00:00 - Introduction 05:37 - Motion and rest 11:04 - Important terms 35:12 - Types of motion 45:57 - Equations of motion 1:46:28 - Motion under gravity 2:11:30 - Graph 2:15:53 - Slope 2:20:30 - Area under the curve 2:23:29 - Equations of motion (Graphical method) 2:36:07 - Graph analysis 3:06:30 - Derivation of equations of motion 4:03:51 -All the best all of you!
Shailendra sir, Kaas apke is tarah ke lecture 11th starting mein hote.... I missed it because on that time every youtube channels was teaching only for jee/neet content. None of one were found like you. 😊😊😊😊 Thanku so much sir. Love from Delhi NCR (Faridabad). ❤❤❤
@@urbestie_army yaa it enough but u need to focus on every line that sir write on the slides because he not repeat same thing again and again. Happy journey 😄😄
Sir there is correction To solve this problem, we'll break it down into two parts: the time it takes for the packet to reach the ground and the velocity at which it reaches the ground. 1. Time to reach the ground: We can use the equation of motion: s = ut + (1/2)gt^2 where: s = displacement (height) = 39.2 m u = initial velocity = 0 (since the packet is dropped) g = acceleration due to gravity = 9.8 m/s^2 t = time (which we want to find) Rearranging the equation to solve for t, we get: t^2 = 2s/g t^2 = 2(39.2)/9.8 t^2 = 8 t = √8 ≈ 2.83 seconds So, it takes approximately 2.83 seconds for the packet to reach the ground. 1. Velocity at which it reaches the ground: We can use the equation: v = u + gt where: v = final velocity (which we want to find) u = initial velocity = 0 g = acceleration due to gravity = 9.8 m/s^2 t = time ≈ 2.83 seconds (found earlier) v = 0 + 9.8(2.83) v ≈ 27.7 m/s So, the packet reaches the ground with a velocity of approximately 27.7 m/s. But you haven't clearly told why height was negative 😔😔
bro, when he says u=9.8 then why you take u=0 and if body moves in downward direction then h should be taken negative and sir is right and your answer is wrong. hope you understand👍
1:27:00 answer is 60km/h Sol- T1=30km /30kmh^-1 = 1hour T2=30/x Vavg=t.d.t/t.t.t 40=60km/total time taken Total time taken=40/60 =1.5hr Now, Let's fine the value of x T1+T2=total time taken 1hr+30/x=1.5 30/x=1.5-1 30/x=0.5 0.5x=30 x=60 answer
Sir I have purchased Arjuna Jee 2025 but jee preparation needs patience and I have it but school has no patience that's why for school preparation 😀 I'm here. Lectures are very helpful for us ❤❤❤❤ Thank you very much sir 😊😊😊
Before studying from him I was afraid so much of phy but he just made my day with his amazing lecture I am finding much difficulties in numerical only as it need more and more practise...
@@Nikhilthakur7466 haan Inka dekh lo aur bas ek topic sir ne Miss kra hai relative velocity aap ise kahi aur se padh lo aur inse pdhne ke baad mind map se pdh lena Lakshman sir ke
sir 1:52:50 wale question mein vertically upwards agar phenk rahe hai ball ko toh pehle ball upar jaegi then 0 velocity attain karne ke baad neeche ayegi toh uss hisaab se tower ki height ke saath voh ek extra distance bhi 2 baar cross kar rhi hai upar jaate waqt and neeche aate waqt
A__________O___________B 1st-take out time interval of A-O Then time interval of A-B Next, Subtract time interval of A-B from A-O Then, take out speed of O-B By putting distance(30km) and the time u just got from 3rd step. And it's done ✔
@@shruti893You didn't understand ? T1= 30/30 ho jayega aur T2=30/x 40(average Speed) = 60( Total Distance ) / 30/30 + 30/ x....T1 me same speed par same distance cover kiya hai toh 1 hojayega 😊
Sir Displacement is a vector quality and distance is scalar soo how we can say that Distance is greater than or equal to displacement I think this is wrong statement may be ye ho sakta hai Distance ≥ | Displacement | iske magnitude ke equal ya greater ho sakta hai
hello sir...sir apne gravity k first part me bola ki object neeche drop ho rhi hogi to positive g hoga...to fir baloon wale question me ut -1/2gt2 kyn???
