MOTION in One Shot - From Zero to Hero || Class 9th

Timestamps -
00:00 - Introduction
01:16 - Difference Between REST and MOTION
03:20 - Physical Quantities
06:45 - Parameters of Motion
11:30 - Questions
18:35 - Uniform & Non-Uniform Motion
25:15 - Concept of Speed
29:46 - Concept of Velocity
35:15 - Parameters: Speed and Velocity
36:03 - Questions
57:09 - Uniformly Accelerated Motion
1:03:44 - Acceleration / Retardation
1:08:16 - Questions
1:15:40 - Equations of Motion
1:23:33 - Questions
1:31:30 - Free Fall (Motion Under Gravity)
1:43:42 - Questions
1:49:00 - Basics of Graph: Slope and Area Under Graph
1:54:45 - Displacement-Time Graph: Slope and Area
1:57:07 - Velocity-Time Graph: Slope and Area
1:59:50 - Graphs on Motion / Rest / Acceleration
2:07:19 - Questions
2:18:53 - Algebraic Derivation of Equations of Motion
2:31:00 - Graphical Derivation of Equations of Motion
2:44:13 - Uniform Circular Motion
2:50:04 - Uniform Circular Motion - Accelerated Motion
2:53:40 - Questions
2:57:23 - End of Lecture

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#Motion #Science #PhysicsWallahFoundation #Physics #Class9th #PhysicsWallah #Class9thScience #Class9thphysics #ZerotoHero

[krishna]

Kal exam hai n??😂😄Legends like karo warna fail hojaoge

[AadarshYadav-rzjk]

Class 9 2024-2025 vale batch like kro❤❤

[-abhumikaparthagoswami]

1:48:55
Answer: t=2.5s
s=31.25
Solution: u=25m/s
t=?
s=?
a= -10m/s²
v=0 ( since it is upwards free fall we consider the final velocity as 0)
1)We have to find t(TIME)
Using first formula of motion
v=u+at
0=25+(-10)t [substituting the variables]
-25=-10t [transposing +25 to LHS because LHS was 0 so it converted into -25]
25=10t [- and - cancel each other]
25/10=t
2.5=t
Therefore TIME=2.5s

Now our UTSAV becomes as follows:
u=25m/s
t= 2.5s
s=?
a= -10m/s²
v=0

2) We need to find s (displacement)
Using the second formula of motion
s=ut+½at²
s=25×2.5+½×(-10)×(2.5)² (substituting the values)
s=62.5+(-31.25) (multiplication khud krlo ya sab kuch mein bataake du?😅😅😂😂)
s=62.5-31.25
s=31.25 m



Smjha toh thoka taali ✋

[priyanshuop]

Summer Vacation wale attendance lagao .❤❤
😂

[Black_Tanmay]

Jo batch 2023 - 2024, 11th m aai h fir bhi 9th ka concept clear krne ke liye aai h, salute u guys and congrats u are serious for your studies 🙂🙃

[parthwashere]

2:18:45
Option B is correct because CAR C is taking less time and covering more distance.

[anuradhagangwar]

Timestamps -
00:00​ - Introduction
01:16​ - Difference Between REST and MOTION
03:20​ - Physical Quantities
06:45​ - Parameters of Motion
11:30​ - Questions
18:35​ - Uniform & Non-Uniform Motion
25:15​ - Concept of Speed
29:46​ - Concept of Velocity
35:15​ - Parameters: Speed and Velocity
36:03​ - Questions
57:09​ - Uniformly Accelerated Motion
1:03:44​ - Acceleration / Retardation
1:08:16​ - Questions
1:15:40​ - Equations of Motion
1:23:33​ - Questions
1:31:30​ - Free Fall (Motion Under Gravity)
1:43:42​ - Questions
1:49:00​ - Basics of Graph: Slope and Area Under Graph
1:54:45​ - Displacement-Time Graph: Slope and Area
1:57:07​ - Velocity-Time Graph: Slope and Area
1:59:50​ - Graphs on Motion / Rest / Acceleration
2:07:19​ - Questions
2:18:53​ - Algebraic Derivation of Equations of Motion
2:31:00​ - Graphical Derivation of Equations of Motion
2:44:13​ - Uniform Circular Motion
2:50:04​ - Uniform Circular Motion - Accelerated Motion
2:53:40​ - Questions
2:57:23​ - End of Lecture

