Live Classes, Video Lectures, Test Series, Lecturewise notes, topicwise DPP, dynamic Exercise and much more on Physicswallah App. Download the App from Google Playstore ( bit.ly/2SHIPW6 ) Physicswallah Instagram Handle : instagram.com/physicswallah/ Physicswallah Facebook Page: facebook.com/physicswallah Physicswallah Twitter Account : twitter.com/PhysicswallahAP?s=20 Physicswallah App on Google Play Store : bit.ly/2SHIPW6 Physicswallah Website: physicswallahalakhpandey.com/ LAKSHYA Batch(2020-21) Join the Batch on Physicswallah App bit.ly/2SHIPW6 Registration Open!!!!
What will you get in the Lakshya Batch? 1) Complete Class 12th + JEE Mains/ NEET syllabus - Targeting 95% in Board Exams and Selection in JEE MAINS / NEET with a Strong Score under Direct Guidance of Alakh Pandey. 2)Live Classes and recorded Video Lectures (New, different from those on UA-cam) 3)PDF Notes of each class. 4)DPP: Daily Practice Problems with each class having 10 questions based on the class of JEE Mains/NEET level. 5)Syllabus Completion by end of January, 2021 with topicwise discussion of Last 10 Years Problems in Boards, JEE Mains/NEET within Lecture. 6)The Complete Course (Video Lectures, PDF Notes, any other Study Material) will be accessible to all the students untill JEE Mains & NEET 2021 (nearly May 2021) 7)In case you missed a live class, you can see its recording. 8)You can view the videos any number of times. 9)Each chapter will be discussed in detail with all concepts and numericals 10)Chapterwise Approach towards JEE Mains/ NEET & Board Exams. ****Test Series for XI & XII**** We provide you the best test series for Class XI,XII, JEE, NEET chapterwise, which will be scheduled for whole year. The test series follows very logical sequence of Basic to Advance questions.& Evaluation of Test and Solution to all the questions at the end of the test. Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 01: Introduction || Average Speed ua-cam.com/video/XIJAZM5G5Fg/v-deo.html Class 11 Chapter 3 Kinematics: Differentiation || Calculus part 01 || Mathematical Tool ua-cam.com/video/UiVYFFRs_BE/v-deo.html Class 11 Physics Chapter 03 : KINEMATICS :Motion in a Straight Line 02 || Instantaneous Velocity || || IIT/ NEET ua-cam.com/video/9nBWFM7XTEw/v-deo.html Class 11 Chap 3: KINEMATICS || INTEGRATION || ||Calculus Part 02 || Mathematical Tools || ua-cam.com/video/_dmHLxqbXYU/v-deo.html Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 03 || Answer Batao Salute Pao #Physicswallah || ua-cam.com/video/-PKUJCIJSfc/v-deo.html Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 04 ||Derivation Of Equations Of Motion Using Integration|| ua-cam.com/video/AxhkozLNDO4/v-deo.html Difference Between Diaplacement and Distance || IIT JEE MAINS AND ADVANCE CONCEPT || ua-cam.com/video/4IkIGhuhuxo/v-deo.html 11 chap 03 : Kinematics 05 | Displacement time Graph -Velocity time Graph - Acceleration time Graph ua-cam.com/video/4OgPa7MKjRA/v-deo.html NEET Best Questions 01 || Motion in a Sraight Line 1 || KINEMATICS NEET || Motion in One Dimension ua-cam.com/video/oq-TjVH-sXc/v-deo.html 11 Chap 03 :Kinematics 06 || Motion Under Gravity || Motion in a Straight Line || Class 11 / JEE || ua-cam.com/video/G0ooGZA_gyc/v-deo.html Relative Velocity || Kinematics|| Motion in a Straight Line 08 || Class 11 Chapter 4 || JEE MAINS ua-cam.com/video/3sLGCWGm610/v-deo.html Best Method For Rain Man Problems | Relative Velocity | Motion in a Plane | Kinematics JEE NEET ua-cam.com/video/tEltpY8vBPM/v-deo.html River Boat Problem || Relative Velocity in 2D || River Man Problem || Motion in a Plane || JEE NEET ua-cam.com/video/2QFRRL50-ds/v-deo.html Projectile Motion 01 || Class 11 chap 4 || Motion in a Plane|| Motion in 2-D || ua-cam.com/video/iUi1M7YkDe4/v-deo.html Projectile Motion 02 || Class 11 chap 4 || Motion in a Plane || Projectile from a Height || ua-cam.com/video/dEqKhmTTXJk/v-deo.html Projectile Motion 03|| Equation Of Trajectory || Derivation Of Equation Of trajectory|| Range Form ua-cam.com/video/js_G3G02AkU/v-deo.html Mock Test 01 | Kinematics Best Questions for IIT JEE NEET | IIT JEE Questions Kinematics ua-cam.com/video/jYLGziSgI-Y/v-deo.html 11 Chap 4 || Circular Motion 01 || Angular Velocity and Angular Displacement || IIT JEE /NEET ua-cam.com/video/JxttUGnNuCU/v-deo.html 11 Chap 4 | Circular Motion 02 | Centripetal and Tangential Acceleration | Angular Acceleration | ua-cam.com/video/pOvYwchAvCA/v-deo.html Circular Motion 03| Centripetal and Centrifugal Force IIT JEE/ NEET | Conical Pendulum |Death Well | ua-cam.com/video/Wt9_bvjCFf0/v-deo.html 11 chap 4 | Circular Motion 04 | Derivation of Centripetal Acceleration or Centripetal Force | ua-cam.com/video/QYbyobdXlrg/v-deo.html 11 chap 4 | Circular Motion 05 | Banking Of Road IIT JEE NEET | Banking of Road with Friction | ua-cam.com/video/C0uVC700EIA/v-deo.html 11 chap 4 || Circular Motion 06 || Motion in a Vertical Circle IIT JEE / NEET || Critical Velocity ua-cam.com/video/e0Tty0V1q2U/v-deo.html 11 chap 4 || Circular Motion 07 || Motion in a Vertical Circle On a Bowl || IIT JEE MAINS / NEET ua-cam.com/video/_WMTtzIZQR8/v-deo.html
Who want Alakh sir as own school teacher ?? 😌😌😌💓💓💓 Edit: it's amazing to see that, students are still watching the old lectures of this "dariwale inshaan 🌝'❤️.... while he has launched so many new batches....this is the power of marker nd white board 😎😎
Sir I joined a coaching recently (fees-3.5 lacs) and the teacher is taking online classes from last 1 month and he is teaching physics too complex to understand . I cried too much as I was not able to understand the concepts then I preferred your video to clear my concepts ...THANK YOU SIR ...you are the GOD of students ...😇
Answer of last question: Given, a = x² We know that a = dv/dt a = dv/dx × dx/dt a = v dv/dx v dv = a dx v dv = x² dx Integrating the above equation; v²/2 = x³/3 + C ... (1) Given when x = 0, v = 0 Putting in Eq. (1), 0 = 0 + C C = 0 Putting in Eq. (1), v²/2 = x³/3 At x = 2 m, v²/2 = 8/3 v² = 16/3 v = ± 4/√3
ALAKH SIR>>>>>>>>>> ITS BEEN 4 AND HALF YEARS SINCE THIS VIDEO AND HE HAS LAUNCHED HIS OWN APP AND OPENED INSTITUTES ALL OVER INDIA ....BUT STILL NO ONE CAN BEAT HIS METHOD OF TEACHING AND EXPLAINING SO WE GET THE FEEL OF THE SUBJECT THANK YOU SO MUCH SIR
@@shephalipatra4356 anyone who knew his channel before he was big would get the feeling. I was subscribed to him since 2017 and these type of videos gives strong nostalgia
After reading the comments from many of our seniors.. I realised that I have chosen a right platform for my neet exam prep. Million tons of thanks to you sir !!
