The shape of the lesson
By the end of the lesson, learners can
- Free fall, strobed: Drop a ball from rest.
- Which equation, and why: Each equation leaves one of the five quantities out.
- Measuring g in the lab.
- List u, v, a, s, t: mark the one you do not need.
- Pick the equation that omits that quantity.
- Free fall: a = g downward; choose a positive direction and keep it.
One picture every idea returns to
For uniformly accelerated motion in a straight line, v = u + at, s = ½(u + v)t, s = ut + ½at² and v² = u² + 2as. They follow directly from the definitions of velocity and acceleration and from the area under a velocity-time graph. Free fall is the special case with a = g.
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, Four equations, one assumption.. When the acceleration is constant, four short equations connect displacement, the two velocities, acceleration and time. 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: Free fall, strobed | Drop a ball from rest. Project tab 2 of the lab (Equations of motion) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Which equation, and why | Each equation leaves one of the five quantities out. | Pairs |
| 21 to 30 min | Teach and check: Measuring g in the lab | Measuring g in the lab. | Corners |
| 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, 17 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 Measuring g in the lab, 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.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
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 |
|---|---|
| Two slips cost marks here. | First, the suvat equations apply only while the acceleration is constant; a journey with separate phases must be split and each phase handled with its own values. Second, mind the signs in free fall: choose one direction as positive and stick to it. A ball thrown upward still has acceleration g acting downward throughout, including at the highest point where the velocity is momentarily zero but the acceleration is not. |
Equations of motion: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
A cyclist accelerates uniformly from rest at 1.5 m s⁻² for 6.0 s. Find the final velocity and the distance travelled.
A train travelling at 40 m s⁻¹ brakes uniformly and stops in a distance of 500 m. Choosing a suitable equation, find the deceleration.
A ball is dropped from rest from a height of 1.8 m. Taking g = 9.8 m s⁻² and ignoring air resistance, find the time to reach the ground and the speed on landing.
A ball is thrown vertically upward at 19.6 m s⁻¹. Taking g = 9.8 m s⁻², find the maximum height reached and the time taken to reach it.
In an experiment to find g, a ball falls from rest through several measured heights, and a graph of fall distance s against time squared t² is plotted. The line is straight through the origin with gradient 4.85 m s⁻². Find g and explain why this graphical method is better than a single drop.
A car covers two stages: 12 s at a constant 20 m s⁻¹, then it accelerates uniformly. Explain why a single suvat equation cannot be applied to the whole journey, and state what you must do instead.
Answer key and marking notes
- v = u + at = 0 + 1.5 × 6.0 = 9.0 m s⁻¹. ✓
- s = ut + ½at² = 0 + ½ × 1.5 × 6.0² = 27 m. ✓
- Known: u = 40, v = 0, s = 500; time not needed, so use v² = u² + 2as. ✓
- 0 = 40² + 2a(500), so a = −1600 / 1000 = −1.6 m s⁻². ✓
- The deceleration is 1.6 m s⁻². ✓
- s = ½gt²: 1.8 = ½ × 9.8 × t², so t² = 0.367, t = 0.61 s. ✓
- v = gt = 9.8 × 0.61 = 6.0 m s⁻¹ (or v = √(2gs) = √(2 × 9.8 × 1.8) = 5.9 m s⁻¹). ✓
- At the top v = 0. Using v² = u² + 2as with a = −9.8: 0 = 19.6² − 2(9.8)s. ✓
- s = 384 / 19.6 = 19.6 m. ✓
- Time to top: v = u + at gives 0 = 19.6 − 9.8t, so t = 2.0 s. ✓
- s = ½g t², so the gradient equals ½g; therefore g = 2 × 4.85 = 9.7 m s⁻². ✓
- Using many points averages out random error and lets anomalies be spotted, giving a more reliable g than one reading. ✓
- The suvat equations require a single constant acceleration, but the acceleration changes between the two stages (zero, then non-zero). ✓
- Treat each stage separately with its own values, then combine the results (for example add the displacements). ✓
17 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.