1:52:36 but in this case two types of acceleration due to gravity are there first is when the ball is thrown up that will be negative g, when ball reaches v=0 then it comes down there positive acceleration of g is applied then why did we only took negative acc for the whole case?
a body travels from A to B at 40 metre per second and from B to A at 60 metre per second calculate the average speed and average velocity? sir in this question 60m/s and 40m/s both are speed so why are you take this as a distance ? Sir please reply my comment
@@Priya-hg3qv Yes, I know the answer..... For eg If you find mean of 3 random numbers you will add all the three numbers and divide it by 3 and you'll get the answer but in case of speed, speed depends upon time as well as distance both, in this question let's say 1:26:44 the 1st 30km is travelled with speed 30km/h , it means it is travels 30 km in 1 hour but 2nd case me speed 60km/ h hai it means vo distance1/2 hr me cover kiya h so time in both cases is different so we can't directly add and find the averag Abr ye uniform motion ka case hota to direct nikal skte the but non uniform motion me average nikalne ke liye hme ye pta hona chahiye ki ye distance kitne time me cover kiya h I hope you understood that 😊
Thanks a lot sir , you teach so well, i have a very weak base in physics and i don't understand physics at all bit today i understood it very well , i was revising this vhapter because tomorrow is my exam , Just praying i score well 🙏🏻🌷✨
Since the helicopter is rising steadily at 2 m/s, the packet will have an initial upward velocity of 2 m/s when released. After 2 seconds, the packet's velocity can be found using the equation: v = u + gt where: v = final velocity u = initial velocity (2 m/s, upwards) g = acceleration due to gravity (9.8 m/s^2, downwards) t = time (2 seconds) v = 2 m/s - (9.8 m/s^2 × 2 s) v = 2 m/s - 19.6 m/s v = -17.6 m/s The negative sign indicates that the packet is now falling downwards. To find the distance between the packet and the helicopter, we need to calculate the distance traveled by both the packet and the helicopter: Distance traveled by packet: s = ut + (1/2)gt^2 = (2 m/s × 2 s) + (1/2 × 9.8 m/s^2 × 2^2) = 4 m + 19.6 m = 23.6 m Distance traveled by helicopter (rising at 2 m/s): s = ut = (2 m/s × 2 s) = 4 m Since the helicopter has risen 4 meters, and the packet has fallen 23.6 meters, the distance between them is: 23.6 m + 4 m = 27.6 m So, the packet is 27.6 meters below the helicopter after 2 seconds.
Pehle sb given likh lo Fir Jo question me poocha hai wo likh lo Or uske baad dekho ki kis equation se sabse jyada variable match ho rhe hai whi lga do.....
From class 9th i was confused in speed, velocity and specially in acceleration Well from this video what i have understood is acceleration is nothing but "non-uniform velocity" Correct me if I'm wrong
Dekho dost mai bhi bhut confused tha fir maine chat gpt se pucha to usne bhut ache se smjhaya hai sab . Us se puch lo .prompt mai bta deta hu ye daalna " teach me Velocity, acceleration ( jo bhi tumhe confusing Lage rha) like I'm a 5 year old . Define in hinglish
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Very nice lecture ❤❤❤❤❤❤❤❤❤❤❤
Proper explanation and hardwork done by sir
2:36:02 Mamla ek dum chakachak hai😅
Sir please provide session pdf
Thankyou so much sir ❤
We want new One shot series for 2024-25 Batches 😢
Manzil series ❤
Yess
UDAY SERIES
My search for the best physics teacher for boards ended here
Lots of love to Shailendra sir❤
Yes❤❤
@@Manshi98hii
Bilkul sahi kaha ❤❤
Same
Bro ya boards ka liya thk hai na
00:00 - Introduction
05:37 - Motion and rest
11:04 - Important terms
35:12 - Types of motion
45:57 - Equations of motion
1:46:28 - Motion under gravity
2:11:30 - Graph
2:15:53 - Slope
2:20:30 - Area under the curve
2:23:29 - Equations of motion (Graphical method)
2:36:07 - Graph analysis
3:06:30 - Derivation of equations of motion
4:03:51 -All the best all of you!
