Worked example · Science · Year 8
DCF: Data and computational thinking → Problem-solving and modelling
A secure Year 8 response, annotated against the success criteria. Answer key: Step 1(d) 45 min, 24 min, 90 min, 0.42 h, 154 min; 18 km/h = 5 m/s. Step 3: section times 24, 30, 45, 30 and 25 minutes; total 30 km in 154 minutes (2.57 h); arrival 12:04; average speed 11.7 km/h. Step 4: O to P 12 km/h, Q to R 9 km/h; change C2 to 12, total 160 minutes. Step 5: D4 should be =B4/C4 and E7 should be =SUM(E2:E6). Step 6: headwind 199 minutes, 12:49; rest stop 174 minutes, 12:24; both 219 minutes, 13:09, which misses the minibus. Accept 0.4167 h for 0.42 h and 11.69 km/h for 11.7 km/h.
(a) Speed = distance ÷ time, so s = d / t.
(b) Multiply both sides by t: d = s × t. Then divide both sides by s: t = d / s.
| In words | In symbols | Spreadsheet formula |
|---|---|---|
| Time in hours = distance ÷ speed | t = d / s | =B2/C2 |
| Time in minutes = time in hours × 60 | minutes = 60 × t | =D2*60 |
| Distance = speed × time | d = s × t | =C2*D2 |
| Average speed = total distance ÷ total time | s = total d / total t | =B7/D7 |
(d) 0.75 × 60 = 45 min. 0.4 × 60 = 24 min. 1.5 × 60 = 90 min. 25 ÷ 60 = 0.42 h. 2 × 60 + 34 = 154 min.
(e) 18 km = 18 000 m. 1 h = 60 × 60 = 3600 s. s = 18 000 m ÷ 3600 s = 5 m/s.
Both rearrangements are correct and the pupil says what was done to each side, rather than relying on a triangle. Each row of the table says the same thing three ways, and the spreadsheet formulae use the right cells. Every conversion is shown with its working, and the m/s conversion changes both units before dividing. Evidences criteria 1 and 2.
(a) The sections are very different: the riders go 18 km/h downhill but only 6 km/h up the steep climb. One speed for the whole ride would hide that, and if the weather changes one section I can change just that one input instead of guessing a new speed for everything.
| Cell | Heading or formula |
|---|---|
| A1 to E1 | Section · Distance (km) · Speed (km/h) · Time (h) · Time (min) |
| D2 | =B2/C2 |
| E2 | =D2*60 |
| B7 | =SUM(B2:B6) |
| E7 | =SUM(E2:E6) |
| B9 | =B7/D7 |
(c) =B4/C4, because the row numbers go up by 1 for each row it is copied down.
The pupil explains decomposition in terms of the ride, not in general. Every heading has a unit, and every formula uses cell references, so changing an input updates the whole model. Evidences criterion 3.
| Row | A: Section | B: Distance (km) | C: Speed (km/h) | D: Time (h) | E: Time (min) |
|---|---|---|---|---|---|
| 2 | 1 | 6 | 15 | 0.40 | 24 |
| 3 | 2 | 4.5 | 9 | 0.50 | 30 |
| 4 | 3 | 9 | 12 | 0.75 | 45 |
| 5 | 4 | 3 | 6 | 0.50 | 30 |
| 6 | 5 | 7.5 | 18 | 0.42 | 25 |
| 7 | Total | 30 | 2.57 | 154 |
(a) t = d / s = 9 km ÷ 12 km/h = 0.75 h. 0.75 h × 60 = 45 min.
(b) 154 min = 2 h 34 min. 09:30 + 2 h 34 min = 12:04.
(c) s = total d / total t = 30 km ÷ 2.57 h = 11.7 km/h. The mean of the five speeds treats every section as if it took the same time, but it doesn't: the group spends 60 minutes on the two slow sections and only 25 minutes on the fast downhill. Slow sections count for more of the time, so the average speed is below 12 km/h.