[shatakshigiri]

Just 10 day of class 9 . Making notes of these onshot lecture just amazing 🥰 trying to complete syllabus before teacher . All the best for those who just entered class 9 .💕👍

[uwlmcgq]

Scaler quantity
The quantity which have only magnitude and no direction
Vector quantity
The quantity which has both magnitude and direction
Distance
The length of actual path travelled by an object
Displacement
The shortest distance between initial and final position
Initial position
The position from which an object starts moving
Uniform motion
If an object covers equal distance in equal interval of time it is said to in uniform motion
Non uniform motion
If an object covers unequal distance in an equal interval of time it is said to be in non uniform motion

[Shyam_kumar]

All numerical :-
11:30 question 1
14:04 question 2
36:04 question 3
45:59 question 4
49:04 question 5
1:08:14 question 6
1:23:34 question 7
1:28:50 question 8
1:43:41 question 9
1:48:45 question 10 homework
2:07:20 question 11
2:16:15 question 12
2:18:09 question 13 homework
2:53:38 question 14
🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥🔥
2:57:22 Thank you 💕 🥰♥️♥️.
All the best my neevians 👍👍

[Official_girl-dq]

Introduction
01:16 - Difference Between REST and MOTION
03:20 - Physical Quantities
06:45 - Parameters of Motion
11:30 - Questions
18:35 - Uniform & Non-Uniform Motion
25:15 - Concept of Speed
29:46 - Concept of Velocity
35:15 - Parameters: Speed and Velocity
36:03 - Questions
57:09 - Uniformly Accelerated Motion
1:03:44 - Acceleration / Retardation
1:08:16 - Questions
1:15:40 - Equations of Motion
1:23:33 - Questions
1:31:30 - Free Fall (Motion Under Gravity)
1:43:42 - Questions
1:49:00 - Basics of Graph: Slope and Area Under Graph
1:54:45 - Displacement-Time Graph: Slope and Area
1:57:07 - Velocity-Time Graph: Slope and Area
1:59:50 - Graphs on Motion / Rest / Acceleration
2:07:19 - Questions
2:18:53 - Algebraic Derivation of Equations of Motion
2:31:00 - Graphical Derivation of Equations of Motion
2:44:13 - Uniform Circular Motion
2:50:04 - Uniform Circular Motion - Accelerated Motion
2:53:40 - Questions
2:57:23 - End of Lecture

Like please

[Aloo_di]

U= 25m/sec
T=?
S=?
A=-10m/sec²
V=0m/sec

According to 1st equation of motion
v= u+at

0m/sec=25m/sec+(-10m/sec)×t
=-25m/sec=-10m/sec ×t
=-25m/sec ÷ -10m/sec = t
=+2.5 sec = t
Hence, time= 2.5 sec

Now, according to 2nd equation of motion

s=ut+1/2at²

s=25m/sec × 2.5sec+1/2(-10m/sec)×(2.5)²
s=25 × 2.5+(-5)×6.25
s= 62.5-31.25
s=31.25m

Hence, the distance covered by it is 31.25m


Thankyou so much sir for teaching us for 3 hours

[manjurathour]

Q- A stone is Reached by it.
Answer - t= 2.5s
And s= 31.25
Explanation-
U=25m/s ; t=? ; s = ? ; a=-10m/s ; v=0
So, from 1st eq of motion
v=u+at
0=25+(-10) t
-10t=25
t=25/10
t=2.5s
Now, from 2nd eq of motion
s=ut+1/2at^2
s=25×2.5+1/2×(-10) ×2.5^2
s=62.5+(-5) ×6.25
s= 62.5-31.25
s=31.25
Sir is the answer and explanation is correct....

