Scene 1

In everyday life, we see some objects at rest
and others in motion. Birds fly, fish swim,
blood flows through veins and arteries, and
cars move. Atoms, molecules, planets, stars
and galaxies are all in motion. We often
perceive an object to be in motion when its
position changes with time. However, there
are situations where the motion is inferred
through indirect evidences. For example, we
infer the motion of air by observing the
movement of dust and the movement of leaves
and branches of trees. What causes the
phenomena of sunrise, sunset and changing
of seasons? Is it due to the motion of the
earth? If it is true, why don’t we directly
perceive the motion of the earth?
An object may appear to be moving for
one person and stationary for some other. For
the passengers in a moving bus, the roadside
trees appear to be moving backwards. A
person standing on the road–side perceives
the bus alongwith the passengers as moving.
However, a passenger inside the bus sees his
fellow passengers to be at rest. What do these
observations indicate?
Most motions are complex. Some objects
may move in a straight line, others may take
a circular path. Some may rotate and a few
others may vibrate. There may be situations
involving a combination of these. In this
chapter, we shall first learn to describe the
motion of objects along a straight line. We
shall also learn to express such  motions
through simple equations and graphs. Later,
we shall discuss ways of describing circular
motion.
Activity ______________ 8.1
•
Discuss whether the walls of your
classroom are at rest or in motion.
Activity ______________ 8.2
•
Have you ever experienced that the
train in which you are sitting appears
to move while it is at rest?
•
Discuss and share your experience.
Think and Act
We sometimes are endangered by the
motion of objects around us, especially
if that motion is erratic and
uncontrolled as observed in a flooded
river, a hurricane or a tsunami. On the
other hand, controlled motion can be a
service to human beings such as in the
generation of hydro-electric power. Do
you feel the necessity to study the
erratic motion of some objects and
learn to control them?
8.1 Describing Motion
We describe the location of an object by
specifying a reference point. Let us
understand this by an example. Let us
assume that a school in a village is 2 km north
of the railway station. We have specified the
position of the school with respect to the
railway station. In this example, the railway
station is the reference point. We could have
also chosen other reference points according
to our convenience. Therefore, to describe the
position of an object we need to specify a
reference point called the origin.
8
M
M
M
M
MOTION
OTION
OTION
OTION
OTION
Chapter
2020-21

Scene 2

The simplest type of motion is the motion
along a straight line. We shall first learn to
describe this by an example. Consider the
motion of an object moving along a straight
path. The object starts its journey from O
which is treated as its reference point
(Fig. 8.1). Let A, B and C represent the
position of the object at different instants. At
first, the object moves through C and B and
reaches A. Then it moves back along the same
path and reaches C through B.

displacement, are used to describe the overall
motion of an object and to locate its final
position with reference to its initial position
at a given time.

•
Walk from one corner of a basket-ball
court to its oppposite corner along its
sides.

•
Measure the distance covered by you
and magnitude of the displacement.

•
What difference would you notice
between the two in this case?

•
Automobiles are fitted with a device
that shows the distance travelled. Such
a device is known as an odometer. A
car is driven from Bhubaneshwar to
New Delhi. The difference between the
final reading and the initial reading of
the odometer is 1850 km.

•
Find the magnitude of the displacement
between Bhubaneshwar and New Delhi
by using the Road Map of India.

The total path length covered by the object
is OA + AC, that is 60 km + 35 km = 95 km.
This is the distance covered by the object. To
describe distance we need to specify only the
numerical value and not the direction of
motion. There are certain quantities which
are described by specifying only their
numerical values. The numerical value of a
physical quantity is its magnitude. From this
example, can you find out the distance of the
final position C of the object from the initial
position O? This difference will give you the
numerical value of the displacement of the
object from O to C through A. The shortest
distance measured from the initial to the final
position of an object is known as the
displacement.
Can the magnitude of the displacement
be equal to the distance travelled by an
object? Consider the example given in
(Fig. 8.1). For motion of the object from O to
A,  the distance covered is 60 km  and the
magnitude of displacement is also 60 km.
During its motion from O to A and back to B,
the distance covered = 60 km + 25 km = 85 km

Fig. 8.1: Positions of an object on a straight line path

while the magnitude of displacement = 35 km.
Thus, the magnitude of displacement (35 km)
is not equal to the path length (85 km).
Further, we will notice that the magnitude of
the displacement for a course of motion may
be zero but the corresponding distance
covered is not zero. If we consider the object
to travel back to O, the final position concides
with the initial position, and therefore, the
displacement is zero. However, the distance
covered in this journey is OA + AO = 60 km +
60 km = 120 km. Thus, two different physical
quantities — the distance and the

8.1.1 A'10'1T0N ALONG A STRAIGHT LINE
The simplest tvoe of motion is the motion
a line. We sli,dl first learn to
d' this by an examv:le, Consu:ler tne
mnnon of an nhmect rnnvme a]nne a srrnw.nt
pal-h. The oblect starts iLS journey from O
w. lich is treated as its reference point
(F [g. 8.1). Let A, B and C represent the
of The Obipc•.t at (litterent Instants. At
first, the ohlect moves rnroupn C anci B and
reaches A. T lien it back along Lile same
pach and reaches C throusli B.
o
while the Of =
'lh11S, the 01 Rrn)
is not equal LO the paun leuetil (65 kun.
Further, we will notice that the of
the d, tor a course at may
be zero but tne corresponflino
covered is not mrO. IT we. consioer the oölent
to travel baek to O. the concuues
with the arid thereture, the
:mernpnt IS zero. However, the •tance
coverea In journey OA + AO = +
60 km 120 Krn, Thus, two different Dhvsical
the distance and the
quantities
B
10
20
30
Fig. 8.1: Positions of an "'•iect a straigkUirL Oath
The total path lenr*h covered by the r '*ft.
This is the covered hy "T
describe distance we need to sp
numerical value and not the -ecuon of
There are certain nuantiti!ts which
are descrtned by spes g only tnelr
numer.eal vanmes. The
pi.vsieal quanücy is its maenitude. FIOL Ch's
can you find out Ice if the
fi.'l C 0! the or•iect trom u. initial
pastnon '[his nil terenoe you rne
numerical value of the Aic p. -lent of the
01 s,ect from O to C throuk' A. The shortest
disl ance measured the inltial to the
or an omet known as tne
displacement.
Can the rr of the disolacement
be equal to he distance travelled bv an
0htect? tne exempl.e given in
(k g. 8.1). For motion of the ontect from O to
A, the distance covered is 60 km and the
mt1Qtitude of is 60 km.
its motion trom O to A and bark to B.
Monon
km
disnlacen. e. used to describe the overall
muuon Of obiect and to locate its final
pus re.crence to its illif.ial pusilion
Activity
8.3
Talee a metre scale and a long
Walk from one corner Of a basKcL-ba_ll
court to its oppposite corner along its
the covered hy you
anu ruaez utude 01 Lile dr:uacement.
What d .erence would you notice
be-Eween the two In tnls case?
Activity
8.4
are With a
car is driven from Bituba_llcsliw-ar to
New The rerenee net-ween the
berween r innesnwar antl Aew uemi

reaches Then it mnvps bark along the same
paLh anå rea_eneS C Ulrougn B.
o
60 km 120 km, Thus, two different phl
the distance and
quantities
B
10
15
20
25
30
Fig. 8.1: Positions of an "'•iect a straig1:lirL Oath
displacen. u e used to describe the o
The total oath rth covered bv the 'Die
and to Ionace Its
rererence to its irnual po
This is the djs cance covered by the Qt. 'To
nescrtrxe (Itstance we need to sp my the
numerical value and not the o. -ecnnn of
There are cert_ain nuanulics
are described by only their
numerical values. The eal valur o a
pnvsuaal quannrvis its maontnvie. Fron rhts
exar1101e. can you find OUL Ice i_he
Lj0Si C Of I oblect fl-uLL1 t,L.
postxÄun O'? This d: u•rence •five you tile
numertoal value ot tne "cev »ent of tne
nmecT trom O to C throne A- Tne shortest
ulNLance rneasnred f. .11 initial to ehc linal
posi€ion of an ohtes+ known as the
displacement.
can the m Ae of the diqnlacernent
be equal to he distance travelled by an
obiect? Consider the exam»le given in
8.1). For motion of the o' from O to
at a -n time,
Activity
8.
T."ke a metre scale and a long rope
W'IIR from one corner Oi
court to its opppusLte corner along
WI easure the distance covered bv
at 01 the
u uat d.. .eaence woulu vuU not
between the two In this case?
Activity
8.
Automobiles are fitted with a dev
gnaws rhe travelleci
Car is driven rrorn Fsnnhanegnwar

