The document covers the concept of motion, detailing types, descriptions, and related terms. It explains distance and displacement, the difference between uniform and non-uniform motion, and introduces scalar and vector quantities. Additionally, it discusses equations of motion and uniform circular motion, providing exercises for reinforcement.

1. MOTION
SUBTOPICS : - 1)TYPES OF MOTION
2) CARTESIAN PLANE
3) DESCRIBING MOTION
4) REST AND MOTION
5) RELATED TERMS
6) SCALAR AND VECTOR
READ NCERT PAGE 98 AND 99 FOR RELATED TOPICS

2. RECAPITULATION

3. CARTESIAN PLANE
+ve y-axis
+ve x-axis
-ve x-axis
-ve y-axis

4. .
DESCRIBING MOTION :- Motion means movement
The motion of an object is perceived when its position changes continuously with respect to some
stationary object .
It is the change in position of a body with time. Motion can be described in terms of the distance
moved or the displacement. ( displacement will be discussed in detail , later in the chapter)
The state of rest and motion are relative to each other
An object may appear to be moving to one person and the same object may appear to be at rest to
another person . For example, when a bus is moving on a road
• the trees appear to move backwards to the passengers sitting inside the bus .
• the person sitting inside the bus finds his fellow passengers at rest.

5. SOME RELATED TERMS : -
• Position : It represents the location of an object .
• Reference point : It is the location from where the position of an object is measured. The reference point is called
the origin .
• For example ,
• Here , kosma’s house is the reference point or the origin and location of shop( object) is specified with respect to
the house .
• MAGNITUDE : The numerical value of a physical quantity is its magnitude .
• DIRECTION : To locate the position of an object, along with the magnitude , its sense of direction is also required .
Example in above case shop is 300m along positive x-axis ( right hand side) from the origin and school is 300m
along negative x axis( left hand side ) from the origin

6. SCALAR AND VECTOR
QUANTITIES
•

7. VIDEO FOR REFERENCE ( WEEK 1 )
https://youtu.be/bYgfwuJnpgU

8. • SUBTOPICS : 1) DISTANCE ( NCERT PAGE 99)
2) DISPLACEMENT ( NCERT PAGE 99)
3) UNIFORM AND NON UNIFORM MOTION ( NCERT PAGE 100)

9. DISTANCE TRAVELLED :- IS THE ACTUAL LENGTH OF THE PATH TRAVELLED BY
AN OBJECT DURING ITS MOTION .
- IT IS A SCALAR QUANTITY ..
e.g, in this case if an object travels
from point O to C , distance travelled
by it will be 25m . Here , O is the
reference point.
During motion of the object , the value of distance can never be zero or negative .
MOTION IN A STRAIGHT LINE

10. DISPLACEMENT
• The displacement of an object in motion is the shortest distance between the initial position ( starting point of the
journey) and the final position ( end point) of the object.
• Displacement does not depend upon the actual path followed by the object.
• It is a vector quantity .
• When final position coincides with initial position , displacement = 0 but distance travelled is not zero.
• The displacement can be positive , zero or negative .
• e.g if a boy starts his journey from his home and first he goes to book shop
to buy a book and then he goes to school then ,
Distance travelled by the boy = 5km + 8km
= 13 km
Displacement = shortest distance between home (initial position) and
school (final position) i.e. hypotenuse of right angle triangle
using Pythagoras theorem,
Displacement = √52
+82
= √25+64
= √89 km
BOOK SHOP
HOME
SCHOOL

11. FEW IMPORTANT POINTS
Both , the distance and displacement are measured in metre or cm or km.
Between two given positions , distance travelled can never be less than the displacement.
Along a straight line motion distance travelled is equal to magnitude of displacement .
VIDEO LINK : https://unacademy.com/lesson/distance-and-displacement/5RZPDBHT