Till date no one can explain better than this ...lakhs of coaching are nothing in front of these short videos...anyone preparing for JEE NEET has to just watch this and practice and excell...best ever teacher in century..in future it is going to be huge huge wonder to students...
The answer of solution would be 4underroot3/3 . a =dv/dt a =dv/ds.ds/dt ( in order to eliminate dt I have done this . a = v.dv/ds Now integrate this .(here a=x², ds =x ) Integration of x².dx = integration of v.dv It would be x³/3 = v²/2 +c Put the value when body initially at rest C=0 Now put x= 2 8/3 = v²/2 Underroot 16/3 = v²/2 Answer would be 4 underoot 3 would come . Thank you very much sir . I am feeling confident . 😊😊
You need to visit pradeep kshetrepal sir's channel 50years+ experienced teacher physics wallah is like a child on the front of the legendary pradeep kshetrapal
Bhaiya mjhe inse samjh ni ata tunning hi nii hoti .. As a junior apse puchna tha mai abhi hi 11th me aya hu konse teacher se pdhu kinematics 🙏 bta dijye
@@prajwalkulkarni4530 Bhaiya maine bhaiya maine kr liya physics galaxy se Aur mai shuru se hi mehnat kr rha full potential .. Results aaye n aaye uska nhi soch rha bs apna full potential de rha . Starting se.. Kia mera ye decision shi h .. (( Jee student - aakash me pdhta hu ))
Last question answer is a=dv/dt dv/dt=x² dv/dx×dx/dt =x² dv/dx×V=x² Vdv=x²dx Integrating both sides V²/2=x³/3+C x=0(given) V=0(given) Applying in EQ.=>C=0 Therefore, V²/2=x³/3 If x=2m V²=(2³/3)×2 V=√(8×2/3) V=√(4×4/3) V=4/√3 m/s OR V= 4√3/3m/s This is the method😊
Answer for the last question: We know that a=dv/dt (a=dv/dt ×ds/ds ) a=dv/ds × ds/dt a=(dv/ds)v {where ds/dt=v} a= vdv/ds Given a=x² We know that a=vdv/ds Integrating on the both the sides vdv/dx=x²dx v²/2=x³/3+c Given that v=0; x=0 On solving the equation (v²/2=x³/3+c) We get c=0 v²/2=x³/3 v²/2=2³/3 v²=2⁴/3 v=root of(2⁴/3)=-+4/root3 I know I am too late but ...still😃
Given, a = x² We know that a = dv/dt a = dv/dx × dx/dt a = v dv/dx v dv = a dx v dv = x² dx Integrating the above equation; v²/2 = x³/3 + C ... (1) Given when x = 0, v = 0 Putting in Eq. (1), 0 = 0 + C C = 0 Putting in Eq. (1), v²/2 = x³/3 At x = 2 m, v²/2 = 8/3 v² = 16/3 v = ± 4/√3
Edit: it's amazing to see that, students are still watching the old lectures of this "dariwale inshaan 🌝'❤️.... while he has launched so many new batches....this is the power of marker nd white board 😎😎
Preparing for neet 2024 Pr abhi tk 11th aur 12th physics clear nhi hua dekhyea aage kya hota hai mera! Abhi toh sir ke wjh se sb clear hota ja rha hai hope I'll crack neet this year or next ,🥺
a=x^2 vdv=x^2dx Integrating both sides with limits 0-v on left side and 0-2 on right side vdv becomes v^2/2 = x^2dx becomes x^3/3 Solving with limits it becomes v^2/2=8/3 v^2 = 16/3 (transposing 2 from left hand side to right hand side (8x2)/3) therefore, v=+- 4/root 3 m/s at x=2m
Answer is 4/√3. Explanation - a = (dv/dx)(dx/dt) a = (v)dv/dx a dx = v dv Integrating both sides, we get, x^3/3 + c1 = v^2 /2 + c2 After applying the condition we will find that c1 = c2 So, x^3/3 = v^2/2 v^2 = 8 * 2 / 3 v^2 = 16/3 v = √16/√3 v = 4/√3 or v = 4√3/3. Thanks.
