Chapter 8: Measurement of Time and Motion

Background idea → Every measurement needs a reference that repeats in a predictable way. Long before people discovered gears, springs and quartz crystals, they watched natural repetitive events and then built a few simple devices that imitate such repetition.

1. Natural events used as clocks

• Day–night cycle: One sunrise to the next was taken as one “day”. At smaller scale, people judged time by the position of the Sun: low in the east (morning), overhead (noon), low in the west (evening).
• Lunar phases: From one full Moon to the next (the lunar month) takes about $$29.5\ \text{days}$$. Farmers and travellers used this as a bigger unit of time.
• Seasons / year: Return of the same season (for example, two successive summers) defined a year, about $$365\ \text{days}$$.
• Stars at night: Sailors observed that particular constellations rise or set at fixed hours; these sightings helped them tell the time after sunset.

2. Devices made before mechanical clocks

1. Sundial
     • A stick (called a gnomon) is fixed upright on a flat, graduated plate.
     • As the Sun moves across the sky, the stick’s shadow moves. The length or direction of the shadow indicates the hour.
     • Limitation – works only in sunlight and must be re–aligned if taken to another latitude.
2. Water clock (clepsydra)
     • A container with a small hole lets water drip at a nearly steady rate.
     • Equal volumes dripping out (or rising in another jar) represent equal intervals of time.
     • With a scale marked on the jar, people could read the elapsed time even on cloudy days or at night.
3. Sand clock (hour-glass)
     • Fine, dry sand flows from an upper bulb to a lower one through a narrow neck.
     • If all the sand takes, say, $$30\ \text{minutes}$$ to empty, turning the glass over starts the next half-hour immediately.
     • Very handy for sailors and cooks who needed short, repeatable periods.
4. Oil lamp / candle clocks
     • A candle or lamp is marked with equally spaced lines. As it burns down to the next line, a fixed time lapses.
     • Useful indoors and in winter but affected by wind and quality of wax/oil.

3. Putting everything together

Thus, before modern clocks and watches, people first relied on the repetitive motions of celestial bodies—Sun, Moon and stars—and then invented simple devices such as sundials, water clocks, sand clocks and graded candles to divide the day into smaller, more manageable intervals.

Step 1 – Recall the class-6 activity

In Grade 6 you tied an eraser to one end of a light thread, fixed the other end to a support and then pulled the eraser slightly to one side and let it go. The eraser moved to and fro about its lowest (mean) position. This to-and-fro motion about a fixed point is called oscillatory motion.

Step 2 – Define a simple pendulum

A simple pendulum consists of

• a small, heavy body (called the bob) of mass $$m$$,
• suspended by a light, inextensible string of length $$L$$,
• from a rigid support.

When the bob is displaced through a small angle $$\theta$$ and released, the force of gravity provides a restoring torque. Consequently the bob executes periodic (to-and-fro) motion. For small angles the time taken for one complete oscillation, called the time-period $$T$$, is

\[T = 2\pi \sqrt{\dfrac{L}{g}}\]
where $$g \approx 9.8\,\text{m s}^{-2}$$ is the acceleration due to gravity.

Step 3 – Compare the eraser set-up with a simple pendulum

The eraser plays the role of the bob and the thread is the suspension string. Therefore the arrangement fulfils every essential condition of a simple pendulum. Minor differences (shape of bob, thickness of string, air resistance, large swing angles) only make the motion slightly less “ideal”; the basic nature of the motion remains the same.

Step 4 – Conclusion

Yes, the eraser-on-a-thread set-up is practically the same as a simple pendulum. Both demonstrate periodic, oscillatory motion that can be studied to measure quantities such as the time-period and even the value of $$g$$.

Step 1 — Recall why “time” alone worked when the distance was the same
If two athletes both run the same distance  $$d$$, the one who takes the shorter time  $$t$$  is the faster, because for that common distance the speed formula
$$\text{speed}=\dfrac{d}{t}$$ simplifies to “smaller t, larger speed”.

Step 2 — Why “time” no longer helps when the distances are different
Suppose Athlete A runs $$100\;\text{m}$$ in $$12\;\text{s}$$ and Athlete B runs $$50\;\text{m}$$ in $$5\;\text{s}$$. Because both the distance and the time differ, we cannot tell directly from the times 12 s and 5 s who was really faster in terms of covering ground.

