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Quick subject briefing
Before this lesson: Consolidation after pupils have met speed calculations, simple graph axes and distance-time graphs. A ruler and calculator are needed; no experiment is required.
A total-distance graph shows how much route has accumulated. Its height is distance; its gradient is speed. For a straight section use changes between two points, even when the section starts away from the origin. A horizontal line means stopped. Whole-journey average speed includes stopped time. Returning still increases total distance.
Words and symbols you will need
- Speed (m/s)
- Distance covered per second; direction is not included.
- Total distance (m)
- The accumulated route travelled, even after a return.
- Elapsed time (s)
- All the time between the stated start and finish, including stops.
- Gradient
- Change in vertical distance value divided by change in horizontal time value.
- Stationary
- Not moving; no additional distance is travelled.
Help pupils choose a method
- Name the interval. Read two points from the same straight graph section.
- Subtract distances and times separately, then divide distance change by time change.
- For a whole-journey average, use the full distance and full elapsed time. Check units and whether the slope matches the motion.
Use the supplied model — pupil page 2
Use the runner model on pupil page 2: 150 m in 30 s gives 5 metres each second. This model starts at zero; question 4 deliberately does not, so both endpoint differences matter.
The revision-note runner travels 150 m in 30 s at constant speed. Find the gradient of the distance-time graph.
Worked answer: Speed = change in distance / change in time = 150 / 30 = 5.0 m/s. A straight sloping line means constant speed.
If pupils are still stuck
Read the relevant Say/Ask/Expect coaching and use the named fresh retry within its replacement time. If it remains insecure, help the pupil write the two changes or the correct whole-journey fraction, then record the specific difficulty for the usual teacher.
KEEP THIS OPEN DURING THE LESSON
Timed teacher run sheet
- Print the complete 8-page pupil booklet in order, one per pupil. Staple each booklet; double-sided printing is fine.
- Read the subject briefing on teacher page 1, then use the timed run sheet on page 2. Keep this guide separate from pupil copies.
- Pupils need pens, a ruler and a calculator. The graph grid, source figure, reading and data are included; no login, devices or apparatus are required.
0-5 minPupil page / Q
1 / 1-3
Read aloud / check: Start from memory on page 1. At 5 minutes, check the three answers.
Next: If insecure, use Try again A in the first 2 reading minutes.
Answers: 1, 2, 3.
Open check 1 and its coaching →
5-15 minPupil page / Q
2 / reading
Read aloud / check: Read the equation table and graph rules. Follow the runner model.
Next: Check Try again A if used; then continue the reading.
15-20 minPupil page / Q
3 / 4
Read aloud / check: Choose a letter and show both endpoint changes. Stop at check 2.
Next: If insecure, Try again B replaces questions 5-6.
Answers: 4.
Open check 2 and its coaching →
20-30 minPupil page / Q
3-4 / 5-10
Read aloud / check: Complete the delivery calculations and Figure 1 questions.
Next: If Try again B was used, leave 5-6 blank. Require units and changes.
Answers: 5, 6, 7, 8, 9, 10.
30-33 minPupil page / Q
5 / 11
Read aloud / check: Pause and complete question 11 independently.
Next: If insecure, Try again C replaces questions 12-13.
Answers: 11.
Open check 3 and its coaching →
33-45 minPupil page / Q
5-7 / 12-19
Read aloud / check: Continue the graph calculations, then draw the different robot graph on page 7.
Next: The grid is supplied. If C was used, leave 12-13 blank; complete 14-19.
Answers: 12, 13, 14, 15, 16, 17, 18, 19.
45-55 minPupil page / Q
8 / 20
Read aloud / check: Correct practice in a different colour. Explain one correction.
Next: Share only practice answers. Keep the exit answer hidden.
Answers: 20.
55-60 minPupil page / Q
8 / 21
Read aloud / check: Cover earlier pages. Complete question 21 independently.
Next: Collect every booklet. Check the exit afterwards; D is next lesson only.
Answers: 21.
