The shape of the lesson
By the end of the lesson, learners can
- Differentiate to go down: Given displacement as a function of time, differentiate once for velocity and again for acceleration. "At rest" (or "instantaneously at rest") means \(v = 0\); "maximum velocity" usually means \(a = 0\).
- Integrate to go up: Integrating loses information, so each integral carries a constant.
- Distance when the velocity changes sign: The integral \(\displaystyle\int_a^b v\,dt\) gives displacement, because area below the axis counts as negative.
- Down: \(v = \dfrac{ds}{dt}\), \(a = \dfrac{dv}{dt}\).
- Up: integrate, then use a known value to find the constant.
- At rest: \(v = 0\). Maximum velocity: \(a = 0\).
- Distance: find where \(v = 0\), split, and add the sizes.
- Stages: the velocity at the end of one formula starts the next.
One picture every idea returns to
\(v = \dfrac{ds}{dt}\) and \(a = \dfrac{dv}{dt}\). Going back, \(v = \displaystyle\int a\,dt\) and \(s = \displaystyle\int v\,dt\), each with a constant found from a known value (usually at \(t = 0\)). Distance travelled is the total area between the velocity graph and the axis, so split the integral wherever \(v = 0\).
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, Differentiate down, integrate up.. When acceleration changes with time the suvat equations fail. 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: Differentiate to go down | Given displacement as a function of time, differentiate once for velocity and again for acceleration. "At rest" (or "instantaneously at rest") means \(v = 0\); "maximum velocity" usually means \(a = 0\). Model the worked example Times at rest on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Integrate to go up | Integrating loses information, so each integral carries a constant. Model the worked example From acceleration to displacement on the board, one line at a time. Your turn: A particle starts from O with velocity \(v = 3\sqrt{t}\ \text{m s}^{-1}\). Find its acceleration when \(t = 4\) and its distance from O at that time. | Pairs |
| 21 to 30 min | Teach and check: Distance when the velocity changes sign | The integral \(\displaystyle\int_a^b v\,dt\) gives displacement, because area below the axis counts as negative. Model the worked example Splitting the integral on the board, one line at a time. | Anticipation / Reaction Guide |
| 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, 29 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 Distance when the velocity changes sign, 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.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
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 |
|---|---|
| \(s(4) - s(0)\) is the DISPLACEMENT. | If the particle turns round in the interval, the distance is bigger: find every time when \(v = 0\) first. |
| Forgetting \(+c\). | If the particle starts with \(v = 3\), dropping the constant gives the wrong velocity at every later time and every answer after it. |
| Integrating straight from 0 to 4 gives \(\tfrac{4}{3}\), because the negative middle piece cancels one positive piece. | A sketch of \(v\) against \(t\) shows you where to split. |
Variable acceleration: using calculus: worksheet
6 questions, 29 marks. Written in the style of the papers; no past paper question is reproduced.
A particle moves so that \(s = 2t^3 - 3t^2 + 5\). Find expressions for \(v\) and \(a\), and their values when \(t = 2\).
A particle starts from O and moves with velocity \(v = 3\sqrt{t}\ \text{m s}^{-1}\). Find its acceleration when \(t = 4\) and its displacement from O at that time.
A particle starts at O with velocity \(3\ \text{m s}^{-1}\) and has acceleration \(a = (4 - 2t)\ \text{m s}^{-2}\). Find its maximum velocity and the time at which it is next at rest, giving the time in exact form.
A particle moves so that \(s = t^3 - 6t^2 + 9t\) m. Find the times when it is at rest and the total distance it travels in the first 4 s.
A particle leaves O with velocity \(v = 8t - 2t^2\ \text{m s}^{-1}\) for \(0 \le t \le 6\). Show that it returns to O at \(t = 6\), and find the total distance travelled.
A particle P starts from rest at O. For \(0 \le t \le 4\) its velocity is \(v = 1.5t^2 - 0.25t^3\ \text{m s}^{-1}\). For \(t \ge 4\) it decelerates uniformly at \(2\ \text{m s}^{-2}\) until it comes to rest.
(a) Show that the speed of P at \(t = 4\) is \(8\ \text{m s}^{-1}\) and that its acceleration at that instant is zero.
(b) Find the total distance travelled by P.
Answer key and marking notes
- \(v = 6t^2 - 6t\); at \(t = 2\), \(v = 24 - 12 = 12\ \text{m s}^{-1}\).