@@aarusharya1544 where is time in the q? See, ({ is integration) a = x² v = {adx => v = x³/3 + C Now, it is given that, x = 0 when v = 0 => 0 = 0³/3 + C => C = 0 To find, v when x = 2, put 2 at x, v = 2³/3 m/s v = 8/3 m/s
Soo till 10th i had my fav teacher on UA-cam was bkp -bhai kii padhaai ...and in 11th it's physics wallah alakh pandey ... really great sirr..hats off to u 🙌🙌✨✨
Sir no words for you... Physics become a easiest subject for me ..on!y because of you... Take my respect sir.. Hats off to you.. What a lectures yrr... 😘😉😊😇🤗☺☺☺☺
@Isha_ actually mera bhi pehle 8/3 hi aa raha tha ans but is ch ki playlist ke nxt video me sir ne iska solution karaya hai video ke end main....usme 4/√3 hi aa raha hai ans
@@taegukie yaa why not look its given that a= x^2 so if we will intergate this than v= x^3/ 3 now ita given x= 2 so putting value of x in velocity so 2^3/3 then v= 8/3 when x is 2
Wrong pta hai kaise ??????? Mai bta ta hu Integration ke time humko x^2 ko totally integrate krna hoga . Na ki pehle hi x ki value put krna hai . Then aakhir me 8/3 +C
Sir u r the best teacher in the world your teaching style is very different for another teacher u r great sir in future we need this type of teacher for every students and in all educational system 👩🏫🤲🤲
The answer is a=x^2 V. (dV/dx) =x^2 Vdv=x^2dx Take Integration of both side V^/2=x^/3+c Put x=0, V=0 You get c=0 V=root over(x^3/3). 2 Put x=2 Ans:4/√3
sir i still watch these old videos of yours because i belive that knowledge is same and the dedication is still the same no matter taught on white board with that sasta caera on on smart board with studio. Hats off to you sir hope this comment reaches alakh sir and reply me
Solution Step 1: Given that: Acceleration in the body; a=x^2 At x=0 , the velocity of the body(u) = 0 Step 2: Calculation of the velocity of the body: In differential form, acceleration (a ) is given as; a=dv/dt Thus, dv/dt=x^2...........(1) Now, the differential form of velocity is v=dx/dt Therefore, from equation (1), we have; dv/dx × dx/dt = x^2 dv/dx × v = x^2 vdv = x^2dx Now, integrating the above for position x=0 to x=2m and for velocity v=0 to v, we have; ∫vdv from v to 0 = ∫x^2dx from 2m to 0 [v^2/2] from v to 0 = [x^3/3] from 2 to 0 [v^2/2 − 0] = [2^3/3 − 0] v^2/2 = 8/3 v^2 = 16/3 v = 4/√3 Thus, The velocity of the body will be = 4m/√3s.
@@akashjaiswar4727 actually i was wrong. Maths Maker is right. watch the next video u will find out yourself and I am in 12 class and waiting for the announcement of board exams.
You're not supposed to disrespect the teacher who're actually spending their time and effort on teaching you. True, they might not be making as much efforts as some other teachers but that doesn't exclude the fact that they actually taught you alot of things which you wouldn't have known otherwise. Also, as for UA-cam teachers, They're also teaching students and have taught many students in past. So it isn't essentially right to label them as 'UA-cam teachers' while comparing them with yet another disturbing term 'Actual teachers'. It's best to not complain and get knowledge from wherever you're able to.
@@ginger0403 Yeah, students who study from UA-cam seem to overlook the fact that there are several teachers whom students are paying for in school and coaching and they're also putting in equal efforts. If students are not able to grasp what those teachers are teaching, they can discontinue their coaching and study solely from UA-cam. But disrespecting those teachers just because these students are not able to understand because of their efforts is indubitably deplorable!
@@ginger0403 Hardwork and good teaching are not so related, even if you work hard but your students dont understand anything then the hard work is in vain...
"True answer with solution " Vfinal square /2 - vinitial square/2 = area under a-x plot or itegration of a w. R. t displacement(x) And given that the initially body is at rest. So vinitial =0 (Vfinal )^2 = 2 (integr.of a= x^2 w.r.t x) Vf ^2 = 2(x^3/ 3 ) =2x^3/3 Vf=√2x^3/3 Or simply v (at x=2m) =√2×8 /3 =+4√3m/s or 4√3/3 m/s
@@chandrasravanthi5123 a=x^2 Then we can write. a=dv/dt And we can also write a=dv/dt= dv/dx multiply by dx/dt(dx/dt=v) Then vdv/dx=x^2 then we can integrate this
Answer=-+4/√3. WE USE a=vdv/ds and displacement in the term of x diya hai to use change krege uske bass cross multiplication expression is /x2dx =/vdv. Uske bad hm 0 to 2integrate karege at x=0 body is at rest then answer is -+4/√3
@@rachitchhillarhey, we use the equation a=vdv/dx toh when you will cross multiply it will give you adx = vdv Isko integrate kro by both sides S adx = S vdv That means S adx = v^2/2 So when you integrate the expression x^2 It will give you X^3/3 +c So c ki value would be zero So x^3/3 = v^2/2 So the ans would be +- 4/ root 3
I have started watching your lectures last month....and really they were better than lectures of my institute ....now I don't have fear of physics anymore...
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LAKSHYA Batch(2020-21)
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What will you get in the Lakshya Batch?
1) Complete Class 12th + JEE Mains/ NEET syllabus - Targeting 95% in Board Exams and Selection in JEE MAINS / NEET with a Strong Score under Direct Guidance of Alakh Pandey.
2)Live Classes and recorded Video Lectures (New, different from those on UA-cam)
3)PDF Notes of each class.
4)DPP: Daily Practice Problems with each class having 10 questions based on the class of JEE Mains/NEET level.
5)Syllabus Completion by end of January, 2021 with topicwise discussion of Last 10 Years Problems in Boards, JEE Mains/NEET within Lecture.
6)The Complete Course (Video Lectures, PDF Notes, any other Study Material) will be accessible to all the students untill JEE Mains & NEET 2021 (nearly May 2021)
7)In case you missed a live class, you can see its recording.
8)You can view the videos any number of times.
9)Each chapter will be discussed in detail with all concepts and numericals
10)Chapterwise Approach towards JEE Mains/ NEET & Board Exams.
****Test Series for XI & XII****
We provide you the best test series for Class XI,XII, JEE, NEET chapterwise, which will be scheduled for whole year.
The test series follows very logical sequence of Basic to Advance questions.&
Evaluation of Test and Solution to all the questions at the end of the test.
Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 01: Introduction || Average Speed
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Class 11 Chapter 3 Kinematics: Differentiation || Calculus part 01 || Mathematical Tool
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Class 11 Physics Chapter 03 : KINEMATICS :Motion in a Straight Line 02 || Instantaneous Velocity || || IIT/ NEET
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Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 03 || Answer Batao Salute Pao #Physicswallah ||
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Class 11 Physics Chapter 03 : KINEMATICS : Motion in a Straight Line 04 ||Derivation Of Equations Of Motion Using Integration||
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Difference Between Diaplacement and Distance || IIT JEE MAINS AND ADVANCE CONCEPT ||
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11 chap 03 : Kinematics 05 | Displacement time Graph -Velocity time Graph - Acceleration time Graph
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NEET Best Questions 01 || Motion in a Sraight Line 1 || KINEMATICS NEET || Motion in One Dimension
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11 Chap 03 :Kinematics 06 || Motion Under Gravity || Motion in a Straight Line || Class 11 / JEE ||
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Best Method For Rain Man Problems | Relative Velocity | Motion in a Plane | Kinematics JEE NEET
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River Boat Problem || Relative Velocity in 2D || River Man Problem || Motion in a Plane || JEE NEET
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Projectile Motion 01 || Class 11 chap 4 || Motion in a Plane|| Motion in 2-D ||
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Projectile Motion 02 || Class 11 chap 4 || Motion in a Plane || Projectile from a Height ||
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Mock Test 01 | Kinematics Best Questions for IIT JEE NEET | IIT JEE Questions Kinematics
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11 Chap 4 || Circular Motion 01 || Angular Velocity and Angular Displacement || IIT JEE /NEET
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11 Chap 4 | Circular Motion 02 | Centripetal and Tangential Acceleration | Angular Acceleration |
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Circular Motion 03| Centripetal and Centrifugal Force IIT JEE/ NEET | Conical Pendulum |Death Well |
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11 chap 4 | Circular Motion 04 | Derivation of Centripetal Acceleration or Centripetal Force |
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11 chap 4 | Circular Motion 05 | Banking Of Road IIT JEE NEET | Banking of Road with Friction |
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11 chap 4 || Circular Motion 06 || Motion in a Vertical Circle IIT JEE / NEET || Critical Velocity
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11 chap 4 || Circular Motion 07 || Motion in a Vertical Circle On a Bowl || IIT JEE MAINS / NEET
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u r my mentor sr
thank u so much sr.
v=4/root3 ...sir...is that answer right
Aap Sahi se donation bhi likhate h
Sir😂😂😂😂,ITNA halwa saval tha.maja nhi Aya..Mera ans Bina pen paper ke a gya
Who want Alakh sir as own school teacher ?? 😌😌😌💓💓💓
Edit: it's amazing to see that, students are still watching the old lectures of this "dariwale inshaan 🌝'❤️.... while he has launched so many new batches....this is the power of marker nd white board 😎😎
Meee
i dont
Meee
Mee
S I wamt
India actually need teachers like you sir !!!🙏🏻🙏🏻🙏🏻
Hey jangid bro
@@apiiiey o tu bhi jangid😃
Yes
Yes bro
what need he is teaching to india
Sir I joined a coaching recently (fees-3.5 lacs) and the teacher is taking online classes from last 1 month and he is teaching physics too complex to understand . I cried too much as I was not able to understand the concepts then I preferred your video to clear my concepts ...THANK YOU SIR ...you are the GOD of students ...😇
Not a god of students but is GOD FOR STUDENTS🤗
Yar paise nhi waste kro bhut regret hota h bad me
3.5lakh fee😂😂😂😂
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T.w left Kar bhai Accha rahe ga Maine bhi left Kiya paise wapas le
Jo wahan Diya tha(3.5 lakhs)uska 10% sir ke account me send karde
This man is single handedly saving my education
Exactly 💯
ur mom is double handedly giving me handjobs
Haa
Who are regularly watching physics wallah lectures nowadays ,btw THANK YOU SIR
Me
Mee
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Answer of last question:
Given, a = x²
We know that a = dv/dt
a = dv/dx × dx/dt
a = v dv/dx
v dv = a dx
v dv = x² dx
Integrating the above equation;
v²/2 = x³/3 + C ... (1)
Given when x = 0, v = 0
Putting in Eq. (1),
0 = 0 + C
C = 0
Putting in Eq. (1),
v²/2 = x³/3
At x = 2 m,
v²/2 = 8/3
v² = 16/3
v = ± 4/√3
Tq
Thnx yrr...I do in the same way...👍👍😊😊
Nice bro👍👍👍 but I am not able to solve it 😩😩😩
Mene bhi krdiya yrr yr😝😝😝
Mahan
From 50 thousand to 5
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Hardwork Pays!
5 million 😁
@ANON THE DOGE yes
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Now a 1b$ company
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17:00
C= -23/6
And Final ans v at 2s= 17/6 ms-¹
wont it be 11/6 ms?
Yes, I also got 17/6 answer.
Yes 17/6 m/s is a right answer
@@zo257 phele integration sikh
yep same
ALAKH SIR>>>>>>>>>>
ITS BEEN 4 AND HALF YEARS SINCE THIS VIDEO AND HE HAS LAUNCHED HIS OWN APP AND OPENED INSTITUTES ALL OVER INDIA ....BUT STILL NO ONE CAN BEAT HIS METHOD OF TEACHING AND EXPLAINING SO WE GET THE FEEL OF THE SUBJECT
THANK YOU SO MUCH SIR
soo true!! 😃
Satish mishra sir se padha hai kabhi ?
My Class Teacher Also Good..
@@aryashree7 ha baut ghatiya padhate hai MR sir jyda accha padhate h
@@adarshsinghbaghel1640 toh mat aa padhne
Actually he changed so much now... These videos give a better nostalgic feeling than his digital classes give...
Board and marker... ❤️
Yea bro
Whyy can u explain me plzzzz🙏🙏🙏
@@shephalipatra4356 bcoz that nostalgia
@@geeteshpaidi200 ohhh😭 it's true ,,,
@@shephalipatra4356 anyone who knew his channel before he was big would get the feeling. I was subscribed to him since 2017 and these type of videos gives strong nostalgia
After reading the comments from many of our seniors.. I realised that I have chosen a right platform for my neet exam prep. Million tons of thanks to you sir !!
don't worry you will fail if you kept spending time in comments!
