NCERT Solutions Class 7 Science Chapter 8 Measurement of Time and Motion

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

CLASS 7 · SCIENCE

NCERT Solutions

Chapter 8: Measurement of Time
and Motion

NCERT Textbook — Curiosity

BOOK PAGES SECTIONS QUESTIONS MEDIUM

105 – 120 13 36 English

Solutions, notes, sample papers & more at 39 pages

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

CLASS 7 · SCIENCE · CURIOSITY

NCERT Solutions — Chapter 8: Measurement of Time
and Motion
Complete NCERT Solutions for Class 7 Science Chapter 8 Measurement of Time and Motion from the NCERT
textbook Curiosity — the chapter that opens with Prerna, the fastest sprinter of her district. Every question is
answered: Activities 8.1 to 8.4, the Think Like a Scientist! box, all in-text questions from pages 106 to 117,
the eleven Let Us Enhance Our Learning exercises (with Tables of Question 7 and 10 worked out), the four
Exploratory Projects and the Holistic Lens task — with every speed, distance and time value recomputed
step by step.

TEXTBOOK BOOK PAGES

Curiosity (Class 7) 105 – 120

SECTIONS QUESTIONS

13 36

MEDIUM

English

In-text Question — Page 106
8.1 Measurement of Time

Q1 How was time measured when there were no clocks and watches?

By watching things that repeat themselves after definite intervals of time. People noticed the
rising and setting of the Sun, the phases of the Moon and the changing seasons, and used
these cycles for timekeeping. First they made calendars, in which a day was defined by one
cycle of sunrise and sunset. Then they built devices to measure the smaller intervals within a
day.

Page 1 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

DEVICE WHAT REPEATS OR FLOWS IN HOW TIME WAS READ
IT

Sundial (Fig. 8.1) The Sun's daily journey across the From the changing position of the shadow
sky of an object cast by sunlight

Water clock, out-flow type Water flowing out of a marked From the falling water level against the
(Fig. 8.2a) vessel markings for time

Water clock, floating-bowl Water entering a bowl through a The bowl filled up in a fixed time and
type (Fig. 8.2b) fine hole at its bottom sank; it was lifted and floated again

Hourglass (Fig. 8.3) Sand flowing from one bulb to From how much sand had run down
another

Candle clock (Fig. 8.4) A candle burning down steadily From the markings on the candle that the
flame had reached

Why it happens: every one of these devices — and every clock made since — needs
one thing: a process that repeats itself, or goes on at a steady rate. Once you have
that, you can cut time into equal pieces and count them.

Did you know? In ancient India the sinking-bowl water clock was called the
Ghatika-yantra. Its hole was made so that the bowl took exactly 24 minutes to fill
and sink. Since 24 h = 1440 min and 1440 ÷ 24 = 60, the day was divided into 60
equal ghatis. Each time the bowl sank, drums, conch shells or a gong announced it.

Activity 8.1: Let us construct — Page 107
8.1 Measurement of Time

ACTIVITY

Q1 Should we make a simple water clock?

Yes — and it needs nothing more than a used transparent plastic bottle. Here is the whole
activity in order.

1. Take a used transparent plastic bottle (½ litre or larger) with its cap.
2. Cut it into two, roughly in the middle [Fig. 8.5a].
3. Using a drawing pin, make a small hole in the cap [Fig. 8.5b].

Page 2 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

4. Place the upper part of the bottle, cap screwed on, inverted over the lower half [Fig. 8.5c].
5. Fill the upper part with water. Add a few drops of ink or colour so that the water level is easy
to see [Fig. 8.5d].
6. The water starts dripping into the lower part. Using a watch, mark the level of water after
every one minute till all the water has dripped down.

Why the marks are not equally spaced: as the water in the top half gets used up,
the push of the water above the hole becomes less, so the dripping slows down a
little. The same difficulty was faced in ancient India — the Arthasastra type of out-
flow water clock was not very accurate because the flow rate decreased as the water
level dropped. That is exactly why the sinking-bowl clock (Ghatika-yantra) was
developed.

Check it yourself: make the hole with a single drawing pin, not a nail. A big hole
empties the bottle in seconds and you get no useful marks.

Q2 Your water clock is ready. Can you now guess how to use it?

Pour the water from the lower part back into the top part and watch the level of water dripping
into the lower part. Every time the level touches a mark you had made, one more minute has
passed.
So the clock is read by counting marks, not by looking at a dial:

Level at the 1st mark → 1 minute over.
Level at the 5th mark → 5 minutes over.
To time something — say how long a friend takes to solve 10 sums — start the water and
note the mark reached when the friend finishes.

Why it works: the water always takes the same time to fall from one mark to the
next as it did while you were making the marks, because the same bottle, the same
hole and the same amounts of water are involved each time. A repeatable process
is the heart of every clock.

Try this: use your water clock to time 10 oscillations of the pendulum you make in
Activity 8.2, and compare it with a wrist watch. You will quickly see why people kept
looking for better clocks.

Page 3 of 39

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Class 7 Science Chapter 8 Measurement of Time and Motion
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In-text Question — Page 109
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Yes — it is the same thing. An eraser hung from a thread and set swinging is a simple
pendulum; only the name and the bob are different.
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Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

GRADE 6 ACTIVITY SIMPLE PENDULUM OF THIS CHAPTER

An eraser tied to a thread A small metallic ball — the bob — tied to a thread

Held in the hand or hung from a hook Hung from a rigid support

Moved to one side and released — it swings to Moved slightly to one side and released — it starts oscillatory
and fro motion

The swinging repeats again and again The motion is periodic: it repeats its path after a fixed interval
of time

Why the pendulum is treated separately: in Grade 6 we only watched the to-and-
fro motion. Here we measure it. A heavy metal bob on a taut thread swings for a
long time without wobbling, so the time for one oscillation — the time period — can
be measured accurately. That is what makes it usable in a clock.

Activity 8.2: Let us experiment — Pages 109 & 110
8.1.1 A simple pendulum

ACTIVITY

Q1 Gently hold the bob, move it slightly to one side and release it. Take care not to
push the bob while releasing it and that the string is taut. Is your pendulum now
oscillating?