[rnztzqo]

H.W QUESTION-
GIVEN=V=0, a=-10, AND u =25, height?,and time.
TO FIND TIME WE WILL USE THE FIRST EQUATION OF MOTION THAT IS( -v=u+at)
0=25+(-10)(t)
0=25+(-10t)
-25=(-10t)
25=10
t=25/10=2.5s
TO FIND HEIGHT WE WILL USE 2nd EQUATION OF MOTION THAT IS -s=ut+1/2 at square.
s=25(2.5)+1/2(-10)(2.5)(2.5)
s=62.5+(-5)(62.5)
s=62.5-31.25
s=31.25
IT IS AN ACCELERATED MOTION SO, UTSAV HAS BEEN FOLLOWED.
SO, T=2.5s AND H=31.25
IS THE EXPLANATION AND ANSWER OF THIS QUESTION CORRECT ?

[abeast]

Sir is scolding children studying 2 months before exam
Me : studying before 1 day😱

[aaravmaheshwari]

To find time
Using first equation
V=u+at
0=25+-10t
25=-10t
25/10=t
2.5seconds Ans

To find displacement
Using 3rd equation
2as=v²-u²
2× -10×s=0-625
-20s =-625
s=625/20
s=31.25m Ans.

[Swati_]

1:48:54 -time-2.5 sec and height or displacement=31.25m

[youtuberedit]