Thankuh bhai
Where is dif.and int. ???
Copy paste 💩💩💩
🙏🙏🙏🙏🙏
Thanks bro
Shailendra sir, Kaas apke is tarah ke lecture 11th starting mein hote.... I missed it because on that time every youtube channels was teaching only for jee/neet content. None of one were found like you. 😊😊😊😊 Thanku so much sir. Love from Delhi NCR (Faridabad). ❤❤❤
Bhai ncert ka level ka cover hojayega na?
@@I_eat_Childrens_hahahhaa Bhai hojaega😊
Hey guys i m not preparing for jee Or neet so for only grade 11 is it enough to score a pretty good marks?
@@urbestie_army yaa it enough but u need to focus on every line that sir write on the slides because he not repeat same thing again and again. Happy journey 😄😄
@@manish_coding ok do i need any reference book to score good or ncert + RUclips teachers are enough to practice question
sir's handwriting is just amazing
agree?👇
yes i also notice it
00:00 - Introduction
05:37 - Motion and rest
11:04 - Important terms
35:12 - Types of motion
45:57 - Equations of motion
1:46:28 - Motion under gravity
2:11:30 - Graph
2:15:53 - Slope
2:20:30 - Area under the curve
2:23:29 - Equations of motion (Graphical method)
2:36:07 - Graph analysis
3:06:30 - Derivation of equations of motion
4:03:51 - Thank You
1:27 :01
Ans : t1= 30/30=1
t2=30/v2
Vavg=60(total distance)/1+(30/v2)=1200/20=60kmh^-1
⏳Timestamps:
00:00 - Introduction
05:37 - Motion and rest
11:04 - Important terms
35:12 - Types of motion
45:57 - Equations of motion
1:46:28 - Motion under gravity
2:11:30 - Graph
2:15:53 - Slope
2:20:30 - Area under the curve
2:23:29 - Equations of motion (Graphical method)
2:36:07 - Graph analysis
3:06:30 - Derivation of equations of motion
4:03:51 - Thank You Bacchon!😊
Copy paste 💩
1:27:16
Ans . The next 30 km is travelled with the speed of 60 km/h .
Sir there is correction
To solve this problem, we'll break it down into two parts: the time it takes for the packet to reach the ground and the velocity at which it reaches the ground.
1. Time to reach the ground:
We can use the equation of motion:
s = ut + (1/2)gt^2
where:
s = displacement (height) = 39.2 m
u = initial velocity = 0 (since the packet is dropped)
g = acceleration due to gravity = 9.8 m/s^2
t = time (which we want to find)
Rearranging the equation to solve for t, we get:
t^2 = 2s/g
t^2 = 2(39.2)/9.8
t^2 = 8
t = √8 ≈ 2.83 seconds
So, it takes approximately 2.83 seconds for the packet to reach the ground.
1. Velocity at which it reaches the ground:
We can use the equation:
v = u + gt
where:
v = final velocity (which we want to find)
u = initial velocity = 0
g = acceleration due to gravity = 9.8 m/s^2
t = time ≈ 2.83 seconds (found earlier)
v = 0 + 9.8(2.83)
v ≈ 27.7 m/s
So, the packet reaches the ground with a velocity of approximately 27.7 m/s.