Scene 3

SCIENCE
100
uestions
1.
An object has moved through a
distance. Can it have zero
displacement? If yes, support
your answer with an example.
2.
A farmer moves along the
boundary of a square field of side
10 m in 40 s. What will be the
magnitude of displacement of the
farmer at the end of 2 minutes 20
seconds from his initial position?
3.
Which of the following is true for
displacement?
(a)
It cannot be zero.
(b)
Its magnitude is greater than
the distance travelled by the
object.
8.1.2 UNIFORM MOTION AND NON-
UNIFORM MOTION
Consider an object moving along a straight
line. Let it travel 5 m in the first second,
5 m more in the next second, 5 m in the
third second and 5 m in the fourth second.
In this case, the object covers 5 m in each
second. As the object covers equal distances
in equal intervals of time, it is said to be in
uniform motion. The time interval in this
motion should be small. In our day-to-day
life, we come across motions where objects
cover unequal distances in equal intervals
of time, for example, when a car is moving
on a crowded street or a person is jogging
in a park. These are some instances of
non-uniform motion.
Activity ______________ 8.5
•
The data regarding the motion of two
different objects A and B are given in
Table 8.1.
•
Examine them carefully and state
whether the motion of the objects is
uniform or non-uniform.
Q
(a)
(b)
Fig. 8.2
Table 8.1
Time
Distance
Distance
travelled by
travelled by
object A in m
object  B in m
9:30 am
10
12
9:45 am
20
19
10:00 am
30
23
10:15 am
40
35
10:30 am
50
37
10:45 am
60
41
11:00 am
70
44
8.2 Measuring the Rate of Motion
2020-21

Scene 4

Look at the situations given in Fig. 8.2. If
the bowling speed is 143 km h–1 in Fig. 8.2(a)
what does it mean? What do you understand
from the signboard in Fig. 8.2(b)?
Different objects may take different
amounts of time to cover a given distance.
Some of them move fast and some move
slowly. The rate at which objects move can
be different. Also, different objects can move
at the same rate. One of the ways of
measuring the rate of motion of an object is
to find out the distance travelled by the object
in unit time. This quantity is referred to as
speed. The SI unit of speed is metre per
second. This is represented by the symbol
m s–1 or m/s. The other units of speed include
centimetre per second (cm s–1) and kilometre
per hour (km h–1). To specify the speed of an
object, we require only its magnitude. The
speed of an object need not be constant. In
most cases, objects will be in non-uniform
motion. Therefore, we describe the rate of
motion of such objects in terms of their
average speed. The average speed of an object
is obtained by dividing the total distance
travelled by the total time taken. That is,

If an object travels a distance s in time t then
its speed v is,

Let us understand this by an example. A
car travels a distance of 100 km in 2 h. Its
average speed is 50 km h–1. The car might
not have travelled at 50 km h–1 all the time.
Sometimes it might have travelled faster and
sometimes slower than this.

Example 8.1 An object travels 16 m in 4 s
and then another 16 m in 2 s. What is
the average speed of the object?

Total distance travelled by the object =
16 m + 16 m = 32 m
Total time taken = 4 s + 2 s = 6 s

Therefore, the average speed of the object
is 5.33 m s–1.

The rate of motion of an object can be more
comprehensive if we specify its direction of
motion along with its speed. The quantity that
specifies both these aspects is called velocity.
Velocity is the speed of an object moving in a
definite direction. The velocity of an object
can be uniform or variable. It can be changed
by changing the object’s speed, direction of
motion or both. When an object is moving
along a straight line at a variable speed, we
can express the magnitude of its rate of
motion in terms of average velocity. It is
calculated in the same way as we calculate
average speed.
In case the velocity of the object is
changing at a uniform rate, then average
velocity is given by the arithmetic mean of
initial velocity and final velocity for a given
period of time. That is,

average velocity = initial velocity + final velocity

where vav is the average velocity, u is the initial
velocity and v is the final velocity of the object.
Speed and velocity have the same units,
that is, m s–1 or m/s.

•
Measure the time it takes you to walk
from your house to your bus stop or
the school. If you consider that your
average walking speed is 4 km h–1,
estimate the distance of the bus stop
or school from your house.

Look at the situations given in Fig. 8.2. If
the ) 43 Rrn n-iin 8,21a)
what does it mean? W hat do you unaersra_nd
from the stenbuard in 8.21b)?
Different Obiects may take different
amounts of time to cover a given distance.
Some of them move tast arid some move
slowlv. The rate at which obieets move can
be dt: rerent. Also. dttterent 0btects can move
at the same rate. One oi the wavs of
measuring the rare Of motion of an oblet':t. is
to iind Lhe distance travclied bv Che ubiect
in unit time. This quantitv is rekerred to as
speed. The SI unit of speed is metre per
second. This is represented by the symnol
m s-l or m,' s. ?ihe orner units of speed
centimetre ner second (cm s-l) ano Kliorne•re
per hour ikm h-l). To the sveed Oi an
obiect, we require only its makunude. The
speed of an ol:gject need not be constant. In
most cases. omects be in non-unuorm
motion. Thererore, we describe the rate of
motion of such obiects in terms of their
averaee soeed. The averaue speed of an 0b; -ct
is oi_ituined by dividivæ the total
travelled by the total time taken. "I'hE is
If an ohiect travels a distance s in tilde t then
Let us understand this bv an e. •un 'le. A
car travels a dist*tnce of 100 km in h. Its
.01 the ume.
Surnetunes it have tria "„iled laster and
and thei_ ano':ber 16 m 2 s. What is
the speed of the object?
Total distence travelled by the object -
Monon
32 m
Therefore, the average speed of the ouject
82.1 SPEED WITH DIRECTION
'lhe rate of motion of an object can be more
conmpreher.sive if we its direction of
mocii wilh its speed. 'The qunntity that
spemttes bi these aspects IS ca 'lect velocity.
VpJ0C.tty rne speed ot an orner n. in a
dennjre dlreeuon. Tne An ohier.t
can be unnorm or variable. t be champed
by char.utne the Obie •t's oecd, direcuon of
mo' ion or both. •n obiect is moving
a stran*kii. Hn a variable speed, we
e -press L •e _ na_nitude of its rate of
ir ot average velocity. It is
uculated th same way as we calculate
In c. ee the velocity of the object is
g at a uniform rate, then average
oc. is given by the arithmetic mean of
'ait; A1 velocity and final velocity for a given
period of tune. That is,
2
Mathernatically, v
2
wyrere v Is the average velocitv. u is the initial
velocilv and V is the velocity 01' the oblcet.
Speed and velocity have the same units,
Activity
8.6
Measure the time it takes vou to walk
from your house to your bus stop or
the scnool. It vou conslrter tnat your
averaue speed Is 4 km h-'
estirr_ ite Che distance of the bus stop
or scnøoi Irom your house.

to out tne (ll•tance Travelled nmect
in unit time. 'I-ms quanrn.v is rererred to as
soecd. The SI unit of soced is metre per
second. This is represented bv the symbol
m s-l or m/s. 'Ihe ot-kier units ot speed ide
centimetre per second (cm s-l) and kilometre
per hour (xm n-l). uzo spemfy the speea 01 an
oDiect, we reauire its maunuude. The
sueed of an oi.neet need not be cunslnnt. In
fnust ohiccts will be in nun-unnorm
motion. "Iloeretore. we describe the rate of
motion of suotl ohiects in terms of their
avernøe_ s,oeed. The avera2e speeci of an Obi -ct
is oi-na_llled bv divruine the total
travelled by the total tune taken. The is,
If an ohiect travels a distance s in ti.ae t then
Let us understand this hv an e. am ne. A
car travels a oi 100 km in h. Its
not h.hve trasrellpd at 50 In -dl the tnne.
Sommtmes It have treziled raster and
someumes slower th n this.
thei_ lher 16 m in 2 s. What is
the speed of the object?
'lhe rate of motion of an oblect can be
cornorenensive if we specriv its directi
mouun ai011E-i wiLh its Sueed. The ouanLiL
Sijecilies i:hese is Ca vel
VeiociLy is the speed of an 01 sier n. avin
definite direction. The velof"+v an
can be uniform or variable. t cap. be Ch.
by changing the •t's direct
mol.on or both. Wnnn •n ompct is
o. a straiøni lin• vananle spee
ir tp- ot avcraue velocitv.
In c. ee the velocity of the obit
ch', g at a unitorm rate, then av
is given by the aritbmenc me
'-nr" veiomry and final velocity for a
period of cime. Inat is,
2
Mathernatically, v
s#ere vm, is the velmit.y, u is the
ve1001tv med Tne nnaL veL001ty 01 rne o
Speea and velocity nave the same
Activity
Measure the time. it takes you to w

Scene 5

The average speed of the car is
50 km h–1 or 13.9 m s–1.

