 Fine. So it looks like some of us are joining a little late. So we'll probably wait for a minute before we start the session itself. But meanwhile, can you all tell me what is going on in your schools? You can tell the name of the school and what is going on. You can type in, you can speak also. Okay. Others? All right. Yashankur Units and Measurements. Differentiation in R&R. Units and Dimensions in PSBB. Okay. Other than PSBB, Raja Jinnagar and Yashankur NPS. Now that is cool. DPS East Motion in one day. Okay. In the Raja Jinnagar, in Raja Jinnagar, they are starting, they have finished Units and Dimensions. In Raja Jinnagar, they have finished Units and Dimensions. Okay. Other than Yashankur, they have started Book 2 as well. That is a little different. Okay. So see some of us probably have done differentiation or integration also in the school. Okay. And even I'm planning to do it. All right. But that is coming probably the next week. So the way I have planned this chapter motion in a straight line or motion in one day, I will do everything that doesn't require integration or integration. Okay. Once we are done with all of those things, then we will do the differentiation, integration and the problems related to that. Understood. So we'll be doing everything. Don't worry. Okay. Somebody is asking about the test. All of you have seen that there is a notification of the test. Have you seen the WhatsApp messages? Yes or no? Some of you have not. So the test will be exactly like how, let's say the notification went for two tests. Right? One was need pattern test. Other was J main pattern test. Okay. Exactly the way need pattern happens. That's the way the test will be for the need pattern test. And same way for the J main also the way J main happens. So that is how exactly it will be. The syllabus of that would be whatever we could cover maybe couple of days before the exam. All right. For example, the seventh of July is the exam. Today is what? Today is 20, 23rd. So we'll be done with the motion in one day. So entire motion one day including the derivatives integration will come. All right. So some of you have joined KVPY group also look like many of us have joined there. Okay. So now why the KVPY group was created was going to understand that. Okay. That KVPY group was created because whatever we are doing in the regular classes that is not sufficient. To complete the KVPY syllabus. All right. However, whatever is happening in the regular classes. That is very much part of the KVPY syllabus. Fine. So whatever is the extra that is required for that we'll be calling you extra. Okay. Now some of you are asking the exact schedule to the extra classes. We will try to put some regime over it, but we cannot say that on every Monday there will be mathematics extra class or every Wednesday there will be chemistry extra class or things like that. All right. But I can tell you roughly about the classes. The KVPY extra classes would be around maybe 8.15 or 8.30 PM. We have to start because your classes for example gets over at 8.30 itself. So on that day you will not be able to attend some other day. Right. So four or five days you'll be coming at around 8.15 or 8.30 for maybe one hour or so. Fine. So that will be the schedule like till maybe November mid fine. And don't worry. We are going to send the KVPY class reminders one day before. Okay. So you'll be knowing that tomorrow there is a class or not. Right. So depending on let's say we need more classes for physics. So maybe more classes of physics will happen. Extra classes. And then suppose more classes of chemistry is required or sorry less class of mathematics is required. So less of mathematics will happen. Okay. So that is how it will be. Understood Anika. Is it clear? All right. So without further ado we can start now. Looks like some of us haven't joined yet. So what we have done in the last session. What are the last thing that we did? Everybody tell me what was the last thing that we did? Let me unmute all of you. You can unmute yourself. Okay. So you can unmute and speak also. Last thing did we do was the derivation of equation of motion. Graphically. Graphically we have derived equation of motion. All of you agree. Okay. So is there any student who is completely new today? No. All right. So great. We have done equation of motion. Have you done the fourth equation of motion or only three question of motion? Fourth also. Fourth also. Okay. Do you remember what was the fourth equation of motion? Distance traveled in the nth second. It was u plus half a. 1 minus 1. Correct. Do you have other equations like this? V is equal to u plus a t. All right. S equal to u t plus all of you write this down. Okay. This these equations will be troubling you for the entire session. So let us pay some respect to these equations. V square equal to u square plus 2 a s. These are the equations that will be applying for situations when what happens? What is the condition when I can use this equation? Type in. Don't speak. What should happen? Then only I can use these equations. Correct. Others quickly type in what should happen? Then only I can use these four equations. Correct. Constant acceleration. If efficient is not constant, we cannot use it. For example, if I say that accession is changing with time, let's say a is equal to t minus t square. So accession depends on time. It is not constant as time progresses. Accession also changes. So for this, we cannot use the equation. But if let's say if I tell you accession is 2 meter per second square. Now it is fixed. So with this, I can use this equation. Okay. I hope it makes sense. Does anyone has any