Key moments:
00:00 Understanding displacement and distance is crucial in physics. Scalar quantities are measured without direction, while vector quantities require direction for measurement.
-Scalar quantities are measured without direction, like mass. Vector quantities, like displacement, require direction for measurement.
-Displacement is the shortest path between two points, regardless of the actual path taken. It remains constant even if the path changes.
-The concept of displacement helps in understanding the shortest distance between two points, irrespective of the actual path taken.
-Distance is the length of the path traveled between two points. It can be different from displacement, which is the shortest path between points.
10:07 Understanding the relationship between displacement, distance, and time intervals is crucial in determining uniform motion in physics.
-Displacement and distance are calculated in meters, emphasizing the importance of understanding these concepts in physics.
-The shortest path is always a straight line, highlighting the significance of straight-line motion in physics.
-Exploring the difference between displacement and distance, and how they relate to uniform motion and time intervals in physics.
-Illustrating uniform motion through examples of objects covering specific distances in set time intervals, emphasizing the concept of uniform motion.
20:09 Speed is the rate at which distance is covered in a specific time unit, typically measured in meters per second. Understanding speed, distance, and time is crucial in physics to calculate various motion-related parameters.
-Speed is typically measured in meters per second, indicating the rate at which an object covers distance in a specific time unit.
-Different units like miles per hour (mph) or kilometers per hour (km/h) can also be used to express speed.
-Understanding the concept of velocity, which includes both speed and direction, is essential in physics.
-Speed, distance, and time are interconnected concepts that play a crucial role in calculating various motion-related parameters in physics.
30:12 Displacement is zero in a round trip scenario where the initial and final points are the same. Average speed is determined by dividing total distance covered by total time taken.
-Displacement is the difference between the initial and final points of motion.
-Average speed is calculated by dividing total distance covered by total time taken.
40:17 Calculating average speed involves dividing total distance by total time. Understanding the relationship between distance, time, and speed is crucial for solving speed-related problems.
-Average speed formula is total distance divided by total time.
-Calculating speed involves multiplying distance and time, then dividing by the appropriate factor.
-Solving speed problems requires understanding the relationship between distance, time, and speed.
-Changing speed or velocity involves calculating the change in speed or velocity.
50:30 Explaining the concept of change in velocity and acceleration with examples, emphasizing the importance of direction in velocity changes.
-Acceleration can lead to changes in speed and direction, affecting velocity.
-Distinguishing between positive and negative acceleration, and the impact on speed.
-Negative acceleration does not always mean speed reduction, but a change in direction.
1:00:35 Acceleration and velocity change are crucial in racing scenarios. Unit conversion from kilometers per hour to meters per second is essential for accurate calculations.
-Acceleration plays a vital role in racing, with quick acceleration leading to speed gains within seconds.
-Unit conversion from kilometers per hour to meters per second is necessary for precise calculations in racing scenarios.
1:10:40 Understanding and correctly applying the three equations of motion is crucial for solving problems related to accelerated motion in physics.
-The importance of using the correct formula for calculating displacement and velocity in physics problems.
-The significance of practicing and mastering the three equations of motion to effectively solve problems related to accelerated motion.
1:20:52 Understanding concepts gradually and practicing regularly is crucial for solving numericals effectively in physics.
-Equations like motion, force, gravity, and work energy are used in physics chapters for solving numericals.
-Regular practice and slow understanding of concepts are essential for academic success.
-The video encourages students to wake up early, practice regularly, and focus on academic preparation for exams.
-The importance of parental support, resources, effort, and dedication in academic success is highlighted.
1:30:48 Understanding free fall involves considering acceleration due to gravity and initial velocity. Displacement and acceleration play crucial roles in determining the motion of an object in free fall.
-Acceleration due to gravity is 9.8 m/s^2 downwards.
-Displacement and acceleration are key factors in free fall motion.
-Initial velocity and direction impact the motion of an object in free fall.
-Understanding the concept of free fall involves considering acceleration and displacement.
1:40:52 Understanding displacement direction, positive and negative slopes, and calculating theta in trigonometry are crucial concepts in physics and mathematics.
-Displacement direction and positive displacement are essential for understanding motion.
-Negative displacement and the importance of following logic to avoid confusion in calculations.
-Explaining the significance of positive and negative slopes in graphs for understanding motion.
-Introduction to trigonometry concepts like theta and slope in graphs for calculating angles and understanding slopes.
1:50:56 Understanding slope in physics involves finding perpendicular distance and area under curve. Displacement is calculated by multiplying time and velocity in graphs for different types of motion.
-Perpendicular distance and dividing by base to find slope in an easy way.
-Calculating displacement by multiplying time and velocity in graphs.
-Understanding the concept of slope in physics using simple language.
-Explaining the process of finding area in graphs by multiplying coordinates.
2:01:00 Understanding velocity-time graphs helps interpret acceleration and displacement. Displacement is represented by the area under the velocity-time graph.
-Acceleration is the rate of change of speed with time. It is represented by the slope of the velocity-time graph.
-Displacement is the area under the velocity-time graph. It indicates the total distance traveled by an object.
-Understanding the relationship between velocity, time, acceleration, and displacement is crucial in analyzing motion.
2:11:05 The video explains the relationship between displacement, time, and speed, highlighting the need to grasp the fundamentals of differentiation before delving into mathematical formulas or theorems.
-Understanding the concept of displacement, time, and speed in a practical scenario.
-Importance of starting with the basics of differentiation before applying formulas or theorems.
-Explaining the significance of having the 'key' to unlock the understanding of mathematical concepts.
2:21:08 Acceleration formulas can be derived using simple mathematical methods, such as the graphical method, which involves practicing derivations multiple times for better comprehension.
-Acceleration formula derived as v=u+at, emphasizing the importance of practicing derivations multiple times for better understanding.
-Introduction to the graphical method for deriving acceleration formulas and the significance of practicing derivations repeatedly for improved comprehension.
2:31:14 Initial velocity affects equations; starting higher helps derive final velocity. Understanding initial velocity is crucial for solving physics problems effectively.
-Starting from a higher point ensures initial velocity is not zero, crucial for equation derivation.
-Deriving equations requires understanding the impact of initial velocity on the final result.
2:41:20 Uniform circular motion involves constant speed but changing direction, impacting velocity and acceleration. Tangents on circular paths determine velocity direction changes.
-Understanding uniform circular motion and its impact on speed and direction.
-Explaining the concept of tangents on circular paths and their relation to velocity direction changes.
-Discussing how velocity changes with direction changes and the concept of acceleration.
-Exploring the factors influencing speed, direction, and acceleration in uniform circular motion.
2:51:25 Understanding circular motion, centripetal acceleration, and displacement calculation is crucial for physics. Continuous learning and practice are essential for exam preparation.
-Centripetal acceleration always points towards the center at 90 degrees. Objects resist changing direction on circular paths due to external forces.
-Displacement calculation involves understanding the radius, revolution, and distance traveled in circular motion. Perimeter and circumference formulas are essential for calculations.

[pranvisinghrajput]

Anybody after board 2024
For base of class 11
👇

[thatsme]

Thank you so much sir this video helped me alot to understand the concept..
Specially UTSAV technique was amazing..
Thank you so much sir ☺️..