Measurement of Time and Motion · Learner pages

Lesson 1 · Time repeating motion carefully

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

Name or initials: ____________   Date: ____________

The SI unit of time is the second (s). A minute contains 60 seconds. A period is the time for one complete repeating cycle.

1. Choose a repeatable event. A clock measures time by counting a process that repeats regularly. A pendulum provides a visible model of repeating motion.
2. Define one complete oscillation. Start at the left extreme. Travel to the right extreme and return to the left extreme: that is one complete oscillation, not two.
3. Time several oscillations. An example record gives 18 seconds for 10 complete oscillations. Divide total time by the number of oscillations: 18 ÷ 10 = 1.8 seconds per oscillation.
4. Repeat and compare. Repeat the timing with the same setup. Small differences such as 18.0 s and 18.4 s can come from reaction time and reading errors. Timing several cycles reduces the relative effect of starting and stopping delay.

Worked example: A supplied data set records 24 s for 12 oscillations. The period is 24 ÷ 12 = 2 s. A second trial gives 24.6 s, so its period is 2.05 s. The small difference is evidence of measurement variation, not proof that one trial should be deleted.

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

   1. left to centre only · 2. left to right only · 3. left to right and back to left

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

   1. 2.2 s · 2. 22 s · 3. 220 s

3. Why time ten cycles instead of only one?

   1. It removes every possible error · 2. Starting/stopping delay is a smaller part of the total time · 3. It changes seconds into metres

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.

Lesson 2 · Calculate speed with meaningful units

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

Name or initials: ____________   Date: ____________

Average speed summarises total distance divided by total elapsed time. A journey may include changes of speed or stops.

1. Record distance and elapsed time. For a journey, record the total path length travelled and the time from start to finish. Name the units before calculating.
2. Divide to find average speed. A toy travels 12 m in 6 s. Average speed = total distance ÷ total time = 12 ÷ 6 = 2 m/s. This describes the whole observed interval.
3. Use consistent units. A model journey covers 180 m in 1 min. Convert 1 min to 60 s before calculating m/s: 180 ÷ 60 = 3 m/s.
4. Use the relationship both ways. At a stated constant speed of 2 m/s for 5 s, distance = speed × time = 10 m. For a 10 m trip at that constant speed, time = distance ÷ speed = 5 s.

Worked example: A classroom model moves 15 m in 5 s, then 9 m in 7 s. Total distance is 24 m and total time is 12 s. Average speed = 24 ÷ 12 = 2 m/s. Keep the full interval, including any pause already included in those times.

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

   1. 100 m/s · 2. 4 m/s · 3. 25 m/s

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

   1. 2 min = 120 s · 2. 2 min = 20 s · 3. 2 min = 200 s

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

   1. 7 m · 2. 0.75 m · 3. 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.

Lesson 3 · Compare journeys using evidence

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

Name or initials: ____________   Date: ____________

A bigger distance does not by itself mean a higher speed. Compare distance and time together.

1. Make a fair comparison. When two journeys take the same time, the one covering the larger total distance has the greater average speed.
2. Compare a shared distance. For the same distance, less elapsed time means greater average speed. Do not decide from a distance or a time alone when both differ.
3. Calculate when both differ. Journey A: 30 m in 15 s gives 2 m/s. Journey B: 40 m in 20 s also gives 2 m/s. The longer journey is not automatically faster.
4. Interpret changing motion. Record distance travelled during successive equal time intervals. Equal distances in all those intervals support uniform speed at that observation scale; unequal distances show non-uniform motion. Coarse records can miss changes between observations.

Worked example: Two robots each travel 36 m. A takes 12 s and B takes 18 s. A averages 36 ÷ 12 = 3 m/s; B averages 36 ÷ 18 = 2 m/s. A has the higher average speed because it covers the same distance in less time.

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

   1. A · 2. Their speeds must be equal · 3. B

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

   1. equal at 3 m/s · 2. A is faster because its trip is shorter · 3. B is faster because its distance is larger

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

   1. uniform motion · 2. non-uniform motion · 3. no motion

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.

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 / 15 s / 450 s

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

   Right to centre / Right to left to right / Right to left only

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

   30 m/s / 60 m/s / 3 m/s

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

   7 s / 32 s / 112 s

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

   no movement / uniform speed at the recorded scale / exactly zero speed

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

   only the fastest second / only the moving time / the full elapsed time, including the stop

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…

   259.2 s / 1.8 s / 9.6 s

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

   Record the differences and consider timing uncertainty / Erase all but the preferred value / Claim clocks never vary

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

   120 m/s / 2 m/s / 480 m/s

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

   1.2 m / 11 m / 30 m

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

   B: 3 m/s / A: 8 m/s / They both average 2 m/s

6. An average speed of 2 m/s proves…

   speed was always exactly 2 m/s / the object never stopped / total distance divided by total time was 2 m/s