Yes. The bob swings from one side to the other and back again, again and again, along the
same path — that is oscillatory motion, and it is periodic because the path repeats after a fixed
interval of time.
Two cautions in the step decide whether the experiment will work:

Do not push the bob. A push gives it extra speed, the swing becomes wide and irregular,
and the readings scatter.
Keep the string taut. A loose string makes the bob dip and jerk instead of swinging along a
smooth arc, and the length of the pendulum keeps changing.

Page 5 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Why it does not go on for ever: air resistance and rubbing at the support take away
a little energy in every swing, so the bob rises less each time and finally stops.
Notice, though, that as the swing becomes smaller the time period hardly changes
— which is precisely what makes a pendulum a good timekeeper.

Q2 Is the time period of your pendulum almost the same every time? What do you
conclude from this observation?

Yes, it is almost the same every time. Here is Table 8.1 with a typical set of readings for a 100
cm pendulum. Time period = (time taken for 10 oscillations) ÷ 10.

TABLE 8.1 TIME PERIOD OF A SIMPLE PENDULUM (LENGTH OF THE STRING = 100 CM)

S.NO. TIME TAKEN FOR 10 OSCILLATIONS (SECONDS) TIME PERIOD (SECONDS)

1. 20.1 20.1 ÷ 10 = 2.01

2. 19.9 19.9 ÷ 10 = 1.99

3. 20.2 20.2 ÷ 10 = 2.02

Time period ≈ 2 s in every trial

Largest difference between trials = 2.02 s − 1.99 s = 0.03 s only

Conclusion: the time period of a simple pendulum of a given length is constant at a place. The
tiny differences between readings come from our reaction time in starting and stopping the
watch, not from the pendulum.

Why we time 10 oscillations and not one: a person's hand takes about 0.2 s to
react. If you time one oscillation of 2 s, that error is a whole 10%. Time 10 oscillations
(20 s) and the same error is spread over ten swings — it becomes 0.02 s per
oscillation, i.e. only 1%. Measuring many repeats and dividing is a standard way of
reducing error.

Tip: count the oscillations as 0, 1, 2, 3 … — start the watch on 'zero', not on 'one'.
Counting from one gives you nine oscillations while you think you have ten.

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Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Think Like a Scientist! — Page 110
8.1.1 A simple pendulum

THINK LIKE A SCIENTIST!

Q1 Suppose you were Galileo experimenting with pendulums, what all would you
investigate? What questions could you ask? Would all pendulums have the same
time period? How would you test this?

A scientist starts by listing everything about the pendulum that could change the time period,
and then changes them one at a time. The things that could matter are:

1. the length of the pendulum (support to the centre of the bob),
2. the mass of the bob,
3. the material of the bob (iron, stone, glass),
4. how far the bob is pulled to one side (the size of the swing),
5. the place where the experiment is done.

Questions worth asking: Does a longer thread swing more slowly? Does a heavier bob swing
faster? Does the swing slow down as it becomes smaller? Is the time period the same in Delhi
and on a hilltop?
Would all pendulums have the same time period? No. Pendulums of different lengths have
different time periods. But all pendulums of the same length have the same time period at a
given place, whatever the bob is made of.

How to test it — the fair-test rule: change only one thing at a time and keep
everything else the same. To test length, use the same bob and only change the
thread. To test mass, keep the same length and only change the bob. If you change
two things together you can never say which one caused the difference.

Did you know? Galileo had no stopwatch. Sitting in a church, he timed a swinging
lamp using his own pulse beat and found that each swing took the same time. That
single observation led, after Huygens, to the pendulum clock.

Q2 Repeat Activity 8.2 using the same bob but with pendulums of two or three
different lengths. Does the time period change? If so, how does the length affect it?

Yes, the time period changes. A typical set of readings, all with the same bob:

Page 7 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

LENGTH OF TIME FOR 10 TIME WHAT WE SEE
PENDULUM OSCILLATIONS PERIOD

25 cm 10.0 s 1.0 s Short pendulum — swings
quickly

50 cm 14.2 s 1.42 s Longer — slower

100 cm 20.0 s 2.0 s Longest — slowest

How length affects it: the longer the pendulum, the greater its time period. A long pendulum
swings slowly; a short one swings fast. Notice also that the time period does not simply double
when the length doubles — going from 25 cm to 100 cm (4 times the length) only doubled the
time period.

Why this is so useful: because the time period depends on length in a fixed way, a
clockmaker can tune a pendulum clock. If Huygens' clock ran slow, the bob was
raised a little (shorter pendulum, smaller time period, faster ticking). Pendulum
clocks still have this adjusting nut under the bob.

Try this: a pendulum of about 1 metre has a time period close to 2 seconds — one
second for each one-way swing. That is why grandfather clocks are tall.

Q3 If changing the length influences the time period, does the bob's mass also matter?
Test this by repeating Activity 8.2 with a fixed pendulum length but with bobs of
different mass. Do you observe any change?

No change is observed. With the length kept fixed at 100 cm and only the bob changed:

BOB USED (LENGTH FIXED AT 100 CM) TIME FOR 10 OSCILLATIONS TIME PERIOD

Small iron ball (light) 20.0 s 2.0 s

Heavy metal ball 20.1 s 2.01 s

Stone of about the same size 19.9 s 1.99 s

The tiny differences are only measurement error. So the conclusion of the box is confirmed: the
time period of a simple pendulum depends on its length but not on the bob's mass. All
pendulums of the same length have the same time period at a given location.

Page 8 of 39

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Class 7 Science Chapter 8 Measurement of Time and Motion
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Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

8.1.2 SI unit of time

ACTIVITY

Q1 Look at the wall clock shown in Fig. 8.9 carefully. What is the smallest interval of
time you can measure with it?

11 12 1
10 2
9 3
8 4
7 6 5

Fig. 8.9, page 112 — redrawn sketch of the wall clock.

One second. One second is the smallest interval of time that we can measure using this clock.
Look at what the dial of Fig. 8.9 has:

The numbers 1 to 12 mark the hours.
Between two numbers there are five small divisions, so the whole dial is divided into 60
small divisions.

Page 10 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

There are three hands — a short hour hand, a longer minute hand, and a thin, fast-moving
second hand.

Second hand crosses the whole dial in 1 min = 60 s

Number of small divisions = 60

Time for the second hand to cross one division = 60 s ÷ 60 = 1 s

Why the smallest division decides this: the smallest interval a measuring device
can give you is the value of one division of its scale — just as an ordinary ruler with
millimetre marks cannot measure less than 1 mm. This clock has no marks between
the small divisions, so nothing shorter than 1 s can be read from it.