Step 3 — Introduce the common comparison quantity: speed
Speed combines distance and time in one number:
$$\text{speed}=\dfrac{\text{distance covered}}{\text{time taken}}$$

Step 4 — Compute and compare the speeds

Because $$v_B(10\;\text{m\,s}^{-1}) > v_A(8.33\;\text{m\,s}^{-1})$$, Athlete B is the faster runner. Comparing the two finishing times alone ($$12\;\text{s}$$ for A and $$5\;\text{s}$$ for B) is misleading — Athlete A spent more time only because he had twice the distance to cover, not because he was slower per metre. Speed packs both distance and time into one number, which is why it is the only fair way to compare races of unequal length.

Step 5 — General rule
Whenever the race distances differ, calculate each competitor’s speed in the same units (for example metres per second or kilometres per hour). The one with the greater speed is the faster.

Thus, for races of different lengths we do not compare times; we compare the speeds, which are obtained from the common formula $$\text{speed}=\dfrac{\text{distance}}{\text{time}}$$.

Step 1 : Recall the meaning of speed

The speed of a runner is the distance he or she covers in unit time.

Mathematically : $$\text{speed} = \dfrac{\text{distance travelled}}{\text{time taken}}$$

If the distance is measured in metres (m) and the time in seconds (s), the speed is obtained in metres per second ( m s^–1).

Step 2 : Define two kinds of motion

• Uniform motion  : An object is said to be in uniform motion along a straight line if it covers equal distances in equal intervals of time, however small those intervals may be.  Thus the speed remains constant.
• Non-uniform motion  : An object is in non-uniform motion if it covers unequal distances in equal time-intervals or, equivalently, its speed keeps changing with time.

Step 3 : Apply the ideas to the marathon runners you observed

• Some runners appeared to run “steady”. If you had taken a stopwatch and measured their time for, say, every 100 m stretch, you would have recorded practically the same time again and again. Their speed therefore stayed the same, so they were in uniform motion.
• The other runners were sometimes quicker and sometimes slower. For them the 100 m timings would not all be equal. Because their speed changed from one interval to the next, their motion was non-uniform.

Step 4 : Summarise the difference

Hence the steady runners were showing uniform motion, whereas the runners who kept speeding up or slowing down were showing non-uniform motion.

Step 1 – Write the data given

• Distance between Swati’s house and school: $$3.6\,\mathrm{km}$$
• Time taken on the bicycle: $$15\,\mathrm{min}$$

Step 2 – Recall the formula for speed

Speed is defined as distance travelled per unit time, that is

$$\text{speed}=\dfrac{\text{distance}}{\text{time}}$$

Step 3 – Convert all quantities to SI units

In the SI system distance must be in metres (m) and time in seconds (s).

• Convert distance: $$1\,\mathrm{km}=1000\,\mathrm{m}$$, therefore

  $$3.6\,\mathrm{km}=3.6\times1000\,\mathrm{m}=3600\,\mathrm{m}$$

• Convert time: $$1\,\mathrm{min}=60\,\mathrm{s}$$, therefore

  $$15\,\mathrm{min}=15\times60\,\mathrm{s}=900\,\mathrm{s}$$

Step 4 – Substitute in the speed formula

$$\text{speed}=\dfrac{3600\,\mathrm{m}}{900\,\mathrm{s}}$$

Step 5 – Carry out the division

$$\dfrac{3600}{900}=4$$

Hence,

So, the bicycle’s speed was $$4\,\mathrm{m/s}$$.

Given data

• Speed of the bus: $$v = 50\;\mathrm{km\,h^{-1}}$$
• Time taken: $$t = 2\;\mathrm{h}$$

Formula for distance

The relation connecting distance, speed and time is

$$\text{Distance} = \text{Speed} \times \text{Time}$$
That is, $$d = v \times t$$.

Substitute the known values

$$d = (50\;\mathrm{km\,h^{-1}}) \times (2\;\mathrm{h})$$

Calculate

Multiply the numbers and observe that the unit hour cancels:

$$d = 50 \times 2\;\mathrm{km} = 100\;\mathrm{km}$$

Conclusion

Raghav’s destination city is 100 km away from his starting point.