- \(a = 12t - 6\); at \(t = 2\), \(a = 18\ \text{m s}^{-2}\).
- \(a = \tfrac{3}{2}t^{-1/2}\), so at \(t = 4\), \(a = \tfrac{3}{2} \times \tfrac{1}{2} = 0.75\ \text{m s}^{-2}\).
- \(s = \displaystyle\int_0^4 3t^{1/2}\,dt = \left[2t^{3/2}\right]_0^4 = 2 \times 8 = 16\) m.
- \(v = 3 + 4t - t^2\).
- \(a = 0\) at \(t = 2\): maximum \(v = 7\ \text{m s}^{-1}\).
- \(v = 0\): \(t^2 - 4t - 3 = 0\), \(t = 2 \pm \sqrt{7}\).
- \(t > 0\), so \(t = 2 + \sqrt{7}\) (about 4.65 s).
- \(v = 3t^2 - 12t + 9 = 3(t - 1)(t - 3)\): at rest at \(t = 1\) and \(t = 3\).
- \(s(0) = 0\), \(s(1) = 4\), \(s(3) = 0\), \(s(4) = 4\).
- Distance \(= 4 + 4 + 4 = 12\) m.
- \(s = 4t^2 - \tfrac{2}{3}t^3\) (since \(s = 0\) at \(t = 0\)); \(s(6) = 144 - 144 = 0\), so it is back at O.
- \(v = 2t(4 - t) = 0\) at \(t = 4\): it turns round there.
- \(s(4) = 64 - \tfrac{128}{3} = \tfrac{64}{3}\).
- Distance \(= \tfrac{64}{3} + \tfrac{64}{3} = \tfrac{128}{3} = 42.7\) m.
- (a) \(v(4) = 24 - 16 = 8\). \(a = 3t - 0.75t^2\), so \(a(4) = 12 - 12 = 0\).
- (b) Stage 1: \(\displaystyle\int_0^4 (1.5t^2 - 0.25t^3)\,dt = \left[0.5t^3 - \tfrac{1}{16}t^4\right]_0^4 = 32 - 16 = 16\) m.
- Stage 2: from 8 to 0 at \(2\ \text{m s}^{-2}\) takes 4 s, distance \(= \tfrac{1}{2}(8 + 0)(4) = 16\) m.
- Total distance \(= 32\) m.
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 particle moves in a straight line so that its displacement from O at time \(t\) s is \(s = 2t^3 - 9t^2 + 12t\) metres. Find the times when the particle is instantaneously at rest.
- A\(t = 1\) and \(t = 2\)
- B\(t = 1.5\) only
- C\(t = 0\) only
- D\(t = 0\), \(t = 1\) and \(t = 2\)
A particle has velocity \(v = 3t^2 - 4t + 1\ \text{m s}^{-1}\) at time \(t\) s. Find its acceleration when \(t = 2\).
- A\(8\ \text{m s}^{-2}\)
- B\(5\ \text{m s}^{-2}\)
- C\(12\ \text{m s}^{-2}\)
- D\(2\ \text{m s}^{-2}\)
A particle starts from rest at O and moves in a straight line with acceleration \(a = 6t\ \text{m s}^{-2}\) at time \(t\) s. Find its displacement from O when \(t = 2\).
- A8 m
- B12 m
- C24 m
- D16 m
A particle moves with acceleration \(a = 12 - 3t\ \text{m s}^{-2}\). When \(t = 0\) its velocity is \(2\ \text{m s}^{-1}\). Find its maximum velocity.
- A\(26\ \text{m s}^{-1}\)
- B\(24\ \text{m s}^{-1}\)
- C\(2\ \text{m s}^{-1}\)
- D\(50\ \text{m s}^{-1}\)
A particle moves in a straight line with velocity \(v = t^2 - 5t + 4\ \text{m s}^{-1}\) at time \(t\) s. Find the total distance it travels from \(t = 0\) to \(t = 5\).
- A\(\tfrac{49}{6}\) m
- B\(\tfrac{5}{6}\) m
- C\(\tfrac{19}{3}\) m
- D\(\tfrac{9}{2}\) m
A particle starts from O and moves in a straight line with velocity \(v = 4\sqrt{t}\ \text{m s}^{-1}\) at time \(t\) s. Find the distance it travels in the first 9 s.
- A72 m
- B162 m
- C54 m
- D12 m
A particle leaves O at time \(t = 0\) and moves in a straight line with velocity \(v = 6t - 3t^2\ \text{m s}^{-1}\). Find the time when it returns to O.
