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Motion Class 9th Notes - Free NCERT Class 9 Science Chapter 8 Notes - Download PDF

Motion Class 9th Notes - Free NCERT Class 9 Science Chapter 8 Notes - Download PDF

Edited By Vishal kumar | Updated on Mar 14, 2024 10:34 AM IST

CBSE Class 9 Science Chapter 8 Motion Notes - Download Free PDF

CBSE Class 9 Science Chapter 8 Motion Notes are a valuable resource not only for Class 10, but also for Classes 11 and 12, and even beyond into mechanics. These motion class 9 notes, curated by Careers360 experts, explain the concept of motion in simple terms, demonstrating how objects move and change position over time. They are more than just textbooks; they serve as the foundation for understanding how the world works. Furthermore, these physics class 9 chapter 8 notes pdf are free to download, so anyone, regardless of internet access, can benefit from learning about the fascinating world of motion and physics.

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NCERT Class 9 Science Chapter 8 Notes

Introduction: Some objects are stationary while others are moving throughout our daily lives. Birds can fly, fish can swim, blood can flow via veins and arteries, and automobiles can move. Atoms, molecules, planets, stars, and galaxies are all moving. When an object's position varies over time, we frequently interpret it as moving. The majority of motions are complicated. Some objects move in a straight line, while others move in a circle. Some of them may rotate, while others may vibrate. There may be times when you need to use a combination of these.

Motion Along a Straight Line

A straight-line motion is the most basic sort of motion. Distance and Displacement are two separate quantities used to represent an object's overall motion and to locate its end position in relation to its initial position at a given moment.

In physics, distance refers to the length of an object's path (the line or curve) through space.

Distance is not dependent on direction. Scalars are physical quantities that can be described completely without the need for direction.

The shortest distance between a body's initial and final positions, as well as its direction, is called displacement as it moves from one location to another.

Because displacement requires both direction and magnitude to be fully described, such physical quantities are referred to as vectors.

A moving body's distance travelled cannot be zero, yet its eventual displacement can be zero.

Uniform Motion and Non-Uniform Motion

A body is said to be in uniform motion if it travels equal distances at equal intervals of time. Non-uniform motion occurs when a body traverses unequal distances in equal intervals or equal distances in unequal intervals.

Speed

The total distance travelled by the object during the time interval in which the motion occurs is defined as its speed. The metric unit of speed is the meter per second. So,

Speed=Distance travelled/time taken=s/t

Here’s is the distance travelled by the object and t is the time taken by the object to travel the distance s. The speed of a body tells us how slowly or quickly it is moving. The average speed of a body is determined by the ratio of the total distance travelled to the total time spent by the body. A body's instantaneous speed is its speed at a given instant. If a body travels the same distance in equal intervals of time, it is said to have constant or uniform speed.

Velocity

The rate at which a body's displacement changes with time is referred to as its velocity.

In SI units, an object's velocity is measured in meters per second.

Velocity is simply the speed of an object moving in a specific direction. An object's velocity can be either uniform or variable. It can be altered by varying the speed, direction, or both of the object's motion. So velocity is a vector quantity, whereas speed is a scalar quantity with only magnitude and no specific direction. When a body travels the same distance in the same period of time in the same direction, it is said to be moving at uniform velocity. If a body covers unequal distances in equal intervals of time and vice versa in a specified direction, or if it changes direction, it is said to be moving with non-uniform velocity.

A body's velocity can be changed in two ways: first, by changing its speed, and second, by changing its direction of motion while keeping its speed constant. In order to change the velocity of the body, both the speed and direction of the body can be changed.

When the velocity of an object changes at a constant rate, the arithmetic mean of the initial and final velocity for a given period of time is used to calculate average velocity. Which is,

average velocity=(u+v)/2

Where u is the initial velocity of the body and v is the final velocity of the body.

Acceleration

The change in velocity of an object per unit time is called acceleration. It is given as,

Acceleration=Change in velocity/time taken

If the velocity of a body varies from an initial value u to a final value v in time t. acceleration ‘a’ is given by,

a = (v-u)/t

If a body moves in a straight path and its velocity increases by the same amount every time interval, it is said to have uniform acceleration. Examples include freely falling bodies, balls rolling down an inclined plane, and so on. If a body's velocity changes by an unequal amount at equal periods of time, it is said to have non uniform acceleration. Positive acceleration occurs when acceleration occurs in the direction of velocity; negative acceleration occurs when acceleration occurs in the opposite direction of velocity; negative acceleration is known as retardation.

Equations of Motion

First Equation of motion

v=u+at, is the first equation of motion, where v is the final velocity and u is the body's initial velocity. The first equation of motion determines the body's velocity at any time t.

We know, Acceleration= Change in velocity time taken

So, a = (v-u)/t

∴ at=v-u

Now, rearranging the above equation we get,

v=u+at

In this manner, we get the first equation of motion.