Example 8.3 Usha swims in a 90 m long
pool. She covers 180 m in one minute
by swimming from one end to the other
and back along the same straight path.
Find the average speed and average
velocity of Usha.
Solution:
Total distance covered by Usha in 1 min
is 180 m.
Displacement of Usha in 1 min = 0 m

During uniform motion of an object along a
straight line,  the velocity remains constant
with time. In this case, the change in velocity
of the object for any time interval is zero.
However, in non-uniform motion, velocity
varies with time. It has different values at
different instants and at different points of
the path. Thus, the change in velocity of the
object during any time interval is not zero.
Can we now express the change in velocity of
an object?

•
At a time when it is cloudy, there may
be frequent thunder and lightning. The
sound of thunder takes some time to
reach you after you see the lightning.

•
Measure this time interval using a
digital wrist watch or a stop watch.

•
Calculate the distance of the nearest
point of lightning. (Speed of sound in
air = 346 m s-1.)
uestions

1.
Distinguish between speed and
velocity.
2.
Under what condition(s) is the
magnitude of average velocity of
an object equal to its average
speed?
3.
What does the odometer of an
automobile measure?
4.
What does the path of an object
look like when it is in uniform
motion?
5.
During an experiment, a signal
from a spaceship reached the
ground station in five minutes.
What was the distance of the
spaceship from the ground
station? The signal travels at the
speed of light, that is, 3 × 108

Example 8.2 The odometer of a car reads
2000 km at the start of a trip and
2400 km at the end of the trip. If the
trip took 8 h, calculate the average
speed of the car in km h–1 and m s–1.

Distance covered by the car,
s = 2400 km – 2000 km = 400 km
Time elapsed, t = 8 h
Average speed of the car is,

Activity
8.7
1000m
At a time when it is cloudv. there may
sound of lhLi11uer takes Lil to
reach you after see the rtnlng.
Cali VUu answer uljs hauucnS?
Measure thls time interval ueing ,
wrist watch or a stoo waLCiL.
te the distance of the nearest
pomt 01 [Speed of sounu in
uestions
50
h
The average speed of the car is
0.0 Usma SWLLUS In a Long
puul. Slie covers 180 m in one minuu-e
by swimming from one end to the other
aptl ba,0K altn the same stra•i Eiht path.
Find the average speed ana average
velocity of Usna.
Tutal distance covered by Ush' 1 min
Displacement of Usha in 1 •ain = O m
l.
8.
mogntüde
an. eqvrÄ.
does $
it
a signal
the
" minutes.
d tance of the
180 m
min
I min
60 s
min
3m s
fro
t the end o:
2400 kl
tnp toc
the ground
traæls at
is, 3 x
tril . ff rne
Average speed ot the car is,
50 km h-
era, -e velocity = Total time taken
60
8.3 R Ite of Chai e of Velocity
During uniform motion of an obieet along a
nr Ime, the velocirv remams consrant
nrne. In this case, the charwe in velocity
of the object for anv tinme fiucrval is zero.
However, in non-unhorm motion, velocity
varies time. It has dj crent values at
di it-rent inatants and at dill.erent points of
tne path. Thus, the cnanee In velocity ot tne
oblect dunng any tlme Inrervai is not zero.
Can we now express me change in velocity of
ScrmvCE

uestions
l.
8.
olution:
and baek Che same natl
Find the average speed and
averag
distance covered by Ush' 1 mi
Displacement of Usha in 1 •ain = O n
mogntüde
an. equ-A.
it
a signal
the
" minutes.
d tance of the
min
3m
1 ruin
60s
fro
at the star
the ground
traæls at
is,
trip
and
Average speed of the car is,
era, -e velocity Lirne La«en
60 s
—0m s-l
The average speed of Usha is 3 m
83 F tte of of V
During uniform motion of an object alon:
str."Ä-11t line, the velocity remains const.%
With time, this case, the cnenge in veloc
or the omect tor any time inTerva1 is ze
However, in non-unitorm mntjnn,
vanes tirne. It has di werent values
duacrent instants and at diner-ent poitll.s

Scene 6

To answer such a question, we have to
introduce another physical quantity called
acceleration, which is a measure of the
change in the velocity of an object per unit
time. That is,

If the velocity of an object changes from
an initial value u to the final value v in time t,
the acceleration a is,

This kind of motion is known as
accelerated motion. The acceleration is taken
to be positive if it is in the direction of velocity
and negative when it is opposite to the
direction of velocity. The SI unit of
acceleration is m s–2 .
If an object travels in a straight line and
its velocity increases or decreases by equal
amounts in equal intervals of time, then the
acceleration of the object is said to be
uniform. The motion of a freely falling body
is an example of uniformly accelerated
motion. On the other hand, an object can
travel with non-uniform acceleration if its
velocity changes at a non-uniform rate. For
example, if a car travelling along a straight
road increases its speed by unequal amounts
in equal intervals of time, then the car is said
to be moving with non-uniform acceleration.

•
In your everyday life you come across
a range of motions in which
(a)
acceleration is in the direction of
motion,
(b)
acceleration is against the
direction of motion,
(c)
acceleration is uniform,
(d)
acceleration is non-uniform.

•
Can you identify one example each
for the above type of motion?

Example 8.4 Starting from a stationary
position, Rahul paddles his bicycle to

attain a velocity of 6 m s–1 in 30 s. Then
he applies brakes such that the velocity
of the bicycle comes down to 4 m s-1 in
the next 5 s. Calculate the acceleration
of the bicycle in both the cases.

In the first case:
initial velocity, u = 0 ;
final velocity, v = 6 m s–1 ;
time, t = 30 s .
From Eq. (8.3), we have

Substituting the given values of u,v and
t in the above equation, we get

In the second case:
initial velocity, u = 6 m s–1;
final velocity, v = 4 m s–1;
time, t = 5 s.

= –0.4 m s–2 .
The acceleration of the bicycle in the
first case is 0.2 m s–2 and in the second
case, it is –0.4 m s–2.

1.
When will you say a body is in
(i) uniform acceleration?  (ii) non-
uniform acceleration?
2.
A bus decreases its speed from
80 km h–1 to 60 km h–1 in 5 s.
Find the acceleration of the bus.
3.
A train starting from a railway
station and moving with uniform
acceleration attains a speed
40 km h–1 in 10 minutes. Find its
acceleration.
Q

To answer such a question, we have to
introduce anotnpr physical quantity called
arceieration, wnich is measure of the
chanue in the velocity of an object per unit
change in velocity
acceleration =
time taken
if the velocity of an obiect changes from
an initial value u to [he finai value v time t,
the acceleration a is,
This kjnd of motion is known as
accelerated motion. The, acceleration is taken
to be posdive ii iL iS in the direction oiveiocny
and negative when it is ooposite to the
direction of velocity. The SI unit of
It an orneer travels in a straiøht line and
its velocity Increases or decreases by eaual
amounus in equal intervals of Lillie. then i .e
acceleration of the object is *Gid to
unitorm. "Ihe motion ot a freely tällir g body
is an example ot unitarmly acne-I •rate I
motion. (m lhe other hand, an 0b can
travel non-uniibrm acceie, if its
velocitv chanees at a non-uniform :ate. For
increases its speed lal amr •ts
in equal Intervals 01 Time. tne car i. sav;
to be moving wun non-unixorm 'lei
Activity
8.8
(a) acceleratk • . is In the directlon of
motion,
(b) accele •atl is against the
die -l motion,
(c) ac. elen is unitorrn.
(dl acce. :raLion is nun-uuiforrn.
Can you Idenl"fy one exanple each
for the above type Ot motion?
puz.uon, Rahul padåles his bicycle to
Monon
attain a velocity of 6 m s-l in 30 s. Then
he brakes such that the veloetty
OJ me Dicycle comes (Inwn to 4 m s-l in
the. next s. Cmcu•ate me acceleration
of the bicycle in both the cases.
initial ve'or•ity, u = O .
From Eq. (8.3), we have
S"bqfituting the given valu o: u,vand
tin cne above equaLior,
iritial •loc ty, u = 6m s
'Ihe accelerai.ion of the bicvcle in the
first case is 0.2 m s-9 and in the second
uestions
vu:rs-
fa. bes
€4 km. trL h-: in S
staring
speed.

