Chapter 8 – Measurement of Time and Motion

How was time measured when there were no clocks and watches?

People watched events in nature that repeat—sunrise / sunset, phases of the Moon, seasons. To mark smaller parts of a day they made simple devices:

Sundial – uses the Sun’s shadow.
Water clock – water drips out (or fills a bowl) in fixed time.
Hour-glass – sand falls from one bulb to another.
Candle clock – a marked candle burns down at a steady rate.

These gave them a way to tell time long before modern clocks.

Activity 8.1 (Build a water-clock)

(The activity itself has no direct question after construction, but it teaches how markings show minutes.)

Is the pendulum similar to the eraser hung on a thread that we saw in Class 6?

Yes. Both are simple pendulums—a small weight (bob) hung by a string that swings back and forth.

8.1.1 A simple pendulum

Think Like a Scientist – “Would all pendulums have the same time period? What affects it?”

Length matters: a longer pendulum swings more slowly, a shorter one more quickly.
Mass does not matter (for small swings).
So pendulums of equal length keep the same time; change the length and the time period changes.

8.1.2 SI unit of time

(No direct question—just facts: SI unit is second, symbol s.)

SCIENCE AND SOCIETY box

For races of different distances how can we fairly compare who was faster?

By using speed—divide the distance each runner covered by the time taken. The larger speed shows who was truly faster.

8.4 Slow or Fast – Speed

How were the marathon runners’ motions different?

Some runners kept almost the same speed all along—that is uniform motion.
Others kept changing speed—speeding up or slowing down—which is non-uniform motion.

Let Us Enhance Our Learning – Exercise Answers

1. Calculate the speed of a car that travels 150 m in 10 s. Express in km/h.

Speed = 150 m ÷ 10 s = 15 m/s
15 m/s = 15 × 3.6 = 54 km/h

2. Two runners finish 400 m in 50 s and 45 s. Who is faster and by how much?

Runner 1 speed = 400 ÷ 50 = 8 m/s.
Runner 2 speed = 400 ÷ 45 ≈ 8.9 m/s.
Runner 2 is faster by 0.9 m/s.

3. A train moves at 25 m/s for 360 km. How much time?

360 km = 360 000 m
Time = distance ÷ speed = 360 000 ÷ 25 = 14 400 s = 4 h.

4. Train travels 180 km in 3 h.

(i) Speed = 180 ÷ 3 = 60 km/h
(ii) 60 km/h = 60 ÷ 3.6 = 16.7 m/s
(iii) In 4 h it covers 60 × 4 = 240 km

5. Fastest horse 18 m/s versus train 72 km/h.

72 km/h = 72 ÷ 3.6 = 20 m/s.
Horse 18 m/s is 2 m/s slower than the train.

6. Uniform vs non-uniform motion (highway vs city).

On an empty straight highway the car keeps almost the same speed → uniform motion.
In city traffic it must brake and speed up often → non-uniform motion.

7. Fill the missing distances for uniform motion

Speed from first two rows: 8 m in 10 s = 0.8 m/s.
Hence every 10 s adds 8 m.

Time (s)	0	10	20	30	50	70
Distance (m)	0	8	16	24	40	56

(Distances at 50 s and 70 s already match 0.8 m/s.)

8. Car covers 60 km, 70 km, 50 km in three hours. Uniform? Average speed?

Distances per hour are different, so non-uniform motion.
Average speed = total distance ÷ total time = (60 + 70 + 50) ÷ 3 = 180 ÷ 3 = 60 km/h

9. Which is more common in daily life, uniform or non-uniform? Give examples.

Non-uniform motion is more common. Examples:

A bus stopping for passengers.
A cyclist slowing for a speed-breaker.
A football kicked across the field (slows then stops).

10. Table of distances vs time – uniform? average speed?

The distances are not equal in equal 10-s gaps, so non-uniform.
Average speed = total distance 60 m ÷ total time 100 s = 0.6 m/s.

11. Vehicle covers 2 km with three speed segments (10 m/s, 5 m/s, unknown). Finish in 200 s. Find missing speed and overall average.

First 500 m at 10 m/s → 50 s.
Next 500 m at 5 m/s → 100 s.
Time left = 200 − (50 + 100) = 50 s.
Remaining distance = 1 km = 1000 m.
Required speed = 1000 ÷ 50 = 20 m/s.
Average speed = total distance 2000 m ÷ 200 s = 10 m/s.