Motion in Physics - Class 9 CBSE Science Notes

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Motion

• An object is said to be in motion when its position changes with time.
• To describe the position of an object, we need a reference point or origin.
• Examples:
  • Movement of dust, leaves, and branches.
  • Phenomena of sunrise, sunset, and changing seasons are due to Earth's motion.

Rest

• A body is said to be in a state of rest when its position does not change with time.
• Examples:
  • Sleeping
  • Sitting
  • Standing
  • Lying

Motion is Relative

A body can be moving for one observer, and at the same time at rest for another observer.

Example: A person sitting in an airplane is at zero velocity relative to the airplane, but is moving at the same velocity as the airplane with respect to the ground.

• Fixation: Refers to the reference point or frame of the observer.
• Direction of Passenger's Motion: Relative to the observer, motion is perceived differently based on the reference frame.
• Relative Motion: Describes how motion is observed differently depending on the frame of reference.

Physical Quantities

Classification of Physical Quantities:

Type	Description	Examples
Scalar	Physical quantities that have magnitude only (value) and no direction involved.	Mass, Volume, Length, Time, Distance, Speed, Work, Energy, Power, etc.
Vector	Physical quantities that have both magnitude and direction.	Force, Velocity, Acceleration, Momentum, etc.

Distance vs Displacement

Aspect	Distance (D)	Displacement (S)
Definition	Actual path length on which an object travels.	Shortest path between two points.
SI Unit	Meter (m)	Meter (m)
Denoted by	Denoted by 'D'	Denoted by 's'
Path	May be curved or irregular.	Always a straight line.

Q. which of the following is universally true? displacement = distance displacement> distance displacement distance displacement = distance

Displacement Problem

Problem:

A farmer moves along the boundary of a square field of side 10m in 40 sec. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?

Solution:

The farmer moves along the boundary of a square field, so each side of the square is 10m. To find the magnitude of displacement after a certain time, we first need to calculate the total number of rounds the farmer completes in 2 minutes 20 seconds (or 140 seconds). One complete round takes 40 seconds. Number of rounds = Total time / Time per round = 140 sec / 40 sec = 3.5 rounds Now, the displacement after 3 full rounds and half of the 4th round will be the diagonal distance of the square (since the farmer reaches a corner diagonally opposite to the starting point). Diagonal of square = √(side² + side²) = √(10² + 10²) = √(100 + 100) = √200 = 10√2 ≈ 14.14m Therefore, the magnitude of the displacement after 2 minutes 20 seconds = 10√2 ≈ 14.14 meters.

Speed

Definition:

Speed is the rate of distance covered in a unit time. It is a scalar quantity.

Formula:

Speed = Distance / Time

For example, if a person travels 100 km in 1 hour, the speed is:

Speed = 100 km / 1 hr = 100 km/h

SI Units:

The SI unit of speed can be m/s (meters per second) or km/h (kilometers per hour).

Velocity

Definition:

Velocity is the rate of change of displacement with respect to time. It is a vector quantity, meaning it has both magnitude and direction.

Formula:

Velocity = Displacement / Time

For example, if an object moves 100 meters to the east in 20 seconds, its velocity would be:

Velocity = 100 m / 20 s = 5 m/s in the eastward direction.

SI Units:

The SI unit of velocity is m/s (meters per second).

Velocity vs Speed

Definition:

• Speed is the rate of distance covered in a unit time. It is a scalar quantity.
• Velocity is the rate of displacement covered in a unit time. It is a vector quantity.

Comparison:

Aspect	Speed	Velocity
Rate of	Distance covered	Displacement covered
Quantity	Scalar	Vector
Denoted by	s	v
SI Unit	km/h or m/s	km/h or m/s
Formula	Speed = Distance / Time	Velocity = Displacement / Time

Change in Speed and Velocity of a Car

Problem:

A car is speeding with 30 km/h towards East and then it returns back with the same speed (30 km/h) along the same path to the initial point. Find the change in speed and the change in velocity of the car?

Solution:

Let the initial speed of the car be 30 km/h towards East.

