Tips for IX - Science - Moving Objects by Ednexa

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Moving Objects
Motion
A body is said to be in motion if it changes its position
with respect to its surroundings and is said to be at
rest if it does not change its position with respect to
its surroundings.
Distance
It is the actual path followed by a body between the
points in which it moves.
Displacement
Displacement is the shortest distance between the
initial and final points of moment.
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Speed
It is related to distance while velocity is related to
displacement
Speed =
Total distance travelled
Total time taken
Velocity =
Displacement
Time Velocity related to speed and direction
Change in speed and Direction is same
1. Change in direction and Speed constant
2. Change in Bothe speed and direction
Unit S.I. :- m/s C.G.S. – Cm/s
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Motion
The motion in which the object covers equal distance
in equal interval of time is known as uniform motion.
The motion in which the object covers unequal
distance in equal interval of time is known as nonuniform motion
Acceleration
The rate of change of velocity is known as time
A = Change in velocity / time = v - u /t
Distance
–
Time
Graphs
Uniformly Accelerated
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for
Uniform
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Velocity time graph for uniform motion
Velocity time graph for uniformly accelerated
motion
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Equations of motions by graphical method
Motion of an object was studied by Newton and then
it was summarized in a set of equations of motion.
The equations analyse rectilinear motion of uniformly
accelerated body. The position of such a body can be
well predicted with the help of a set of three
equations, called kinematical equations.
A body is moving along a straight line with initial
velocity ‘u’. After time interval ‘t’ it attains final velocity
‘v’ due to an acceleration ‘a’. Then the set of three
equations is given as:
v = u + at _________ (1)
It represents velocity time relation.
1 2
s = ut + at _________ (2)
2
It describes position time relation.
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v2 = u2 + 2as _________ (3)
It represents relation between position and velocity.
These equations can be derived by graphical
method. (As shown in fig.)
1. Equation for velocity - time relation.
An object having non zero
initial velocity u starts from
point P. It’s velocity goes
on increasing with
respective to time and
becomes v when it reaches
point Q. The change in velocity is at uniform rate.
change in velocity
a=
time
a=
QP
t
∴ QP = at _______ (a)
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To find change in velocity, draw PR parallel to OS.
QS = QR + RS
= QR + PT
Put QS = v, PT = u, and QR = at, we get.
v = at + u
v = u + at
The velocity - time relation is obtained here. This is
the first kinematical equation.
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2. Equation for position - time relation.
Let distance travelled by the
object be ‘s’ in the time
interval
‘t’
under
uniform
acceleration ‘a’. In figure, the
distance travelled is given by
area enclosed within TPQS.
s = Area of quadrilateral TPQS.
s = area of rectangle PRST + area of triangle PRQ.
s = (TS × TP) +
1
(PR × QR)
2
But TP = u, TS = PR = t and QR = at
s = ut +
1
a t2
2
This is the second kinematical equation.
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3. Equation for position Velocity relation:
The distance‘s’ travelled by
object
in
time
‘t’
under
uniform acceleration ‘a’ is
Given by area enclosed
within quadrilateral TPQS.
s = Area of trapezium TPQS.
s=
1
(QS + TP) × TS
2
Put TP = u, QS = v and OS = t
∴s=
1
(v + u) × t
2
__________(a)
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From velocity time relation we have
a=
v−u
v −u
i.e. t =
t
a
(v − u)
1
s = (v + u) ×
2
a
∴ 2as = v2 - u2
v2 = u2 + 2as
This is third kinematical equation
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Answer the following
Q.1Distinguish between
i.
Distance and displacement
Ans:
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ii. Speed and velocity
Ans:
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Q.2 Give reasons
i.
An object at rest can be considered to have
uniform motion
Ans:
•
An object is said to be in uniform motion
when it has constant speed.
•
For an object at rest, its speed is always zero.
•
Hence, an object at rest can be considered to
have uniform motion.
ii. When a body falls freely to the ground, its motion
has a uniform acceleration
Ans:
•
When a body falls freely, the only force acting
on it is earth’s gravitational force.
•
This force is practically uniform near the
earth’s surface.
•
Hence, when a body falls freely to the ground,
its motion has a uniform acceleration.
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Solved Examples
1. A person swims 100 m in 45 second at the start.
Then he covers 80 m in the next 45 second and
in the last 10 second he covers 250 m. Calculate
the average speed.
Ans:
Given:
Total distance
= 100 m + 80 m +25 m = 205 m
Total time taken = 45 s + 45 s +10 s = 100 s
Average speed =?
total distance covered
Average speed =
total time taken
=
205
100
= 2.05 m/s
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2. An athlete runs on a circular track of length 400 m
in 25 second and returns to the starting point.
Calculate average speed and average velocity.
Ans:
Given:
Total distance covered
Total displacement
= 400 m
= 0 m/s as he returns to
Starting point
Total time taken
= 25 second
Average speed
= ?
Average velocity
=?
Average speed
=
total distance covered
total time taken
400
= 25
= 16 m/s
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total distance
= total time taken
Average velocity
0
= 25
= 0 m/s
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3. A motorboat starts from rest and moves with
uniform acceleration. If it reaches to a velocity of
15 m/s in 5 second, calculate acceleration and
distance covered by it in given time.
Ans:
Given:
Initial velocity
= 0 m/s
Final velocity
= 15 m/s
Time taken
= 5 second
Acceleration
=?
Distance covered = ?
Acceleration
=
v−u
t
=
15 − 0
5
= 3 m/s2
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By second kinematical equation,
s
= ut +
1 2
at
2
s
=0+
1
× 3 × 25
2
=
75
2
= 37.5 m
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•
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