Did you know? A sports stopwatch reads to one-hundredth or even one-thousandth
(a millisecond) of a second. That is how the winner is decided when two sprinters
seem to cross the finish line together — the very thing that amazes Prerna at the
start of this chapter.

In-text Questions — Pages 112 & 113
8.2 Slow or Fast · 8.3 Speed

Q1 For races covering the same distance, we can tell who was faster by measuring
time. But how can we tell that when comparing races for different distances?

By comparing the distance each runner covers in unit time — that is, by comparing their
speeds, not their times.
Time alone is useless when the distances differ. Compare these two runs:

RUNNER DISTANCE TIME DISTANCE IN 1 SECOND

Prerna 100 m 12.5 s 100 ÷ 12.5 = 8 m/s

Her friend 400 m 62.5 s 400 ÷ 62.5 = 6.4 m/s

The friend ran for a much longer time and covered a much longer distance, yet Prerna was the
faster of the two — because she covered more distance in each second.

Page 11 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Why speed is the fair comparison: dividing distance by time strips away how long
the race was. Whatever the length of the race, speed answers the same question —
'how much ground per second?' — so two completely different races can be placed
side by side.

Q2 What do we mean when we say something is moving fast or slow?

We mean a comparison of the distances moved in a given interval of time. Something is
moving fast if it covers more distance in the same time than another object; it is slow if it covers
less.

Bus in 1 h → 40 km

Cyclist in 1 h → 15 km

Same time, more distance ⇒ the bus is faster

'Fast' and 'slow' are therefore never absolute words. A cyclist is fast compared with a person
walking and slow compared with a bus.

Why the time must be the same: if you compare 40 km in 1 h with 15 km in 20 min,
you cannot say anything just by looking — 15 km in 20 min is really 45 km in an hour,
so the cyclist would be the faster one. Only when the interval of time is the same
does 'more distance' mean 'faster'.

Page 12 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Q3 All the players begin from the starting line together but after sometime they are
not running together (Fig. 8.10). How do you decide who is running faster amongst
them?

Starting line

Fig. 8.10, page 113 — redrawn sketch: boys running a race on a straight track.

The one who is ahead of the others at that instant is running faster. All the runners started
together from the same starting line, so they have all been running for the same time. The one
who is ahead has therefore covered more distance in the same time.

Start

A

B

C

Positions of three runners after the same time
All three started together. Runner A is ahead, so A has covered the most distance in that time and is
the fastest.

Page 13 of 39

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Class 7 Science Chapter 8 Measurement of Time and Motion
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Since speed is distance ÷ time, its unit is a unit of length divided by a unit of time. The SI unit of
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Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

IF DISTANCE IS …AND TIME IS …THE UNIT OF USED FOR
IN… IN… SPEED IS

metre (m) second (s) m/s (the SI unit) Runners, falling objects, physics
problems

kilometre (km) hour (h) km/h Buses, trains, speedometers

Converting: 1 m/s = 1 × 3600 m in 1 h = 3600 m/h = 3.6 km/h

So: m/s → km/h : multiply by 3.6 · km/h → m/s : divide by 3.6

Check it yourself: a train at 72 km/h moves at 72 ÷ 3.6 = 20 m/s, and a runner at 8
m/s runs at 8 × 3.6 = 28.8 km/h.

Activity 8.4: Let us calculate — Pages 114 & 115
8.3 Speed

ACTIVITY

Q1 Calculate the speed of the train between the two stations and record it in Table 8.2.

Use the railway timetable: note the distance to the next stopping station and the difference
between the departure time from your station and the arrival time at the next one, then divide.

Speed of the train = Distance till the next station ÷ Time taken till the next station

Sample answer (nearest railway station: Bhopal Junction) — fill your own timetable values in
the same way:

Page 15 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

TABLE 8.2: FINDING THE SPEED OF TRAINS

S.NO. NAME NAME OF DISTANCE TIME TAKEN SPEED OF THE
OF THE THE NEXT TILL THE TILL THE TRAIN BETWEEN
TRAIN STATION NEXT NEXT THESE TWO
STATION (KM) STATION (H) STATIONS (KM/H)

1. Vande Jhansi 291 3.00 291 ÷ 3.00 = 97 km/h
Bharat
Express

2. Shatabdi Jhansi 291 3.50 291 ÷ 3.50 = 83.1 km/h
Express

3. Punjab Mail Bina 139 2.25 139 ÷ 2.25 = 61.8 km/h
(Express)

4. Intercity Vidisha 54 0.75 54 ÷ 0.75 = 72 km/h
Express

5. Passenger Vidisha 54 1.50 54 ÷ 1.50 = 36 km/h
train

Watch the time carefully: a timetable gives clock times, not durations. If a train
leaves at 06:20 and reaches at 07:50, the time taken is 1 h 30 min = 1.5 h, not 1.30 h.
Always turn the minutes into a fraction of an hour (30 min = 0.5 h, 45 min = 0.75 h,
15 min = 0.25 h).

Q2 Compare the speeds of the trains. Which is the fastest train?

The train which has covered the maximum distance in unit time is the fastest train — that
is, the one with the highest speed. In the sample table above, that is the Vande Bharat
Express at 97 km/h.
Arranged from fastest to slowest: Vande Bharat (97 km/h) > Shatabdi (83.1 km/h) > Intercity (72
km/h) > Punjab Mail (61.8 km/h) > Passenger train (36 km/h).

Why the longest journey is not automatically the fastest: Punjab Mail travelled
139 km while the Intercity travelled only 54 km, yet the Intercity is the faster of the
two. A big distance in a big time can still mean a small speed. Only distance per hour
settles the matter.

Page 16 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Did you know? Superfast trains stop at fewer stations, and every stop costs slowing
down, waiting and speeding up. That is a large part of why a passenger train on the
same track is so much slower.

In-text Questions — Page 116
8.4 Uniform and Non-uniform Linear Motion

Q1 I once watched a part of marathon on a straight road stretch. I noticed that some
people seemed to be running at the same speed during that distance while some
people would speed up or slow down. How were their motion different?

The two groups of runners were in two different kinds of linear motion.

