The shape of the lesson
By the end of the lesson, learners can
- Scalars and vectors: In the lab, give the particle a positive initial velocity and a negative acceleration, and run it for long enough that it turns round.
- Displacement-time graphs: The gradient of a displacement-time graph is the velocity.
- Velocity-time graphs: The gradient of a velocity-time graph is the acceleration, and the area under it is the displacement.
- Sketch the velocity-time graph, marking every time and velocity you know.
- Gradient of \(v\)-\(t\) is acceleration; of \(s\)-\(t\) is velocity.
- Area under \(v\)-\(t\) is displacement: triangles, rectangles, trapezia.
- Sign: area below the axis is negative displacement but adds to distance.
- Two particles: write each displacement in terms of \(t\) and set them equal.
One picture every idea returns to
Gradient of a displacement-time graph \(=\) velocity. Gradient of a velocity-time graph \(=\) acceleration. Area between a velocity-time graph and the time axis \(=\) displacement, with area below the axis counting as negative displacement.
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, Gradients and areas.. Two graphs tell the whole story of motion in a straight line. 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: Scalars and vectors | In the lab, give the particle a positive initial velocity and a negative acceleration, and run it for long enough that it turns round. Project tab 1 of the lab (Kinematics lab, linear motion and graphs) and let a student drive it while the class predicts. Model the worked example Distance versus displacement on the board, one line at a time. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Displacement-time graphs | The gradient of a displacement-time graph is the velocity. Model the worked example Reading a displacement-time graph on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: Velocity-time graphs | The gradient of a velocity-time graph is the acceleration, and the area under it is the displacement. Model the worked example Reading Fig. 1 on the board, one line at a time. Your turn: A bus starts from rest and accelerates at \(1.5\ \text{m s}^{-2}\) for 10 s, travels at constant speed for 40 s, then decelerates at \(2.5\ \text{m s}^{-2}\) to rest. Find the total distance. | Frayer Model |
| 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, 26 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 Velocity-time graphs, 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.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
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 |
|---|---|
| Average speed is total DISTANCE divided by time, not the average of the start and end speeds (that only works for constant acceleration). | Return to Scalars and vectors: ask a pair to state the correct version and show it on the lab before moving on. |
| On a displacement-time graph the HEIGHT is position, not speed. | A particle can be far from the start (high on the graph) and still be at rest (flat line). |
| Area below the time axis is negative DISPLACEMENT but positive DISTANCE. | Read the question: "distance travelled" adds the areas; "displacement" subtracts. |
Displacement, velocity and acceleration graphs: worksheet
6 questions, 26 marks. Written in the style of the papers; no past paper question is reproduced.
A runner completes one lap of a 400 m track in 80 s, finishing where she started. Find her distance travelled, displacement, average speed and average velocity.
A displacement-time graph consists of straight lines joining \((0, 0)\), \((5, 30)\), \((8, 30)\) and \((12, -10)\), with \(t\) in seconds and \(s\) in metres. Find the velocity in each stage, the total distance travelled and the final displacement.
A tram starts from rest and accelerates at \(1.5\ \text{m s}^{-2}\) for 10 s. It then travels at constant speed for 40 s before decelerating at \(2.5\ \text{m s}^{-2}\) to rest. Sketch the velocity-time graph and find the total distance.
A car starts from rest and accelerates uniformly to a speed \(V\ \text{m s}^{-1}\) in 12 s. It travels at this speed for 30 s and then decelerates uniformly to rest in 8 s. The total distance is 1000 m. Find \(V\) and the acceleration.
A particle moves in a straight line. Its velocity decreases uniformly from \(8\ \text{m s}^{-1}\) at \(t = 0\) to \(-6\ \text{m s}^{-1}\) at \(t = 7\). Find (a) the time at which it is instantaneously at rest, (b) the distance travelled in the 7 s, (c) its displacement at \(t = 7\).
Car A starts from rest at a junction O and accelerates at \(0.8\ \text{m s}^{-2}\) for 20 s, then continues at constant speed. At the instant A starts, car B passes O travelling in the same direction at a constant \(12\ \text{m s}^{-1}\). Find the time when A overtakes B and the distance from O.