the acceleration a is,
This kind of motion is known as
accelerated motion. The accelerption is t*ikpn
to De postrtve in me an-ecnon ot veinetty
and neeacive when it is opposite to the
directiun of velocity. The SI unit of
accelerai.ion is m s-2
an oblect travels in a straitfht line and
its velocity increases or decrea«es by equal
amounts in equal Intervnls ot nrne, then v .e
accelera Lion Of the oblect is said to
uniiorm. The motion of a freely fallin body
is an example of uniformly acoel •rate I
motion. On the other hand, an 0b, can
travel wjrh non-unnorm accele, if Its
veiociTV chanpes at a nnn-ururnrm -'ate. For
exarno/e, if a car Craven' II", a stra_iv.lt
roau Hlereases its •
in equal of dime, Gw-n Lhe car i.
to be moving with non-uniiorm 'lei t.lon
Activity
a range of munons in . Ch
8.8
(a) acre]c-ratie. is In the directlon of
(b) accele •atl is against
. motion,
(c) ac lion is unilorrn.
(d) acce. -'ration is non-uniinrm.
Can you Identify one example
each
From Eq. (8.3), we have
SubstitutinQ the Qiven valu 's o: u, c
tm tne above equauor,
•loc cy, u = 6 m s-l
final •looty, u = 4 m s-l
5s
arreleraTion of the hlovele in
first case is 0.2 m and in lhe
uestions
fa. bes
€4 km. trL h-l in.
staring

Scene 7

Graphs provide a convenient method to
present basic information about a variety of
events. For example, in the telecast of a
one-day cricket match, vertical bar graphs
show the run rate of a team in each over. As
you have studied in mathematics, a straight
line graph helps in solving a linear equation
having two variables.
To describe the motion of an object, we
can use line graphs. In this case, line graphs
show dependence of one physical quantity,
such as distance or velocity, on another
quantity, such as time.

The change in the position of an object with
time can be represented on the distance-time
graph adopting a convenient scale of choice.
In this graph, time is taken along the x–axis
and distance is taken along the y-axis.
Distance-time graphs can be employed under
various conditions where objects move with
uniform speed, non-uniform speed, remain
at rest etc.

Fig. 8.3: Distance-time graph of an object moving
with uniform speed

We know that when an object travels equal
distances in equal intervals of time, it moves
with uniform speed. This shows that the

distance travelled by the object is directly
proportional to time taken. Thus, for uniform
speed, a graph of distance travelled against
time is a straight line, as shown in Fig. 8.3.
The portion OB of the graph shows that the
distance is increasing at a uniform rate. Note
that, you can also use the term uniform
velocity in place of uniform speed if you take
the magnitude of displacement equal to the
distance travelled by the object along the
y-axis.
We can use the distance-time graph to
determine the speed of an object. To do so,
consider a small part AB of the distance-time
graph shown in Fig 8.3. Draw a line parallel
to the x-axis from point A and another line
parallel to the y-axis from point B. These two
lines meet each other at point C to form a
triangle ABC. Now, on the graph, AC denotes
the time interval (t2 – t1) while BC corresponds
to the distance (s2 – s1). We can see from the
graph that as the object moves from the point
A to B, it covers a distance (s2 – s1) in time
(t2 – t1). The speed, v of the object, therefore
can be represented as

We can also plot the distance-time graph
for accelerated motion. Table 8.2 shows the
distance travelled by a car in a time interval
of two seconds.

Table 8.2: Distance travelled by a
car at regular time intervals

8.4
1 :cal of
Graohs provide a convenient method to
prcsent b.] 'öic inform-anon about a vaciety of
events. For example, in the telecast ot a
one-day cricket maT0h, vertical bar graphs
show the run rate of a team in each over, As
you have studied in '_:lucn.a€ies. a strai[iht
h. in sol',iug a linear equacion
having two variames.
To describe the motion of an ohiect, we
can use line urauhs. In this case. line uraons
show of one physical quami€,i.v,
such as distance or velocity, on another
quantity, such as time.
8.4.1 DISTANCE-TIME GRAPHS
me charge in the position of an ohiect with
time can be represented on the distance-time
a convenient scale chntre.
In VI-don, Orne is t_dKeE1 alone the Y—axis
and distance is taken a10i1Q the y-axis
various I-ions where ohiects move
ununrrn speed, non-unnonn speed, .-en-
10
o
Fig. 8.3: Di*$ance-tt. •e groph of an object moving
wetn unyorm speed
We know that when an ohipc•t travels eanal
distances in equal illtervais 01 time, it moves
urgilorrn speed. This shows that the
die,tance travQ]ed by the ohiect is directly
prnnoruonal co rune Taken. for
sueed, a eraDh of distance travelled ayeu11St
The portion ot the graph shows that the
distance is increasjng at a 1 mnuorrn rate. Note
that, you can atso use the term uniiorm
veiocuv in Oiaee of soeed if vou take
the of equal to the
dlQtance travelled by tne Object along tne
we can use the distance-time granh to
determine the speed of an obiect. To do so,
consiEIer a small part. AB of the distence-Cime
gr:ann snoum mg 8.3. Draw a line
to rne x-axis from point A and ann her line
parallel to [he y-axis irom point B. The -e two
the rune interval [t — t,) w, le corresnonds
to tb' distance {s? — S I, •an see from [he
graoh LAa€as the ves frum the point
c-,-n also plot the diqtance-time graph
'01 elera_tpql motion. ;le. 8.2 shows the
travel lea DY a car in a time interval
of two seconds.
e 8.2: D stance trave e y
car at regular time intervals
Time in seconds Distance in metres
10
12
16
25
36
ScrmvCE

vine two variables.
"l"o describe the motion of an obtect, we
use Grnnns.
depenuence Of one pnvslcal quanu1V,
ch as disuance or velocity, on anotuer
antity, such as time.
Ll DISTANCE-TIME GRAPHS
e change. in the position of an ohiect with
can be reoreserued on the Giseanee-time
a convenient scale of choice.
ume is taken alone Ehe y-axis
d dlsranc:e iS taxen along rne y-ar•s
stanee-U_Lne vraons can be ernoloved
riuus witere obiects move
speed, non-unitonn speed, .-en-
8.3: Distance-h. •e aranh of an object moving
We can use the distance-time graph
determine the, speed ot an oblect. To do
cumsider a small oart AB of the distance-til
g, shown in Fik 8.3. Draw a lir•e para.
to the x-axis trom point A and ann her Ij
parallel to rhp y-ans rrom point B. _e t
lines meet each 0Lner at Poir L to
ABC. NOW, on thc denu
the Unje interval (t w, le L Q correspon
to th' distance (52 — s i', We •an see trom
grant. rhe ves Irom Vhe po
tt B. iL covers db. •ance (sm— S,) in
v the object, theref(
W also plot the distance-time gra
C *stance trav tllcd by a car in a time inter
ot two seconds.
car at re ular time intervals
Time in seconds
Distance in metre

Scene 8

The distance-time graph for the motion
of the car is shown in Fig. 8.4. Note that the
shape of this graph is different from the earlier
distance-time graph (Fig. 8.3) for uniform
motion. The nature of this graph shows non-
linear variation of the distance travelled by
the car with time. Thus, the graph shown in
Fig 8.4 represents motion with non-uniform
speed.

The variation in velocity with time for an
object moving in a straight line can be
represented by a velocity-time graph. In this
graph, time is represented along the x-axis

Fig. 8.4: Distance-time graph for a car moving with
non-uniform speed

Fig. 8.5: Velocity-time graph for uniform motion of
a car

and the velocity is represented along the
y-axis. If the object moves at uniform velocity,
the height of its velocity-time graph will not
change with time (Fig. 8.5). It will be a straight
line parallel to the x-axis. Fig. 8.5 shows the
velocity-time graph for a car moving with
uniform velocity of 40 km h–1.
We know that the product of velocity and
time give displacement of an object moving
with uniform velocity. The area enclosed by
velocity-time graph and the time axis will be
equal to the magnitude of the displacement.
To know the distance moved by the car
between time t1 and t2 using Fig. 8.5, draw
perpendiculars from the points corresponding
to the time t1 and t2 on the graph. The velocity
of 40 km h–1 is represented by the height AC
or BD and the time (t2 – t1) is represented by
the length AB.
So, the distance s moved by the car in
time (t2 – t1) can be expressed as
s = AC × CD
= [(40 km h–1) × (t2 – t1) h]
= 40 (t2– t1) km
= area of the rectangle ABDC (shaded
in Fig. 8.5).
We can also study about uniformly
accelerated motion by plotting its velocity–
time graph. Consider a car being driven along
a straight road for testing its engine. Suppose
a person sitting next to the driver records its
velocity after every 5 seconds by noting the
reading of the speedometer of the car. The
velocity of the car, in km h–1 as well as in
m s–1,  at different instants of time is shown
in table 8.3.