THE RUNNERS WHO KEPT THE SAME THE RUNNERS WHO SPED UP AND SLOWED
SPEED DOWN

Uniform linear motion — moving along a straight Non-uniform linear motion — moving along a straight line
line with a constant speed with a speed that keeps changing

Cover equal distances in equal intervals of time Cover unequal distances in equal intervals of time (250 m in
(say 200 m in every minute) one minute, 150 m in the next)

Their speed at any moment equals their average Their speed at a moment may be more or less than their
speed average speed

Both groups were running along the same straight road, so both motions are linear. What
separates them is only whether the speed stays constant.

Why 'linear' and 'uniform' are two separate ideas: 'linear' describes the path — a
straight line. 'Uniform' describes the speed — unchanging. A car going round a
circular track at a steady 40 km/h has an unchanging speed but not a straight path,
so it is not uniform linear motion.

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Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Q2 Do you remember learning about linear motion in the chapter ‘Measurement of
Length and Motion’ in the Grade 6 Science textbook Curiosity?

Yes. In Grade 6 we learnt that when an object moves along a straight line, its motion is
called linear motion.
That chapter separated the common types of motion:

Linear motion — along a straight line: a train on a straight track, a marching band on a
straight road, a ball rolling down a straight slope.
Circular motion — along a circular path: the hands of a clock, a merry-go-round.
Oscillatory motion — to and fro about a fixed position: a swing, and the pendulum of
Section 8.1.1 of this chapter.

This chapter takes linear motion one step further and asks a new question about it: is the speed
constant or changing? That is what divides linear motion into uniform and non-uniform.

Tip: the train of Fig. 8.11 shows both in one journey — non-uniform from A to B
(speeding up), uniform from B to C (constant speed), non-uniform again from C to D
(slowing down to a halt).

In-text Question — Page 117
8.4 Uniform and Non-uniform Linear Motion

Q1 Which of the two trains is in uniform linear motion between 10:00 AM and 11:00
AM?

Train X. Train X covers equal distances in equal intervals of time, so it is in uniform linear
motion, while Train Y is in non-uniform linear motion.
Reading the distance column of Table 8.3 for each 10-minute interval:

Page 18 of 39

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Class 7 Science Chapter 8 Measurement of Time and Motion
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INTERVAL (10 MIN EACH) TRAIN X — DISTANCE (KM) TRAIN Y — DISTANCE (KM)

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Position (km)

80

60 agl

co m
40
m .
m as e
.co 20 a g l
se m
g l a
a 0
c
10:00 10:10 10:20 10:30 10:40 10:50 11:00
m .
m a s e
. co agl
Time (AM)
e m
g l as
a
Distance–time graph of Table 8.3. Train X gives a perfectly straight line — uniform motion. Train Y's
line bends, because its speed keeps changing.

co m
m .
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.co


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Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Why the same average speed does not make them alike: both trains started at 0
km and reached 120 km in one hour, so both have an average speed of 120 km/h.
But Train Y reached there by running slow for a while and fast at other times.
Average speed tells you about the whole journey; it hides what happened inside it.
The graph does not hide it — a straight line means uniform, a bent line means non-
uniform.

Let Us Enhance Our Learning — Pages 118 & 119
Chapter exercises

LET US ENHANCE OUR LEARNING

Q1 Calculate the speed of a car that travels 150 metres in 10 seconds. Express your
answer in km/h.

Speed = Total distance covered ÷ Total time taken
= 150 m ÷ 10 s

= 15 m/s

Now change m/s into km/h:

150 m = 150 ÷ 1000 km = 0.15 km

10 s = 10 ÷ 3600 h

Speed = 0.15 km ÷ (10/3600) h = 0.15 × 360 = 54 km/h

The short way gives the same answer:

1 m/s = 3.6 km/h

15 m/s = 15 × 3.6 = 54 km/h

Page 20 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Where 3.6 comes from: in 1 second an object at 1 m/s covers 1 m. In 1 hour (3600 s)
it covers 3600 m = 3.6 km. So 1 m/s is the same speed as 3.6 km/h — multiply by 3.6
going one way, divide by 3.6 coming back.

Q2 A runner completes 400 metres in 50 seconds. Another runner completes the same
distance in 45 seconds. Who has a greater speed and by how much?

First runner: speed = 400 m ÷ 50 s = 8 m/s

Second runner: speed = 400 m ÷ 45 s = 8.89 m/s (8 8/9 m/s) ≈ 8.9 m/s

The second runner has the greater speed — the one who took less time for the same
distance.

Difference = 8.89 m/s − 8 m/s = 0.89 m/s (exactly 8/9 m/s ≈ 0.9 m/s)

In km/h the numbers come out neatly:

8 m/s = 8 × 3.6 = 28.8 km/h
8.89 m/s = 8.89 × 3.6 = 32 km/h

Difference = 32 − 28.8 = 3.2 km/h

Why less time means more speed: the distance is the same for both, so speed
depends only on the time in the denominator. A smaller denominator gives a bigger
answer. Cutting 5 s off a 400 m run raised the speed by about 0.9 m/s.

Q3 A train travels at a speed of 25 m/s and covers a distance of 360 km. How much time
does it take?

Total time taken = Total distance covered ÷ Speed

Page 21 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

The speed is in m/s, so first put the distance in metre:

360 km = 360 × 1000 m = 3,60,000 m

Time = 3,60,000 m ÷ 25 m/s = 14,400 s

14,400 s ÷ 60 = 240 min = 240 ÷ 60 = 4 h

Check by working in km/h instead:

25 m/s = 25 × 3.6 = 90 km/h

Time = 360 km ÷ 90 km/h = 4 h ✔

Tip: mixing units is the commonest mistake in this chapter. Before dividing, look at
the two quantities and ask: 'are they in the same family — m with s, or km with h?'

Q4 A train travels 180 km in 3 h. Find its speed in: (i) km/h (ii) m/s (iii) What distance will
it travel in 4 h if it maintains the same speed throughout the journey?

(i) In km/h

Speed = 180 km ÷ 3 h = 60 km/h

(ii) In m/s

180 km = 1,80,000 m · 3 h = 3 × 3600 s = 10,800 s

Speed = 1,80,000 m ÷ 10,800 s = 16.67 m/s (that is 50/3 m/s ≈ 16.7 m/s)

Check: 60 ÷ 3.6 = 16.67 m/s ✔

(iii) Distance in 4 h at the same speed

Total distance covered = Speed × Total time taken

= 60 km/h × 4 h = 240 km

Page 22 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Why part (iii) needs the words 'same speed throughout': the formula distance =
speed × time uses one single speed for the whole journey. That is allowed only if the
motion is uniform — otherwise you would need the average speed for those
particular 4 hours, which the question does not give.