Answer key and marking notes
- Distance 400 m; displacement 0.
- Average speed \(= 400 \div 80 = 5\ \text{m s}^{-1}\).
- Average velocity \(= 0 \div 80 = 0\).
- Velocities: \(\dfrac{30}{5} = 6\), then 0, then \(\dfrac{-10 - 30}{4} = -10\ \text{m s}^{-1}\).
- Distance \(= 30 + 40 = 70\) m.
- Final displacement \(= -10\) m (10 m behind the start).
- Top speed \(= 1.5 \times 10 = 15\ \text{m s}^{-1}\); deceleration time \(= 15 \div 2.5 = 6\) s.
- Graph: up to 15 at \(t = 10\), flat to \(t = 50\), down to 0 at \(t = 56\).
- Distance \(= 75 + 600 + 45 = 720\) m.
- Area \(= \tfrac{1}{2}(12)V + 30V + \tfrac{1}{2}(8)V = 40V\).
- \(40V = 1000\), so \(V = 25\).
- Acceleration \(= 25 \div 12 = 2.08\ \text{m s}^{-2}\).
- Acceleration \(= \dfrac{-6 - 8}{7} = -2\ \text{m s}^{-2}\), so \(8 - 2t = 0\) at (a) \(t = 4\).
- Areas: \(\tfrac{1}{2}(4)(8) = 16\) m above, \(\tfrac{1}{2}(3)(6) = 9\) m below.
- (b) Distance \(= 16 + 9 = 25\) m.
- (c) Displacement \(= 16 - 9 = 7\) m.
- While accelerating, A's displacement is \(0.4t^2\). Setting \(0.4t^2 = 12t\) gives \(t = 30\), which is after the 20 s, so A has not caught B by then.
- At \(t = 20\), A has travelled \(0.4 \times 400 = 160\) m and has speed 16 \(\text{m s}^{-1}\).
- For \(t \ge 20\): \(160 + 16(t - 20) = 12t\), so \(4t = 160\), \(t = 40\) s.
- Distance \(= 12 \times 40 = 480\) m from O.
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.
Part of a particle's velocity-time graph lies below the time axis. What does the total area between the graph and the time axis, with every part counted as positive, represent?
- AThe distance travelled
- BThe displacement from the start
- CThe acceleration
- DThe average velocity
The displacement-time graph of a particle moving in a straight line is shown. Find the velocity of the particle between \(t = 8\) and \(t = 14\).
- A\(-5\ \text{m s}^{-1}\)
- B\(5\ \text{m s}^{-1}\)
- C\(2.14\ \text{m s}^{-1}\)
- D\(0\ \text{m s}^{-1}\)
A particle moves 12 m due east in 4 s, then 20 m due west in 6 s. Find its average speed and its average velocity for the whole 10 s.
- A\(3.2\ \text{m s}^{-1}\) and \(0.8\ \text{m s}^{-1}\) due west
- B\(3.2\ \text{m s}^{-1}\) and \(3.2\ \text{m s}^{-1}\) due west
- C\(0.8\ \text{m s}^{-1}\) and \(3.2\ \text{m s}^{-1}\) due west
- D\(3.17\ \text{m s}^{-1}\) and \(0.8\ \text{m s}^{-1}\) due west
The velocity-time graph of a car is shown. Find the total distance travelled by the car.
- A180 m
- B240 m
- C156 m
- D120 m
A car starts from rest and accelerates uniformly to \(25\ \text{m s}^{-1}\) in 10 s. It travels at \(25\ \text{m s}^{-1}\) for \(T\) s, then decelerates uniformly to rest in 15 s. The total distance is 1000 m. Find \(T\).
- A27.5
- B15
- C22.5
- D40
A particle's velocity decreases uniformly from \(8\ \text{m s}^{-1}\) at \(t = 0\) to \(-4\ \text{m s}^{-1}\) at \(t = 6\). Find the distance travelled in the 6 s.
- A20 m
- B12 m
- C36 m
- D16 m
Car A passes a point O at a constant \(18\ \text{m s}^{-1}\). At that instant car B starts from rest at O and accelerates at \(3\ \text{m s}^{-2}\) until it reaches \(27\ \text{m s}^{-1}\), then continues at that speed. Find the time after passing O when B overtakes A.