But you haven't clearly told why height was negative 😔😔
bro, when he says u=9.8 then why you take u=0 and if body moves in downward direction then h should be taken negative and sir is right and your answer is wrong. hope you understand👍
@@arpitsrivastav1118 Brother I said there is correction ig
u=0 because object is dropped/released
@@reetasingh9672 I think you should search this question on google
@@arpitsrivastav1118 I've checked on Meta AI😅 dono hi way glt ni h but thnks
Hi
2024-2025 batch attention pls😅
😢@@MaahiAadi-bc4pn
😮
🤔🤔🤔
uss question ka answer is 30km/h
❤❤❤❤❤
The teacher is here to change the full RUclips history🤫🤫💥
Like 👍 students watching this lecture in the academic session 2024-25
2:11:32 ----- 17.6 m÷s and -19.6
Same ans19.6
Same
Shouldn't it be -17.6 m/s as acc it - ?
Yes
@@NEWZ_EDITZ19.6
भाइयों कभी हार मत मानना क्योंकि आपकी मां आपके जीत का आसाय लगा बैठी है ❤❤❤❤😢❤😢❤😢😢😢😢
❤❤❤❤❤❤❤❤❤❤❤ aap kinse school me ho
Kaha se ho sister , tumhari baatein ❤
🥹🥹🥹🥹🥹🥹🥹🥹🥹❤❤❤❤❤❤❤
Papa bhi
Thanks a lot sis❤
1:27:00 answer is 60km/h
Sol- T1=30km /30kmh^-1 = 1hour
T2=30/x
Vavg=t.d.t/t.t.t
40=60km/total time taken
Total time taken=40/60 =1.5hr
Now,
Let's fine the value of x
T1+T2=total time taken 1hr+30/x=1.5
30/x=1.5-1
30/x=0.5
0.5x=30
x=60 answer
Wrong the answer is 30km per hr
Yes write ans
Answer is right
My answer exactly same
Yes@@NOELSE
Kon kon 2024. 2025 batch wala hai Like karo❤❤❤❤❤
Hi
Hey, jo ye sir padha rhe isme deleted syllabus nai haina ?
Nhi hai deleted topics@@I_ll_win
Me
⏳Timestamps:
00:00 - Introduction
05:37 - Motion and rest
11:04 - Important terms
35:12 - Types of motion
45:57 - Equations of motion
1:46:28 - Motion under gravity
2:11:30 - Graph
2:15:53 - Slope
2:20:30 - Area under the curve
2:23:29 - Equations of motion (Graphical method)
2:36:07 - Graph analysis
3:06:30 - Derivation of equations of motion
4:03:51 - Thank You likeeeeee
Copy paste 💩
Teaching style wow
Explanation wow ❤
Hair cut and sir you looking a wow
Explanation wow wow wow 😅
शिक्षा से बड़ा कोई ज्ञान नहीं और गुरु से मिले आशीर्वाद इससे बड़ा कोई सम्मान नहीं ❤❤❤❤😢😊😊😊 ______ love you sir ♥️🥰
Ha
You are absolutely right 🙏
Hello ji navya
Yes
Itna gyaan kyo de rahe ho
2:10:00. (I)= -17.6
--------------(ii)= 19.6
Anybody from 2024-2025 batch >>>>>>>>>>>>>>>>>
Here
Me
X=60 km track
X=30kmh speed
Train travel next.
60/40= 1.5 t.D.t/ t.t.t
1.5=we right (2).
30/2
=60 m/s speed Hp
Sir I have purchased Arjuna Jee 2025 but jee preparation needs patience and I have it but school has no patience that's why for school preparation 😀 I'm here.
Lectures are very helpful for us ❤❤❤❤
Thank you very much sir 😊😊😊
same here bro...
@@KhushiSahani-hk4rq Ek thali ke chate bate hi hain :)
Sir, 2:09:57, mujhe displacement 4+19.6=23.6 mila, par doosro ko 15.6 mila? Sir, shouldn't g be +9.8m/s
Before studying from him I was afraid so much of phy but he just made my day with his amazing lecture I am finding much difficulties in numerical only as it need more and more practise...
Bhai inke one shot dekh kr scl exams mei acche marks score kr skte hai kya?
