We come across thousands of moving things on a daily basis. How are these moving objects different from one another? Class 9 Motion chapter describes this in detail. In this blog, you get to learn about Class 9 motion in a simple yet convenient language. Learn everything starting from the definition of motion, to its types, solved examples and question answers with these exclusive notes!
This Blog Includes:
- What is Motion?
- Graphical Representation of Motion
- Class 9 Motion Important Equations
- Deriving the Equations of Motion
- Class 9 Motion NCERT PDF
- Class 9 Motion Notes PPT
- Class 9 Motion NCERT Important Q&A
- One-Word Questions on Class 9 Motion
- MCQs on Class 9 Motion
What is Motion?
If the position of an object varies over time, the object is considered to be in motion:
Motion on the Straight Line
Distance – The distance covered by the object is defined as the complete length of the path covered by the object between the two endpoints.
Distance is a numeric number. We may not specify the direction in which the object moves when mentioning the distance travelled by the object. The distance in Physics is the length of the direction (line or curve) defined by an object travelling through space. The distance is independent of the course. Thus, certain physical quantities that do not need guidance for their full definition are called scalars. When a body moves from one position to another, the shortest distance between the body’s original and final position, along with its path, is known as displacement.
Displacement has both orientation and magnitude for its full definition, and thus these physical quantities are called Vectors. The distance travelled by a moving body cannot be zero, but the final displacement of a moving body may be Zero.
Displacement is denoted by Δx
Δx = xf − x0
xf = Final position on the object
x0 = Initial position of the object
Must Read: Here are NCERT Class 9 Science Solutions!
Difference Between Distance and Displacement
|Distance provides the complete details of the path taken by the object||Displacement does not provide the complete details of the path taken by the object|
|Distance is always positive||Displacement can be positive, negative or zero|
|It is a scalar quantity||It is a vector quantity|
|The distance between two points may not be unique||The displacement between two points is always unique|
Uniform vs Non-Uniform Motion
If the body covers equal lengths in equal time intervals, so it is assumed to have a Uniform Motion. If the body covers unequal distances at equal intervals or equal distances at unequal intervals, so the body is said to have Non-uniform Motion.
Speed is defined as the total distance travelled by the object in the time interval during which the motion takes place. SI unit of speed is meter per second.
Speed = Distance Travelled / Time Taken
The velocity of a body is known as the rate of change of displacement of a body with the passage of time. The velocity of a moving object is measured in meters per second in SI units.
Acceleration is a measure of the change in the velocity of an object per unit of time. SI unit of acceleration is ms-2. The equation can mathematically be written as:
The body is said to have a uniform acceleration if it is going on a straight path and the velocity shifts (increases or decreases) by equal proportions at equal time intervals.
A body is said to have a non-uniform acceleration if the velocity shifts (increases or decreases) by unequal proportions at unequal time frames.
Graphical Representation of Motion
Distance – Time Graphs
Distance – Time Graphs represents a change in position of the object with respect to time. The graph in case the object is stationary (means the distance is constant at all time intervals) – Straight line graph parallel to x = axis
For a body at rest, as the slope is zero, so the speed of the body is zero
For a body moving with uniform speed
For an accelerated motion., the slope of the graph is increasing with time
For decelerated (speeding down) motion., the slope of the graph is decreasing with time
Velocity- Time Graphs
Constant velocity is a straight line graph, velocity is always parallel to the x-axis. When a body is moving with a uniform velocity, the slope of AB indicates zero acceleration
When a body starts from rest and moves with uniform acceleration, then it is greater is the slope of the v-t graph, greater will be the acceleration
When a body is moving with uniform acceleration and its initial velocity is not zero
When a body is moving with increasing acceleration, the slope gradually increases with time
When a body is moving with decreasing acceleration, the slope decreases with time
When a body is moving with uniform retardation and its initial velocity is not zero, as θ > 90°, the graph has a negative slope
Must Read: Structure Of An Atom – Class 9 Science Notes
Class 9 Motion Important Equations
The equations of motion represent the relationship between an object’s acceleration, velocity and distance covered if and only if,
- The object is moving on a straight path
- The object has a uniform acceleration
Here are the three important equations of the class 9 Motion chapter:
- Equation for Velocity – Time Relation
v = u + at
- Equation for Position – Time Relation
s = ut + 1/2 at2
- Equation for the Position – Velocity Relation
2a s = v2 – u2
u: initial velocity
a: uniform acceleration
v: final velocity
s: distance travelled in time t
Deriving the Equations of Motion
Let’s now look at the derivation of these three equations of Class 9 Motion chapter:
Deriving the Equation for Velocity – Time Relation
Acceleration = Change in velocity / time taken
Acceleration = (final velocity – initial velocity) / time
a = (v – u)/t
so, at = v – u
v = u + at
Deriving Equation for Position – Time Relation
We know that, distance travelled by an object = Area under the graph
So, Distance travelled = Area of OPNR = Area of rectangle OPQR + Area of triangle PQN
s = (OP * OR) + (PQ * QN) / 2
s = (u * t) + (t * (v – u) / 2)
s = ut + 1/2 at2 [because at = v – u]
Deriving the Equation for Position – Velocity Relation
We know that, distance travelled by an object = area under the graph
So, s = Area of OPNR = (Sum of parallel sides * height) / 2
s = ((PO + NR)* PQ)/ 2 = ( (v+u) * t)/ 2
2s / (v+u) = t [equation 1]Also, we know that, (v – u)/ a = t [equation 2]On equating equations 1 and 2, we get,
2s / (v + u) = (v – u)/ a
2as = (v + u) (v – u)
2 as = v2 – u2
Here are some important study notes on Linear Equation in Two Variables
Class 9 Motion NCERT PDF
Class 9 Motion Notes PPT
Class 9 Motion NCERT Important Q&A
Given that the farmer traverses the full square field’s perimeter in 40 seconds, the farmer’s total distance travelled in 40 seconds is 4*(10) = 40 metres.