Open check 4 and its coaching →
Estimated 60 minutes. Practice: 10 minutes, a 3-minute check and 12 minutes, including the supplied-grid graph. A uses reading time; B replaces 5-6; C replaces 12-13; D is next lesson only. Timings are not classroom-trialled.
Check the entry, hinge and mid-practice responses at their marked pauses. Share only practice answers at 45 minutes. Keep question 21 and the next-lesson answers hidden until submission.
Download teacher guide & answers ↓If several pupils are stuck
Use the matching coaching and page 2 method. Direct the named retry within its replacement time. If it still fails, help pupils set up the distance/time fraction and return the exact difficulty to the usual teacher.
Estimated 60 minutes. Practice: 10 minutes, a 3-minute check and 12 minutes, including the supplied-grid graph. A uses reading time; B replaces 5-6; C replaces 12-13; D is next lesson only. Timings are not classroom-trialled.
ENTRY CHECK
Check 1: Question 1, Question 2, Question 3
Look for: Question 1: 3.0 m/s; question 2: 90 s; question 3: stationary, 0 m/s because distance is unchanged.
If secure: Continue to the reading and model on pupil page 2.
If: Question 1 multiplies distance and time or gives an incorrect unit; question 2 does not multiply minutes by 60.Open coaching
Likely issue: Equation choice or the time unit is insecure.
- Say
- Speed is distance covered in one second, so divide distance by time. A minute contains 60 seconds.
- Ask
- For questions 1-2, what are 18 divided by 6, and 1.5 multiplied by 60?
- Expect
- 3.0 m/s and 90 s. Metres divided by seconds give m/s.
Next: Use the equation/units table on page 2. Direct Try again A on page 1 in the first 2 reading minutes, then check the fresh work.
Pupil page 2, equations and units; Try again A on page 1.
If: Question 3 calls the speed 12 m/s or says the object keeps moving.Open coaching
Likely issue: Graph height has been confused with gradient.
- Say
- The height gives the total distance so far. During a horizontal section no extra distance is travelled, so speed is zero.
- Ask
- How much additional distance is travelled from 8 s to 14 s in question 3?
- Expect
- Zero metres, so the object is stationary at 0 m/s. Its existing 12 m total is not a speed.
Next: Use page 2, the horizontal-line rule. Direct Try again A on page 1 in reading time and require the stationary reason.
Pupil page 2, A graph section needs two changes; Try again A on page 1.
HINGE CHECK
Check 2: Question 4
Look for: B, with distance change 12 m and time interval 4 s, giving 3.0 m/s. A correct letter alone is insufficient.
If secure: Start questions 5-10. Require both endpoint differences to establish the method.
If: A uses 17 divided by 6.Open coaching
Likely issue: The pupil divides final coordinates instead of using changes across this section.
- Say
- This section starts after zero time and distance. Use the distance added during it and the time it takes, not the final coordinates from the origin.
- Ask
- For question 4, what are 17 minus 5 and 6 minus 2?
- Expect
- 12 m and 4 s. Dividing gives 3.0 m/s, option B.
Next: Use the endpoint method on page 2, then Try again B on page 3 instead of questions 5-6. Check both changes.
Pupil page 2, A graph section needs two changes; Try again B on page 3.
If: C reports only the distance change, or D divides time change by distance change.Open coaching
Likely issue: The meaning or order of the gradient ratio is insecure.
- Say
- A speed needs distance per unit time. Divide metres by seconds; a distance change alone is not a speed.
- Ask
- The changes are 12 m and 4 s in question 4. Which division has unit m/s?
- Expect
- 12 / 4 = 3.0 m/s. The reversed division has unit s/m.
Next: Use the speed equation and endpoint method on page 2. Complete Try again B on page 3 instead of questions 5-6.
Pupil page 2, speed and gradient; Try again B on page 3.
If: B is chosen without both changes or with incorrect working.Open coaching
Likely issue: A correct choice may be a guess.
- Say
- The letter needs a checkable method. Write the two endpoint differences before dividing.
- Ask
- Explain how the two given points produce your speed in question 4.
- Expect
- Distance rises by 12 m in 4 s, so speed is 3.0 m/s. B follows from those changes.