- A\(t = 3\)
- B\(t = 2\)
- C\(t = 1\)
- D\(t = 6\)
A particle starts from rest. For \(0 \le t \le 2\) its velocity is \(v = 3t^2\ \text{m s}^{-1}\). After \(t = 2\) it decelerates uniformly at \(4\ \text{m s}^{-2}\) until it comes to rest. Find the total distance travelled.
- A26 m
- B44 m
- C30 m
- D18 m
Quiz key
\(v = 6t^2 - 18t + 12 = 6(t - 1)(t - 2)\), so \(v = 0\) when \(t = 1\) and \(t = 2\). The tempting \(t = 1.5\) solves \(a = 12t - 18 = 0\), which is when the velocity is least, not zero. \(t = 0\) is when \(s = 0\), the particle is at O but moving at \(12\ \text{m s}^{-1}\), so it is not at rest there.
\(a = \dfrac{dv}{dt} = 6t - 4\), so at \(t = 2\), \(a = 8\ \text{m s}^{-2}\). The tempting 5 is the VELOCITY at \(t = 2\); acceleration is its rate of change. 12 drops the \(-4\) term when differentiating, and 2 is \(\displaystyle\int_0^2 v\,dt\), which is a displacement.
\(v = 3t^2 + c\), and \(v = 0\) at \(t = 0\), so \(c = 0\). \(s = t^3 + k\), and \(s = 0\) at \(t = 0\), so \(k = 0\). At \(t = 2\), \(s = 8\) m. The tempting 24 m uses \(s = \tfrac{1}{2}at^2\) with \(a = 12\), but suvat is not valid because the acceleration changes. 12 m is the velocity at \(t = 2\), and 16 m integrates \(6t\) as \(6t^2\).
\(a = 0\) when \(t = 4\). \(v = 12t - 1.5t^2 + c\) with \(c = 2\), so \(v(4) = 48 - 24 + 2 = 26\ \text{m s}^{-1}\). The tempting 24 forgets the \(+2\) from the initial velocity. 2 integrates \(-3t\) as \(-3t^2\), and 50 drops the \(-1.5t^2\) term altogether.
\(\displaystyle\int_0^1 v\,dt = \tfrac{11}{6}\), \(\displaystyle\int_1^4 v\,dt = -\tfrac{9}{2}\), \(\displaystyle\int_4^5 v\,dt = \tfrac{11}{6}\). Distance \(= \tfrac{11}{6} + \tfrac{9}{2} + \tfrac{11}{6} = \tfrac{49}{6}\) m (8.17 m). The tempting \(\tfrac{5}{6}\) is the size of \(\displaystyle\int_0^5 v\,dt\), the displacement, where the pieces cancel. \(\tfrac{19}{3}\) stops at \(t = 4\), and \(\tfrac{9}{2}\) splits only at \(t = 1\).
\(s = \displaystyle\int_0^9 4t^{\frac{1}{2}}\,dt = \left[\tfrac{8}{3}t^{\frac{3}{2}}\right]_0^9 = \tfrac{8}{3} \times 27 = 72\) m. The velocity is never negative, so this is also the distance. The tempting 162 m multiplies by \(\tfrac{3}{2}\) instead of dividing. 54 m uses \(\tfrac{1}{2}(0 + 12) \times 9\), a constant-acceleration formula that does not apply, and 12 is the velocity at \(t = 9\).
\(s = 3t^2 - t^3 + c\) with \(c = 0\) (it starts at O). \(s = t^2(3 - t) = 0\) gives \(t = 3\) (other than \(t = 0\)). The tempting \(t = 2\) is when \(v = 0\): the particle turns round there, at its furthest point from O, and then needs another second to come back. \(t = 1\) is when \(a = 0\).
Stage 1: \(\displaystyle\int_0^2 3t^2\,dt = 8\) m, and \(v(2) = 12\ \text{m s}^{-1}\). Stage 2: \(0 = 12^2 - 2 \times 4 \times s\), so \(s = 18\) m. Total \(= 26\) m. The tempting 44 m drops the 2 in \(v^2 = u^2 + 2as\) (giving 36 m for stage 2). 30 m uses \(\tfrac{1}{2}(0 + 12) \times 2 = 12\) m for stage 1, wrongly treating it as constant acceleration, and 18 m is stage 2 alone.
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.