Second Equation of motion:

s=ut+at2/2 is the second equation of motion. ‘u’ is the initial velocity here,‘s’ is distance travelled by the body in time ‘t’ and ‘a’ is uniform acceleration.

Now consider the average velocity formula, when the velocity of an object changes at a constant rate. That is,

average velocity=(u+v)/2

Distance travelled by an object is,

s= (u+v)t/2

The first equation of motion is, v=u+at

Substituting the first equation of motion in the above equation, we get

s=(u+u+at)t/2

=(2u+at)t/2

=(2ut+at2)/2

Rearranging the above equation we get,

s=ut+at2/2

Third Equation of motion:

v2=u2+2as is the third equation of motion. ‘u’ is the initial velocity here, ‘v’ is the final velocity. ‘s’ is distance travelled by body in time ‘t’ and ‘a’ is uniform acceleration.

This equation gives the body's velocity as it travels a distance of s.

v=u+at And s=ut+at2/2 are the first and second equations of motions respectively.

Rearranging the first equation we get,

t=(v-u)/a

Substituting this in the second equation of motion we get,

s=u((v-u)/a + a((v-u)/a)2/2

s=(2uv-2u2+v2+u2-2uv)/2a

Rearranging the above equation we get,

v2=u2+2as

The following three crucial points should be remembered while addressing uniformly accelerated motion problems using these three equations of motion.

  • u=0 is the initial velocity of a body starting from rest.

  • When a body comes to a complete stop, its final velocity is equal to zero.

  • The acceleration of a body moving at a constant velocity is zero.

Graphical representation of motion

A graph depicts the relationship between two sets of data, one of which contains dependent variables and the other of which contains independent variables.

Line graphs can be used to depict the motion of an object. Line graphs in this case show the relationship between one physical quantity, such as distance or velocity, and another physical quantity, such as time.

Distance Time Graphs

On a distance-time graph, the change in an object's position over time can be represented.

The x-axis in this graph represents time, while the y-axis represents distance.

Because they precisely depict velocity, distance time graphs of a moving body can be utilized to compute the speed of the body.

As illustrated in Figure a, the distance time graph for a body moving at a constant speed is always a straight line since the distance travelled by the body is directly proportional to time.

The distance time graph for a body travelling at a non-uniform speed is depicted in figure b below as a curve.

1647345467156

When the object is at rest, the distance time graph is parallel to the time axis as shown in below figure a),

1647345465611

To determine the body's speed from a distance time graph [figure d)], say at point A, first draw a perpendicular AB on the time axis and a perpendicular AC on the distance axis, so that AB represents the body's distance travelled in time interval OB and we know,

Speed=Distance travelled/time taken=s/t=AB/OB

Velocity time graphs

A velocity-time graph depicts the variation in velocity with time for an object travelling in a straight line. Time is shown on the x-axis, while velocity is displayed on the y-axis in this graph. The displacement of an object travelling with uniform velocity is given by the product of velocity and time. The magnitude of the displacement will be equal to the area contained by the velocity-time graph and the time axis.

If a body moves at a steady velocity, its velocity time graph will be a straight line parallel to the time axis, as seen in Figure (e).

1647345466912


Figure f) depicts a straight-line velocity time graph with uniformly changing velocity. The velocity-time graph can be used to determine the value of acceleration.

Draw a perpendicular RP from point R to calculate acceleration at the time corresponding to point R, as shown in figure f).

Acceleration=Change in velocity/time taken

The time taken is equal to OR, and the change in velocity is indicated by PR.

Acceleration=PR/OR

This is the same as the velocity time graph's slope. As a result, we can deduce that the slope of a moving body's velocity-time graph determines its acceleration.

As shown in Figure f), the distance travelled by a moving body in a given time is equal to the area of the triangle OPR.

Distance travelled=Area of triangle OPR

=0.5Area of rectangle ORPQ

Thus, Distance travelled=(OQ)(OR)/2

When a body's velocity varies in an irregular pattern, the body's velocity-time graph is a curved line.

Equations of motion by graphical method

When an object moves in a straight path with uniform acceleration, we already understand equations of motion. We already know how to calculate them, but a graphical way can also be used.

Equation for velocity time relation:

Consider the velocity-time graph of an object moving with uniform acceleration, as seen in Figure a).

The initial velocity of the object is u (at point A) and rises to v (at point B) in time t, as shown by this graph. The velocity is changing at a constant rate of a.

Again, the time‘t’ is represented by OC, the initial velocity u by OA, and the final velocity of the body after time t by BC in the figure.

1647345466427

The graph in figure a) clearly shows that BC=BD+DC=BD+OA. As a result,

We have v = BD+u ……………………………… (1)

We should now determine the value of BD. The object's acceleration from the velocity-time graph is given by,

Acceleration=Change in velocity time taken=BDAD=BDOC=BDt

We get, BD=at

Substituting the value of BD in eqn (1) we get,

v=u+at

Which is the first equation of motion and velocity time relation.