Table 8.3: Velocity of a car at
regular instants of time

Time
Velocity of the car
  (s)
(m s–1)
(km h
–1)

0
0
0
5
2.5
9
10
5.0
18
15
7.5
27
20
10.0
36
25
12.5
45
30
15.0
54

35
Fig. 8.4: Distance-time aranhfor a car moving with
The distance-time graph for the motion
of the car is shown in 2, 24.4. Nnt_e
Shane: rnis JS €11 tne
disLance-tuue graph (k 8.3) for u Luorn_
The neoure of this shm s nor-
lilltear varia at' the disiuuee tra. ell. oy
tne ear time. Thus, tne
in
Fip 8.4 represents monon witn n0i. -unnorm
The variation in velocity with G. T (e •r an
ohlect moving in a stratdtlt line "-an be
renresenred bv a veiooitv-urv In this
graph, tune is represen'0J3 tile x-axis
Time (h)
• . : Velæity-ärne graphfor uniform rnotion q/
mg 8.5
Monon
and the velocity is represented along the
rhe onlec•t moves ururnrm
the he121it oi iLS velocnv-L'i11me not
time Q. 8.5). It a stra. , :
velocity-tjme graph tör a car moving witn
urn101-m Velocir.v Of 40 km h-l
We know that the muduct of velocity and
tiple of an obic.ct moviu.g
tu„lttorrn veqmtty, area eneJ0€æd by
veinetry-ume grann and tne: ome axts Wili ne
equal to the rnaelütuue 01 tile disoLaeemer1t.
To know the disiance moved bv the car
bet*.repn time t, and t, Fig. 8.5, dr•s.w
the crxrregponrtlng
to r ne Time and on The pranh "he veloeuy
of 40 km h-L is revresentcd DV he AC
or BD and the thne (G— 4) •s cpacsen€ccl by
So, the distance mm •d by the car in
tir .e (ta — tl) can be r xpre. -sed as
arec of he rectangle ABDC (shaded
also studv ahout uniformly
acc (ert '-ed motion by .g its ve-100ity-
ann. Consu-lera car nerno driven along
- St' dll_• FLL road lor cesüné its en21_ne. SuoDose
a pet-sun next to die driver recu,d3 its
ot tne sneedometer ot tne car. 'lhe
ve•oc1W of the, car, in km h-l as well as in
m s-l, at rerent insi_alits of time is shown
in table 8.3.
re ular instants of time
Time
(s)
10
15
20
25
30
Velocity of the car
2.5
5.0
7.5
10.0
12.5
15.0
(km KI)
18
27
36
45
54

10
o
Fig. 8.4: Distance-firrte a car moving with
n01t-Lutyorrrt swed
The distance-time m-anh for the motion
of the car is shown in Q. 8.4. Note tl
(listanee-ttrne graph (Wig. 8.3) tor v Illorn_
rnnrnnn. The nature Of oranh shm nor
hncar valiuuun of the tra. oy
che car time. 'Illus. the
FÄg 8.4 represents motion wiLh noi. -unilorm
8.4.2 VELOCITY-TIME
The variation in velocity with t. •r an
oblect movinz in a strai2i1t line "-an be
represented a velocity-tim s, •aph, In this
equal to the m:" 'it"de of the
To know the disrance movea ny rl«
between time t, aud t, usitlé Fi±. 8.3.
perpendir-ulars from the poini.s correspu
to the time and t2 on the s'raph 'he ve
ot 1 km h-l is representea Dy he
or BD and the time (ta— 4) 's
the
So, the dista_nce s mm •d by the (
• (t40 kn. x (t, —t)
arec of he rectangle ABDC (sl
a.so study about unifc
acc lert '-ed moLion by its vel•
- st- ail Flt road tor testing its ev ftine. Suj
a person next (inver reenr
veLucWv aiccr cverv 5 secunds bv notin
reaG, of the speedome€er of the cal
veloc:ty or the ear, in kill as well
m s-l, at cil"erent instants of time IS s
in table 8.3.
e oc yo a cara
gular instants of time
Velocity of the car
Time
(s)
10
2.5
5.0
(km h -
18

Scene 9

In this case, the velocity-time graph for the
motion of the car is shown in Fig. 8.6. The
nature of the graph shows that velocity
changes by equal amounts in equal intervals
of time. Thus, for all uniformly accelerated
motion, the velocity-time graph is a
straight line.

Fig. 8.6: Velocity-time graph for a car moving with
uniform accelerations.

You can also determine the distance
moved by the car from its velocity-time graph.
The area under the velocity-time graph gives
the distance (magnitude of displacement)
moved by the car in a given interval of time.
If the car would have been moving with
uniform velocity, the distance travelled by it
would be represented by the area ABCD
under the graph (Fig. 8.6). Since the
magnitude of the velocity of the car is
changing due to acceleration, the distance s
travelled by the car will be given by the area
ABCDE under the velocity-time graph
(Fig. 8.6).
That is,
s = area ABCDE
= area of the rectangle ABCD + area of
the triangle ADE

In the case of non-uniformly accelerated
motion, velocity-time graphs can have any
shape.

Fig. 8.7: Velocity-time graphs of an object in non-
uniformly accelerated motion.

Fig. 8.7(a) shows a velocity-time graph
that represents the motion of an object whose
velocity is decreasing with time while
Fig. 8.7 (b) shows the velocity-time graph
representing the non-uniform variation of
velocity of the object with time. Try to interpret
these graphs.

•
The times of arrival and departure of a
train at three stations A, B and C and
the distance of stations B and C from
station A are given in table 8.4.

Table 8.4: Distances of stations B
and C from A and times of arrival
and departure of the train

Station
Distance
Time of
Time of
from A
arrival
departure
(km)
(hours)
(hours)

A
0
08:00
08:15
B
120
11:15
11:30
C
180
13:00
13:15

•
Plot and interpret the distance-time
graph for the train assuming that its
motion between any two stations is
uniform.

In this case, the velry•jty-time graph for Cbe
monon ot rne car is snnwn 111 too. 8.6. Toe
g 2r
nature of the graoh shows tuat velocity
clu bv equal uluounts in equal intervals
ot Thus, tor all accelerated
motton, tne velocity-time graph is a
straignt line
t}
10
Fig. 8.6: Velrrit!'-time grrwh for a car moving with
You can also determine the diqt* nce
the ear from its velocity-time
The area under the velocity-time
the di•.tance of di:4] •ment)
If tne car would have bee ' m with
mmilorm velocrtv. the distaxre vruvehed b
would be represented by the area
maonitude or tne vmocjty of the ear is
charip]np due. to acceleration. t' -stance s
bv the car wdl be 'en the area
ATÆCLFE under the velocit) -t graph
area ABCDI
area of i -ecaangle ABCD + area of
2
In the case of non-uniformly accelerated
motion, velocity-time graphs can have any
10
sc
ec
3
4
Time (s)
5
o
Fig. 8 Velryy:ty-ttme granhs of an object tn non-
unyonruy a •cet •t motton.
*ig. 713) sb. •vs .rplncitv-time
tx leureserts n. •tiun 01 ail ut.)leei, wnuse
city is ore with ti±ne waile
reoresenti-u T ne. non-unllorm Variation of
velucih ot object with Lime. Try
Activity
8.9
e
me. times of arrival and denariure of a
train at three. A, B and C and
the cilsrance_ Qi sranong B C from
s ances o s a Ions
and C from A and times of arriv
d departure
Station Distance
from A
120
180
th trai
Time of Time of
arrival departure
(hours)
08:00
11:15
13:00
(hours)
08:15
1 1:30
13:15
F\at and interpret t.ne dietance-tjme
for the train assu111J112 that its
motmn between any two stations is
uniaorm.
ScrmvCE