12. TEST YOUR
KNOWLEDGE
S.no Rounds Displacement Distance
1 1
2 1.5
3 2
4 2.5
Q1) A particle is moving in a circle of diameter 200 m . Fill the table given below
Q2) A ball thrown vertically upwards reaches a maximum height h. It then returns to ground. Calculate the distance
travelled and the displacement .
Q3) Vector quantities need both a magnitude and a _____.
Q4) How far an object has moved in comparison to where it began is called _______________________.
Q5) A car drives 84 meters in forward direction. It's displacement and distance would be the same. ( TRUE/FALSE)

13. UNIFORM MOTION AND NON-
UNIFORM MOTION
UNIFORM MOTION : When an object travels equal distances in equal intervals of time, howsoever small the interval
may be , the motion of object is said to be uniform .For example, movement of hands of a watch.
1ST
SEC 2ND
SEC 3RD
SEC
Here, the yellow car travels 50m in each
second i.e. it is in uniform motion .
NON UNIFORM MOTION : When an object travels unequal distances in equal intervals of time , howsoever small the interval
may be , the motion of object is said to be non- uniform
Here, the red car travels
50m in 1st
sec
100m in 2nd
sec
150m in 3rd
sec
50m in 4th
sec i.e. unequal distance in equal intervals
of time. Hence, car is in non – uniform motion.
https://youtu.be/qRR_1Gj6Kzw

14. WORK TO DO
1) Read pages 99 and 100 of NCERT thoroughly.
2) View the ppt and record the notes in your physics register .
3) A Youtube video link has been given along with each topic for concept clarification .
4) Solve ‘ Test your knowledge ‘ in your physics register .
5) Solve NCERT intext questions given on page no. 100 in your register.
.

15. e.g. a car running at a constant speed of 10km/h i.e. e.g. a freely falling object
It covers 10 km every hour.
RECAPITULATI
ON

16. UNIFORM AND NON-UNIFORM MOTION
SPEED :- OF A BODY IS THE DISTANCE TRAVELLED BY THE BODY PER UNIT TIME.
DISTANCE
SPEED =
TIME
IF A BODY TRAVELS A DISTANCE ‘S‘ IN TIME ‘T ‘ THEN ITS SPEED ‘V’ IS
S
V =
T
THE SI UNIT OF SPEED IS METER PER SECOND M/S OR MS -1
SINCE SPEED HAS ONLY MAGNITUDE IT IS A SCALAR QUANTITY.
AVERAGE SPEED :- IS THE RATIO OF THE TOTAL DISTANCE TRAVELLED TO THE TOTAL
TIME TAKEN .
TOTAL DISTANCE TRAVELLED
AVERAGE SPEED =
TOTAL TIME TAKEN

18. SPEED WITH DIRECTION
THE RATE OF MOTION OF A BODY IS MORE MEANINGFUL IF WE SPECIFY ITS DIRECTION OF MOTION
ALONG WITH SPEED. THE QUANTITY WHICH SPECIFIES BOTH THE DIRECTION OF MOTION AND SPEED IS
VELOCITY.
I) VELOCITY :- OF A BODY IS THE DISPLACEMENT OF THE BODY PER UNIT TIME.
DISPLACEMENT
VELOCITY =
TIME TAKEN
SINCE VELOCITY HAS BOTH MAGNITUDE AND DIRECTION, IT IS A VECTOR QUANTITY.
II) AVERAGE VELOCITY :- IS THE RATIO OF THE TOTAL DISPLACEMENT TO THE TOTAL
TIME TAKEN.
TOTAL DISPLACEMENT
AVERAGE VELOCITY =
TOTAL TIME TAKEN
AVERAGE VELOCITY IS ALSO THE MEAN OF THE INITIAL VELOCITY ‘U’ AND FINAL VELOCITY ‘V’.
INITIAL VELOCITY + FINAL VELOCITY U + V
AVERAGE VELOCITY = VAV =
2 2
SPEED AND VELOCITY HAVE THE SAME UNITS M/S OR MS -1