@@Nikhilthakur7466 haan Inka dekh lo aur bas ek topic sir ne Miss kra hai relative velocity aap ise kahi aur se padh lo aur inse pdhne ke baad mind map se pdh lena Lakshman sir ke
@@Nikhilthakur7466 mind map kevl 50 min max rhega vo bhi 2× me krke dekhoge to kaam ban jaayega
@@atul-vh3zv hey can you provide me with the link.
@@STUDYNEZUKO-js2wr what kind of link??
Who is seeing this video on November
Me
1:26:56 x = 60
t1=30km/30km per h= 1 hour
t2=30/x
Vavg=40km per hour
40=60/1+30/x
solve for x we get x=60
Wrong hai bhai
Answers is correct but process is wrong
Homework question 1:27:20 answer is 60km/h
Kaise bata do
Is me kya find karna hai kaise
💯 done sir thanks for the amazing session JAI HIND SIR 🎉🎉🎉🎉❤❤❤❤❤❤❤❤❤
Are you a NDA ASPIRANT...,?
2:11:22
Velocity= (+17.6ms-1)
S=(+19.6m)
1:27:18 solution 60km per hour
sir 1:52:50 wale question mein vertically upwards agar phenk rahe hai ball ko toh pehle ball upar jaegi then 0 velocity attain karne ke baad neeche ayegi toh uss hisaab se tower ki height ke saath voh ek extra distance bhi 2 baar cross kar rhi hai upar jaate waqt and neeche aate waqt
Who is watching in September 😅
Me
Me 😅😅
@@Shekhawatakshu all the best 👍
@@Anas_khan_69 thnx✨
Who is watching in October
27oct..🙌
31 oct
Today ☠️ for first time
@@somu_sempai matlab 11th barbaad ho jayega
Matt krna yarr
1:27:14 ANSWER =60km/h.. sir equation of motion agyi smj thanks for amazing lectures ❤❤❤❤ *KOP*KING OF PHYSICS ❤❤
How can you explain??
A__________O___________B
1st-take out time interval of A-O
Then time interval of A-B
Next, Subtract time interval of A-B from A-O
Then, take out speed of O-B
By putting distance(30km) and the time u just got from 3rd step.
And it's done ✔
@@Uzma-7-8-6 ok thank you
@@shruti893You didn't understand ? T1= 30/30 ho jayega aur T2=30/x
40(average Speed) = 60( Total Distance ) / 30/30 + 30/ x....T1 me same speed par same distance cover kiya hai toh 1 hojayega 😊
@@Reaper-px6lw ouu Okie got it thenku😃👍🏻
Sir AAP jitna bhi time le lo 7-12 ghante but pura concept padha do questions k sath koi bhi topic skip mat Karo 🥰🥰🥰 love you sir
Hlo
1:27:09 hour ka question ka ans 60km/h..
Sir Displacement is a vector quality and distance is scalar soo how we can say that Distance is greater than or equal to displacement I think this is wrong statement may be ye ho sakta hai Distance ≥ | Displacement | iske magnitude ke equal ya greater ho sakta hai
Yes bro
You are right 👍
Bro distance can be greater or equal to displacement this is the correct statement
2:11:45 sir first answer is 20.4m/s
2nd answer is 22.4m below from helicopter 🚁
Send total solution
Fastly
I think it's wrong
Is this lecture enough for class 11 boards???
Sir instantaneous speed instantaneous velocity kaha hai sir😢😢
Whi n yar
@@BHUMIKASINGH-sb4mt sir ne skip kyu kr diaa
SIR I AM IN CLASS 11TH FROM 2024-25 BATCH.. AND I WANT TO PREPARE FOR SCHOOL LEVEL EXAMS.. IS THIS VIDEO ENOUGH FOR DETAILED EXPLANATION OF NCERT?
U were a very much needed teacher❤
2:00:39 sir isme shuru me equation of motion ki 2nd equation bhi to lag sakti thi ?? Any help??
Haa laga skte hai
1:38:40 imp ques
1:49:53 motion under gravity questions start
Sir your lecture is awesome I have no doubt for this chapter but that DPP is provide has relative velocity questions so how can we solve that question
hello sir...sir apne gravity k first part me bola ki object neeche drop ho rhi hogi to positive g hoga...to fir baloon wale question me ut -1/2gt2 kyn???
