Measurement of Time and Motion · Answer guide

Lesson 1 · Time repeating motion carefully

Distinguish a full oscillation from half a swing and calculate a period from repeated cycles.

1. One full cycle starting at the left extreme is…

   Answer: left to right and back to left. A complete cycle returns to the starting position and phase of motion.

2. Ten complete oscillations take 22 s. The period is…

   Answer: 2.2 s. 22 seconds shared across 10 cycles gives 22 ÷ 10 = 2.2 s per cycle.

3. Why time ten cycles instead of only one?

   Answer: Starting/stopping delay is a smaller part of the total time. The same reaction delay is usually a smaller fraction of a longer measured interval. It does not remove all error.

Discuss: Use the supplied records 20.0 s, 20.4 s and 19.8 s for ten cycles. Calculate each period. If you observe a real pendulum, use a teacher-secured soft object, small swings and clear space; the data activity alone is sufficient.

Teaching note: A journey from the left extreme to the right extreme is half an oscillation; returning to the same starting extreme completes the cycle. Open responses need discussion; they are not automatically marked.

Lesson 2 · Calculate speed with meaningful units

Calculate average speed, distance and time while keeping units consistent.

1. A toy travels 20 m in 5 s. Its average speed is…

   Answer: 4 m/s. Average speed = 20 ÷ 5 = 4 m/s.

2. Which conversion is needed for 120 m in 2 min, expressed in m/s?

   Answer: 2 min = 120 s. Each minute is 60 seconds, so 2 min = 120 s; the speed is 1 m/s.

3. At a constant 3 m/s for 4 s, the distance is…

   Answer: 12 m. Distance = 3 m/s × 4 s = 12 m.

Discuss: Invent a safe toy journey with a whole-number distance and time. Calculate its average speed and explain what the average cannot tell you about individual moments.

Teaching note: 12 m in 6 s gives an average of 2 m/s. It does not prove the object travelled at exactly 2 m/s at every instant. Open responses need discussion; they are not automatically marked.

Lesson 3 · Compare journeys using evidence

Use fair comparisons and equal-interval records to describe speed.

1. Both toys move for 10 s. A travels 15 m; B travels 20 m. Which has higher average speed?

   Answer: B. For the same elapsed time, the larger distance gives the larger average speed.

2. A covers 24 m in 8 s; B covers 30 m in 10 s. Their average speeds are…

   Answer: equal at 3 m/s. 24 ÷ 8 = 3 and 30 ÷ 10 = 3, so the averages match.

3. Distances during three equal time intervals are 2 m, 3 m and 5 m. This record shows…

   Answer: non-uniform motion. The distances differ even though the time intervals are equal.

Discuss: Create three equal-interval records for a toy journey. Make one set show equal distances and one show unequal distances. Explain exactly what each set supports.

Teaching note: A single average cannot show whether a journey was smooth, stop-start or variable. It summarises the total trip. Open responses need discussion; they are not automatically marked.

Final check · Set 1

Try one set after the lessons. Use the other set after reviewing any errors.

1. Fifteen oscillations take 30 s. The period is…

   2 s. 30 ÷ 15 = 2 seconds per oscillation.

2. Which is one full oscillation from the right extreme?

   Right to left to right. The bob returns to the starting extreme after travelling in both directions.

3. A cart travels 45 m in 15 s. Average speed?

   3 m/s. 45 ÷ 15 = 3 m/s.

4. At constant 4 m/s, how long does 28 m take?

   7 s. Time = distance ÷ speed = 28 ÷ 4 = 7 s.

5. Equal 5 s intervals have distances 10 m, 10 m, 10 m. This supports…

   uniform speed at the recorded scale. Equal distances in equal recorded intervals support uniform speed at that scale.

6. A journey includes a 5 s stop. Average speed over the whole trip uses…

   the full elapsed time, including the stop. Average speed for the whole journey uses all elapsed time.

Final check · Set 2

Try one set after the lessons. Use the other set after reviewing any errors.

1. Twelve oscillations take 21.6 s. The period is…

   1.8 s. 21.6 ÷ 12 = 1.8 seconds per cycle.

2. Repeated timings differ a little. A good response is…

   Record the differences and consider timing uncertainty. Measurement variation should be reported and investigated.

3. A model travels 240 m in 2 min. Average speed in m/s?

   2 m/s. 2 min = 120 s, so 240 ÷ 120 = 2 m/s.

4. At constant 5 m/s for 6 s, distance is…

   30 m. Distance = 5 × 6 = 30 m.

5. A: 16 m in 8 s. B: 27 m in 9 s. Which average is higher?

   B: 3 m/s. A averages 2 m/s and B averages 3 m/s.

6. An average speed of 2 m/s proves…

   total distance divided by total time was 2 m/s. An average is the ratio for the whole interval; the speed within it can vary.