Next: Ask for both endpoint differences. If insecure, use the page 2 method and Try again B instead of questions 5-6.
Pupil page 2, endpoint method; Try again B on page 3.
MID PRACTICE CHECK
Check 3: Question 11
Look for: Rest from 4-8 s and whole-journey average 12 / 8 = 1.5 m/s, including the stop.
If secure: Continue questions 12-19, including the new graph on page 7.
If: Gives 3 m/s by using 4 s for the whole-journey average.Open coaching
Likely issue: Stopped time has been removed from the denominator.
- Say
- Average speed over a whole journey uses the whole elapsed time. The clock still runs while the object is stopped.
- Ask
- In question 11, what is the total elapsed time from the beginning to the stated finish?
- Expect
- 8 s, including the 4 s stop. Average = 12 / 8 = 1.5 m/s.
Next: Read the whole-journey rule on page 2. Direct Try again C on page 5 instead of questions 12-13, then continue 14-19.
Pupil page 2, Whole journeys and returns; Try again C on page 5.
If: Treats the constant 12 m height as a nonzero speed, or misses the rest interval.Open coaching
Likely issue: Accumulated distance and speed are still being confused.
- Say
- The total stays at 12 m during 4-8 s. Staying at the same accumulated distance means no motion in that interval.
- Ask
- How much distance is added after 4 s in question 11, and what is the rest interval?
- Expect
- No distance is added; rest is 4-8 s. For the complete run, 12 / 8 = 1.5 m/s.
Next: Use page 2, the horizontal-line rule and whole-journey definition. Complete Try again C instead of questions 12-13.
Pupil page 2, graph rules and whole-journey rule; Try again C on page 5.
EXIT CHECK
Check 4: Question 21
Look for: Rest 2-5 s; moving speeds 4.0 and 2.0 m/s with the first faster; whole average 1.75 m/s.
If secure: Record the objectives demonstrated by the independent work, then identify any precise follow-up.
If: Uses the final coordinates for the last section, or chooses the highest section as fastest.Open coaching
Likely issue: The endpoint-gradient or height-versus-speed distinction needs follow-up.
- Say
- Compare the changes during each section. The last stage starts at 8 m and 5 s, so its distance and time changes must both be subtracted.
- Ask
- In question 21, what distance and time are added during 5-8 s? Compare its speed with 0-2 s.
- Expect
- 6 m in 3 s gives 2.0 m/s; the first stage is 8 m in 2 s, or 4.0 m/s, so the first is faster.
Next: Next lesson, revisit the endpoint method and graph rules on page 2, then complete Try again D independently on page 8.
Pupil page 2, endpoint gradient; Try again D on page 8, next lesson only.
If: Misses the 2-5 s rest, excludes it from the average or averages the moving-stage speeds.Open coaching
Likely issue: Whole elapsed time and stationary intervals need follow-up.
- Say
- The unchanged 8 m reading identifies a stop. For average speed, use the full 14 m and all 8 s; do not take an unweighted mean of stage speeds.
- Ask
- Which interval adds time but no distance in question 21? What division gives the whole average?
- Expect
- 2-5 s adds stopped time. Whole average = 14 / 8 = 1.75 m/s.
Next: Next lesson, use the whole-journey rule on page 2 before Try again D on page 8. Keep that follow-up outside today’s 60 minutes.
Pupil page 2, whole-journey rule; Try again D on page 8, next lesson only.
Answers keyed to the pupil handout
- Question 1
Pupil page 1 - 18 / 6.0 = 3.0 m/s.
- Question 2
Pupil page 1 - 1.5 × 60 = 90 s.
- Question 3
Pupil page 1 - Stationary: total distance does not change. Speed = 0 m/s. The height 12 m is a distance, not a speed.
- Question 4
Pupil page 3 - B: distance change = 17 - 5 = 12 m; time interval = 6 - 2 = 4 s; speed = 12 / 4 = 3.0 m/s.
- Question 5
Pupil page 3 - t = distance / speed. First: 12 / 1.5 = 8.0 s. Second: 12 / 2.0 = 6.0 s.