🙃
@@aryanantil4796 I see you are a hypocrite unless you really want to fail
@@harshadvikram7458 yes i am a hypocrite
@@aryanantil4796 tu bhi yhi kr rha h bhai...spending time in comments
@@aryanantil4796 Bro thinks just spending 3-5 hrs in a day doing nthing does anything ;-;
I provide here solution 👇
a = x^2
dv/dt = x^2
(dv/dx).(dx/dt) = x^2
(dv/dx) . v= x^2
vdv = x^2 dx
Int ( vdv = x^2 dx)
v^2 /2 =x^3 /3 +c
at x=0 the particle at rest hence v=0
Hence c=0
v^2 /2 =x^3 /3
v^2 =x^3 ( 2/3)
At x= 2
v^2 = 8.2/3
v= √16/3
v =4/√3
😅
The video is still trending in October 2019!!!!!
Who said
Even now also this video is trending....
In November also
February 2020
@@saswatisaha8155 today also means on 3 feb but I know it will soon become trending video in April or may of 2020 similarly 21,22,23......
Till now and will forever
Till date no one can explain better than this ...lakhs of coaching are nothing in front of these short videos...anyone preparing for JEE NEET has to just watch this and practice and excell...best ever teacher in century..in future it is going to be huge huge wonder to students...
😑
It's true,
Homework answer
8/3 ms^-1
Thank you sir ❤❤❤
Correct, I did it in the same way..😊
Thats actually incorrect
@@curiousity003it is incorrect bro
@@aperson4543 yes, I watched lec 4 then I realised that my answer is incorrect.
from 50k to 10 million salute to you sir jii !!!!!!!!!!!!!!!!!!!!!!!!!!🧔🧔🧔🧔🧔🧔🧔🧔
गुरुर्ब्रह्मा ग्रुरुर्विष्णुः गुरुर्देवो महेश्वरः । गुरुः साक्षात् परं ब्रह्म तस्मै श्री गुरवे नमः ॥
A big salute to you bro
@@ManuChauhan7 thanks bro it;s mean alought
In the era of trolling teachers by making fun of them this is so warm😃❤
0= answer
Please give the meaning
ANSWER OF QUESTION:- V=3m/s or V=2.66m/s.
Mere bhi yahi aara h..
@@himanshupokhriyal6883ooh bhai...tum bhi meri raftaar se chal rahe ho😅😅😅
Han same aa ra. Shi h kya?
Mera bhi yahi aaya bhai
Sir i never know that physics teacher's are also like you...you are a masterpiece😇💞
i never knew ur mom was so hot bro wtf
(17:54) Answer is 8/3...Thankyou sir awsome lecture 🤘🏼🤘🏼
Bro u have to put unit (m/s)
no its, 4/√ 3
@@AkankshaGupta-pr6ovwrong its 8/3 m/s
you are absolutely correct!!8/3m/s
Yes you are correct
Thanks for ur dedication sir
Ans)8/3 m/s
Right answer bro
Right 💯💯
Galat answer bro
Right 👍
@@Anuradhakumari97927 galat hai yr mera bhi same aaya tha lekin galat hai
Best
Physics
Teacher.
Who
Thinks
The
Same
👇
👇
👇
Like wala bhikari
Thanks
@@NoobGaming-hx9jq sahi khaa ab hum sab dislike karenge
You are right bro
Big dasssss______hai sala bikhari credit leta hai
Best teacher ever.. Thank you very much sir !.. Your level of teaching is simply awesome.. 🔥
Did your goal completed?
The answer of solution would be 4underroot3/3 .
a =dv/dt
a =dv/ds.ds/dt ( in order to eliminate dt I have done this .
a = v.dv/ds
Now integrate this .(here a=x², ds =x )
Integration of x².dx = integration of v.dv
It would be
x³/3 = v²/2 +c
Put the value when body initially at rest
C=0
Now put x= 2
8/3 = v²/2
Underroot 16/3 = v²/2
Answer would be 4 underoot 3 would come .
Thank you very much sir .
I am feeling confident .
😊😊
How tf xdx and vdv can be simultenously integrated?
@@KattarHinduShivendra YES both should be integrated {adt={dv
{=integration
Why ds=x??
I really don't fear physics anymore..
You are the best sir
Yaaa
oi
Uff course
Ya I really don't fear physics because I am the student of Alakh sir
You need to visit pradeep kshetrepal sir's channel 50years+ experienced teacher physics wallah is like a child on the front of the legendary pradeep kshetrapal
17:44 --- V =2.66 m/s
X displacement h velocity nhi h bhai
@@hackerdoctor5041 Tera bhi Galt hai vha a function of x de raha hai t nhi hai
मेरा भी यही आया hai
Mera bhi yahi a Raha hai
Po
Preparing for JEE 2023 and still going for his video's explanation because of his amazing teaching skills 🧡🧡👏
Same Brother Best of Luck
Same bro
Bhaiya mjhe inse samjh ni ata tunning hi nii hoti .. As a junior apse puchna tha mai abhi hi 11th me aya hu konse teacher se pdhu kinematics 🙏 bta dijye
@@prajwalkulkarni4530
Bhaiya maine bhaiya maine kr liya physics galaxy se
Aur mai shuru se hi mehnat kr rha full potential .. Results aaye n aaye uska nhi soch rha bs apna full potential de rha . Starting se.. Kia mera ye decision shi h .. (( Jee student - aakash me pdhta hu ))
Preparing for neet 2023 .. Doing the same... Inke alawa kisi s smjh hi nhi ata
6 yrs and still counting✨…
How much more years this lecture will be as useful and as powerful as 6years ago and now ❤
Sir aap mere liye TEACHER Nahi ek GOD hain.
Thanks for the videos...