- A13.5 s
- B12 s
- C6 s
- D9 s
The velocity-time graph of a particle for \(0 \le t \le 10\) is shown. Find the displacement of the particle from its starting point at \(t = 10\).
- A37.5 m
- B40.5 m
- C46.5 m
- D33 m
Quiz key
Area under a velocity-time graph gives displacement, with area below the axis counting as negative. Counting every part as positive adds up the lengths of all the movements regardless of direction, which is the distance travelled. The tempting "displacement" would SUBTRACT the area below the axis. Acceleration is the gradient, not the area.
Gradient \(= \dfrac{0 - 30}{14 - 8} = -5\ \text{m s}^{-1}\): the particle moves back towards its start at \(5\ \text{m s}^{-1}\). The tempting \(5\ \text{m s}^{-1}\) is the speed; velocity carries the sign of the direction. 2.14 divides 30 by the whole 14 s, and 0 confuses this stage with the flat stage from \(t = 5\) to \(t = 8\), when the particle is at rest.
Distance \(= 12 + 20 = 32\) m, so average speed \(= 32 \div 10 = 3.2\ \text{m s}^{-1}\). Displacement \(= 20 - 12 = 8\) m due west, so average velocity \(= 8 \div 10 = 0.8\ \text{m s}^{-1}\) due west. The tempting 3.17 averages the two stage speeds (\(3\) and \(3.33\)), which is wrong because the stages last different times. Swapping the two values mixes up distance and displacement.
Area \(= \tfrac{1}{2} \times 4 \times 12 + 10 \times 12 + \tfrac{1}{2} \times 6 \times 12 = 24 + 120 + 36 = 180\) m (or use the trapezium: \(\tfrac{1}{2}(10 + 20) \times 12 = 180\)). The tempting 240 m is \(12 \times 20\), treating the whole journey as if at top speed. 156 m leaves out the first triangle, and 120 m counts only the constant-speed stage.
Area \(= \tfrac{1}{2} \times 10 \times 25 + 25T + \tfrac{1}{2} \times 15 \times 25 = 125 + 25T + 187.5\). Setting this equal to 1000 gives \(25T = 687.5\), so \(T = 27.5\). The tempting 15 forgets the halves in both triangles (or treats the whole journey as a rectangle of length \(T + 25\)). 22.5 forgets the half in the first triangle only, and 40 is \(1000 \div 25\), the time at top speed for the whole distance.
The velocity falls by \(2\ \text{m s}^{-1}\) each second, so \(v = 0\) at \(t = 4\). Area above the axis \(= \tfrac{1}{2} \times 4 \times 8 = 16\); area below \(= \tfrac{1}{2} \times 2 \times 4 = 4\). Distance \(= 16 + 4 = 20\) m. The tempting 12 m is the DISPLACEMENT \(16 - 4\) (also what the trapezium formula gives). 36 m uses \(\tfrac{1}{2}(8 + 4) \times 6\), and 16 m ignores the motion after the particle turns round.
B reaches \(27\ \text{m s}^{-1}\) at \(t = 9\), when B has gone \(\tfrac{1}{2} \times 3 \times 81 = 121.5\) m and A has gone \(162\) m, so B has not yet caught up. After that, \(121.5 + 27(t - 9) = 18t\), giving \(9t = 121.5\) and \(t = 13.5\) s. The tempting 12 s solves \(1.5t^2 = 18t\), which assumes B keeps accelerating past \(27\ \text{m s}^{-1}\). 6 s is when the speeds are equal (B is furthest behind then), and 9 s is when B stops accelerating.
Areas: \(\tfrac{1}{2} \times 3 \times 6 = 9\), \(4 \times 6 = 24\), \(\tfrac{1}{2} \times 2 \times 6 = 6\) above the axis and \(\tfrac{1}{2} \times 1 \times 3 = 1.5\) below. Displacement \(= 9 + 24 + 6 - 1.5 = 37.5\) m. The tempting 40.5 m adds the area below the axis: that is the distance travelled, not the displacement. 46.5 m treats \(-3\) as \(+3\) in a trapezium for the last stage, and 33 m ignores the last 3 s.
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.