ppt on motion class 9th

2. Introduction
• We use the word ‘rest’ very often. For example, when someone
is doing no work or lying on the bed, we often say that the
person is resting. This means that the person is not moving.
Scientifically as well, the word ‘rest’ has a similar meaning.
• Scientifically, we say an object is at rest when the position of
the object does not change with time, with respect to its
surroundings.
• Similarly, motion is defined as the change of position of an
object with time, with respect to its surroundings.

3. Acceleration
• Acceleration is definedas the rate of change of velocity.
It is a vector quantity and its direction is given by the direction of the
force causing the acceleration. Mathematically, acceleration is given
as: Accelerations= Change in velocity
• Suppose the velocity of a car is u at time t1. Later, at time t2, its
velocity becomes v.
• Change in velocity = (v − u), time interval = t2 − t1

4. Uniform and Non Uniform
Acceleration
• UniformAcceleration- If the rate of change of velocity remains
constant, then the acceleration is uniform. Examples of uniform
acceleration include a ball under free fall, a ball rolling on an
inclined plane and a car accelerating on a straight, traffic-free road.
Non-UniformAcceleration- If the rate of change of velocity changes
with time, then the acceleration is non-uniform. An example of non-
uniform acceleration is a car accelerating on a straight road with
traffic.

5. Circular motion
• A body is said to be in circular motion when it rotates about a fix
point.
• In circular motion, the velocity can never be constant, but the speed
of the moving body can be constant.
• A body moving in a circular path at a constant speed is said to be in
uniformcircular motion.
• In Uniform Circular motion, the object in one revolution moves 2 r in
T seconds

6. Uniform Circular Motion
Uniform circular motion is the
motion of an object traveling
at a constant (uniform) speed
on a circular path.
Period T is the time required
to travel once around the
circle, that is, to make one
complete revolution.
T
r
v
2


7. Example : A Tire-Balancing Machine
The wheel of a car has a radius of r = 0.29m and is
being rotated at 830 revolutions per minute (rpm) on
a tire-balancing machine. Determine the speed (in
m/s) at which the outer edge of the wheel is moving.
The speed v can be obtained directly from ,
but first the period T is needed. It must be expressed
in seconds.
T
r
v
2


8. 830 revolutions in one minute
T=1.2*10-3 min, which corresponds to 0.072s
sm
s
m
T
r
v /25
072.0
)29.0(22



9. Uniform circular motion emphasizes that
1. The speed, or the magnitude of the velocity
vector, is constant.
2. Direction of the vector is not constant.
3. Change in direction, means acceleration
4. “Centripetal acceleration” , it points toward the
center of the circle.

10. 5) Graphical representation of motion :-
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)
20 – 10 10
= --------- = ----
10 – 5 5
= 2 ms -1
A
B
10
20
30
t1 t2
s1
s2
C
Time (s)
Distance(m)
X
Y
5 10 15 20
Distance – time graph for a body moving with uniform speed
0

11. II) THE DISTANCE – TIME GRAPH FOR NON
UNIFORM MOTION IS NON LINEAR. THIS IS
BECAUSE IN NON UNIFORM SPEED A BODY
TRAVELS UNEQUAL DISTANCES IN EQUAL
INTERVALS OF TIME.
20
40
Time (s)
Distance(m)
X
10
30
50 10 15 20
Distance – time graph for a body moving with non uniform speed
Y