After returning, the car is still traveling with a speed of 30 km/h in the opposite direction (West). Hence, the final speed is 30 km/h.

Change in Speed (ΔS):

Change in speed is the difference between the final speed and the initial speed.

ΔS = final speed - initial speed

ΔS = 30 km/h - 30 km/h = 0 km/h

Change in Velocity (ΔV):

Velocity is a vector quantity, and since the direction changes, we have to consider the sign of the velocity.

Let the initial velocity be 30 km/h towards East, and the final velocity be -30 km/h towards West.

ΔV = final velocity - initial velocity

ΔV = -30 km/h - 30 km/h = -60 km/h

Speed and Velocity of a Car

Problem:

A car travels a distance of 200 km from Delhi to Ambala towards North in 5 hours. Calculate (i) speed, and (ii) velocity, of the car for this journey.

Solution:

Given:

• Distance travelled by the car = 200 km
• Time taken = 5 hours
• Direction = North

(i) Speed:

Speed is the rate at which distance is covered. It is a scalar quantity.

Formula: Speed = Distance / Time

Speed = 200 km / 5 hours = 40 km/h

(ii) Velocity:

Velocity is the rate at which displacement is covered in a given direction. It is a vector quantity and takes into account both the distance and the direction.

Since the car is traveling in a straight line towards the North, the displacement is equal to the distance covered, and the direction is North.

Velocity = Displacement / Time

Velocity = 200 km / 5 hours = 40 km/h towards North.

Acceleration and Retardation

Definition:

Acceleration (or Retardation) is the rate of change in velocity per unit time. It is a vector quantity as it has both magnitude and direction.

Formula:

Acceleration is denoted by a and is given by:

a = (V - u) / t

• u = initial velocity
• V = final velocity
• t = time taken for the change in velocity

SI Unit:

The SI unit of acceleration is m/s² or km/h² depending on the units of velocity and time.

Retardation

Definition:

Retardation is a term used to describe the decrease in velocity with time. It is a type of non-uniform motion where the velocity decreases over time. It has the same definition as acceleration but occurs when velocity decreases.

Retardation is also known as deacceleration.

In this case, the final velocity v is less than the initial velocity u, and the change in velocity or acceleration is negative.

Example:

An example of retardation is when an airplane lands and reduces its speed until it comes to a stop.

Uniform vs Non-Uniform Acceleration

Parameters	Uniform Acceleration	Non-Uniform Acceleration
Meaning	Equal amount of velocity increases in equal intervals of time.	Velocity changes by unequal amounts in equal intervals of time.
Velocity-Time Graph	Straight line	Curved line
Example	A free-fall object	Circular motion types where speed is constant but direction changes at every point.

Velocity-Time Graph

The graph for uniform acceleration shows a straight line, while for non-uniform acceleration, it shows a curved line.

Graphical Representation of Motions - Distance-Time Graph

In a Distance-Time Graph, time is represented along the x-axis, and distance is represented along the y-axis.

In this graph:

• The x-axis represents time, which is the independent quantity.
• The y-axis represents distance, which is the dependent quantity.

Important Points:

In a distance-time graph, the slope of the graph indicates the speed or velocity of the object. The steeper the slope, the faster the motion.

If the graph is a straight line, it indicates uniform motion. If the graph is curved, it indicates non-uniform motion.

Equations of Motion

1. Equation for Velocity-Time Relation

For an object moving with constant acceleration, the velocity-time relation is:

V = u + at

• V: Final velocity
• u: Initial velocity
• a: Acceleration
• t: Time

2. Equation for Displacement-Time Relation

The displacement-time relation for uniformly accelerated motion is:

S = ut + (1/2)at²

• S: Displacement
• u: Initial velocity
• a: Acceleration
• t: Time

3. Equation for Displacement-Velocity Relation

The relation between displacement, velocity, and acceleration is:

v² = u² + 2as

• v: Final velocity
• u: Initial velocity
• a: Acceleration
• s: Displacement

Algebraic Derivation of Equations of Motion

1. Derivation of Velocity-Time Equation

We know that acceleration a is the rate of change of velocity:

a = (V - u) / t

Multiplying both sides by t, we get:

at = V - u

Rearranging this equation:

V = u + at

2. Derivation of Displacement-Time Equation

To find the displacement S, we use the formula for average velocity.