1. 8.6: Velocity-time graph for a car moving with
uregörrrt accelerations.
You can also determine the disü nce
ved bv the car trom its velority-time
e area unaer the ve1001tv-nrnp Q, •w.
distance (inauL1iLude Of •ment)
wed bv Lhe car in a kiv-en of chne.
the ear would have bee 11 m w:lth
veloejty, the 'Ijstali?e VT D'
uder the graph (F W. 8.6). Sin,e t he
IQi1itude of the velocity of the ear is
due to acceleration, t' Le 'stance s
a oy the car oe tne area
under the velocnyt:me graph
Q. 8.6).
area or r te -ee.nngle ABCD + area of
8
Fig. 8 7; Velocttu-ttme aranhs of an object tn
a •cet _ motion.
reltreserrs
n. OT an ontect
prelu Cu y is a, ore Ising wit h time wh
F 2. 8.7 (b) s'•ows the veloeitv-tiille era
f he non-unitorm varial ion
or . ae oojectwlth ume. Try to Interp
Activity
8.9
1
me times Of arrival and departure of a
train at l_nree sEanons A, b and C and
i
and C from A and times of arriv'
and de arture of the train
Station Distance
from A
(km)
120
180
Time of Time o
arrival departu
(hours)
08:00
11:15
13:00
(hours)
08:15
1 1:30
13:15

Scene 10

8.5 Equations of Motion by
Graphical Method

When an object moves along a straight line
with uniform acceleration, it is possible to
relate its velocity, acceleration during motion
and the distance covered by it in a certain
time interval by a set of equations known as
the equations of motion. For convenience, a
set of three such equations are given below:
v = u + at
(8.5)
s = ut + ½ at2
(8.6)
2 a s = v2 – u2
(8.7)
where u is the initial velocity of the object
which moves with uniform acceleration a for
time t, v is the final velocity, and s is the
distance travelled by the object in time t.
Eq. (8.5) describes the velocity-time relation
and Eq. (8.6) represents the position-time
relation. Eq. (8.7), which represents the
relation between the position and the velocity,
can be obtained from Eqs. (8.5) and (8.6) by
eliminating t. These three equations can be
derived by graphical method.

Consider the velocity-time graph of an object
that moves under uniform acceleration as

•
Feroz and his sister Sania go to school
on their bicycles. Both of them start at
the same time from their home but take
different times to reach the school
although they follow the same route.
Table 8.5 shows the distance travelled
by them in different times

Table 8.5: Distance covered by Feroz
and Sania at different times on
their bicycles

Time
Distance
Distance
travelled
travelled
by Feroz
by Sania
(km)
(km)

•
Plot the distance-time graph for their
motions on the same scale and
interpret.

1.
What is the nature of the
distance-time graphs for uniform
and non-uniform motion of an
object?
2.
What can you say about the
motion of an object whose
distance-time graph is a straight
line parallel to the time axis?
3.
What can you say about the
motion of an object if its speed-
time graph is a straight line
parallel to the time axis?
4.
What is the quantity which is
measured by the area occupied
below the velocity-time graph?

Fig. 8.8: Velocity-time graph to obtain the equations
of motion

Activity
8.10
Feroz and his sister Snni.o go to sphool
on rnelr meveles, tsnrh 01 rnem srarr. at
the same tune from thuir holile but take
nrerent. runes to reach r.ne sennol
H.5 rue 'iterance Eravened
able 8.5: Distance covered by Fero
and Sania at different times o
their bicycles
0.8
1.6
2.3
3.0
3.6
Time
8:00 am
8:05 am
8:10 am
8:15 am
8:20 am
8:25 am
Distance
travelled
by Feroz
(km)
1.0
1.9
2.8
3.6
Distance
travelled
by Sania
(km)
Plot the distance-time graph for tneir
motions on the same scale and
uestions
l.
4.
of the
graphsfor unconn
and non-un.Vorm rnotion of
What can you say
of an object
dismrzce-tfrne
une parallel
What can goo,
nuåon Qf
true occtu.péc•d,
veioci*th'-f$ne
8.5 Equations of hCotion by
When an obiect moves along a straight line
relate its veinejty, acceleration dn_ring motion
anci Lhe ciisv_ance covered bv it in a ceruain
tirne interval by a set of ations known as
the era 'Atlons of motion. For conventpnce, a
set oi unree such equauons are given below:
where u is the initial velocitv of the obiect
w] lich moves with utij tnrm acceleration ator
distance travelled by the 2t i- tin.e t
Eq. (8.5) describes the velo -tune rela tion
ann EA. t8.5J represenr= e •,osjrjon-tlme
retia Lion. Eq. (8.7), v lich. reoresenLS tile
rel the ,.JOSL veloci
irnir..ktinp t. 'aes t.nree. equauons can be
8.5.1 EQUA'i10N FOR VELOCITY-TIME
the velocity-time Qraoh of an obiect
moves unüer uniiui'ill accelerauon as
Time t,)
Fig. 8.8: graph to obtain the equations
Monon

Time
8:00 am
8:05 am
8:10 am
8:15 am
8:20 am
8:25 am
ana sama at
their bicycles
0.8
1.6
2.3
3.0
3.6
amerent times o
Distance
Distance
travelled
travelled
by Feroz
by Sania
(km)
(km)
1.0
1.9
2.8
3.6
where u is the initial velncitv of the
tittie t, V is the fif].11 s
travelled by t,ne i.
Eq. lb.5) deserwes the velo -tune re:
and Eq. (8.6) represent= e -.,osN.ion
re' anon t)eiureen the •on Ineve]
can he obtainea fro. = B, (S, 5) and
uenvea memod.
8.5.1 ION FOR VELOCITY-
the velocity-time graph of an
Inuves unaer unnorlll acceleraUl
Hot tne distance-time graph "or tneir
motions on the saule scale arid
uestions
l.
of the
graphsfor unconn
and non-un.Vorm rnotion of
What can you say
of an object
dismrzce-tfrne
une parallel
What can goo,
nution Qf

Scene 11

Substituting OA = u, OC = AD = t and BD = at,
we get

From the velocity-time graph shown in
Fig. 8.8, the distance s travelled by the object
in time t, moving under uniform acceleration
a is given by the area enclosed within the
trapezium OABC under the graph. That is,
s = area of the trapezium OABC

Substituting OA = u, BC = v and OC = t,
we get

From the velocity-time relation (Eq. 8.6),
we get

Example 8.5 A train starting from rest
attains a velocity of 72 km h–1 in
5 minutes. Assuming that the
acceleration is uniform, find (i) the
acceleration and (ii) the distance
travelled by the train for attaining this
velocity.

shown in Fig. 8.8 (similar to Fig. 8.6, but now
with u ≠ 0). From this graph, you can see that
initial velocity of the object is u (at point A)
and then it increases to v (at point B) in time
t. The velocity changes at a uniform rate a.
In Fig. 8.8, the perpendicular lines BC and
BE are drawn from point B on the time and
the velocity axes respectively, so that the
initial velocity is represented by OA, the final
velocity is represented by BC and the time
interval t is represented by OC. BD = BC –
CD, represents the change in velocity in time
interval t.
Let us draw AD parallel to OC. From the
graph, we observe that
BC = BD + DC = BD + OA
Substituting
BC = v and OA =  u,
we get
v
= BD + u
or
BD = v – u
(8.8)
From the velocity-time graph (Fig. 8.8),
the acceleration of the object is given by

Using Eqs. (8.8) and (8.9) we get
v = u + at

8.5.2 EQUATION FOR POSITION-TIME
RELATION

Let us consider that the object has travelled
a distance s in time t under uniform
acceleration a. In Fig. 8.8, the distance
travelled by the object is obtained by the area
enclosed within OABC under the velocity-time
graph AB.