Given

• Speed of the train  = $$90 \\, \mathrm{km\,h^{-1}}$$
• Distance to be covered  = $$360 \\, \mathrm{km}$$

Formula used

The relation connecting distance, speed and time is:

$$\text{Speed} = \dfrac{\text{Distance}}{\text{Time}}$$

To find time, we rearrange the formula:

$$\text{Time} = \dfrac{\text{Distance}}{\text{Speed}}$$

Substitution of values

Both the distance and speed are already in compatible units (kilometres and kilometres per hour), so we can substitute directly:

$$\text{Time} = \dfrac{360 \\, \mathrm{km}}{90 \\, \mathrm{km\,h^{-1}}}$$

Simplification

$$\text{Time} = \dfrac{360}{90} \\, \mathrm{h}$$

Divide numerator and denominator by 90:

$$\text{Time} = 4 \\, \mathrm{h}$$

Result

The train will take 4 hours to cover a distance of 360 km.

Step 1: Write the formula for speed.

Speed $$v$$ is defined as the distance travelled divided by the time taken:

Step 2: Substitute the given values (in SI units).

Distance $$d = 150 \, \mathrm{m}$$; Time $$t = 10 \, \mathrm{s}$$.

Therefore,

$$v = \frac{150 \, \mathrm{m}}{10 \, \mathrm{s}} = 15 \, \mathrm{m/s}$$

Step 3: Convert $$15 \, \mathrm{m/s}$$ to $$\mathrm{km/h}$$.

To change metres to kilometres, divide by 1000; to change seconds to hours, multiply by 3600:

$$15 \, \mathrm{m/s} = 15 \times \frac{1 \, \mathrm{km}}{1000 \, \mathrm{m}} \times \frac{3600 \, \mathrm{s}}{1 \, \mathrm{h}}$$

$$= 15 \times \frac{3600}{1000} \, \mathrm{km/h} = 15 \times 3.6 \, \mathrm{km/h}$$

$$= 54 \, \mathrm{km/h}$$

Hence, the cars speed is $$54 \, \mathrm{km/h}$$.

Step 1 Write the given data

• Distance covered by each runner: $$d = 400\,\text{m}$$
• Time taken by Runner A: $$t_A = 50\,\text{s}$$
• Time taken by Runner B: $$t_B = 45\,\text{s}$$

Step 2 Recall the formula for speed

$$\text{Speed} = \frac{\text{Distance}}{\text{Time}}$$

Step 3 Calculate each runners speed

Runner A: $$v_A = \frac{d}{t_A} = \frac{400\,\text{m}}{50\,\text{s}} = 8\,\mathrm{m\,s^{-1}}$$

Runner B: $$v_B = \frac{d}{t_B} = \frac{400\,\text{m}}{45\,\text{s}} = 8.89\,\mathrm{m\,s^{-1}}$$ (rounded to two decimal places)

Step 4 Find the difference

$$\Delta v = v_B - v_A = 8.89\,\mathrm{m\,s^{-1}} - 8\,\mathrm{m\,s^{-1}} = 0.89\,\mathrm{m\,s^{-1}}$$

Step 5 State the result

Runner B is faster than Runner A by approximately $$0.9\,\mathrm{m\,s^{-1}}$$.

Given: Speed $$v = 25\,\mathrm{m/s}$$, distance $$s = 360\,\mathrm{km}$$.

Step 1 — Convert distance to metres

$$1\,\mathrm{km} = 1000\,\mathrm{m}$$

$$s = 360 \times 1000\,\mathrm{m} = 360\,000\,\mathrm{m}$$

Step 2 — Use the speed formula

$$\text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \;\Longrightarrow\; \text{Time} = \dfrac{\text{Distance}}{\text{Speed}}$$

Step 3 — Substitute the values

$$t = \dfrac{360\,000\,\mathrm{m}}{25\,\mathrm{m/s}}$$

$$t = 14\,400\,\mathrm{s}$$

Step 4 — Convert seconds to hours

$$1\,\mathrm{hour} = 60\,\text{min} = 3600\,\mathrm{s}$$

$$t = \dfrac{14\,400}{3600}\,\mathrm{h} = 4\,\mathrm{h}$$

Therefore, the train takes 4 hours to cover 360 km at 25 m/s.

Data given

• Horse: $$v_\text{horse}=18\,\mathrm{m\,s^{-1}}$$
• Train: $$v_\text{train}=72\,\mathrm{km\,h^{-1}}$$

To compare the two speeds we first express them in the same unit.