Nice
👍
Sab mere hi sahare rho
17:41 The answer is 8/3 m/s. But sir we are not able to take your salute because you are the great legend
True answer is 4/√3
Yes
@@pritighosh851 I saw that it wrong 😔
@@bhartirajan9727 hm me vhi 😞
Bhai answer chahe galt aya ho ya aye lekin bhai try hi sab kuch dilati hai
Right answer
Last question answer is
a=dv/dt
dv/dt=x²
dv/dx×dx/dt =x²
dv/dx×V=x²
Vdv=x²dx
Integrating both sides
V²/2=x³/3+C
x=0(given)
V=0(given)
Applying in EQ.=>C=0
Therefore, V²/2=x³/3
If x=2m
V²=(2³/3)×2
V=√(8×2/3)
V=√(4×4/3)
V=4/√3 m/s
OR
V= 4√3/3m/s
This is the method😊
How can you solve this question 🙃
Mera to 8/3aa RHA h
Ha@@Isha123sharmayy
@@Isha123sharmayymera bhi same aa rha hai
@@Isha123sharmayy mera bhi same aa rha hai
17:33 8/3 m/s
maza a gaya
11/6
Answer for the last question:
We know that a=dv/dt
(a=dv/dt ×ds/ds )
a=dv/ds × ds/dt
a=(dv/ds)v {where ds/dt=v}
a= vdv/ds
Given a=x²
We know that a=vdv/ds
Integrating on the both the sides
vdv/dx=x²dx
v²/2=x³/3+c
Given that v=0; x=0
On solving the equation (v²/2=x³/3+c)
We get c=0
v²/2=x³/3
v²/2=2³/3
v²=2⁴/3
v=root of(2⁴/3)=-+4/root3
I know I am too late but ...still😃
Hey babes , what is the value of v ..( after put the value of t = 1 nd v = _2
Please answer 🙂🙂
V = 2.66 m/s 🙂🙂
@@Swastik_Debnath correct mera bhi same aya hai😂 8/3
Your answer is wrong
4️⃣➗🌱3️⃣
Hu wants and like to join alakh sir offline classes imagine if sir open their offline classes and five million student come to attend class 😉🤔🤔
Haha 😄🤣
Opened
Maths wallah 🤣😂😂
Khul gyi bro
Hu or who😂🤣
India deserves teachers like you sir❤️🔥❤️
17:21 answer is : 8/3=2.66 sir Bipul das
Acceleration ko velocity mei integrate Kiya uske baad c = 0
And x ki value 2 put kar dee
Yahi process Kiya na ??
@@GoodMan98rf haa
@@GoodMan98rf Limits are from 0-2 then why do we need c? It's a definite interaction
@@GoodMan98rf x meter me diya hai bhai .
2:13, my father's reaction after My result.
😂🤣
🤣🤣🤣
😂🤣🤣
Aree bhai bhai bhai🤣🤣🤣🤣🤣🤣
Lol
Finally we get Jeetu bhaiya!!!
Jeetu bhaiya baad me aye h sir 4 saal se padha rhe h yt p
Yes bro
Given, a = x²
We know that a = dv/dt
a = dv/dx × dx/dt
a = v dv/dx
v dv = a dx
v dv = x² dx
Integrating the above equation;
v²/2 = x³/3 + C ... (1)
Given when x = 0, v = 0
Putting in Eq. (1),
0 = 0 + C
C = 0
Putting in Eq. (1),
v²/2 = x³/3
At x = 2 m,
v²/2 = 8/3
v² = 16/3
v = ± 4/√3
How a= dv/dx*dx/dt.??😟
Thank you harshit.....
@@suhanisahoo5461 its Harshil
@@harshilpatel7171 sry sry it's by mistake 😅
@@suhanisahoo5461 Ok no problem 😂❤️
answer is ------ 4/√2 m/s
PHYSICS, CHEMISTRY,AND BIOLOGY ARE EASY WHEN ALLAKH PANDEY SIR...MRITUNJAY SIR AND RITU RATTEWAL MAM...IS TEACHER...respectively...
Hiii
Neha mam for maths ❤️❤️
@@fanofabd7613. Meeru ante unnollu bro
@@corzoncorzon7011 ??? Neha mam classes free eyy bro yt lo vuntai
@@fanofabd7613 meme bro nee dp choosi ala anna🤣
Sir you make whole physics an interesting subject
Bhai tum galatfaimi ke sikar ho gy ho
Edit: it's amazing to see that, students are still watching the old lectures of this "dariwale inshaan 🌝'❤️.... while he has launched so many new batches....this is the power of marker nd white board 😎😎
From 50k to a unicorn company amazing journey a baig big salute to yo respected sir thanks for teaching us so grea❤❤😊
ANS FOR LAST QUESTION IS 4✓3 M/S THANK YOU SIR!!!!
@@ksoh1oe276 pata hai 4√3 hai answer
how?
Preparing for neet 2024
Pr abhi tk 11th aur 12th physics clear nhi hua dekhyea aage kya hota hai mera! Abhi toh sir ke wjh se sb clear hota ja rha hai hope I'll crack neet this year or next ,🥺
17:06 8/3ms^-1
same
right
Yes bro
It will be 17/16 ms-¹ check ur answer
@@captainnk7633 isne last question ka answer btaya h
a=x^2
vdv=x^2dx
Integrating both sides with limits 0-v on left side and 0-2 on right side
vdv becomes v^2/2 = x^2dx becomes x^3/3
Solving with limits
it becomes v^2/2=8/3
v^2 = 16/3 (transposing 2 from left hand side to right hand side (8x2)/3)
therefore, v=+- 4/root 3 m/s at x=2m
Is 8/3 m/s wrong?
Answer is 4/√3.
Explanation -
a = (dv/dx)(dx/dt)
a = (v)dv/dx
a dx = v dv
Integrating both sides, we get,
x^3/3 + c1 = v^2 /2 + c2
After applying the condition we will find that c1 = c2
So,
x^3/3 = v^2/2
v^2 = 8 * 2 / 3
v^2 = 16/3
v = √16/√3
v = 4/√3
or
v = 4√3/3.
Thanks.
❤️❤️❤️ right
Thanks
Can you explain wtapp plz
Reply me plz
@@jakkalashobha6526 yes
Answer = 8/3 m/s or 2.67m/s
x ko integrate kaise kiya wrt time
???