The average velocity v_avg is:

v_avg = (u + V) / 2

Now, displacement S is:

S = v_avg * t

Substituting v_avg = (u + V) / 2 into the equation:

S = ((u + V) / 2) * t

Substitute V = u + at into the equation:

S = ((u + (u + at)) / 2) * t

Now simplify:

S = (2u + at) * t / 2

S = ut + (1/2)at²

3. Derivation of Displacement-Velocity Equation

We start with the equation V = u + at:

Squaring both sides:

V² = (u + at)²

Expanding the right side:

V² = u² + 2uat + a²t²

From the equation S = ut + (1/2)at², solve for at²:

at² = 2S

Substitute this into the equation:

V² = u² + 2aS

This is the required equation:

V² = u² + 2aS

Graphical Derivation of Equations of Motion

1. Equation for Velocity-Time Relation (v = u + at)

The velocity-time graph is a straight line if the acceleration is constant. Let's denote:

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• t = Time taken

Steps for Graphical Derivation:

1. Plot a velocity-time graph. The slope of this graph represents acceleration.

2. The slope of a line is calculated as the ratio of the change in velocity to the change in time. This is given by:

a = (v - u) / t

3. Rearranging, we get:

v = u + at

This is the required equation for velocity-time relation.

Graphical Representation:

On a velocity-time graph, the line from point A to point C represents constant acceleration. The slope of the line, denoted as a, is the rate of change of velocity with respect to time.

The distance between the points on the graph can be interpreted as the change in velocity over time.

Thus, the equation v = u + at is derived graphically.

2. Derivation of Equation using the Slope of Velocity-Time Graph

Consider the following points:

• P = Initial point
• Q = Final point
• BD = Change in velocity (v - u)
• CD = Change in time (t)

From the graph, the slope is:

Slope = (BD - CD) / time

Thus, we get:

a = (v - u) / t

Rearranging, we get:

v = u + at

This is again the same equation for velocity-time relation, confirming that the graphical derivation matches the algebraic formula.

Graphical Derivation of Equations of Motion-2 (S = ut + ½at²)

1. Area Representation in Velocity-Time Graph

The equation for displacement \( S \) in uniformly accelerated motion can be derived by considering the area under the velocity-time graph.

For a velocity-time graph, the area under the graph represents displacement. Let's break the graph into areas:

• The area of triangle \( \triangle ABC \) represents the distance covered under uniform acceleration.
• The area of rectangle \( LOACD \) corresponds to the distance covered with constant velocity \( u \).

The total area under the graph is the sum of these two areas, which gives the displacement \( S \).

Steps for Derivation:

1. The area of triangle \( ABC \) is given by:

Area of \( ABC \) = \( \frac{1}{2} \times BC \times AB \)

2. The area of rectangle \( LOACD \) is given by:

Area of \( LOACD \) = \( BC \times OD \)

3. The total area under the graph is:

S = Area of \( ABC \) + Area of \( LOACD \)

S = \( \frac{1}{2} \times BC \times AB + BC \times OD \)

Substituting the values:

Now, \( AB \) represents the time \( t \), \( BC \) represents the change in velocity \( (v - u) \), and \( OD \) represents the time \( t \), so we have:

S = \( \frac{1}{2} \times (v - u) \times t + u \times t \)

4. Simplifying the expression:

S = ut + \frac{1}{2} (v - u) \times t

5. Since \( v = u + at \), we substitute this into the equation:

S = ut + \frac{1}{2} (u + at - u) \times t

S = ut + \frac{1}{2} at^2

This is the required equation for displacement during uniformly accelerated motion.

Conclusion:

Hence, we have derived the equation for displacement:

S = ut + \frac{1}{2} at^2

Graphical Derivation of Equation of Motion-3 (2as = v² - u²)

1. Area Representation in Velocity-Time Graph

In this derivation, we use the velocity-time graph to derive the equation \( 2as = v^2 - u^2 \), where:

• \( v \) = final velocity
• \( u \) = initial velocity
• \( a \) = acceleration
• \( s \) = displacement

The area under the velocity-time graph represents displacement. By considering the areas, we can derive the equation.