Step 1 – Convert train speed to m/s

Unit facts: $$1\,\mathrm{km}=1000\,\mathrm{m}$$ and $$1\,\mathrm{h}=3600\,\mathrm{s}$$

$$v_\text{train}=72\,\mathrm{km\,h^{-1}}=72\times\frac{1000\,\mathrm{m}}{3600\,\mathrm{s}}=72\times\frac{5}{18}\,\mathrm{m\,s^{-1}}$$

$$72\div18=4\;\Rightarrow\;v_\text{train}=4\times5=20\,\mathrm{m\,s^{-1}}$$

Step 1 : Recall the definitions

• Uniform motion  When a body travels equal distances in equal intervals of time, howsoever small those intervals are chosen, its speed remains the same throughout. Its speed is said to be constant.
• Non-uniform motion  When a body travels unequal distances in the same time intervals (or, equivalently, its speed keeps changing), the motion is non-uniform.

The mathematical way to see this is to look at the speed $$v = \frac{\text{distance}}{\text{time}}$$ during each chosen time interval. If the value of $$v$$ is the same for every interval, the motion is uniform; if the values differ, it is non-uniform.

Step 2 : Example A — a car on an empty straight highway

Suppose the car is set on cruise control at 60 km h⁻¹. In class-7 units we convert

$$60\;\text{km h}^{-1} = 60 \times \frac{1000\;\text{m}}{3600\;\text{s}} = 16.7\;\text{m s}^{-1}$$.

Now observe it for successive 5-second intervals:

The distance columns are equal, the speed column has the same value each time; hence the motion is uniform.

Step 3 : Example B — the same car in city traffic

Traffic lights, turns and braking make the speed keep changing. Measure over the same 5-s intervals:

The distances (40 m, 10 m, 65 m) are all different, and so are the speeds (8, 2, 13 m s⁻¹). Therefore the motion is non-uniform.

Step 4 : Stating the distinction clearly

• On the highway the car’s speed stays constant; equal distances are covered in every equal time-slot ⇒ uniform motion.
• In city traffic the car alternately accelerates, slows or stops; the distances covered in equal time-slots differ ⇒ non-uniform motion.

Thus, the same car can show either type of motion depending on the driving conditions.

Step 1 : Find the constant speed of the object

For uniform motion, the ratio $$\dfrac{\text{distance}}{\text{time}}$$ is the same for every interval.

Between 0 s and 10 s:

$$v = \dfrac{8\,\text{m} - 0\,\text{m}}{10\,\text{s} - 0\,\text{s}} = \dfrac{8}{10} = 0.8\;\text{m\,s}^{-1}$$

Check with another known pair (10 s to 30 s):

$$v = \dfrac{24\,\text{m} - 8\,\text{m}}{30\,\text{s} - 10\,\text{s}} = \dfrac{16}{20} = 0.8\;\text{m\,s}^{-1}$$

The speed is indeed constant: $$v = 0.8\;\text{m\,s}^{-1}$$.

Step 2 : Work out the missing time entries

The given times go 0 s, 10 s, 20 s, 30 s, ?, 50 s, ?, 70 s. They rise in equal jumps of 10 s, so the gaps must be

• 40 s after 30 s
• 60 s between 50 s and 70 s

Step 3 : Calculate the missing distances

• At 20 s: $$s = v \times t = 0.8 \times 20 = 16\;\text{m}$$
• At 40 s: $$s = 0.8 \times 40 = 32\;\text{m}$$ (already supplied, confirms consistency)
• At 60 s: $$s = 0.8 \times 60 = 48\;\text{m}$$

Step 4 : Completed table

All missing values are now filled, and every pair of successive readings keeps the same speed of $$0.8\;\text{m\,s}^{-1}$$, confirming uniform motion.

Step 1. Recall the definition of uniform motion.
If a body covers equal distances in equal intervals of time, its motion is called uniform. Otherwise, the motion is non-uniform.

Step 2. Compare the distances covered in each hour.
First hour: $$60\,\mathrm{km}$$
Second hour: $$70\,\mathrm{km}$$
Third hour: $$50\,\mathrm{km}$$

The three distances are not equal. Because the car does not cover the same distance in each equal time-interval of one hour, its motion is non-uniform.

Step 3. Find the total distance travelled.
\[\text{Total distance}=60\,\mathrm{km}+70\,\mathrm{km}+50\,\mathrm{km}=180\,\mathrm{km}\]

Step 4. Find the total time taken.
There are three one-hour intervals, so
$$\text{Total time}=1\,\mathrm{h}+1\,\mathrm{h}+1\,\mathrm{h}=3\,\mathrm{h}$$

Step 5. Calculate the average speed.
Average speed $$=\dfrac{\text{total distance}}{\text{total time}}$$
\[\text{Average speed}=\dfrac{180\,\mathrm{km}}{3\,\mathrm{h}}=60\,\mathrm{km\,h^{-1}}\]

Conclusion.
The motion is non-uniform, and the average speed of the car for the whole journey is $$60\,\mathrm{km\,h^{-1}}$$.

