = 0\).
- \(t = \dfrac{18 \pm \sqrt{124}}{10}\), giving \(t = 0.686\) and \(t = 2.914\).
- Time above 10 m \(= \dfrac{2\sqrt{124}}{10} = 2.23\) s.
- Distance in the \(n\)th second \(= s(n) - s(n-1) = u + a\left(n - \tfrac{1}{2}\right)\).
- 2nd second: \(u + 1.5a = 20\). 4th second: \(u + 3.5a = 32\).
- Subtract: \(2a = 12\), so \(a = 6\ \text{m s}^{-2}\) and \(u = 20 - 9 = 11\ \text{m s}^{-1}\).
- Stage 1: \(t = 6 \div 0.5 = 12\) s; \(s = \tfrac{1}{2}(0 + 6)(12) = 36\) m.
- Stage 2: remaining \(300 - 36 = 264\) m at \(6\ \text{m s}^{-1}\) takes 44 s.
- Total time \(= 12 + 44 = 56\) s.
- First ball: \(20 = 5t^2\), so it lands at \(t = 2\) s.
- The second ball therefore takes \(2 - 1 = 1\) s.
- Down positive: \(20 = V(1) + \tfrac{1}{2}(10)(1)^2 = V + 5\).
- \(V = 15\).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A car travelling at \(20\ \text{m s}^{-1}\) brakes uniformly and comes to rest in a distance of 50 m. Find the deceleration and the time taken to stop.
- A\(4\ \text{m s}^{-2}\) and 5 s
- B\(8\ \text{m s}^{-2}\) and 2.5 s
- C\(4\ \text{m s}^{-2}\) and 2.5 s
- D\(0.4\ \text{m s}^{-2}\) and 50 s
In which of these situations can the constant acceleration (suvat) equations be used?
- AA ball thrown vertically upwards, with air resistance ignored
- BA particle whose acceleration is \(3t\ \text{m s}^{-2}\) at time \(t\)
- CA particle whose velocity is \(4t^2\ \text{m s}^{-1}\) at time \(t\)
- DA skydiver whose air resistance increases with speed
A ball is thrown vertically upwards with speed \(15\ \text{m s}^{-1}\). Find the greatest height it reaches above the point of projection.
- A11.25 m
- B22.5 m
- C33.75 m
- D7.5 m
A particle has initial velocity \(2\ \text{m s}^{-1}\) and constant acceleration \(3\ \text{m s}^{-2}\). Find the distance it travels during the 4th second of its motion.
- A12.5 m
- B32 m
- C14 m
- D11 m
A ball is thrown vertically upwards at \(12\ \text{m s}^{-1}\) from a point 16 m above the ground. Find the time it takes to reach the ground.
- A3.35 s
- B2.99 s
- C1.79 s
- D0.954 s
A ball is projected vertically upwards at \(20\ \text{m s}^{-1}\). For how long is it more than 15 m above the point of projection?
- A2 s
- B1 s
- C3 s
- D4 s
Particle A is released from rest at the top of a tower 60 m high. At the same instant particle B is projected vertically upwards at \(30\ \text{m s}^{-1}\) from the ground directly below. Find when the particles meet and their height above the ground then.
- AAfter 2 s, 40 m above the ground
- BAfter 2 s, 20 m above the ground
- CAfter 2 s, 60 m above the ground
- DAfter 3.46 s, at ground level
A car passes point A with speed \(u\ \text{m s}^{-1}\) and accelerates uniformly. It passes point B, 150 m beyond A, 6 s later with speed \(32\ \text{m s}^{-1}\). Find \(u\) and the acceleration.
- A\(u = 18\), \(a = 2.33\ \text{m s}^{-2}\)
- B\(u = 25\), \(a = 1.17\ \text{m s}^{-2}\)
- C\(u = 18\), \(a = 5.33\ \text{m s}^{-2}\)
- D\(u = 18\), \(a = 4.67\ \text{m s}^{-2}\)
Quiz key
\(0 = 20^2 + 2a \times 50\) gives \(a = -4\), a deceleration of \(4\ \text{m s}^{-2}\). Then \(50 = \tfrac{1}{2}(20 + 0)t\) gives \(t = 5\) s. The tempting 2.5 s is \(50 \div 20\), which assumes the car keeps its full speed all the way; its average speed is only \(10\ \text{m s}^{-1}\). \(8\ \text{m s}^{-2}\) loses the 2 in \(2as\), and 0.4 divides 20 by 50.
Ignoring air resistance, a projected ball has acceleration \(g = 10\ \text{m s}^{-2}\) downwards at every instant, rising or falling, so suvat applies. An acceleration of \(3t\) changes with time, and \(v = 4t^2\) gives \(a = 8t\), so both need calculus. The skydiver's resultant force changes as the speed changes, so the acceleration is not constant either.
\(0 = 15^2 - 2 \times 10 \times s\), so \(s = \dfrac{225}{20} = 11.25\) m. The tempting 22.5 m is \(15 \times 1.5\): it uses the time to the top (1.5 s) at full speed throughout, or drops the 2 in \(v^2 = u^2 + 2as\). 33.75 m takes \(a = +10\) in \(s = ut + \tfrac{1}{2}at^2\), and 7.5 m is \(15 \div 2\).
\(s(4) = 2 \times 4 + 1.5 \times 16 = 32\) m and \(s(3) = 2 \times 3 + 1.5 \times 9 = 19.5\) m, so the distance in the 4th second is \(32 - 19.5 = 12.5\) m. The tempting 32 m is the distance in the first 4 seconds. 14 and 11 are the velocities at \(t = 4\) and \(t = 3\), not distances (their average, 12.5, is correct because the acceleration is constant).
\(-16 = 12t - 5t^2\) gives \(5t^2 - 12t - 16 = 0\), so \(t = \dfrac{12 + \sqrt{464}}{10} = 3.35\) s (the other root is negative). The tempting 2.99 s adds the 1.2 s to the top to the time for a 16 m drop from rest, forgetting that the ball first rises 7.2 m higher. 1.79 s ignores the upward throw entirely, and 0.954 s is the size of the negative root.
\(5t^2 - 20t + 15 = 0\) gives \(t^2 - 4t + 3 = 0\), so \(t = 1\) or \(t = 3\). The ball is above 15 m between these times, for \(3 - 1 = 2\) s. The tempting 1 s is only the first time it reaches 15 m (on the way up), and 3 s is when it passes 15 m on the way down. 4 s is the time until it returns to the point of projection.
Setting \(60 - 5t^2 = 30t - 5t^2\) gives \(t = 2\) (the \(5t^2\) terms cancel). Height \(= 30 \times 2 - 5 \times 4 = 40\) m. The tempting 20 m is the distance A has FALLEN, not its height above the ground. 60 m ignores gravity on B (\(30 \times 2\)), and 3.46 s is when A would hit the ground if they never met.
\(150 = \tfrac{1}{2}(u + 32) \times 6\) gives \(u + 32 = 50\), so \(u = 18\). Then \(a = \dfrac{32 - 18}{6} = 2.33\ \text{m s}^{-2}\). The tempting \(u = 25\) is the AVERAGE speed \(150 \div 6\), not the initial speed. 5.33 is \(32 \div 6\), ignoring \(u\), and 4.67 drops the 2 in \(v^2 = u^2 + 2as\).
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.