- Question 6
Pupil page 3 - Average speed = 12 / 5.0 = 2.4 m/s. No: the average exceeds 1.5 m/s, so the speed must exceed the limit at some time. An average below a limit would not prove that every instantaneous speed was below it.
- Question 7
Pupil page 4 - 7.5 m. Accept 7.0-8.0 m for a clear reading from the supplied graph.
- Question 8
Pupil page 4 - 25 m: halfway between 15 m at 20 s and 35 m at 30 s.
- Question 9
Pupil page 4 - 10-20 s and 40-50 s. The total distance is unchanged in both intervals.
- Question 10
Pupil page 4 - (15 - 0) / (10 - 0) = 1.5 m/s.
- Question 11
Pupil page 5 - Rest: 4-8 s. Whole-journey average = 12 / 8 = 1.5 m/s. Include the stopped time; 12 / 4 = 3 m/s describes only the moving stage.
- Question 12
Pupil page 5 - (35 - 15) / (30 - 20) = 20 / 10 = 2.0 m/s. 35 / 30 gives an average from the origin, not this section’s speed.
- Question 13
Pupil page 6 - (40 - 35) / (40 - 30) = 5 / 10 = 0.50 m/s.
- Question 14
Pupil page 6 - Total distance / total elapsed time = 40 / 50 = 0.80 m/s. The complete journey lasts 50 s, including its stops; excluding stops answers a different question.
- Question 15
Pupil page 6 - 20-30 s: this section has the greatest gradient. Graph height gives accumulated distance, not speed; the highest horizontal section is stationary.
- Question 16
Pupil page 6 - Travelling west still adds to the total distance travelled. The vertical axis is accumulated distance, not position or signed displacement, so it increases during this movement.
- Question 17
Pupil page 7 - Time / s horizontally; total distance travelled / m vertically. Plot (0,0), (3,6), (5,6), (9,18), using the supplied consistent scales. Join consecutive points with straight segments. Use the supplied teacher graph to check plotting.
- Question 18
Pupil page 7 - Rest: 3-5 s. The fastest section is 5-9 s: (18 - 6) / (9 - 5) = 12 / 4 = 3.0 m/s. The first section is 6 / 3 = 2 m/s.
- Question 19
Pupil page 7 - 18 / 9 = 2.0 m/s. Include the 2 s stop; do not take an unweighted mean of the three section speeds.
- Question 20
Pupil page 8 - Accept a correct answer plus the reason for the original error. For the all-correct route: height gives accumulated distance; gradient gives speed. Figure 1 is highest at 40-50 s, but horizontal there, so speed is zero; 20-30 s has the greatest gradient.
- Question 21
Pupil page 8 - (a) 2-5 s. (b) First: 8 / 2 = 4.0 m/s. Last: (14 - 8) / (8 - 5) = 6 / 3 = 2.0 m/s. The first section is faster. (c) 14 / 8 = 1.75 m/s, including the stop.
- Try again A
- (a) 4.0 m/s. (b) 30 s. (c) Stationary, 0 m/s: no distance is added.
- Try again B
- Distance change = 20 m; time interval = 5 s; speed = 4.0 m/s.
- Try again C
- Rest: 4-8 s. Average speed = 16 / 8 = 2.0 m/s. The whole time includes the stop.
- Try again D - next lesson
- Rest: 2-4 s. Speeds: 6 / 2 = 3.0 m/s and (12 - 6) / (8 - 4) = 1.5 m/s. Whole-journey average = 12 / 8 = 1.5 m/s.
Worksheet selection and marking
The selected worksheet parts total 16 marks. The adapted note graph/application adds 5 practice marks: axis quantities/units and plotting (2), rest and greatest speed (2), whole average (1). Scales are supplied. Accept graph-reading tolerance and valid method/follow-through. No tangents or velocity-time areas are assigned.
Selection: Printed worksheet Q1(c,d), Q2(a-i), pages 1-3.
Open the maintained worked solutions →Collect and follow up: Collect all named booklets. Use the independent exit to record whether section gradients, stationary intervals and whole-journey averages are secure. Tell the usual teacher the specific error and who needs Try again D.