Every value in the table is correct, the working for section 3 shows the formula, the substitution with units and the conversion, and the arrival time is found by changing minutes into hours and minutes first. The pupil explains why average speed is not the mean of the speeds. Evidences criteria 2 and 3.
(a) O to P is steepest, so that is where they were fastest: they covered the most distance for each minute. Between P and Q the line is flat, so the distance did not change for 10 minutes. They had stopped, fixing the chain at Pont Haearn.
(b) O to P: 6 km in 30 min = 0.5 h, so s = 6 km ÷ 0.5 h = 12 km/h. Q to R: 10.5 − 6 = 4.5 km in 70 − 40 = 30 min = 0.5 h, so s = 4.5 km ÷ 0.5 h = 9 km/h.
(c) Section 2 matches the plan exactly (9 km/h). Section 1 is wrong: the plan says 15 km/h but the real riders managed 12 km/h, so the model says 24 minutes when it took 30. She should change C2 from 15 to 12. D2 becomes 6 ÷ 12 = 0.5 h = 30 min, so the total becomes 154 + 6 = 160 minutes, arriving at 12:10. Only 4 riders did the practice, so the whole year group may be slower still.
The pupil reads both the slope and the flat section correctly, works out the real speeds from the graph and uses them to test the model one section at a time. The pupil names the exact input to change and notices the test data came from a small group. Evidences criterion 4.
(a) D4 says =C4/B4, which is speed ÷ distance = 12 ÷ 9 = 1.33. That is not a time: km/h ÷ km gives “per hour”, not hours. It also cannot be right, because at 12 km/h they would ride 12 km in 1 hour, so 9 km must take less than 1 hour. Correct formula: =B4/C4, which gives 0.75 h.
(b) =SUM(E2:E5) stops at row 5, so it leaves out section 5 (25 minutes). Correct formula: =SUM(E2:E6).
(c) D7 = 3.15 h × 60 = 189 min, but E7 says 164 min. The sheet disagrees with itself, so something must be wrong. Elis could add a check cell, =D7*60-E7, which should always be 0, and check one row by hand against the answer to Step 3(a).
Both errors are found, explained in terms of what the formula actually calculated, and fixed. The pupil uses units and a common-sense check to prove D4 is wrong, and designs a check cell that would catch a missing row automatically. Evidences criterion 5.
| What if… | Cell changed or added | New total time | New arrival time |
|---|---|---|---|
| 1. Headwind halves speed on section 3 | C4 changed from 12 to 6. D4 = 9 ÷ 6 = 1.5 h = 90 min | 154 − 45 + 90 = 199 min (3 h 19 min) | 12:49 |
| 2. 20 minute rest stop | New column F, Stop (min). F4 = 20 | 154 + 20 = 174 min (2 h 54 min) | 12:24 |
| 3. Both together | C4 = 6 and F4 = 20 | 199 + 20 = 219 min (3 h 39 min) | 13:09 |
Heading for F1: Stop (min). New E7: =SUM(E2:E6)+SUM(F2:F6).
No. With the headwind and the stop they arrive at 13:09, 9 minutes after the minibus leaves. If they start at 09:15 instead, 219 minutes later is 12:54, which leaves 6 minutes to spare. Cutting the stop to 10 minutes would give 12:59, but 1 minute is too close.
Each change starts from the correct model, names the exact cell, and the new totals and times are right. The pupil extends the model with a new column instead of fudging the total, and tests two fixes before choosing the safer one. Evidences criterion 6.
The group rides at the speed of its slowest rider, and that rider will get more tired towards the end, so sections 4 and 5 will probably be slower than planned and the arrival time later. The model also has no time for road crossings, gates or punctures, like the chain stop in the practice ride; each of those adds minutes the model does not count. I would add a column of extra minutes for each section, using real data from the practice ride.
Both assumptions come from evidence on the stimulus sheet, each says which way it pushes the prediction, and the pupil suggests a way to improve the model. Evidences criterion 6.