Compound Units

Activities to try

1. 1

   Rate workshop: best-buy, two-leg journey, and km/h↔m/s converter

   Why it works: A rate is 'how much for exactly one', computed by dividing the right way round; average speed is total distance ÷ total time (not the mean of speeds) because more time spent at the slower speed pulls the average down; conversions are a tracked chain, not a single memorised factor; and fair comparisons need a common per-unit basis.

   1. Best-buy station: compare two shelf tags (e.g. 400 g for £1.80 vs 250 g for £1.20). Put both onto a shared 'cost per 100 g' line and state which is cheaper.
   2. Two-leg journey station: assemble a journey of 60 km at 60 km/h then 60 km at 40 km/h, and — before doing any arithmetic — predict the average speed for the whole journey.
   3. Now compute it: write down the time each leg takes (60÷60 = 1 h and 60÷40 = 1.5 h), then total distance (120 km) and total time (2.5 h), then average speed (120 ÷ 2.5 = 48 km/h). Compare this with your prediction and explain the gap.
   4. Converter station: turn a km/h figure into m/s as two visible steps — ×1000 (km→m), then ÷3600 (per hour→per second) — tracking units so they cancel on screen.
   5. Sanity-check magnitude: 36 km/h should come out as about 10 m/s — a fast cycling/sprinting pace, well above a walk — not 0.6 or 600.
2. 2

   Average-speed performance task: work the journey on paper

   Why it works: Watching the pupil compute (rather than recite a rule) reveals whether they use total÷total or average the speeds; because equal distances are covered at unequal speeds, the correct 48 cannot be reached by the mean method, so the method is visible in the number produced. The equal-TIMES counterexample shows the mean is correct only when equal times are spent at each speed.

   1. Work out: a cyclist rides 60 km at a steady 60 km/h, then another 60 km at a steady 40 km/h. Find the average speed for the whole journey, showing every step on paper.
   2. Find each leg's time first (60÷60 = 1 h; 60÷40 = 1.5 h), then total distance (120 km), total time (2.5 h), and average speed (120 ÷ 2.5 = 48 km/h).
   3. Answer probes: which leg took more time, so should the average be nearer 60 or nearer 40 — does 50 fit? Then redo with equal TIMES (1 h at 60, 1 h at 40) and explain why the average is 50 this time.
   4. Convert your 48 km/h into metres per second, talking through each step and the units.
   5. Compare two products (e.g. 400 g for £1.80 vs 250 g for £1.20): put both on a cost-per-100 g basis (45p vs 48p), say which is the better buy, and explain what would go wrong if you just compared their prices.
3. 3

   Short check on rates, best-buy and conversion

   Why it works: The mean-of-speeds error and wrong-way-round division make sharp, diagnostic wrong options, so a quick check can surface whether these specific slips persist after teaching.

   1. Work out and type the cost per 100 g for a product given its price and weight.
   2. Choose the better buy between two differently sized packs where the smaller pack is actually cheaper per unit, so that 'the bigger pack must be cheaper per unit' is a tempting but wrong option.
   3. Choose the average speed of a two-leg equal-distance journey, with the mean-of-the-two-speeds value (e.g. 50) offered as the key trap.
   4. Give a km/h→m/s conversion as a typed value, then check its magnitude.