doubts here? Anyone has any doubts? No. Okay. Okay. Now one final thing I want to discuss with you before jumping on to all the kind of numericals. Have you seen that in grade 9 when you were using these equations, sometimes you write accession to be positive. Sometimes you write accession negative. Sometimes velocity is positive. Sometimes negative. Have you observed that? Yes or no? Everyone, have you observed that or not? Right? So have I discussed about the signs in the last session? I haven't, right? So I'll discuss it quickly. So write down the sign convention. When to take what positive and negative. By the way, till grade 9th, okay, we have remembered many things because we were not aware of the few basics. That is why we remembered some thumb rules that if it is deceleration, we'll always take it negative. All right. So that way we have remembered some things. What I'm telling you is you don't need to remember any of that. Okay. In fact, if you don't remember all of that, it will help you. Okay. So we are going to follow a sign convention going forward, which looks like this. We are dealing with vector quantities. All of you understand, right? We are dealing with vectors, which have directions. For example, velocity, displacement, acceleration, all three have directions, right? So what we will do is that we will assume all of you write down, assume one direction to be positive. You can say that this direction is positive. What will happen to the opposite direction, all of you? Opposite direction would be correct. Very good. Opposite direction would be negative. All right. So let me write this down here. Now, all of you listen. Let's say there is a numerical. Okay. Now, God hasn't come and told us that, okay, every time right-hand side, if velocity is there, take it positive. Every time upward, acceleration is there, take that as positive. No, it's nothing like that. Okay. So in a problem, we are going to assume, let's say right-hand side positive. That is what we are assuming. So if velocity is on the right-hand side, then the value of velocity is positive. If acceleration is on the left-hand side, then the value of velocity is negative value. Fine. Similarly, let's say if you assume downward to be positive, if downward is positive and if accession is downward, so will you write position positive or negative? You'll write it positive, right? Because in that direction, which you have assumed, accession is there. So if velocity is upward, velocity is upward, you will assume it to be positive or negative. No. You had assumed downward positive, right? Velocity is upward, so velocity becomes negative. Velocity becomes negative, right? So we are going to throw all these garbage from our head that every time one direction is permanently positive, one direction is permanently negative. All that is not true. Okay. You'll get the correct answer only, don't worry. Fine. So we don't need to break our head what is positive, what is negative. And one more thing, there is nothing called as deceleration. Okay. There is nothing called deceleration. Acceleration is there. If acceleration is in the direction of velocity, you call it acceleration only. And if acceleration is in opposite direction of velocity, you call it deceleration. Okay. Let's not use the word deceleration going forward. It is acceleration only, but in opposite direction of the velocity. Clear? Okay. So now most of these things gets cleared when we solve numericals only. If you don't solve numerical, all of these things will just keep our mind occupied with a lot of doubts. Okay. Somebody is asking, can we use the word deceleration? No. Only acceleration is there. Only acceleration is there. For example, let's say this is the initial velocity. Okay. This is the initial velocity. U is equal to 5 meter per second this way. All right. And you have acceleration that way. Acceleration is equal to 2 meter per second square. Fine. So if you are saying that U is equal to 5 meter per second, what do you have to write acceleration as? Everyone? If this is plus 5, acceleration is minus 2 meter per second square. Minus 2 meter per second square. And if you write U as minus 5 meter per second, what will be the acceleration? Plus 2 meter per second square. So in this case, in this case, we assumed that direction to be positive. Here, we are assuming that direction to be positive. Nothing is incorrect. That is all proper only. Okay. Is it clear? I hope it is clear. Let us solve numericals so that things become even more clear. Okay. Somebody is saying in case 2 minus 5, in case 2 minus 5 is retardation. Minus 5? Aditya, you can speak up. Anu can speak. Aditya Anan? Yes. Speak. Okay. No problem. No, what are you asking? I'm not understanding. But in the case 2 minus 5 is retardation. Minus 5 is velocity. This is velocity. See, it is going to be, you know, because till 9, we assume that, you know, we assume that retardation means negative acceleration. Yes or no? All of us assume that retardation means negative acceleration. Right? But here I'm telling you that even though it is retardation, it can be positive. Positive negative depends on which direction you're assuming positive. Getting it. Retardation, in a way, if you really want to retardation or deceleration, it only means that acceleration and velocity are of opposite signs. That's all. Okay. So keep your head open. As in keep your mind open. Think in a very, let's say open-minded way. Then only you'll be able to solve numerical like as if you're solving a puzzle. Fine. So let us take a couple of simple straightforward questions to understand what I have written here, right? Then it will be a little bit more clear. So let's do that. Whatever I'm scribbling, you need to note it down because it will not show when I'll share this PPT with you and the group. Fine. So let's say there is this tower of height 10 meters. 