Q5 The fastest galloping horse can reach the speed of approximately 18 m/s. How does
this compare to the speed of a train moving at 72 km/h?

Bring both speeds to the same unit before comparing.

Train: 72 km/h = 72 ÷ 3.6 = 20 m/s

Horse: 18 m/s = 18 × 3.6 = 64.8 km/h

IN M/S IN KM/H

Galloping horse 18 64.8

Train 20 72

The train is faster. It moves 2 m/s (that is 7.2 km/h) faster than the fastest galloping horse. Put
another way, the horse's speed is 18 ÷ 20 = 0.9, i.e. 90% of the train's speed — remarkably
close for an animal.

Why the comparison is impossible without converting: the number 72 looks four
times as big as 18, which wrongly suggests the train is four times faster. The units
are different, so the raw numbers cannot be compared at all. Always convert first,
then compare.

Page 23 of 39

as e
Class 7 Science Chapter 8 Measurement of Time and Motion
a g l AglaSem · NCERT Solutions

co m
m.
Distinguish between uniform and non-uniform motion using the example of a car
e
Q6

m l as
.co
moving on a straight highway with no traffic and a car moving in city traffic.

a g
se m
g l a
a

POINT OF CAR ON AN EMPTY STRAIGHT CAR IN CITY TRAFFIC

co m
. ag
DIFFERENCE HIGHWAY

em
g l as
a
Type of motion Uniform linear motion Non-uniform linear motion

Speed Stays constant, say 60 km/h, Changes constantly — slows at red

co m
m.
because nothing forces the driver lights and crossings, speeds up when

se
to change it the road clears

om
Distance .incequal
l a
g in one minute, then only
emof time
Equal: 1 km in every minute akm
Unequal: 1

a s
agl
intervals 300 m in the next

Speedometer reading Needle stays near one mark Needle keeps moving up and down,

m a s
.co agl
and falls to zero at signals

se m
l a
Distance–time graph A straight line A bent, uneven line, flat wherever the

a g car is stopped

co m
Speed at any instant vs Same as the average speed Sometimes more, sometimes less than
average speed the average speed
m .
o m l a se
g
.c are moving along a straight road, so both are in linearamotion.
m
Both cars The difference lies

l a se in whether the speed stays constant.
only
ag
Why real journeys are almost never uniform: even the highway car must slow at a
se m
com g l a
.
toll plaza or a curve. Uniform linear motion is an idealisation — that is exactly why
m a
ase
the book says we have to use average speeds.

agl

co m
Data for an object covering distances in different intervals of time are given in the
m .
e
Q7

m l as
.co
following table. If the object is in uniform motion, fill in the gaps in the table.
a g
se m
g l a TIME (S) 0 10 20 30 ___ 50 ___ 70
a c
m .
s e
DISTANCE (M) 0 8 ___ 24 32 40 ___ 56

m a
e m . co agl
g l as
a
First find the speed from a pair of values that is completely given, then use it for every gap.

co m
m .
m as e
.co


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Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Speed = 8 m ÷ 10 s = 0.8 m/s

Check with another complete pair: 24 m ÷ 30 s = 0.8 m/s ✔ and 40 m ÷ 50 s = 0.8 m/s ✔

Now fill the gaps using distance = 0.8 × time and time = distance ÷ 0.8:

Distance at 20 s = 0.8 × 20 = 16 m

Time for 32 m = 32 ÷ 0.8 = 40 s

Seventh column (between 50 s and 70 s) = 60 s, and distance = 0.8 × 60 = 48 m

TIME (S) 0 10 20 30 40 50 60 70

Distance (m) 0 8 16 24 32 40 48 56

The values filled in are shown in orange: 16 m, 40 s, 60 s and 48 m.

56
48
Distance (m)

32

16

0
0 20 40 60 70
Time (s)
All eight readings lie on one straight line through the origin — the mark of uniform motion. The
orange points are the values filled in.

Page 25 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Why one straight line is enough: in uniform motion the object covers 0.8 m in
every second, no matter when you start counting. So distance and time rise together
in a fixed ratio, and the graph can only be a straight line through the origin. Any gap
can then be read straight off that line.

Q8 A car covers 60 km in the first hour, 70 km in the second hour, and 50 km in the third
hour. Is the motion uniform? Justify your answer. Find the average speed of the car.

No, the motion is not uniform.
Justification: an object in uniform motion covers equal distances in equal intervals of time.
Here the intervals are equal (one hour each) but the distances are not:

1st hour → 60 km · 2nd hour → 70 km · 3rd hour → 50 km

60 km ≠ 70 km ≠ 50 km ⇒ non-uniform motion

Average speed:

Total distance covered = 60 + 70 + 50 = 180 km

Total time taken = 1 + 1 + 1 = 3 h

Average speed = 180 km ÷ 3 h = 60 km/h
In SI units: 60 ÷ 3.6 = 16.67 m/s

Why we add first and divide once: average speed is total distance ÷ total time —
never the average of the three separate speeds. Here the two happen to agree
((60+70+50)/3 = 60) only because the three intervals are equal. If the car had covered
60 km in 1 h and 70 km in 2 h, averaging the speeds would give a wrong answer.

Did you know? The car's average speed of 60 km/h is exactly the speed it had in the
first hour — yet at no single moment of the second or third hour was it moving
'averagely'. Average speed is a summary of the journey, not a description of any
instant in it.

Page 26 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Q9 Which type of motion is more common in daily life—uniform or non-uniform?
Provide three examples from your experience to support your answer.

Non-uniform motion is far more common in daily life. Uniform linear motion is an
idealisation; we seldom find objects moving with a constant speed over long distances or for
long intervals of time.

1. A school bus on its morning route. It speeds up on an open stretch, slows for speed
breakers, and stops completely at every pick-up point and red light. In one minute it may
cover 600 m, in the next only 100 m.
2. Walking to school or to the market. You start slowly, walk fast when you are late, slow
down at a crossing, and stop to greet a friend. The distance covered in each minute keeps
changing.
3. A cricket ball after it is hit. It leaves the bat very fast, slows down in the air, and slows still
more after it hits the ground and rolls to the boundary.

Two more from a train journey: the train of Fig. 8.11 is non-uniform between A and B while
gathering speed, and again between C and D while braking — uniform only in the middle
stretch B to C.