19. RATE OF CHANGE OF VELOCITY
During uniform motion of a body in a straight line the velocity remains constant with time. In this case the change in
velocity at any time interval is zero ( no change in velocity).
During non uniform motion the velocity changes with time. In this case the change in velocity at any time interval is not
zero. It may be positive (+ ve) or negative (- ve).
The quantity which specifies rate of change in velocity is acceleration.
Acceleration :- is the change in velocity of a body per unit time.( Or the rate of change of velocity.)
Change in velocity
Acceleration = time
If the velocity of a body changes from initial value ‘u’ to final value ‘v’ in time ‘t’, then acceleration ‘a' is
v - u
a = t
The SI unit of acceleration is ms - 2
Uniform acceleration :- if the change in velocity is equal in equal intervals of time it is uniform acceleration.
Non uniform acceleration :- if the change in velocity is unequal in equal intervals of time it is non uniform acceleration

20. a) Distance – Time graphs :-
The change in the position of a body with time can be represented on the distance time graph. In this graph distance is taken
on the y – axis and time is taken on the x – axis.
i) The distance time graph for uniform speed is a straight line ( linear ). This is because in uniform speed a body travels equal
distances in equal intervals of time.
We can determine the speed of the body from the distance – time graph.
For the speed of the body between the points A and B, distance is (s2 – s1) and time is (t2 – t1).
s (s2 – s1)
v = ---- v = -----------
t (t2 – t1)
200 – 100 100
= --------- = ----
4 – 2 2
= 50 ms -1
(S)
(M)
A
B

21. • Subtopics : 1) Distance time graphs
2) Velocity time graphs
(NCERT pages 104 to 107)

22. Graphical representation of motion
Distance – Time graphs :-
The change in the position of a body with time can be represented on the distance time graph. In this graph distance is taken on the
y – axis and time is taken on the x – axis.
i) The distance time graph for uniform speed is a straight line ( linear ). This is because in uniform speed a body travels equal
distances in equal intervals of time.
We can determine the speed of the body from the distance – time graph.
For the speed of the body between the points A and B, distance is (s2 – s1) and time is (t2 – t1).
here, A (t1 , s1 )
s (s2 – s1) or
v = ---- v = ----------- A(2,100) and
t (t2 – t1) B(t2 , s2 )
OR
200 – 100 100 B(4,200)
= --------- = ----
4-2 2
= 50 ms -1
Hence, slope of distance – time graph gives speed.
(s)
(m)
A
B

23. VELOCITY – TIME GRAPHS

24. WHEN BODY IS MOVING WITH UNIFORM ACCELERATION

25. • Velocity – time graph is also used to calculate the acceleration of the body
• Slope of velocity time graph gives the acceleration .
• In graph ,

27. VELOCITY –TIME GRAPH SUMMARY

28. LINK FOR DISTANCE TIME GRAPH
https://youtu.be/aemCJKt1fG0

29. LINK FOR VELOCITY – TIME GRAPH
https://youtu.be/FFGj6ODbeA8

30. TEST YOUR KNOWLEDGE
• Q1) What does area under velocity time graph represent?
• Q2) A body has zero acceleration . What will be the nature of distance time graph ?
• Q3) From the given v-t graph, it can be inferred that the object is
a) moving with uniform acceleration c) at rest
b) in uniform motion d) non- uniform motion
Q4) The speed - time graph of a car is given . Using the data in the graph
calculate the total distance covered by the car.
a) 1250 m c) 850 m
b) 1500 m d) 875m
• Q5) A car of mass 1000 kg is moving with a velocity of 10 m/s. If the velocity-time graph for this car is a horizontal line parallel to
the time axis, then the velocity of the car at the end of 25 s will be:
(a) 40 m/s (C) 10 m/s
(b) 25 m/s (d) 250 m/s

31. WORK TO DO
• 1) Read pages 104 to 107 of NCERT.
• 2) Complete notes in physics registers .Draw all the graphs given in ppt .( with pencil n scale )
• 3) Attempt questions given on page 107 in NCERT.
• 4) Solve ‘TEST YOUR KNOWLEDGE’ IN PHYSICS REGISTERS.
.

