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Quick subject briefing
Before this lesson: Consolidation after velocity, acceleration, straight-line graphs and uniform-acceleration calculations have been taught. Both tiers follow the same selected route; no graph areas or tangents are assigned.
Velocity includes direction; acceleration measures its change per second. On a velocity-time plot height gives velocity and gradient gives acceleration. A falling line above zero is slowing positive-direction motion, not backwards motion. A flat nonzero line means moving with zero acceleration. For uniform forward motion, v²=u²+2as gives squared speed, so take a square root.
Words and symbols you will need
- Velocity, u or v (m/s)
- Speed with direction; u is initial and v final in the equation.
- Acceleration, a (m/s²)
- Change in velocity each second; can be positive, negative or zero.
- Gradient / elapsed time
- Velocity change divided by the time change between the same two points.
- Displacement, s (m)
- Distance along the chosen straight-line direction for these forward tests.
- Uniform / v²
- Uniform acceleration stays constant. v² is squared velocity and needs a square root to recover speed.
Help pupils choose a method
- Read axis quantity and sign convention; use velocity to identify direction.
- For acceleration use both endpoint differences, later minus earlier, and divide by elapsed time.
- For uniform forward u,a,s data, calculate v²=u²+2as, then take the positive square root. Check units and plausibility.
Use the supplied model — pupil page 2
Use pupil page 2. The first test gains 12 m/s in 6 s, so acceleration is 2 m/s². The separate second test gives squared speed 16, so final speed is 4 m/s; the intermediate 16 is not the speed.
A velocity changes uniformly from 4.0 to 16 m/s in 6.0 s. Find acceleration. In a separate straight-line test, u=2.0 m/s, a=1.5 m/s² and s=4.0 m, with uniform acceleration forward. Find final speed.
Worked answer: Acceleration = (16-4.0)/6.0 = 2.0 m/s². Separate test: v² = u²+2as = 2.0²+2×1.5×4.0 = 16 m²/s². Final speed v = √16 = 4.0 m/s. Take the square root after calculating v².
If pupils are still stuck
Use the matching Say/Ask/Expect and named independent retry in its replacement time. If still insecure, help write the two differences or the v² substitution with labels, then record the specific step needing follow-up. Keep next-lesson support outside today’s 60 minutes.
KEEP THIS OPEN DURING THE LESSON
Timed teacher run sheet
- Print the complete 8-page pupil booklet, one per pupil, stapled in order. Double-sided printing is fine.
- Read teacher page 1, then use the timed run sheet on page 2. Keep the 10-page teacher guide separate from pupil copies.
- Pupils need pens, a ruler and a calculator. Reading, data and figures are supplied. No login, pupil devices or practical apparatus are required.
0-5 minPupil page / Q
1 / 1-3
5-15 minPupil page / Q
2 / reading
Read aloud / check: Read quantities and the two models. Explain why a square root is needed.
Next: Check A if used; then continue reading.
15-20 minPupil page / Q
3 / 4
Read aloud / check: Choose a letter with both changes and a motion reason; stop.
Next: If insecure B replaces 5-6. The remaining source questions use 0-20 s.
Answers: 4.
Open check 2 and its coaching →
20-30 minPupil page / Q
4 / 5-9
Read aloud / check: Complete the graph, then read velocities and gradients.
Next: If B was used leave 5-6 blank. Require signed acceleration.
Answers: 5, 6, 7, 8, 9.
30-33 minPupil page / Q
5 / 10
33-45 minPupil page / Q
5-7 / 11-16
Read aloud / check: Finish graph explanations, then apply both equations.
Next: If C was used leave 11-12 blank; complete 13-16. Final speed needs square root.
Answers: 11, 12, 13, 14, 15, 16.
45-55 minPupil page / Q
8 / 17
Read aloud / check: Correct practice in a different colour and explain a correction.
Next: Share only practice answers. Refer to the checked teacher graph for Q5.
Answers: 17.
55-60 minPupil page / Q
8 / 18
Read aloud / check: Cover earlier pages; complete question 18 independently.
Next: Collect booklets and check afterwards. D is next lesson only.
Answers: 18.
Open check 4 and its coaching →
Estimated 60 minutes. Practice: 10 minutes, a 3-minute check and 12 minutes, including the two note applications. A uses reading time; B replaces 5-6; C replaces 11-12; D is next lesson only. Timings are not classroom-trialled.
Check entry, hinge and mid-practice at the marked pauses. At 45 minutes share only practice answers. Keep the independent exit and next-lesson answers hidden until submission.
Download teacher guide & answers ↓If several pupils are stuck
Use the matching Say/Ask/Expect and named independent retry in its replacement time. If still insecure, help write the two differences or the v² substitution with labels, then record the specific step needing follow-up. Keep next-lesson support outside today’s 60 minutes.
Estimated 60 minutes. Practice: 10 minutes, a 3-minute check and 12 minutes, including the two note applications. A uses reading time; B replaces 5-6; C replaces 11-12; D is next lesson only. Timings are not classroom-trialled.