Why uniform motion is so rare: to keep a speed exactly constant, everything that
slows a body down — friction, air resistance, traffic, curves, gradients — must be
balanced perfectly, second after second. In everyday situations that almost never
happens. The nearest examples are things like the tip of a clock's second hand or a
ceiling fan running at a steady setting.

Q10 Data for the motion of an object are given in the following table. State whether
the speed of the object is uniform or non-uniform. Find the average speed.

TIME (S) 0 10 20 30 40 50 60 70 80 90 100

DISTANCE (M) 0 6 10 16 21 29 35 42 45 55 60

Work out the distance covered in each 10-second interval by subtracting successive distances.

Page 27 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

TIME (S) 0 10 20 30 40 50 60 70 80 90 100

Distance (m) 0 6 10 16 21 29 35 42 45 55 60

Covered in that interval — 6 4 6 5 8 6 7 3 10 5
(m)

The intervals of time are all equal (10 s each), but the distances covered in them — 6, 4, 6, 5, 8, 6,
7, 3, 10, 5 m — are not equal.
So the speed of the object is non-uniform.

Average speed = Total distance covered ÷ Total time taken

= 60 m ÷ 100 s

= 0.6 m/s

In km/h: 0.6 × 3.6 = 2.16 km/h

actual motion
60

average 0.6 m/s
Distance (m)

40

20

0
0 30 60 90
Time (s)
The red line joining the readings is uneven — the speed keeps changing. The dashed grey line is the
straight path the object would have followed had it moved uniformly at its average speed of 0.6 m/s.

Page 28 of 39

ase
Class 7 Science Chapter 8 Measurement of Time and Motion
a g l AglaSem · NCERT Solutions

co m
m.
Why the graph is the quicker test: you do not even have to subtract. If the plotted

m l a se
points lie on one straight line, the motion is uniform; if the line bends, it is non-
o
c the line is steepest between 80 s and 90 s (10 m inag10 s = 1 m/s, the
uniform. .Here
m
se stretch) and flattest between 70 s and 80 s (3 m in 10 s = 0.3 m/s, the slowest).
l a
ag
fastest

o m
e
. c
m and covers a distance of 2 km. In the first 500 ag
s
A vehicle moves along a straight line
lam/s and in the next 500 m, it moves with a speed of
Q11

a g
m, it moves with a speed of 10
5 m/s. With what speed should it move the remaining distance so that the journey

m
is complete in 200 s? What is the average speed of the vehicle for the entire
co
m.
journey?

o m l a se
ANSWER .c a g
m
e time used up in the first two stretches, then see how much time and distance are left.
sthe
l a
ag
Find

m a s
.co agl
Total distance = 2 km = 2000 m

se m
g l a
a
First stretch: time = 500 m ÷ 10 m/s = 50 s

Second stretch: time = 500 m ÷ 5 m/s = 100 s
com
m .
m as e
l
Time used so far = 50 + 100 = 150 s

m .co a g
l a se
ag Remaining distance = 2000 − (500 + 500) = 1000 m

se m
Remaining time = 200 s − 150 s = 50 s
com g l a
m . a
ase
Required speed = 1000 m ÷ 50 s = 20 m/s (= 72 km/h)

agl
Average speed for the whole journey:

co m
m .
m as e
.co l
Average speed = Total distance ÷ Total time
a g
a s em = 2000 m ÷ 200 s = 10 m/s (= 36 km/h)

agl
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


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Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

10 m/s 5 m/s 20 m/s (required)

500 m 500 m 1000 m

50 s 100 s 50 s

Whole journey: 2000 m in 200 s

The three stretches of the journey. The bar widths are drawn in proportion to the distances.

Why the average is not (10 + 5 + 20) ÷ 3 = 11.67 m/s: the three speeds were held
for different lengths of time — 50 s, 100 s and 50 s. The slow 5 m/s stretch lasted
twice as long as either of the others, so it pulls the average down. Average speed
must always be worked out as total distance ÷ total time.

Check it yourself: add the three times, 50 + 100 + 50 = 200 s ✔, and the three
distances, 500 + 500 + 1000 = 2000 m ✔. Both totals agree with the question, so the
answer is consistent.

Exploratory Projects — Pages 119 & 120
Interdisciplinary projects

EXPLORATORY PROJECTS

Q1 Construct a floating bowl-type water clock. Experiment by using bowls of different
sizes and making holes of different sizes in them so that the sinking time of the
bowl can be close to 24 minutes.

What to build: a copy of the ancient Ghatika-yantra — a bowl with a fine hole at the bottom,
floated on water in a bucket. Water enters slowly, the bowl fills and finally sinks. The bowl is
lifted out and floated again, and each sinking marks one ghati.
Method

1. Take a large tub or bucket of water and a light metal or plastic bowl (a katori works well).
2. Make one fine hole exactly at the centre of the bottom with a pin or a thin nail.

Page 30 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

3. Float the bowl gently on the water so that no water enters over the rim. Start a watch at that
moment.
4. Note the time at which the bowl sinks. Record it in a table.
5. Repeat with a bigger hole, a smaller hole and bowls of different sizes.

TRIAL BOWL HOLE MADE SINKING WHAT TO DO NEXT
WITH TIME

1 Small Thin nail (large hole) 2 min Sinks far too fast — use a smaller
katori hole

2 Small Drawing pin 9 min Still fast — use a bigger bowl
katori

3 Large bowl Drawing pin 21 min Very close — try a slightly finer hole

4 Large bowl Sewing needle ≈ 24 min Target reached — this is one ghati

Why the sinking time changes the way it does: a bigger bowl needs more water
to sink, so it takes longer. A bigger hole lets water in faster, so it sinks sooner. To
lengthen the time, make the bowl bigger or the hole smaller — and change only one
of the two at a time, so you know which change did what.

Did you know? 24 minutes is exactly 1/60 of a day (24 h = 1440 min; 1440 ÷ 60 = 24).
That is why the ancient Indian day had 60 ghatis. Every sinking was announced with
drums, conch shells or a gong.

Q2 Design an activity for measuring the pulse rate (number of times the pulse of a
person beats in 1 minute) of your friends. Think of an activity where you can use
your pulse to measure time and develop a story over that idea.

