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Here you will find Motion Class 9 Notes.

## Motion Class 9 Notes

Motion is a change in position of an object over time. Motion is described in terms of displacement , distance, velocity, acceleration, time and speed.

If the position of a body is not changing with respect to a given time , the body is said to be at rest, stationary or in constant position.

Everything naturally wants to move and change. In the world of mechanics, there are four basic types of motion.
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Rotary Motion

Rotary motion is anything that moves in a circle. This type of motion was among the first discovered in ancient times. Think of a spinning wheel on which people spun wool. A car’s engine works the same way. Like linear cylinders, rotary actuators are used across a wide range of industries and come in electric, pneumatic and hydraulic options.

Oscillating Motion

Something that oscillates moves back and forth. Anything that repeats the motion cycle after a certain period is considered to be oscillating. This type of motion is found everywhere in our world: a sprinkler system, the pendulum of a clock or even sound waves.

You may be thinking that a rotary actuator functions as an oscillating device, and for that matter, so does a linear one when it repeats a continuous movement. When it comes to actuators, linear and rotary can be viewed as oscillating.

Linear Motion

Simple enough, linear motion is anything that moves in a straight line, like our linear actuators. Time, as far as we know, moves in a linear fashion. Just like rotary devices, you can find linear cylinders in electric, pneumatic or hydraulic options. They have driven the field of automation, manufacturing, robotics and others into a new age, because in the past, rotary motion was the only means to create motion.
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Irregular motion

It is motion which has no obvious pattern to its movement. It is also called Zig Zag motion. Example: a flying bee.

UNIFORM MOTION AND NONUNIFORM MOTION

When a body covers equal distance in equal interval of times then the body is said to describe Uniform motion.
When a body covers unequal distance in equal interval of times or vice – verse, then the body is said to describe No uniform motion.

For an example:

Car A

 Time in second 0 5 10 15 20 25 30 35 Distance Travel in Meter 0 10 20 30 40 50 60 70

Car B

 Time in second 0 5 10 15 20 25 30 35 Distance Travel in Meter 0 5 15 20 40 42 60 65

#### Measuring the Rate of Motion

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,

Average Speed = Total Distance Travelled / Total Time Taken

If an object travels a distance s in time t then its speed v is,
$V = {s \over t}$

Example:

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

Solution:

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

Average Speed = Total Distance Travelled / Total Time Taken

Average Speed $= {{32m} \over {6\sec }} = 5.3m/\sec$
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Velocity

velocity is defined as the rate of change of position. It is a vector physical quantity; both speed and direction are required to define it.

Relative velocity

Relative velocity is a measurement of velocity between two objects as determined in a single coordinate system.

Average Velocity $= {{InitialVelocity + FinalVelocity} \over 2}$

V  Avg $= {{u + \upsilon } \over 2}$

Example:

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.

Solution:

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

V Avg  $= {s \over t}$
V Avg $= {{400Km} \over {8h}} = 50km/h$
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Example: 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

Average Velocity $= {{InitialVelocity + FinalVelocity} \over 2}$

Average Velocity $= {{180m} \over {60\sec }} = 3m/\sec$

Velocity  $= {{Displacement} \over {Time}} = {{0m} \over {60\sec }} = 0m/\sec$

#### Acceleration

Acceleration is a measure of the change in the velocity of an object per unit time.

$Acceleration = {{ChangeInVelocity} \over {TimeTaken}}$

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

$A = {{u - \nu } \over T}$

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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.

Example
: Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m /s in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m /s in the next 5 s. Calculate the acceleration of the bicycle in both the cases.

Solution:
In the first case:
initial velocity, u = 0 ;
final velocity, v = 6 m /s ;
time, t = 30 s .
we have

$a = {{u - \nu } \over T}$

$a = {{60 - 0} \over {30}} = 2m/s$

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

$a = {{4 - 6} \over 5}$

= -0.4m 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.
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Quick Review :-

• 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 three equations, namely
v = u + at
s = ut + ½ at2
2as = v2 – u2
• 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 three equations, namely
v = u + at
s = ut + ½ at2
2as = v2 – u2

You were reading Motion Class 9 Notes. Read Science notes for Class 9.