ENTRY CHECK
Check 1: Question 1, Question 2, Question 3
Look for: Q1 +2.0 m/s²; Q2 zero acceleration while moving; Q3 speed 7.0 m/s.
If secure: Continue to page 2 reading and models.
If: Uses 10/3 rather than (10-4)/3, or says constant velocity means stationary.Open coaching
Likely issue: Velocity and change in velocity are confused.
- Say
- Acceleration depends on the change, not the final velocity. If velocity stays nonzero and constant, the object still moves but does not accelerate.
- Ask
- For question 1 what is 10-4? In question 2 does 6.0 m/s describe rest?
- Expect
- Change 6.0 m/s gives 2.0 m/s²; 6.0 m/s means moving, with acceleration zero.
Next: Use the relevant method on page 2; complete Try again A on page 1 during the first 2 reading minutes.
Pupil page 2, acceleration and flat-line rules.
If: Reports 49 m/s as the answer to question 3.Open coaching
Likely issue: Squared speed has not been converted to speed.
- Say
- v² is speed multiplied by itself. Use the positive square root for the speed asked here.
- Ask
- In question 3, which positive number multiplied by itself gives 49?
- Expect
- 7, so final speed is 7.0 m/s, not 49 m/s.
Next: Use the relevant method on page 2; complete Try again A on page 1 during the first 2 reading minutes.
Pupil page 2, uniform-acceleration method.
HINGE CHECK
Check 2: Question 4
Look for: A, with velocity change -6 m/s, elapsed time 4 s and acceleration -1.5 m/s²; motion east and slowing.
If secure: Begin questions 5-9. Require both endpoint differences and a separate direction reason.
If: B uses final coordinates 2/6.Open coaching
Likely issue: The relevant graph interval has not been isolated.
- Say
- Subtract the two velocities and the two times. This section does not start at the origin.
- Ask
- In question 4, what are 2-8 and 6-2?
- Expect
- -6 m/s and 4 s, giving -1.5 m/s².
Next: Use page 2 endpoint/sign methods; complete Try again B on page 3 instead of questions 5-6.
Pupil page 2, graph-section method.
If: C calls the motion west because acceleration is negative.Open coaching
Likely issue: Acceleration direction has been confused with travel direction.
- Say
- The velocity values remain positive, so travel is east. Their decrease gives negative acceleration and slowing.
- Ask
- In question 4 are the velocities above or below zero throughout the section?
- Expect
- Above zero: eastward motion while slowing.
Next: Use page 2 endpoint/sign methods; complete Try again B on page 3 instead of questions 5-6.
Pupil page 2, motion-direction rule.
If: D reverses the velocity difference, or A is chosen without working.Open coaching
Likely issue: The sign is lost or the correct letter may be a guess.
- Say
- Keep the same chronological order in both differences: later minus earlier. Then describe motion using velocity, not gradient.
- Ask
- Show both differences and explain east/slowing for question 4.
- Expect
- (2-8)/(6-2)=-1.5 m/s²; positive velocity means east, decreasing velocity means slowing.
Next: Use page 2 endpoint/sign methods; complete Try again B on page 3 instead of questions 5-6.
Pupil page 2, endpoint and sign rules.
MID PRACTICE CHECK
Check 3: Question 10
Look for: Q10 increasing acceleration 1.5 m/s²; flat acceleration zero while moving at 9.0 m/s.
If secure: Continue 11-16 with the same quantity distinctions.
If: Uses 9/4 or an incorrect velocity change for the increasing stage.Open coaching
Likely issue: Final velocity has replaced change in velocity.
- Say
- The first 3 m/s was already present. Only the additional 6 m/s is gained during these 4 seconds.
- Ask
- In question 10, how much is added to 3 m/s to reach 9 m/s?
- Expect
- 6 m/s added in 4 s, so acceleration is 1.5 m/s².
Next: Use page 2 acceleration/flat-line methods. Complete Try again C on page 5 instead of questions 11-12.
Pupil page 2, a=change in velocity/time.
If: Calls the flat section stationary or gives acceleration 9 m/s².Open coaching
Likely issue: Graph height and gradient are confused.
- Say
- Height gives velocity; gradient gives acceleration. At height 9 m/s with no slope, the cart moves steadily with zero acceleration.
- Ask
- In question 10, does velocity change during the flat section, and is its velocity zero?
- Expect
- It does not change, so acceleration is zero; velocity is 9 m/s, so it is moving.
Next: Use page 2 acceleration/flat-line methods. Complete Try again C on page 5 instead of questions 11-12.
Pupil page 2, velocity versus acceleration.
EXIT CHECK
Check 4: Question 18
Look for: Q18 gradient -2.0 m/s² with eastward slowing; separate final speed 6.0 m/s; constant 4 m/s gives zero acceleration while moving.
If secure: Record independently secure gradient, motion-description and equation methods.
If: Wrong endpoint gradient/sign or describes westward travel in part (a).Open coaching
Likely issue: Endpoint changes or direction versus acceleration needs follow-up.
- Say
- Use later minus earlier for velocity and time separately. Positive velocity still indicates eastward travel while it decreases.
- Ask
- For question 18(a), what are 4-12 and 5-1? Are both velocities positive?