Part 1 — Measuring the pulse rate

1. Place the first two fingers (not the thumb) of your right hand on the inside of a friend's left
wrist, below the base of the thumb, and press very gently until you feel the throbbing.
2. Count the beats for 30 s with a watch and multiply by 2. Repeat three times and take the
average.
3. Record every friend's reading, first while sitting quietly, then after 20 skips or a short run.

Page 31 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

NAME BEATS IN 30 S (AT PULSE RATE (PER PULSE RATE AFTER
REST) MIN) EXERCISE

Sample: 39 78 116
Meera

Sample: Arun 36 72 124

Part 2 — Using the pulse as a clock. If your pulse rate is 72 per minute, then one beat ≈ 60 ÷
72 = 0.83 s. Use it to time things where a stopwatch is not allowed: how long a paper plane stays
in the air, how many beats a friend takes to run 50 m, how many beats a pendulum needs for 10
oscillations.
Sample story: 'The Boy Who Carried a Clock in His Wrist'. Kabir forgets his watch on the day of
the school sports. His friend is about to run the 100 m and there is no one to time her. Kabir
remembers this chapter, puts two fingers on his wrist and counts — 10 beats from the whistle to
the finish line. Later, at home, he measures his pulse rate as 72 per minute, works out 10 × 0.83
= 8.3 s, and realises his friend has beaten the school record. But he also learns why Galileo's
method was replaced: when he was excited at the race his own heart was beating faster, so his
'clock' was running fast.

Why the pulse is a poor clock, scientifically: its rate changes with excitement,
exercise, illness and age, so the 'unit' itself keeps changing. A good clock needs a
process that repeats at a rate which does not depend on what is happening around
it — which is exactly the advantage a pendulum has over a heartbeat.

Page 32 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Q3 What might be the reasons for the slight differences in the time periods of a
pendulum of a given length in different readings taken in Activity 8.2. Think of ways
to control those and repeat the activity to check if the difference in readings is
reduced.

POSSIBLE REASON FOR THE HOW TO CONTROL IT
DIFFERENCE

Reaction time — our hand starts and Time 20 or 30 oscillations instead of 10 and divide; start the watch at the
stops the watch a little late count 'zero' when the bob passes the mean position, where it moves fastest
and the instant is easiest to judge

Miscounting the oscillations Count only the passes through the mean position in one direction; have a
second student count aloud

Push while releasing the bob, or too Hold the bob, let the thread become taut, and release without any push;
wide a swing keep the swing small

Slack or stretching thread, or a Use a thin, strong, non-elastic thread; clamp it firmly; measure the length
slipping knot — the length keeps up to the centre of the bob each time
changing

A shaky support, or the bob swinging Fix the support to a heavy table or a rigid stand; release the bob in one
in a circle instead of a plane plane only

Draughts of air from a fan or Switch off the fan and close the window while taking readings
window

Least count of the watch (a wall Use a stopwatch or a mobile phone stopwatch that reads to 0.01 s
clock reads only to 1 s)

Repeat and compare. Take five readings with the old method and five with the controls in
place, and put them side by side:

Before: 19.6, 20.4, 19.8, 20.5, 20.1 s for 10 oscillations → time period 1.96 to 2.05 s, spread

= 0.09 s

After (30 oscillations, stopwatch, fan off): 60.1, 60.0, 60.2, 60.1, 59.9 s → time period 2.003

to 2.007 s, spread = 0.004 s

Page 33 of 39

ase
Class 7 Science Chapter 8 Measurement of Time and Motion
a g l AglaSem · NCERT Solutions

co m
m.
Why the spread shrinks: a fixed error of about 0.2 s in starting and stopping is

m l a se
spread over 30 oscillations instead of 10, so its effect on one oscillation becomes
o
.c smaller. Removing draughts and a shaky support takes
a g away the causes
m
three times
a e the pendulum itself behave differently from trial to trial.
smade
ag l
that

o m
e
. c
m Measure the time taken by a swing for 10 ag
s
Visit a playground with a few swings.
a period. Repeat it a few times with children of
Q4

agl
oscillations and calculate its time
different weights to find out if its time period is almost the same. Repeat this with

m
swings of different lengths. Find out how the time period changes with increasing
co
m.
length of the swings. Is the swing also an example of a pendulum?

o m l a se
ANSWER .c a g
m
s—e a swing is a large pendulum. The chains are the thread, the seat with the child on it is
l a
ag
Yes
the bob, and the frame overhead is the rigid support.

a s
cotomthe same side) and divide by 10. agl
What to do: give the swing one gentle push, wait for the motion to settle, then time 10
.
em
oscillations (each oscillation = out and back

a s
l THE
SWING CHILDagON TIME FOR 10 TIME
SWING OSCILLATIONS PERIOD

. com3.16 s
e m
as
Long swing (about 2.5 m) Light child 31.6 s

Long.c om(about 2.5 m) a g l
m
ase
swing Heavier child 31.7 s 3.17 s

agl Long swing (about 2.5 m) Two children together 31.5 s 3.15 s

se m
Medium swing (about 1.6 Light child
com
25.4 s 2.54 s
g l a
m . a
ase
m)

Short swing (about 1.0 m)
agl
Light child 20.0 s 2.00 s

om
Findings

With children of different weights on the same swing, the timeem
. c
m a s period stays almost the

.csame agl8.2.
o — just as the bob's mass made no difference in Activity
m With swings of different lengths, the time period clearly changes: the longer the swing, the
l a se
ag greater the time period. A tall swing swings lazily; a short one swings quickly.
.c
s e m
. c om
Why a swing is not a perfect pendulum: the 'bob' is a child, not a small heavy ball,
a g la
s e
and its centre keeps shifting mas the child leans back and forward — which is exactly
how children make a g la go higher by themselves. For clean readings, ask the
a swing
rider to sit still.

co m
m .
m ase
.co


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Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Safety tip: stand well clear of the swinging seat, and do the timing from the side,
never from the front.

Holistic Lens — Page 120
Interdisciplinary task

HOLISTIC LENS

Q1 Gather the timings of the winners of the races — 100 m, 200 m, and 400 m for men
and women in the last two Olympic games. Calculate and compare their speeds. In
which event is the speed the fastest?

Method: for each event, speed = distance ÷ winning time. Look up the official timings on the
Olympics website or in a newspaper archive; the sample below uses the winning times of the
Tokyo 2020 and Paris 2024 Games, rounded to two decimal places.