Thus, the distance s travelled by the object
is given by

s = area OABC (which is a trapezium)
= area of the rectangle OADC + area of
the triangle ABD

shown in Fig, 8.8 (eirnilar to Fio. 8.6, but now
From ernnh, you can See 'hat
iruual velocitv of the oi_neet is u (at bornt A)
then it increases to v in tilne
t- The velocity at a rate a.
Ftff. 8.8, the pernentljcular lines BC and
BE, are drawn from pomt B on tne time and
the velocity axes resvectivelv, so that the
initial velueity is represc.nted by ON the filial
vevv•itv iS by BC and tne ne
inrerval t is renre.senred by OC. HD = BC •
CD. represents the change in velocity in time
Let us draw AD parallel to OC. From the
Subsl:lkutiug BC = v and OA :
(8.8)
or
From the velociLv-tixne graph (Fig. 8.8).
the acceleration of the oleieet is given by
AD OC
Substituting OC = t, we get
= OAX OC + (AD x BD)
Substituting OA = u, OC = AD = t and BD = at
2
2
8.5.3 EQUATION FOR POSITION-VELOCITY
From the velocitv-time graph shown in
FiQ. 8.8, the distance s travelk.d b: obiect
in time t, moving under unitorrz „-ration
a is given by the area enc10k v itmn tne
trapezium OAHU unner e •apn. 'Inat is,
2
Substituting
•rcnaa
2
the velocity-time relation (Eq. 8.6),
or
BD = at
Using Eqs. (8.8) and (8,9) we
8.5.2 EQUATION FOR POSITION-UME
Let us consider that the ohiec, has travelled
distance in timr„ t under uniiorm
acceleration a, In c.d, the distance
enmosea •tvnnm : unaer tne velocity-time
Thus, the distance s travelled by the object
= area OARC (which is a trapezium)
area of Lhe rectangle OÄDC + alea of
the
Using Eqs. (8.11) and (8.12) we have
attains a velocity of 72 km h-l in
5 minutes. Assuming that the
acceleration is unitnrm. find K) the
acceleratinn and (li) the djstanee
travelled by the train for attaining this
veiociuy
ScrmvCE

represents the change in velocity in time
Let us draw AD Darallel to OC. From the
SuosLILuting BC v and OA =
or
Fram the veloc1TV-time graph (F irk. 8.5).
acceleration of the ooject is given by
Chan r. Ln velocity
AD OC
bstituting OC = t, we get
BD = at
ing Eqs. (8.8) and (8,9) we
5.2 EQUATION FOR POSITION-UME
t us consider that the oblec, has travelled
under unitorm
(l*stance s in timr„
oeierarton a, In c.d, the distance
bv the oi is 'bl.ained bv che arca
closed wooin •AL n. I-unlel tile velocity-tnne
Thus, the distance s travelled by the object
8.5.3 EQUATION FOR POSITION-VELOCI
From the velocity-time grnnh shown
F 12.8.8, the disvance s Lraveueu
in Lime t, movinu under unhurr- --raLi
a is eiven bv the area enclu Eii.liin
trapezaum OABC under 'V e k •avn. Tl'tat i:
(8.
2
the velocity-time relation (Eq. 8.
(8.
Using Eqs. (8.11) and (8.12) we have
s=—
or
Lxalllnne 0.0 star L Lile mroru re:
atLåins velocity of 72 km h-l
5 minutes. Assuming that th

Scene 12

MOTION
109
Solution:
We have been given
u = 0 ; v = 72 km h–1 = 20 m s-1 and
t = 5 minutes = 300 s.
(i) From Eq. (8.5) we know that
(
)
v – u
a =
t
–1
–1
–2
20 m s
– 0 m s
=
300s
1
=
m s
15
(ii) From Eq. (8.7) we have
2 a s = v2 – u2 = v2 – 0
Thus,
2
–1 2
–2
=
2
(20 m s
)
=
2×(1/15) m s
v
s
a
= 3000 m
= 3 km
The acceleration of the train is 
1
15 m s– 2
and the distance travelled is 3 km.
Example 8.6 A car accelerates uniformly
from 18 km h
–1 to 36 km h
–1 in 5 s.
Calculate (i) the acceleration and (ii) the
distance covered by the car in that time.
Solution:
We are given that
u
=
18 km h–1  = 5 m s–1
v
=
36 km h–1  = 10 m s–1 and
t
=
5 s .
(i)   From Eq. (8.5) we have
v – u
a =
t
= 
-1
-1
10 m s  – 5 m s
5s
= 1 m s–2
(ii) From Eq. (8.6) we have
s = u t + 1
2
a t 2
= 5 m s–1 × 5 s + 
1
2 × 1 m s–2 × (5 s)2
= 25 m + 12.5 m
= 37.5 m
The acceleration of the car is 1 m s
–2
and the distance covered is 37.5 m.
Example 8.7 The brakes applied to a car
produce an acceleration of 6 m s-2 in
the opposite direction to the motion. If
the car takes 2 s to stop after the
application of  brakes, calculate the
distance it travels during this time.
Solution:
We have been given
a = – 6 m s–2  ; t = 2 s and v = 0 m s–1.
From Eq. (8.5) we know that
v = u + at
0 = u + (– 6 m s–2) × 2 s
      or u = 12 m s–1 .
From Eq. (8.6) we get
s = u t + 
1
2  a t 2
= (12 m s–1 ) × (2 s) + 
1
2   (–6 m s–2 ) (2 s)2
= 24 m – 12 m
= 12 m
Thus, the car will move 12 m before it
stops after the application of brakes.
Can you now appreciate why drivers
are cautioned to maintain some
distance between vehicles while
travelling on the road?
uestions
1.
A bus starting from rest moves
with a uniform acceleration of
0.1 m s-2  for 2 minutes. Find (a)
the speed acquired, (b) the
distance travelled.
Q
2020-21

Scene 13

2.
A train is travelling at a speed
of 90 km h–1. Brakes are applied
so as to produce a uniform
acceleration of – 0.5 m s-2. Find
how far the train will go before it
is brought to rest.
3.
A trolley, while going down an
inclined 
plane, 
has 
an
acceleration of 2 cm s-2. What will
be its velocity 3 s after the start?
4.
A racing car has a uniform
acceleration of 4 m s-2. What
distance will it cover in 10 s after
start?
5.
A stone is thrown in a vertically
upward direction with a velocity
of 5 m s-1. If the acceleration of the
stone during its motion is 10 m s–2

in the downward direction, what
will be the height attained by the
stone and how much time will it
take to reach there?

When the velocity of an object changes, we
say that the object is accelerating. The change
in the velocity could be due to change in its
magnitude or the direction of the motion or
both. Can you think of an example when an
object does not change its magnitude of
velocity but only its direction of motion?

Let us consider an example of the motion
of a body along a closed path. Fig 8.9 (a) shows
the path of an athlete along a rectangular
track ABCD. Let us assume that the athlete
runs at a uniform speed on the straight parts
AB, BC, CD and DA of the track. In order to
keep himself on track, he quickly changes
his speed at the corners. How many times
will the athlete have to change his direction
of motion, while he completes one round? It
is clear that to move in a rectangular track
once, he has to change his direction of motion
four times.
Now, suppose instead of a rectangular
track, the athlete is running along a
hexagonal shaped path ABCDEF, as shown
in Fig. 8.9(b). In this situation, the athlete
will have to change his direction six times
while he  completes one round. What if the
track was not a hexagon but a regular
octagon, with eight equal sides as shown by
ABCDEFGH in Fig. 8.9(c)?  It is observed that
as the number of sides of the track increases
the athelete has to take turns more and more
often. What would happen to the shape of
the track as we go on increasing the number
of sides indefinitely? If you do this you will
notice that the shape of the track approaches
the shape of a circle and the length of each of
the sides will decrease to a point. If the athlete
moves with a velocity of constant magnitude
along the circular path, the only change in
his velocity is due to the change in the
direction of motion. The motion of the athlete
moving along a circular path is, therefore, an
example of an accelerated motion.
We know  that the circumference of a circle
of radius r is given by π
2 r . If the athlete takes
t seconds to go once around the circular path
of radius r, the speed v is given by

When an object moves in a circular path
with uniform speed, its motion is called
uniform circular motion.

(a) Rectangular track
(b) Hexagonal track

(d) A circular track
(c) Octagonal shaped track

Fig. 8.9: The motion of an athlete along closed tracks
of different shapes.