Steps for Derivation:

1. The area of the trapezoid \( COABD \) under the velocity-time graph is given by:

Area of \( COABD \) = \( \text{sum of the sides} \times h \) = \( 8 \)

Step-by-Step Explanation:

2. The equation for displacement can be written as:

\( (DA + BD) \times OD = 8 \)

3. Expanding the terms, we get:

\( (DA + BD) \times OD = (v + u) \times t \)

4. Now, \( v - u = at \), so:

\( v - u = at \)

5. Substituting this into the equation, we get:

\( (v + u)(v - u) = 2as \)

Final Equation:

Thus, the derived equation becomes:

\( 2as = v^2 - u^2 \)

Conclusion:

Hence, we have derived the equation \( 2as = v^2 - u^2 \), which relates the acceleration, displacement, initial velocity, and final velocity of an object in uniformly accelerated motion.

Free Fall (Motion Under Gravity)

Definition:

Free fall motion under gravity is when an object moves solely under the influence of gravity, without any other external forces acting on it. This type of motion is called acceleration due to gravity.

Key Points:

• The object experiences a uniform acceleration towards the Earth's surface.
• The rate of acceleration is constant at a value of approximately 9.8 m/s² (on the surface of the Earth), and is denoted as g.
• During free fall, the only force acting on the object is gravity (ignoring air resistance).
• If an object is dropped from a height, its initial velocity is zero. As it falls, its velocity increases due to the constant acceleration.

Mathematical Representation:

For an object in free fall, we can use the equations of motion with acceleration \( g \) to describe its position and velocity over time.

• Velocity after time \( t \): v = u + gt, where \( u \) is the initial velocity (0 if dropped), \( v \) is the final velocity, and \( g \) is the acceleration due to gravity.
• Displacement after time \( t \): s = ut + (1/2)gt², where \( u \) is the initial velocity (0 if dropped), and \( s \) is the displacement.

Example:

If an object is dropped from a height, the initial velocity \( u = 0 \). The object will experience an acceleration of \( g = 9.8 \, \text{m/s}^2 \). Using the equations above, we can calculate its velocity and displacement at any point during its fall.

Conclusion:

Free fall motion under gravity is a fundamental concept in physics, and it helps us understand the behavior of objects under the influence of gravity alone. The constant acceleration \( g \) plays a crucial role in determining the velocity and displacement of objects in free fall.

Car Acceleration Problem

Problem:

A car starts from rest, and accelerates at a constant rate. After 0.25 minutes, it is moving with a velocity of 25 m/s. At what rate is the car accelerating?

Solution:

We know the following information:

• Initial velocity, \( u = 0 \) (since the car starts from rest)
• Final velocity, \( v = 25 \, \text{m/s} \)
• Time, \( t = 0.25 \, \text{min} = 15 \, \text{seconds} \) (since 1 minute = 60 seconds)

To find:

The acceleration, \( a \), using the equation of motion:

v = u + at

Substitute the known values:

25 = 0 + a(15)

So, solving for \( a \):

a = 25 / 15 = 1.67 m/s²

Answer:

The rate of acceleration of the car is 1.67 m/s².

Truck Retardation Problem

Problem:

A truck came to rest when the brake was applied for 4 seconds to get a retardation of 3 m/s². Calculate how far the truck would have travelled after applying the brake.

Solution:

We know the following information:

• Initial velocity, \( u \) (before applying brake) = unknown
• Final velocity, \( v = 0 \) (since the truck comes to rest)
• Time, \( t = 4 \, \text{s} \)
• Retardation, \( a = -3 \, \text{m/s}^2 \) (negative because it is a deceleration)

To find:

The distance travelled by the truck while it is decelerating. We will use the equation of motion:

s = ut + (1/2)at²

However, we do not know \( u \) (the initial velocity). To find it, we will use another equation of motion:

v = u + at

Substitute the known values:

0 = u - 3(4)

Solving for \( u \):

u = 12 \, \text{m/s}

Now, calculate the distance:

Substitute \( u = 12 \, \text{m/s} \), \( a = -3 \, \text{m/s}^2 \), and \( t = 4 \, \text{s} \) into the first equation:

s = (12)(4) + (1/2)(-3)(4)²

Now, calculate the value:

s = 48 + (1/2)(-3)(16)

s = 48 - 24 = 24 \, \text{meters}

Answer:

The truck would have travelled 24 meters after applying the brake.