Page 35 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

EVENT GAMES WINNING TIME SPEED = DISTANCE ÷ TIME SPEED
(S) (M/S) (KM/H)

100 m, men Tokyo 9.80 100 ÷ 9.80 = 10.20 36.7
2020

100 m, men Paris 2024 9.79 100 ÷ 9.79 = 10.21 36.8

200 m, men Tokyo 19.62 200 ÷ 19.62 = 10.19 36.7
2020

200 m, men Paris 2024 19.46 200 ÷ 19.46 = 10.28 37.0

400 m, men Tokyo 43.85 400 ÷ 43.85 = 9.12 32.8
2020

400 m, men Paris 2024 43.40 400 ÷ 43.40 = 9.22 33.2

100 m, Tokyo 10.61 100 ÷ 10.61 = 9.43 33.9
women 2020

100 m, Paris 2024 10.72 100 ÷ 10.72 = 9.33 33.6
women

200 m, Tokyo 21.53 200 ÷ 21.53 = 9.29 33.4
women 2020

200 m, Paris 2024 21.71 200 ÷ 21.71 = 9.21 33.2
women

400 m, Tokyo 48.36 400 ÷ 48.36 = 8.27 29.8
women 2020

400 m, Paris 2024 48.17 400 ÷ 48.17 = 8.30 29.9
women

Comparison: the average speed is highest in the 100 m and 200 m sprints — they come out
almost equal, at about 10.2 m/s for men and 9.3 m/s for women — and it is clearly lowest in
the 400 m (about 9.2 m/s and 8.3 m/s). So the speed is fastest in the short sprints, and it falls as
the race gets longer.

Why the longer race is slower: a sprinter cannot keep top speed for very long. In
the 100 m almost the whole race is run at full speed; in the 400 m the runner must
hold something back to avoid tiring before the finish. Notice also that the 200 m
average can equal or beat the 100 m average — because in the 100 m a large part of
the time goes in getting up to speed from a standing start, while in the 200 m that
slow start is spread over twice the distance.

Page 36 of 39

Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Do it with fresh data: use the timings from the most recent Games and redo the
division yourself — the pattern (100 m ≈ 200 m > 400 m) will stay the same even
though the numbers change.

Chapter at a glance
People first kept time by watching events in nature that repeat after definite intervals — the
rising and setting of the Sun, the phases of the Moon, the changing seasons. From these
they made calendars, and then devices for smaller intervals within a day: sundials, water
clocks, hourglasses and candle clocks.
India has a rich record of timekeeping. Shadow-based measurement appears in Kautilya's
Arthasastra, and Varahamihira gave an accurate expression around 530 CE. The sinking-bowl
water clock or Ghatika-yantra, first mentioned by Aryabhata, took 24 minutes to fill and
sink — so a day was divided into 60 ghatis. The Samrat Yantra at Jantar Mantar, Jaipur, is 27
m tall and can measure intervals as short as 2 seconds.
A simple pendulum is a small metallic ball (the bob) hung from a rigid support by a long
thread. Starting from the mean position O, it completes one oscillation when it goes O → A
→ B → O. The time taken for one oscillation is its time period.
The time period of a simple pendulum of a given length is constant at a place. It depends
on the length of the pendulum but not on the mass of the bob. This is why pendulums, and
every other clock, work on a process that repeats itself.
The SI unit of time is the second (s). Larger units are the minute (min) and hour (h): 60 s =
1 min and 60 min = 1 h. Unit symbols are written in lower case, in the singular, with a space
after the number and no full stop — 's', 'min', 'h' (never 'sec' or 'hrs').
Speed = total distance covered ÷ total time taken. Its SI unit is metre/second (m/s); it
may also be expressed in km/h. Rearranging: distance = speed × time, and time = distance
÷ speed. 1 m/s = 3.6 km/h.
Because an object seldom keeps exactly the same speed throughout, the speed we
calculate is really the average speed. In this book, 'speed' means average speed.
An object moving along a straight line with constant speed is in uniform linear motion — it
covers equal distances in equal intervals of time. If its speed keeps changing, the motion is
non-uniform linear motion. In Table 8.3, Train X is uniform (20 km every 10 min) and Train
Y is non-uniform.

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Class 7 Science Chapter 8 Measurement of Time and Motion AglaSem · NCERT Solutions

Quick revision

TERM WHAT IT MEANS EXAMPLE / DETAIL POINT TO REMEMBER
FROM THE CHAPTER

Sundial A device in which time is read The Samrat Yantra at It shows local or 'solar
from the changing position of Jantar Mantar, Jaipur — 27 time', so a correction is
the shadow of an object cast by m tall, shadow moves needed to get Indian
sunlight about 1 mm per second Standard Time

Water clock A clock that uses the flow of The out-flow type (Fig. The out-flow type slows
water out of, or into, a vessel to 8.2a) and the floating- down as the water level
measure time bowl type or Ghatika- drops — that is why the
yantra (Fig. 8.2b) sinking bowl was developed

Ghatika (ghati) The time unit measured by one The hole was made so 24 h ÷ 24 min = 60, so a day
sinking of the bowl of the that the bowl took 24 was divided into 60 equal
Ghatika-yantra minutes to fill and sink ghatis

Simple A small metallic ball (bob) Fig. 8.7a — rigid support, At rest the bob hangs at the
pendulum suspended from a rigid support long thread, bob mean position O
by a long thread

Oscillation One complete to-and-fro journey Fig. 8.7b, with A and B the Going only from O to A and
of the bob: O → A → B → O (or two extreme positions back to O is half an
A → B → A) oscillation

Time period The time taken by a pendulum to Measured in Activity 8.2 as For a given length it is
complete one oscillation (time for 10 oscillations) ÷ constant at a place; it does
10 not depend on the bob's
mass

Periodic Motion that repeats its path after The oscillatory motion of Every clock — old or
motion a fixed interval of time a pendulum modern — is built on some
process that repeats

Second (s) The SI unit of time 60 s = 1 min; 60 min = 1 h Write '5 s', not '5 sec.';
symbols stay lower case
and singular

Speed The distance covered by an Speed = total distance SI unit m/s; also km/h. 1
object in unit time covered ÷ total time taken m/s = 3.6 km/h

Average speed Total distance covered divided A car covering 60 km, 70 It is not the average of the
by the total time taken for the km and 50 km in three separate speeds
whole journey successive hours has an
average speed of 60 km/h

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