Me BHI yah I soch Raha hu Abhishek sir me negative bataya hai😢😢
1:52:36 but in this case two types of acceleration due to gravity are there first is when the ball is thrown up that will be negative g, when ball reaches v=0 then it comes down there positive acceleration of g is applied then why did we only took negative acc for the whole case?
at 2:04 hr ballon question why are you taking acceleration in negative sign....you are wrong here
Yess
Sir bhut aacha lecture tha love from uttarakhand
Sir please ek shedule bataye ki kis tarah se padhe
Bhai ase koi ni btata pdna to suru kr apne ap samajhne me a jayega
Sir pls ek week m 3 lecture physics ka plz sir 12 jan se exam hai plz plz sir ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤
How was your exam? 😊
velocity ... 17.6 m per second and the packet is 19.6 m away from the helicopter.
Can solved it
a body travels from A to B at 40 metre per second and from B to A at 60 metre per second calculate the average speed and average velocity? sir in this question 60m/s and 40m/s both are speed so why are you take this as a distance ? Sir please reply my comment
Sir please answer my doubt
@@Priya-hg3qv yeh video purani hain ..google karo answer mil jayga
Google par dusra answer bta rha hai
@@Priya-hg3qv Yes, I know the answer.....
For eg If you find mean of 3 random numbers you will add all the three numbers and divide it by 3 and you'll get the answer but in case of speed, speed depends upon time as well as distance both, in this question let's say 1:26:44 the 1st 30km is travelled with speed 30km/h , it means it is travels 30 km in 1 hour but 2nd case me speed 60km/ h hai it means vo distance1/2 hr me cover kiya h so time in both cases is different so we can't directly add and find the averag
Abr ye uniform motion ka case hota to direct nikal skte the but non uniform motion me average nikalne ke liye hme ye pta hona chahiye ki ye distance kitne time me cover kiya h
I hope you understood that 😊
Sir class was amazing
But plz ab initio method sai bi kijiye differentiation plz..
.
Thanks a lot sir , you teach so well, i have a very weak base in physics and i don't understand physics at all bit today i understood it very well , i was revising this vhapter because tomorrow is my exam ,
Just praying i score well 🙏🏻🌷✨
How much did you scored ?
90+ ?
Since the helicopter is rising steadily at 2 m/s, the packet will have an initial upward velocity of 2 m/s when released.
After 2 seconds, the packet's velocity can be found using the equation:
v = u + gt
where:
v = final velocity
u = initial velocity (2 m/s, upwards)
g = acceleration due to gravity (9.8 m/s^2, downwards)
t = time (2 seconds)
v = 2 m/s - (9.8 m/s^2 × 2 s)
v = 2 m/s - 19.6 m/s
v = -17.6 m/s
The negative sign indicates that the packet is now falling downwards.
To find the distance between the packet and the helicopter, we need to calculate the distance traveled by both the packet and the helicopter:
Distance traveled by packet:
s = ut + (1/2)gt^2
= (2 m/s × 2 s) + (1/2 × 9.8 m/s^2 × 2^2)
= 4 m + 19.6 m
= 23.6 m
Distance traveled by helicopter (rising at 2 m/s):
s = ut
= (2 m/s × 2 s)
= 4 m
Since the helicopter has risen 4 meters, and the packet has fallen 23.6 meters, the distance between them is:
23.6 m + 4 m = 27.6 m
So, the packet is 27.6 meters below the helicopter after 2 seconds.
Why u had taken acceleration as negative in first question?😅
Amazing lecture 🙏☺️ thank u sir jiii
is this lecture of ch 2 enough for mid-term exam
It's all depends upon you keep practicing numericals then it will be enough for Exam !