As a result, the farmer’s average distance travelled in one second is 40m/40 = 1m.
Two minutes and 20 seconds is equal to 140 seconds. In this interval, the farmer has gone a total distance of 1 m * 140 = 140m.
Because the farmer is travelling along the square field’s perimeter, the total number of laps he will accomplish is 140m/40 = 3.5 laps.
Now, the farmer’s total displacement is determined by his starting position. If the farmer’s beginning position is in one corner of the field, his final position will be in the other corner (since the field is square).
In this situation, the farmer’s total displacement will be equal to the length of the diagonal line connecting the square’s opposing corners.
Applying the Pythagoras theorem, the length of the diagonal can be obtained as follows: √(102+102)= √200= 14.14m, which is the maximum possible displacement of the farmer. If the farmer starts at the midpoint between two neighbouring square corners, his net displacement will be equal to the square’s side, which is 10m. This is the minimum displacement possible. If the farmer starts at any position around the square’s perimeter, his net displacement after 140 metres will be between 10 and 14.14 metres.
The distance between the spacecraft and the ground station is equal to the entire distance travelled by the signal since it travels in a straight line.
5 minutes = 5*60 seconds = 300 seconds.
Speed of the signal = 3 × 108 m/s.
Therefore, total distance = (3 × 108 m/s) * 300s
= 9*1010 meters.
Uniform Acceleration: In this sort of acceleration, the body travels in a straight line with a consistent velocity increase or decrease (it changes at a constant rate in any constant time interval).
Non-Uniform Acceleration: In this sort of acceleration, the body moves in a straight line but its velocity varies at a non-uniform pace (it changes at a different rate for a given constant time interval).
Given, initial velocity (u) = 90 km/hour = 25 m.s-1
Terminal velocity (v) = 0 m.s-1
Acceleration (a) = -0.5 m.s-2
As per the third motion equation, v2-u2=2as
Therefore, distance traveled by the train (s) =(v2-u2)/2a
s = (02-252)/2(-0.5) meters = 625 meters
Hence, the train must travel 625 meters at an acceleration of -0.5 ms-2 before it reaches the rest position.
Given, initial velocity of the ball (u) = 0 (since it began at the rest position)
Distance traveled by the ball (s) = 20m
Acceleration (a) = 10 ms-2
As per the third motion equation,
2*(10ms-2)*(20m) + 0
v2 = 400m2s-2
Therefore, v= 20ms-1
The ball hits the ground with a velocity of 20 meters per second.
As per the first motion equation,
Therefore, t = (v-u)/a = (20-0)ms-1 / 10ms-2 = 2 seconds
Therefore, the ball reaches the ground after 2 seconds.
(a) It’s possible; an object launched into the air accelerates at a constant rate owing to gravity. Its velocity, on the other hand, is zero when it reaches its maximum height.
(b) It’s impossible; acceleration denotes a change in speed, whereas uniform speed denotes a speed that remains constant throughout time.
(c) It is feasible because the acceleration of an object moving on a circular trajectory is perpendicular to the object’s path.
Class 9 Atoms And Molecules Study Notes for students!
One-Word Questions on Class 9 Motion
1. __________________ is defined as the total distance traveled by the body in the time interval during which the motion takes place.
2. __________________ unit of speed is m/s
3. The ratio of total distance to total __________________ taken by the body gives its average speed.
4. The rate of change of displacement of an object with the passage of time is known as __________________ of the object.
5. velocity of a body is a vector quantity involving both distance and __________________.
MCQs on Class 9 Motion
- If the displacement of an object is proportional to the square of time, then the object moves with:
A. Uniform Velocity
B. Uniform acceleration
C. Increasing acceleration
- Suppose a boy is enjoying a ride on a marry-go-round which is moving with a constant speed of 10 m/s. It implies that the boy is:
A. At rest
B. Moving with no acceleration
C. In accelerated motion
- A particle is moving in a circular path of radius r. The displacement after half a circle would be:
- Which of the following can sometimes be ‘zero’ for a moving body?
i. Average velocity
ii. Distance travelled
iii. Average speed
A. Only (i)
B. (i) and (iv)
C. (i) and (ii)
- Which of the following statement is correct regarding the velocity and speed of a moving body?
A. Velocity of a moving body is its speed in a given direction
B. Velocity of a moving body is always higher than its speed
C. Speed of a moving body is always higher than its velocity
These were class 9 Motion study notes! Hope you found answers to all your questions in this blog. Selecting the right stream after class 10th plays a great role in making a successful career. Reach out to Leverage Edu experts and make the right career choice! Book a FREE counselling session today!