The equation s = ut + ½at² may be applied only when:
- Athe velocity is constant
- Bthe object starts from rest
- Cthe acceleration is constant
- Dthere is no gravity
An object starts from rest and accelerates at 3.0 m s⁻² for 5.0 s. Its final velocity is:
- A8.0 m s⁻¹
- B15 m s⁻¹
- C1.7 m s⁻¹
- D30 m s⁻¹
A body moving at 2.0 m s⁻¹ accelerates uniformly at 4.0 m s⁻² for 3.0 s. The distance covered is:
- A18 m
- B12 m
- C36 m
- D24 m
A body starts from rest and accelerates at 2.0 m s⁻² over 9.0 m. Its final speed is:
- A6.0 m s⁻¹
- B3.0 m s⁻¹
- C18 m s⁻¹
- D36 m s⁻¹
Which equation of motion does not contain displacement s?
- As = ut + ½at²
- Bv² = u² + 2as
- Cv = u + at
- Ds = ½(u + v)t
Which equation of motion does not contain time t?
- Av = u + at
- Bv² = u² + 2as
- Cs = ut + ½at²
- Ds = ½(u + v)t
An object is dropped from rest. Taking g = 9.8 m s⁻², its speed after 3.0 s of free fall is about:
- A15 m s⁻¹
- B44 m s⁻¹
- C3.3 m s⁻¹
- D29 m s⁻¹
Taking g = 9.8 m s⁻², how far does an object fall from rest in the first 1.0 s?
- A4.9 m
- B9.8 m
- C2.0 m
- D19.6 m
An object is thrown vertically upward. At the highest point of its motion:
- Aboth velocity and acceleration are zero
- Bthe acceleration is zero but the velocity is maximum
- Cthe velocity is zero but the acceleration is g downward
- Dthe acceleration points upward
In an experiment to find g, fall distance s is plotted against time squared t². The gradient of the straight line equals:
- Ag
- B½g
- C2g
- Dg²
A car decelerates uniformly from 20 m s⁻¹ to rest in 50 m. The magnitude of its acceleration is:
- A8.0 m s⁻²
- B2.0 m s⁻²
- C0.4 m s⁻²
- D4.0 m s⁻²
For uniformly accelerated motion, the average velocity equals:
- A(u + v) / 2
- Bu + v
- C(v − u) / 2
- D½ v
A ball falls freely from rest. The ratio of the distance fallen in the first 2.0 s to that in the first 1.0 s is:
- A2 : 1
- B1 : 1
- C4 : 1
- D8 : 1
Taking g = 10 m s⁻², the time for an object to fall 45 m from rest is:
- A4.5 s
- B3.0 s
- C2.1 s
- D9.0 s
Rearranging v = u + at to make t the subject gives:
- At = (v + u) / a
- Bt = a(v − u)
- Ct = v / (u + a)
- Dt = (v − u) / a
Using s = ½(u + v)t, a body that speeds up uniformly from 4.0 to 16 m s⁻¹ over 6.0 s travels:
- A60 m
- B120 m
- C30 m
- D20 m
A stone is thrown straight down at 5.0 m s⁻¹. Taking g = 9.8 m s⁻², its speed after 2.0 s is:
- A19.6 m s⁻¹
- B14.6 m s⁻¹
- C24.6 m s⁻¹
- D9.8 m s⁻¹
Quiz key
The suvat equations assume a constant acceleration throughout the interval.
v = u + at = 0 + 3.0 × 5.0 = 15 m s⁻¹.
s = 2.0 × 3.0 + ½ × 4.0 × 3.0² = 6 + 18 = 24 m.
v² = 0 + 2 × 2.0 × 9.0 = 36, so v = 6.0 m s⁻¹.
v = u + at links u, v, a and t but not s.
v² = u² + 2as relates u, v, a and s with no t.
v = gt = 9.8 × 3.0 = 29 m s⁻¹.
s = ½ × 9.8 × 1.0² = 4.9 m.
At the top the velocity is momentarily zero, yet gravity still acts, so the acceleration is g downward.
Since s = ½g t², the gradient of s against t² is ½g, so g is twice the gradient.
0 = 20² + 2a(50), so a = −400/100 = −4.0 m s⁻²; the magnitude is 4.0 m s⁻².
With uniform acceleration the average velocity is the mean of the initial and final velocities, (u + v)/2.
Since s = ½gt², distance is proportional to t², so the ratio is 2² : 1² = 4 : 1.
45 = ½ × 10 × t², so t² = 9.0 and t = 3.0 s.
From v = u + at, v − u = at, so t = (v − u)/a.
s = ½(4.0 + 16) × 6.0 = 10 × 6.0 = 60 m.
v = u + gt = 5.0 + 9.8 × 2.0 = 24.6 m s⁻¹.
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.