3.
4.
gféJfJ
dcumwtx•d
t,'uz
When the velarltv of an ohiect chanoes,
Sav [bat (_n_nect is aceelejaLi1iQ. Tue. eir _1iQe
in the velocity coul'] be d' [e to cha its
or the nn•ecnnn of the n or
bolil. Can VOU think Of an e_Xd111Dle an
olneet does not its de of
velocity but ony its directiC,) .uouon?
(a) Rectangular track
(c) Octagonal shaped track (d] A circular track
Fig. 8.9: mnHon Qfan athlete along closed tracks
OJ dwuent shapes.
Let us consider an example of the motion
ofa body along a cloaed path. Fig 8.9 (a) snows
the path ot an athlete a]nng a rectancfl'lar
track Let us assnme tnat T.ne aTn•ece
on the straiont parts
AB. BC. Ci") and LIA or the trar•k. In orner to
keep himself on track, he qurc:K1v chanpes
his sneed at the corners. How mariv lirnes
wili t-he atolete have to chanQe his direcuon
of motion. while he conn.JieLes one round? It
is clear that to move in a rectaneular track
once, he has to change his direction of motion
Now, suppose instead of a rectangular
track, the athlete is running •long a
hexagonal shaped path as hown
in 8.9(b). In this situatic A, thc ath'ete
will have to cha_nge his direc or six times
wnlle he competes one noun. Wnat ir me
traci was not _apm. but
reQ"lar
with eirlht 'ides as shown by
jn R.S- It js observed that
as "he. numbe - s Of the track increases
thr• atheneLe : 'as ._aKe turns more and more
Aten. •tld haouen to the shaDe of
the tra . S we eo on uncreasinQ dle nulilber
of in .1eiii1iLclv? If vou do diis vou will
nc -j' .e tha the shave of the track av['roaches
he of a circle and the lenr:2U1 of each of
tk. sides will decrease to a point. If the ai hlete
moves with a velocity ot constant 103 gnitude
along the circular path, the only change in
his velocity is due to the change in the
direction ot motion. me morjon ot tne athlete
movmg along a circular parh is, thererore, an
examnl.e ot an accelerated motion.
We know that tne ctrcumTerence of a circle
of radius ris given by 21tr . It the athlete takes
t seconds to 20 once around the circular path
of radius r, the speed v is given by
21tr
When an object moves in a circular path
with nmrorm speed. ics motion is caned
umiorrn circular motion.
ScrmvCE

dcumwtx•d
t,'uz
'len the veloe!ty of an chenges,
the is Cn-
Ille Velocii,v COULU be due to ii IKS
or the direction of the n or
tn. Can you 01 an an
leet does not cnanoe. its mavr•irs de Of
.ueity but only its dueetic,) . uution?
(a) Rectangular track
is tl'»at to rrl.nve in a
once, he. has to change his direcuon of moti
four times.
Now, suppose instead of a rectangu
track, the athlete is running • 1011*
hexagonal shaped path ABCDER as ho
in F 8.90)). In this situatic A, thL athl
will have to change his direr; or six tin
while he complelcs one nun. What if t
tracl was not a he: -arm. but a reemx
with eifht sides as shown
. in Fig. R.SO? It is observed tt
as •ne numm - or the track increas
thr• atnelete : •as 'n turns more and
,gren. wnet uld nappen to the snane
the tra . •s we go on Increasing rhe numi
of sine. in If vou do cnis you
nc -ire the shane. the, track ar»proael-
'he B: lane of a circle and lhe lern of eacr.
tk. Slues will decrease LO a Donn. 11 Lhe athil
moves with a veloeiuv of constant maQ1iiLu
along the circular path, the Oidv chanüc::
his velocity is due to the change in
direction of motion. me motion of tne athl
moving along a circular path is, therefore,
exam Vle of an accelerated motion.
Wc. know that the circumference of a cir
of radius ris given by 21tr . If ihe athlete
t seconds to once around the circular
of radius r, the speed v is given by
21tr
(8.

Scene 14

MOTION
111
If you carefully note, on being released
the stone moves along a straight line
tangential to the circular path. This is
because once the stone is released, it
continues to move along the direction it has
been moving at that instant. This shows that
the direction of motion changed at every point
when the stone was moving along the circular
path.
When an athlete throws a hammer or a
discus in a sports meet, he/she holds the
hammer or the discus in his/her hand and
gives it a circular motion by rotating his/her
own body. Once released in the desired
direction, the hammer or discus moves in the
direction in which it was moving at the time
it was released, just like the piece of stone in
the activity described above. There are many
more familiar examples of objects moving
under uniform circular motion, such as the
motion of the moon and the earth, a satellite
in a circular orbit around the earth, a cyclist
on a circular track at constant speed
and so on.
Activity _____________8.11
•
Take a piece of thread and tie a small
piece of stone at one of its ends. Move
the stone to describe a circular path
with constant speed by holding the
thread at the other end, as shown in
Fig. 8.10.
Fig. 8.10: A stone describing a circular path with
a velocity of constant magnitude.
•
Now, let the stone go by releasing the
thread.
•
Can you tell the direction in which the
stone moves after it is released?
•
By repeating the activity for a few times
and releasing the stone at different
positions of the circular path, check
whether the direction in which the
stone moves remains the same or not.
What
you have
learnt
•
Motion is a change of position; it can be described in terms of
the distance moved or the displacement.
•
The motion of an object could be uniform or non-uniform
depending on whether its velocity is constant or changing.
•
The speed of an object is the distance covered per unit time,
and velocity is the displacement per unit time.
•
The acceleration of an object is the change in velocity per unit
time.
•
Uniform and non-uniform motions of objects can be shown
through graphs.
•
The motion of an object moving at uniform acceleration can be
described with the help of the following equations, namely
v = u + at
s = ut + ½ at2
2as = v2 – u2
2020-21

Scene 15

SCIENCE
112
where u is initial velocity of the object, which moves with
uniform acceleration a for time t, v is its final velocity and s is
the distance it travelled in time t.
•
If an object moves in a circular path with uniform speed, its
motion is called uniform circular motion.
Exercises
1.
An athlete completes one round of a circular track of diameter
200 m in 40 s. What will be the distance covered and the
displacement at the end of 2 minutes 20 s?
2.
Joseph jogs from one end A to the other end B of a straight
300 m road in 2 minutes 30 seconds and then turns around
and jogs 100 m back to point C in another 1 minute. What are
Joseph’s average speeds and velocities in jogging (a) from A to
B and (b) from A to C?
3.
Abdul, while driving to school, computes the average speed for
his trip to be 20 km h–1. On his return trip along the same
route, there is less traffic and the average speed is
30 km h–1. What is the average speed for Abdul’s trip?
4.
A motorboat starting from rest on a lake accelerates in a straight
line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the
boat travel during this time?
5.
A driver of a car travelling at 52 km h–1 applies the brakes and
accelerates uniformly in the opposite direction. The car stops
in 5 s. Another driver going at 3 km h–1 in another car applies
his brakes slowly and stops in 10 s. On the same graph paper,
plot the speed versus time graphs for the two cars. Which of
the two cars travelled farther after the brakes were applied?
6.
Fig 8.11 shows the distance-time graph of three objects A,B
and C. Study the graph and  answer the following questions:
Fig. 8.11
2020-21

Scene 16

(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
7.
A ball is gently dropped from a height of 20 m. If its velocity
increases uniformly at the rate of 10 m s-2, with what velocity
will it strike the ground? After what time will it strike the
ground?
8.
The speed-time graph for a car is shown is Fig. 8.12.

(a) Find how far does the car travel in the first 4 seconds.
Shade the area on the graph that represents the distance
travelled by the car during the period.
(b) Which part of the graph represents uniform motion of
the car?
9.
State which of the following situations are possible and give
an example for each of these:
(a) an object with a constant acceleration but with zero velocity
(b) an object moving with an acceleration but with uniform
speed.
(c) an object moving in a certain direction with an
acceleration in the perpendicular direction.
10.
An artificial satellite is moving in a circular orbit of radius
42250 km. Calculate its speed if it takes 24 hours to revolve
around the earth.

(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
A ball is gently dropped from a of 20 m. If its velocity
increases urc.r.ormly at the rate 01 10 m so, with what velocity
it strike tne ground? Alter what tlrne Wlil It strike tne
'Ihe speed-time graph for a car is shown is Fig. 8.12.
8.
9.
10
Monon
8-
g
4
o
2
4
6
(a) Find how far hes car ir (he first 4 seennds.
Shade Lhe +he erauh th. •t I epresenLs the dL5L_ance
travelled -oy •ar dur1LIC Lhe period.
(b) Which pav ot the graph represents uniform motion of
State. of the fob Wii g situations are possible and give
(a) an object wit a acceleration but with zero velocity
(bj an object mov.ng with an acceleration but with uniform
(c) moving in certain direction with an
ace at-ion in the perpendicular direction.
qrtfffcfal satellite is moving in a circular orbit of radius
km. Calculate its speed if it takes 24 hours to revolve
around the ear

8
g
4.
4
6
8
10
9.
10.
Snarle the ar, Ohe szraph th •t epresents the dis
travened -oy •ar dunne the period.
(b) Which pav or the graph represents uniform motl
State of the g situations are possible axu
(a) an ubiect a acceleration but with zero ve
(b) an Obiect mov.ng vnth an acceleration but with
(c) moving in certain direction wit:
ace ation in the perpendicular direction.
artifieial satellite is moving tn a circular orbit of
J km. Catemuate its speed if it takes 24 hours to re
around Lhe earth.

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