- Expect
- -8 m/s in 4 s gives -2.0 m/s². Both velocities are positive, so travel is east, slowing.
Next: Next lesson revisit the corresponding page 2 methods, then complete Try again D independently on page 8.
Pupil page 2, endpoint and direction methods.
If: Reports 36 m/s in (b), or calls the flat 4 m/s stage stationary.Open coaching
Likely issue: Squared speed or height versus gradient remains insecure.
- Say
- The calculation in (b) gives v²; take its square root. A nonzero flat velocity is steady motion, not rest.
- Ask
- For question 18, what is √36, and does a velocity fixed at 4 m/s change or become zero?
- Expect
- Final speed 6.0 m/s. Fixed 4 m/s has zero acceleration but is still moving.
Next: Next lesson revisit the corresponding page 2 methods, then complete Try again D independently on page 8.
Pupil page 2, squared-speed and flat-line methods.
Answers keyed to the pupil handout
- Question 1
Pupil page 1 - (10-4.0)/3.0 = +2.0 m/s².
- Question 2
Pupil page 1 - Acceleration = 0 m/s² because velocity is unchanged. It is moving, not stationary.
- Question 3
Pupil page 1 - v = positive square root of 49 = 7.0 m/s. 49 is the squared speed.
- Question 4
Pupil page 3 - A. a=(2.0-8.0)/(6.0-2.0)=-6/4=-1.5 m/s². Velocity stays positive, so the object travels east while slowing.
- Question 5
Pupil page 4 - Horizontal line from (20 s, 1.0 m/s) to (24 s, 1.0 m/s), as in the supplied teacher graph.
- Question 6
Pupil page 4 - 2.0 m/s east. The velocity is positive and constant during 4-8 s.
- Question 7
Pupil page 4 - a = (2.0-0)/(4.0-0) = +0.50 m/s².
- Question 8
Pupil page 4 - a = (0-2.0)/(12-8) = -0.50 m/s². The acceleration points west while the robot slows travelling east. Deceleration magnitude is 0.50 m/s², but the signed answer is negative.
- Question 9
Pupil page 4 - Velocity stays constant at 2.0 m/s east. Acceleration is zero because velocity is unchanged; the robot is moving, not stationary.
- Question 10
Pupil page 5 - Increasing section: (9.0-3.0)/4.0 = +1.5 m/s². Flat section: acceleration 0 m/s²; cart still moves at +9.0 m/s.
- Question 11
Pupil page 5 - 12-16 s, when velocity is zero throughout.
- Question 12
Pupil page 6 - Velocity increases linearly from 0 to 1.0 m/s east. Acceleration is constant and positive because gradient is constant: (1.0-0)/(20-16) = +0.25 m/s².
- Question 13
Pupil page 6 - The robot has a constant acceleration of +0.50 m/s²; its velocity increases linearly from 0 to 2.0 m/s east. The robot accelerates; its velocity increases.
- Question 14
Pupil page 6 - Velocity remains positive until reaching zero at 12 s, so motion is east, slowing. The falling line shows negative acceleration, not negative velocity. Its sign above/below the time axis establishes direction.
- Question 15
Pupil page 7 - a=(v-u)/t=(15-3.0)/6.0=+2.0 m/s².
- Question 16
Pupil page 7 - v²=u²+2as=1.0²+2×2.0×6.0=25 m²/s². v=√25=5.0 m/s, the positive root for the stated forward motion.
- Question 17
Pupil page 8 - Accept a corrected answer with the reason for the error. All-correct route: in 8-12 s velocity is positive until the object stops, so it moves east; its falling velocity gives negative acceleration. The sign of velocity sets motion direction, not the sign of gradient.
- Question 18
Pupil page 8 - (a) a=(4.0-12)/(5.0-1.0)=-8/4=-2.0 m/s²; moving east while slowing. (b) v²=2²+2×2×8=36 m²/s²; v=6.0 m/s. (c) Acceleration 0 m/s²; it still moves at 4.0 m/s east.
- Try again A
- (a) (14-2)/4=3 m/s². (b) 0 m/s², still moving. (c) √81=9 m/s.
- Try again B
- a=(2-10)/(7-3)=-8/4=-2 m/s². Positive velocity means moving east; it slows because velocity decreases.
- Try again C
- Increasing: (11-5)/3=2 m/s². Constant: 0 m/s², still moving at 11 m/s.
- Try again D
- Gradient=(5-15)/(7-2)=-2 m/s²; moving east and slowing. Separate cart: v²=1+2×1×12=25, so v=5 m/s. At constant velocity 3 m/s acceleration is zero, but the cart still moves.
Worksheet selection and marking
Nine source parts total 13 marks; maintained note practice 1 and 5 add 6 marks. Accept correct signed working and equivalent wording. The uniform-acceleration equation is shared at both tiers.
Selection: Q3(a-i), source pages 3-4; incomplete Figure 2.
Open the maintained worked solutions →Collect and follow up: Collect every named booklet with working and graph completion. Use question 18 to identify endpoint/sign, direction, square-root or flat-line errors, then report the exact step and who needs Try again D.