Step 1 – Recall the definitions

• Uniform motion: An object covers equal distances in equal intervals of time. Mathematically, its speed v stays constant, so distance $$s$$ is proportional to time $$t$$: $$s = v\,t$$.
• Non-uniform motion: The distance covered in equal time intervals keeps changing; the speed $$v = \dfrac{s}{t}$$ is not constant. The object may speed up or slow down (i.e. it has acceleration).

Step 2 – Compare with daily life

Almost everything around us starts, stops, turns or moves on rough ground. Forces such as friction, air resistance and the push or pull we apply keep changing the speed. Therefore the motion we usually see is non-uniform. Truly uniform motion (for example, a satellite in outer space or a train running on a long straight track at steady speed) is comparatively rare in ordinary surroundings.

Step 3 – Three personal examples of non-uniform motion

1. A city bus ride: When a bus moves through traffic it accelerates after a stop, slows down at turns and brakes at traffic lights, so the speed keeps varying.
2. Cycling on an uneven road: While pedalling up a slope I go slower, then speed up while going downhill or on a flat stretch.
3. Running in the playground: I start from rest, run fast to catch a ball, and finally decelerate to stop—different speeds in different intervals.

All three clearly show that non-uniform motion dominates our day-to-day experience.

Step 1 : Calculate the speed in every 10 s interval

The speed varies from one interval to the next, ranging between 0.3 m s⁻¹ and 1.0 m s⁻¹ (a few intervals happen to share the same value, but most differ).

Conclusion 1: The object moves with non-uniform speed.

Step 2 : Average speed for the whole journey

Total distance travelled in 100 s:

$$s_{\text{total}} = 60\;\text{m}$$

Total time taken:

$$t_{\text{total}} = 100\;\text{s}$$

Average speed $$v_{\text{avg}}$$ is

$$v_{\text{avg}} = \frac{s_{\text{total}}}{t_{\text{total}}} = \frac{60\;\text{m}}{100\;\text{s}} = 0.6\;\text{m\,s}^{-1}$$

Conclusion 2: The average speed of the object is 0.6 m s⁻¹.

Given data

• Total distance to be covered: $$D = 2\,\text{km} = 2000\,\text{m}$$
• First part: $$s_1 = 500\,\text{m}$$ with speed $$v_1 = 10\,\text{m/s}$$
• Second part: $$s_2 = 500\,\text{m}$$ with speed $$v_2 = 5\,\text{m/s}$$
• Total time allowed for the whole journey: $$T = 200\,\text{s}$$

Step 1 ― Time taken for the first 500 m

The time, $$t_1$$, is calculated from $$\text{speed} = \dfrac{\text{distance}}{\text{time}}$$:

$$t_1 = \dfrac{s_1}{v_1} = \dfrac{500}{10} = 50\,\text{s}$$

Step 2 ― Time taken for the next 500 m

$$t_2 = \dfrac{s_2}{v_2} = \dfrac{500}{5} = 100\,\text{s}$$

Step 3 ― Time left for the remaining distance

Total time already spent:

$$t_\text{spent} = t_1 + t_2 = 50 + 100 = 150\,\text{s}$$

Time still available:

$$t_3 = T - t_\text{spent} = 200 - 150 = 50\,\text{s}$$

Step 4 ― Remaining distance to be covered

Distance already covered: $$s_1 + s_2 = 500 + 500 = 1000\,\text{m}$$

Remaining distance:

$$s_3 = D - (s_1 + s_2) = 2000 - 1000 = 1000\,\text{m}$$

Step 5 ― Required speed for the remaining distance

Using $$v = \dfrac{\text{distance}}{\text{time}}$$ for the third part:

$$v_3 = \dfrac{s_3}{t_3} = \dfrac{1000}{50} = 20\,\text{m/s}$$

Step 6 ― Average speed for the whole journey

Average speed $$\bar{v}$$ is defined as $$\bar{v} = \dfrac{\text{total distance}}{\text{total time}}$$

$$\bar{v} = \dfrac{D}{T} = \dfrac{2000}{200} = 10\,\text{m/s}$$

Final results

• Speed required for the last 1000 m: $$20\,\text{m/s}$$
• Average speed for the whole 2 km journey: $$10\,\text{m/s}$$