32. MOTION
• SUBTOPICS :- EQUATIONS OF MOTION ( NCERT PAGE 107-109)
UNIFORM CIRCULAR MOTION (NCERT PAGE 110, 111)

33. EQUATIONS OF MOTIONS BY GRAPHICAL
METHOD :-
The motion of a body moving along a straight line with uniform acceleration can be
described with the help of three equations called equations of motion.
The equations of motion are :-
i) v = u + at
ii) s = ut + ½ at2
Iii) 2as = v2
– u2
where u - is the initial velocity
v - is the final velocity
a - is acceleration
t - is the time
s - is the distance travelled by the object in time ‘t’.

34. A) EQUATION FOR VELOCITY – TIME RELATION ( v = u + at ) :-
• Consider a velocity – time graph for a body moving with uniform acceleration ‘a’. The initial velocity is u (at
A )and final velocity is v at (B) in time t. The velocity changes at uniform rate ‘a’.
• Perpendicular lines BC and BE are drawn from point B to the time and velocity axes respectively, so that the
initial velocity is OA and final velocity is BC and time interval is OC. Draw AD parallel to OC.
• Here , BD = BC – CD i.e. change in velocity in time ‘t’ .
• We observe that
• OC = t
BC = BD + DC = BD + OA
Substituting BC = v and OA = u
We get v = BD + u
or BD = v - u
change in velocity
Acceleration = ---------------------------
time
BD BD v - u
a = ----- = ----- or a = ---------
AD OC t
v – u = at or v = u + at

35. B) EQUATION FOR POSITION – TIME RELATION (s = ut + ½ at2
) :-
Consider a velocity – time graph for a body moving with uniform acceleration ‘a’ travelled a distance s in time t.
The distance travelled by the body between the points A and B is the area OABC under vel – time graph AB
i.e. S = area OABC ( which is a trapezium )
= area of rectangle OADC + area of triangle ABD
1
= OA X OC + --- ( AD X BD )
2
substituting OA = u, OC = AD = t,
BD = v – u = at
we get ,
1
s = u x t + -- ( t x at )
2
or s = ut + ½ at2

36. C) EQUATION FOR POSITION – VELOCITY RELATION (2as = v2
–u2
) :-
Consider a velocity – time graph for a body moving with uniform acceleration ‘a’ travelled a distance s in
time t.
The distance travelled by the body between the points A and B is the area OABC.
S = area of trapezium OABC
(OA + BC) X OC
S = ----------------------
2
SUBSTITUTING OA = u, BC = v AND OC = t
( u + v ) x t
we get s = -----------------
2
From velocity – time relation
( v – u )
t = -----------
a
( v + u ) x ( v – u )
s = ----------------------- or 2as = v2
– u 2
2a

37. UNIFORM CIRCULAR MOTION
The motion of a body in a circular path is called circular motion.
UNIFORM CIRCULAR MOTION :- If a body moves in a circular path with uniform speed, its motion is
called uniform circular motion.
Uniform circular motion is accelerated motion because in a circular motion a body continuously
changes its direction.
The circumference of a circle of radius r is given by 2лr. If a body takes time t to go once around the
circular path, then the speed v is given by
2лr
V = ----
t

38. TEST YOUR KNOWLEDGE
• Q1) A car increases its speed from 20km/h to 50km/h in 10 seconds. What is its acceleration ?
• Q2) A racing car has uniform acceleration of 4m/s². What distance will it cover in 10 seconds after
start ?
• Q3) A train starting from rest moves with a uniform acceleration of 0.2m/s² for 5 minutes. Calculate the speed
acquired and distance travelled in this time .
• Q4) A cyclist goes once round a circular track of diameter 105 m in 5 minutes. Calculate his speed .
• Q5) A bus was moving with a speed of 54km/h. On applying brakes, it stopped in 8 seconds. Calculate the
acceleration and the distance travelled before stopping .

39. WORK TO DO
COMPLETE NOTES IN PHYSICS REGISTERS .
• DO PRACTICE OF DERIVATION OF EQUATIONS OF MOTION USING GRAPH METHOD .
• SOLVE ‘ TEST YOUR KNOWLEDGE ‘ IN PHYSICS REGISTERS .
• READ NCERT PAGES 107- 111.
• TRY TO ATTEMPT NCERT EXERCISE .