12. T H E C H A N G E I N T H E V E L O C I T Y O F A B O D Y W I T H T I M E C A N B E R E P R E S E N T E D O N T H E V E L O C I T Y
T I M E G R A P H . I N T H I S G R A P H V E L O C I T Y I S T A K E N O N T H E Y – A X I S A N D T I M E I S T A K E N O N T H E X
– A X I S .
I ) I F A B O D Y M O V E S W I T H U N I F O R M V E L O C I T Y , T H E G R A P H W I L L B E A S T R A I G H T L I N E P A R A L L E L
T O T H E X – A X I S . T H I S I S B E C A U S E T H E V E L O C I T Y D O E S N O T C H A N G E W I T H T I M E .
T O D E T E R M I N E T H E D I S T A N C E T R A V E L L E D B Y T H E B O D Y B E T W E E N T H E P O I N T S A A N D B
W I T H V E L O CI T Y 2 0 K M H - 1
V = ---
S = V X T
V = 20 KM H-1 = AC OR BD
T = T2 – T1 = DC
= AC (T2 – T1)
S = AC X CD
S = AREA OF THE RECTANGLE ABDC
b) Velocity– time graphs :-
20
40
Time (s)
Velocity(kmh-1)
X
10
30
50 10 15 20
t1 t2
A B
C D
Velocity – time graph for a body moving with uniform velocity
Y

13. II) IF A BODY WHOSE VELOCITY IS INCREASING WITH TIME, THE GRAPH IS A
STRAIGHT LINE HAVING AN INCREASING SLOPE. THIS IS BECAUSE THE VE LOCITY
INCREASES BY EQUAL AMOUNTS WITH EQUAL INTERVALS OF TIME.
THE AREA UNDER THE VELOCITY – TIME GRAPH IS THE DISTANCE (MA GNITUDE
OF DISPLACEMENT) OF THE BODY.
THE DISTANCE TRAVELLED BY A BODY BETWEEN THE POINTS A AND E IS THE
AREA ABCDE UNDER THE VELOCITY – TIME GRAPH.
S = AREA ABCDE
= AREA OF RECTANGLE ABCD
+ AREA OF TRIANGLE ADE
S = AB X BC + --- ( AD X DE )
A
B
10
20
30
t1 t2
C
Time (s)
Velocity(ms-1)
X
Y
10 20 30 40
Velocity – time graph for a body moving with uniform acceleration
D
E
0

14. III) IF A BODY WHOSE VELOCITY IS DECREASING WITH TIME,
THE GRAPH IS A STRAIGHT LINE HAVING AN DECREASING
SLOPE. THIS IS BECAUSE THE VELOCITY DECREASES BY
EQUAL AMOUNTS WITH EQUAL INTERVALS OF TIME.
IV) IF A BODY WHOSE VELOCITY IS NON UNIFORM, THE
GRAPH SHOWS DIFFERENT VARIATIONS. THIS IS BECAUSE
THE VELOCITY CHANGES BY UNEQUAL AMOUNTS IN EQUAL
INTERVALS OF TIME.
20
40
Time (s)
Velocity(ms-1)
X
10
30
5
0
10 15 20
20
40
Time (s)
Velocity(ms-1)
X
10
30
50 10 15 20
Velocity – time graph for a uniformly
decelerated motion
Velocity – time graph for
non uniform acceleration
Y
Y

15.  To Derivate- v=u+at
Initial velocity[u]= AO = CD
Final velocity[v]= BC
Time[t]= OC = AD
As BC=BD+CD
v=BD+u
BD= v-u------------- [i]
As AB = slope of acceleration
So, AB = BC-CD
AD

16. AB= BD
OC
a = BD
t
BD = at
Put the value of BD in [i],
we get,
at = v-u
v = u+at Hence, derivated..........

17. Let s = displacement in time [t]
Area of OABC = Area of OADC + Area of ABD
Area of OADC = OA * OD
Area of ABD = ½ * (AD * BD)
Area of OABC = [OA * OD] [½ * (AD * BD)]

18. Substituting
OA=v
AD = OC = t
From the 1st equation, BD = at
S = u*t + ½ (t*at)
S = ut + ½ at
Hence, derived............

19. S = area of trapezium OABC
Area = ½ [OA+BC] OC
Substituting OA = u
BC = v and OC = t
By the 1st eqn. of motion
v = u+at
v-u = a
t

20. S = ½ [u+v] * v-u
a
S = [v+u] [v-u]
2a
2as = v - u
Hence, derived.........

21. C