@@aarusharya1544 where is time in the q?
See, ({ is integration)
a = x²
v = {adx
=> v = x³/3 + C
Now, it is given that,
x = 0 when v = 0
=> 0 = 0³/3 + C
=> C = 0
To find, v when x = 2, put 2 at x,
v = 2³/3 m/s
v = 8/3 m/s
@@blaablaa-jx7hy are sorry...ye mai nhi mera bhai tha...abhi vo class 8th mei hai. To use pta nhi tha.
Aur mai nda crack kr chuka hu.
To sorry to my younger bro
Use mai smjha dunga.
Iska correct Ans +-4/underroot3 hoga
Abhi solve kiya maine.
anyone is learning now ?
Me😂
Me tooo@@RishikPatro-ou7qr
Me😂
Yesss!
Yesss😅
Yeh mere liye super teacher hai sir ko sat sat naman 🙏🙏🙏🙏🙏🙏🙏 agar meri physics acchi hai toh bas pura credit inkho jata hai ❤❤
Soo till 10th i had my fav teacher on UA-cam was bkp -bhai kii padhaai ...and in 11th it's physics wallah alakh pandey ... really great sirr..hats off to u 🙌🙌✨✨
Same here
Yehh same
Same to you 😀😀
Woah
Same thing is applicable wid meh 😃
Omg!! Same🤩🥰😍❤
Sir no words for you...
Physics become a easiest subject for me ..on!y because of you...
Take my respect sir..
Hats off to you..
What a lectures yrr...
😘😉😊😇🤗☺☺☺☺
17:50 answer is 8/3 m/s. Sir you are the best teacher in the world
How ??? Explain
no it should be 4/√3 m/s
@Isha_ actually mera bhi pehle 8/3 hi aa raha tha ans but is ch ki playlist ke nxt video me sir ne iska solution karaya hai video ke end main....usme 4/√3 hi aa raha hai ans
@Isha_ maine bhi same kiya h
pr sabka ans kuch or kyu h 💀😶
@Isha_ arey hn!! sir ne agli video k last m bta diya tha🕺
Sir ap mere liye god of physics and god of chemistry dono ho, thanks for fabulous teaching 🙏🙏🙏🙏🙏🙏
Same here
Moreover, the way u teach is wonderful.. Easy to understand
U r so lucky; u got ❤️ from alakh sir 4 years ago!!!
@@Tanmayyy28 Shai bat hai bhai
17:26 answer is 8/3 m/s
Same bro! Idk how come some ppl are saying 4 root3😕
@@pragattii_show can u explain pls?
@@taegukie yaa why not look its given that a= x^2 so if we will intergate this than v= x^3/ 3 now ita given x= 2 so putting value of x in velocity so 2^3/3 then v= 8/3 when x is 2
@@pragattii_s thank you so much
@@pragattii_s C kaha gaya
Class 11th 2024-25 attendance
😅
yeh video mene 10th ma dekhli thi T_T
last question answer
integrate both sides and equate them
answer will be root 2^4/3 or 4/root 3
Correct. Same here.
:D
Ry8😀
This is the last step
Plz tell me what is the ans of Second last question
@@HarmandeepKaur- + - 4/√3
4/√3 is the ans
a= vdv/dx
vdv/dx=x^2
vdv= x^2dx
N then integration...
Wrong ans completely wrong
Correct
Wrong hai
Last me ja kar thodi se mistake kr di
Wrong pta hai kaise
???????
Mai bta ta hu
Integration ke time humko
x^2 ko totally integrate krna hoga .
Na ki pehle hi x ki value put krna hai . Then aakhir me 8/3 +C
@@jvlogs6878 but in the next vdo sir declared 4/√3 as correct
Answer for last q . Is =8/3m/sec or 2.67 m/s
18 may 22
No unnecessary theory
Very nice understanding concepts
Answer of end question=2.67m/s
wrongggggg
Sir exactly 3:57 your face expression is osm i repeated almost 5 times 😂😂😂😂😂😂
Hi
Yeah same here😂😂😂😂
Hii
@@NITian_Sayantika yes 😊
Yes same here........
I just come to comment the same thing here and i get this
V=8m/3sec or V=2.67m/sec
Sir u r the best teacher in the world your teaching style is very different for another teacher u r great sir in future we need this type of teacher for every students and in all educational system 👩🏫🤲🤲
answer:
v = ± (4 / √ 3) m/s
I am following Sir since he had 50k subscribers and I was in 9th. Now it's 5.71 million And I am in 12th. Time flies so fast.
how was your preparation
9th m 11th ka kya padh rha tha bhai
And after 1 yr 9.13million
Me in class 11 with Alakh sir having 10.1M sub
@@IXADIBYANSHUJHA now 10.3 M 🎉
17:29 Correct ans is 8/3 m/s or 2.6666....... m/s
Best teacher in the world
NICE LECTURE SIR, i really enjoyed it
my ans of the last ques is 8/3 m per sec
same
yeah i also got this answer but is it the right answer someone confirm please
@@thissidesoniya no it is 4/root 3
@@joeldsilva3030 yeah I got it...thank you
@@thissidesoniya Yup
The answer is
a=x^2
V. (dV/dx) =x^2
Vdv=x^2dx
Take Integration of both side
V^/2=x^/3+c
Put x=0, V=0
You get c=0
V=root over(x^3/3). 2
Put x=2
Ans:4/√3
Yes this is correct
same here
Bhai ye V. (dV/dx) formula kaha se mila??
@@apurboghosal4875 a=dv/dt
a=dv/dt *dx/dx
a=dv/dx *dx/dt
as we know dx/dt = v
therefore , a=dv/dx * V
Bhai velocity root me nhi aaygi so v=2.7m/s
Given: a=x^2
We know that v•dv÷dx=x^2
When we integrate it and solve it properly ,Then we get v=4/√3 m
I'm gettin v=2m/s by same method 💀💀
Nah ! On the same you directed your solution I solved it properly and I got 2.6 m/s.
Thank you so much sir
Answer is 8/3 m
8/3m per second
Yes
Is it 8/3+c m
@@Vishal132tno answer is 4/√3 as v square= 8/3 so after taking under root ans is 4/√3
Ans is 8/3 .....firstly integrate acceleration to find velocity,then put the value of displacement in the equation.