Stone Thrown Vertically Upwards Problem

Problem:

A stone is thrown in a vertically upward direction with a velocity of 5 m/s. If the acceleration of the stone during its motion is 10 m/s² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Solution:

Given the following data:

• Initial velocity, \( u = 5 \, \text{m/s} \)
• Acceleration due to gravity, \( a = -10 \, \text{m/s}^2 \) (negative because it acts downward)
• Final velocity at the highest point, \( v = 0 \, \text{m/s} \) (because the stone stops before falling back down)

To Find:

We need to calculate the maximum height attained by the stone and the time taken to reach this height. We will use the following equations of motion:

1. To calculate the time taken to reach the highest point:

v = u + at

Substitute the values:

0 = 5 + (-10)t

Solving for \( t \):

t = 5 / 10 = 0.5 \, \text{seconds}

2. To calculate the maximum height attained:

s = ut + (1/2)at²

Substitute the values:

s = (5)(0.5) + (1/2)(-10)(0.5)²

Now, calculate the value:

s = 2.5 - 1.25 = 1.25 \, \text{meters}

Answer:

The stone will reach a maximum height of 1.25 meters and it will take 0.5 seconds to reach there.

Uniform Circular Motion

Definition:

The motion of an object along a circular path, covering equal distances along the circumference in the same interval of time, is known as uniform circular motion.

Examples:

• Motion of the Moon
• Motion of the Earth
• A cyclist on a circular track

Velocity in Uniform Circular Motion:

The velocity \( v \) for a circular path with radius \( r \) and time \( t \) is given by:

v = 2πr / t

In uniform circular motion, the speed remains constant, but the direction of the velocity changes continuously.

Particle Motion on Circular Path

Problem:

A particle is moving in a circular path with a radius of 35m. What is the distance traveled and displacement after completing half a revolution?

Given:

• Radius (R) = 35 m
• Half revolution

Solution:

Distance traveled: The distance covered after half a revolution is half of the circumference of the circular path. The formula for the circumference of a circle is \( C = 2 \pi R \). So, the distance traveled in half revolution is:

Distance = \( \frac{1}{2} \times 2 \pi R = \pi R \)

Distance = \( \pi \times 35 \approx 110 \, \text{m}

Displacement: The displacement is the straight-line distance between the initial and final positions of the particle, which is the diameter of the circle. Therefore, the displacement is:

Displacement = \( 2R = 2 \times 35 = 70 \, \text{m}

Conclusion:

Therefore, after completing half a revolution:

• Distance traveled = 110 m
• Displacement = 70 m

Artificial Satellite Speed Calculation

Problem:

An artificial satellite is moving in a circular orbit of radius 42,250 km. Calculate its speed if it takes 24 hours to revolve around the Earth.

Given:

• Radius of the orbit, \( r = 42,250 \, \text{km} \)
• Time taken for one revolution, \( t = 24 \, \text{hours} \)

Solution:

The speed of the satellite can be calculated using the formula for the speed in uniform circular motion:

Speed \( v = \frac{2 \pi r}{t}

Where:

• \( r \) is the radius of the orbit (42,250 km)
• \( t \) is the time for one revolution (24 hours)
• \( \pi \approx 3.1416 \)

First, we need to convert the time to seconds:

24 hours = 24 × 60 × 60 = 86,400 seconds

Now, substitute the values into the formula:

v = \( \frac{2 \times 3.1416 \times 42,250,000}{86,400} \)

Calculating this gives:

v ≈ 3,071.5 \, \text{km/s}

Conclusion:

The speed of the satellite is approximately 3,071.5 km/s.