Equation for position time relation:

Assume the body has travelled s distance in time t with uniform acceleration a. The area encompassed within OABC under the velocity-time graph AB in Fig. (a) is used to calculate the distance travelled by the body.

Thus the object travels a distance, which is given by,

s = Area of OABC (trapezium).

s = area of the rectangle OADC + area of the triangle ABD.

s=OA x OC+(AD)(BD/2

Putting OA= u, OC = t and BD = at, we have

s=(u x t)+(t x at)/2

Then we get,

s=ut+at2/2

This is an equation of position-time relation.

Equation for position velocity relation:

Consider the graph in Figure (a). We know that the area under line AB, which is the area of trapezium OABC, gives the distance s travelled by a body in time t.

Thus, we have.

distance travelled by object= s=Area of trapezium OABC

s=(sum of parallel sides x height)/2=(OA+CB) x OC/2

we have, OA+CB=u+ v and OC=t, so we have

s=(u+v)t/2

By velocity time relation, t=(v-u)/a

Substituting this ‘t’ in equation for ‘s’ we have,

s=(u+v)(v-u)t/2a

Therefore we have,

v2=u2+2as

This is an equation for position velocity relation.

Uniform circular motion

The motion of an object is called uniform circular motion when it moves along a circular route at a constant speed.

Many examples of circular motion may be found in our daily lives, such as vehicles driving around a circular track and many others.

In addition, the earth and other planets orbit the sun in roughly circular orbits.

Even if the speed of a particle moving in a circular motion remains constant, the particle accelerates due to the continually changing direction of the velocity. A uniform circular motion occurs when an object moves in a circular route at a constant speed.

In circular motion, we use angular velocity instead of velocity as we did in linear motion. The force required to move a body along a circular route is referred to as centripetal force.

We know that 2πr equals the circumference of a circle of radius r. If a body takes t seconds to move once around a circular path of radius r, the velocity v is given by v =2πrt.

Although uniform linear motion is not accelerated, uniform circular motion is.

Significance of NCERT class 9 Science chapter 8 notes

Revision Aid: CBSE class 9 physics ch 8 notes on motion are useful tools for reviewing key concepts covered in the chapter.

Simplified Explanations: These ch 8 physics class 9 notes are intended to provide straightforward explanations, allowing students to understand complex topics without undue difficulty.

Step-by-Step Approach: The motion class 9 notes provide a systematic breakdown of concepts, guiding students through each step to improve comprehension.

Aligned with CBSE Syllabus: The class 9 physics chapter 8 notes are closely aligned with the CBSE Science syllabus for Class 9, covering all important topics for exams.

Offline preparation: physics class 9 chapter 8 notes pdf, allowing students to study offline at their leisure without requiring internet access.

Enhanced Accessibility: Students can work through the material at their own pace, strengthening their understanding of motion and improving their academic performance.

Foundation for Future Learning: By providing a solid understanding of motion, these ch 8 physics class 9 notes lay the groundwork for further studies in science and related fields.

NCERT Solutions of Class 9 Subject-Wise

NCERT Class 9 Exemplar Solutions for Other Subjects:

Class 10 Chapter Wise Notes

Frequently Asked Question (FAQs)

1. What are the main points covered in NCERT notes for class 9 Science chapter 8?

 NCERT class 9 Science chapter 8 notes covers the following topics in depth:

  • Motion along a straight line

  • Speed

  • Velocity

  • Acceleration

  • Equations of uniformly accelerated Motion

  • Graphical representation of motion

  • Distance Time Graph

  • Velocity Time Graph

  • Equations of motion by graphical method

  • Uniform Circular Motion

2. How will CBSE class 9 Science chapter 8 notes benefit students?

Subject experts have created Class 9th Science chapter 8 notes that will give you further information on the topics. You can reinforce your foundation with these class 9 science chapter 8 notes. Important concepts are well-explained here. Every aspect of Motion is covered in these notes.

3. State the difference between speed and velocity.

 According to Class 9th Science chapter 8 notes,

Speed

Velocity

It is the distance travelled by an object during a period of time.

It refers to the movement of an object over a period of time.

Direction is not needed to describe it.

Direction is needed to describe it.

Speed cannot be negative.

Velocity can be negative.

Speed = Distance/time

Velocity = Displacement/time

4. According to Class 9th Science chapter 8 notes, what is Uniform and non-uniform acceleration?

A body is said to be in uniform motion if it travels equal distances at equal intervals of times.

Non-uniform motion occurs when a body traverses unequal distances in equal intervals or equal distances in unequal intervals.

5. What is Uniform Circular Motion?

According to Class 9th Science chapter 8 notes, the motion of an object is called uniform circular motion when it moves along a circular route at a constant speed. These topics can also be obtained from Class 9 Science chapter 8 notes pdf download.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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