Yes same as mine answer
Bhai ans galat hai...U can't integrate bcz variable are different..
sir i still watch these old videos of yours because i belive that knowledge is same and the dedication is still the same no matter taught on white board with that sasta caera on on smart board with studio. Hats off to you sir hope this comment reaches alakh sir and reply me
a=x²
Find v when x is 2m , given at x=0, body is at rest. 17:44
2.6m/s, am I right?
Solution
Step 1: Given that:
Acceleration in the body;
a=x^2
At x=0 , the velocity of the body(u) = 0
Step 2: Calculation of the velocity of the body:
In differential form, acceleration (a ) is given as; a=dv/dt
Thus,
dv/dt=x^2...........(1)
Now, the differential form of velocity is
v=dx/dt
Therefore, from equation (1), we have;
dv/dx × dx/dt = x^2
dv/dx × v = x^2
vdv = x^2dx
Now, integrating the above for position x=0 to x=2m and for velocity v=0 to v, we have;
∫vdv from v to 0 = ∫x^2dx from 2m to 0
[v^2/2] from v to 0 = [x^3/3] from 2 to 0
[v^2/2 − 0] = [2^3/3 − 0]
v^2/2 = 8/3
v^2 = 16/3
v = 4/√3
Thus,
The velocity of the body will be = 4m/√3s.
🎉
finally someone answered
Is it necessary to integrate the dv in LHS ??
Yaha tak aagye aage bhi jaaoge ,
Physics wallah ke saath Dhoom machaoge .....
😁😁
Sir on applying the laws of motion the ans comes to be4/√3 m/s
As per v²-u²=2as
Very simple sir!!!
@@Lets_crack_it0806 in which class u r ?
@@akashjaiswar4727 actually i was wrong. Maths Maker is right. watch the next video u will find out yourself and I am in 12 class and waiting for the announcement of board exams.
@@Lets_crack_it0806 date aa gai he didi 1-7-2021
Last answer is C= -23/6 .... And V=29/6
Why are UA-cam teachers always better than the actual ones ...
Why can't this kind of amazing ones actually be teaching us in person
You're not supposed to disrespect the teacher who're actually spending their time and effort on teaching you. True, they might not be making as much efforts as some other teachers but that doesn't exclude the fact that they actually taught you alot of things which you wouldn't have known otherwise. Also, as for UA-cam teachers, They're also teaching students and have taught many students in past. So it isn't essentially right to label them as 'UA-cam teachers' while comparing them with yet another disturbing term 'Actual teachers'. It's best to not complain and get knowledge from wherever you're able to.
@@ginger0403 Yeah, students who study from UA-cam seem to overlook the fact that there are several teachers whom students are paying for in school and coaching and they're also putting in equal efforts. If students are not able to grasp what those teachers are teaching, they can discontinue their coaching and study solely from UA-cam. But disrespecting those teachers just because these students are not able to understand because of their efforts is indubitably deplorable!
School teachers ko troll kar deti hu kool lagugi huehuehue
@@ginger0403 bangali.
@@ginger0403 Hardwork and good teaching are not so related, even if you work hard but your students dont understand anything then the hard work is in vain...
17:52 Velocity is 8/3 m/s
riight
Right
"True answer with solution "
Vfinal square /2 - vinitial square/2 = area under a-x plot or itegration of a w. R. t displacement(x)
And given that the initially body is at rest. So vinitial =0
(Vfinal )^2 = 2 (integr.of a= x^2 w.r.t x)
Vf ^2 = 2(x^3/ 3 )
=2x^3/3
Vf=√2x^3/3
Or simply v (at x=2m) =√2×8 /3
=+4√3m/s or 4√3/3 m/s
Same here🖐️
Still confused...😬 Thanks anyway
@@thasneemamohammed4671 what is your confusion ?
Please ask 🙏🏻🙏🏻
But limit is not given ryt? So how r u solving this with initial nd final velo?
@@wheresthatbirdie given that at x=0 body is at rest which indicates that initial velocity is zero.
13 millon Subscriber's complete🎉❤
❤❤❤
end ke question ka ans hai (8/3or 2.66)
Yes's.
Answer of last question is 4/√3 m/s. This is Very easy questionfor me sir because of you❤✨
17:05 .....17/6
Correct
@@class11yayv 6.3 aaya Mera
Answer:8/3m/s=v
The last question is simple just use the formula
a= v.dv/dx
Then,
X^2=v.dv/dx
vl uu explain wth including steps ..?
@@chandrasravanthi5123 sir ne explain kiya hai ye question iske baad wala video dekho
@@chandrasravanthi5123 a=x^2
Then we can write. a=dv/dt
And we can also write a=dv/dt=
dv/dx multiply by dx/dt(dx/dt=v)
Then vdv/dx=x^2 then we can integrate this
Answer=-+4/√3. WE USE a=vdv/ds and displacement in the term of x diya hai to use change krege uske bass cross multiplication expression is /x2dx =/vdv. Uske bad hm 0 to 2integrate karege at x=0 body is at rest then answer is -+4/√3
Bro 8/3 tak aagya uske baad kaise 4/√3 aya can u explain
@@rachitchhillarhey, we use the equation a=vdv/dx toh when you will cross multiply it will give you adx = vdv
Isko integrate kro by both sides
S adx = S vdv
That means
S adx = v^2/2
So when you integrate the expression x^2
It will give you
X^3/3 +c
So c ki value would be zero
So x^3/3 = v^2/2
So the ans would be +- 4/ root 3
I have started watching your lectures last month....and really they were better than lectures of my institute ....now I don't have fear of physics anymore...
17:53 answer
4/√3
I don't know who are these people who have disliked your video mannnn
sir thankk you so much , you are life saver btw last Q ans is 8 by 3
It's right 👍👍👍👍
Sir answer is 4/√3 mls your old lectures are very use full❤
In this why just we use a=vdv/dx instead of a=dv /dx .. plz explain 😢
17:32 ans: v=4/√3m/s
Your teaching method was awsmmm......finally I got a good platform of physics you made physics really interesting..... Thanx sir for your hard work.