2:11:10
i) -17.6 m/s
ii)-19.6 m
Thank you very much sir ji 👍👍 for this fantastic lec very nice explanation crystal clear sir ji ❤️ am neet aspirant from Arjuna batch ❤️❤️❤️❤️❤️❤️❤️
This is sufficient video for us,
Thank you sir
We all proud of you
Does this lecture also cover the deleted syllabus for 2024-25 batch?
Yes
SIR AAP KI HAIRSTYLE KING JONG KI TARAH HAI😘😘
Mam class 11th school exam walo ka liya full year batch bhi launch kar dena agala saal please hame bhi pass hona hai 🥺🥺
thanks you sir your videos always helps a lot, i am preparing for 11th by seeing all your videos
1:55:08 sir mene 2nd method bhi kar liya
thank you very much sir
this has really build my basic concepts
Are u preparing for boards from this video or rivision
hi
Sir please cover the syllabus fast my exams are near
00:00 - Introduction
05:37 - Motion and rest
11:04 - Important terms
35:12 - Types of motion
45:57 - Equations of motion
1:46:28 - Motion under gravity
2:11:30 - Graph
2:15:53 - Slope
2:20:30 - Area under the curve
2:23:29 - Equations of motion (Graphical method)
2:36:07 - Graph analysis
3:06:30 - Derivation of equations of motion
4:03:51 - Thank You Bacchon!
Sir you skip instantaneous velocity and instateneous acceleration
Sir kaise pta chalega ki kb konsi equation of motion lgani hai
Pehle sb given likh lo
Fir Jo question me poocha hai wo likh lo
Or uske baad dekho ki kis equation se sabse jyada variable match ho rhe hai whi lga do.....
Why you right ds upon dt in derivation by calculus in 3 rd equation
In the end we all become stories ✨❤️🩹💐
1:25:44 if we solve 2AB/AB ,, it will be AB not 2..... 💔💔💔
AB is numerator and denominator
Thank you so much sir the lecture is awesome ❤❤❤❤❤❤❤❤❤❤
2:10:30 ans for (i) 21.6 m/s (ii) 8m is it correct?
It is enough for CBSE class 11th physics exam
1:27:11 - homwwork answer- 60 km /hour
sir according to quantum mechanics it is possible that a object present at different point
Yeah this is true
Sir aap time ka tension mt lo please lecture ko lamba krdo
1:27:20 ans = 60 km/h
Same
@@AdityaPratapsingh-h5k same
your teaching is marvelous
:D
Who is watching in October😅❤
From arjuna neet cz of backlogs for half yearly 🤧🙃🍃
From class 9th i was confused in speed, velocity and specially in acceleration
Well from this video what i have understood is acceleration is nothing but "non-uniform velocity"
Correct me if I'm wrong
Dekho dost mai bhi bhut confused tha fir maine chat gpt se pucha to usne bhut ache se smjhaya hai sab . Us se puch lo .prompt mai bta deta hu ye daalna " teach me Velocity, acceleration ( jo bhi tumhe confusing Lage rha) like I'm a 5 year old . Define in hinglish
Thx devdutt :)
Thanku so much sir for this amazing lecture ;;; love from kashmir ❤❤❤
21.6 =velocity,and 47.2 m
1:34:00 and 2:10:00 best question to solve
Answers kitne aaye
1:27:17
Answer is 60kmh-1
Sir your handwriting is very very good 👍
Shalindar sir ko kaun kaun dil se pasand karta h
are behen padho 😅
Hmm
@@devshukla9440 haa gjb 😊😊
Sir acha nhi bhut shi pada te hai
Thik padate h
1:27:00 question ka answer 60m/sec aaya hai sir
@@Nancy___11921 oh bro Thank you for correcting my mistake ☺️☺️☺️
Sir, is Derivation important in class 11 of equations of motion?
Sir non uniform acceleration
Concept of differentiation and integration mare syllabus mai hai 😢😢
Your strategy of way of learning is unique and awesome ❤❤
sir aap bhoutttt accha padhate hoo.. thanku soo much humare liye itni mehnat karne ke liye
So beautiful so elegant just 😎 teaching like wow ✨ wow