10 meters is the height. Okay. And you have a ball here. This is a ball. Okay. This ball is thrown up with a initial velocity of 2 meter per second. Okay. As you throughout that, the acceleration is downwards of 10 meter per second. It's choir. Fine. You need to find out the time. Time taken to reach the ground. Find out all of you. Which equation do you think you'll be using? Okay. Right. Right. Right. Everyone try it quick. Those who are joining late. Can you tell me why you're joining late? What is there any thing that you cannot avoid? Wifi got disconnected. Alarm went off. Okay. All right. So I can see some of your answering. Great. Shall I, shall I discuss now? Shall I discuss now? Okay. Many of you got the answer, but then all of you getting different, different answers. Let me tell you. How the particle will move everyone. How do you move? It'll go up. Reach a maximum height and then go downwards like this. It'll travel like that. Okay. Now. This is my initial point. And this is my final point. Do I know the value of displacement? What is the displacement? Everyone. Displacement is down. Displacement is a vector, right? It is down. See, it is 10. 10 meters displacement. Right. So if I have to write the variables with the sign, I will first have to assume one direction positive. Which direction do you think should I assume positive? Although it doesn't matter. You'll get the correct answer only. Which direction should we assume positive? Everyone. It doesn't matter though, but what do you have taken? You can tell me that. I'm sure all of you have taken upward positive. Or downward positive, some of you taken. Great. Okay. Good to see that. Try something new. Anyways, so we'll take upward positive and downward. Then we'll become negative. So acceleration is minus 10. Displacement is minus 10. Initial velocity is plus two. All of you agree with all of this, right? Now we'll use s equal to ut plus half 80 square. Okay. Somebody asking how displacement minus 10. This is your initial position. This is your final position. So displacement is what? A line connecting initial to the final. So this is the displacement from here. Till there initial to final. So displacement is minus 10. Now I know why there's a discomfort. I'll discuss that. Don't worry. S U is plus two into T. Acceleration is minus 10 half. 10 into T square. So the equation you'll be getting is five T square. Minus two T minus 10 is equal to zero. Has anybody got this equation? Okay. All right. So this will give you the value of time. When you solve this quality equation, I'm not solving it because finance doesn't matter here. So this is the way you solve it. Now some of you might be wondering. Plus 10. No, it is minus 10 only. This will go on the left hand side. Displacement is minus 10. Okay. Now some of you might be wonder how many of you wondering that it, we are not considering the fact that it is going up and then coming down. So this time I have not considered how many of you thinking like that? It goes up and then comes down. So what about this time? I know you will be worried about that. But let me tell you it is automatically considered. Okay. For example, if you solve it like this, what will happen is when it goes up, when it goes up and comes back. Okay. Then same velocity two meter per second, which is upward right now. When it comes back, becomes downward. All of you agree? Yes or no? This two, which was upward. Why it is coming down becomes same value, but downward. So instead of plus two, you will have to then substitute minus two there. Then if you want to consider this time is T1, this time is T2 and that time is T3. Total answer is T1 plus T2 plus T3. Then in order to find T3, you need to consider velocity as minus two. Right now you're considering as plus two. Do you all understand what I'm doing here? Type in. Okay. So here all of you know it, but then we are not accepting. What are we not accepting? We are not accepting that S is displacement. We are not accepting that S is, we're not accepting that XS, which is the displacement is the distance between initial and final point. We want to consider this also, but will that add up any displacement going up and coming down displacement? Will it add up anything? Not add up anything, right? Displacement would be 10 meters downward only. Doesn't affect displacement and this equation S equal to UT plus half it is squared. This is displacement. So it does not matter where it goes. You'll get a correct answer only. Those who are still wondering, no, no, it should be a different answer. Do one thing. Try this way and then try that way also splitting the motion into three parts. Find out T1, T2 and T3 and then compare whether you get the same answer. You will get the same answer. Fine. But you can see directly we're getting the answer without splitting it. Okay. So let us solve more and more questions in order to understand this concept. This we have done. Have you done this question last time around? Do you remember doing this? No. Okay. Do this quick. This is not really the equation of motion. Definition of average acceleration it is. Find out quickly what is the answer. Done. Anybody? What is the answer? Okay. Different answers people are getting. All right. Okay. Some of you are getting zero also. Okay. So let's do this. Car is traveling. Let's say this direction. This way it is going with. 20 meter per second. And after 20 seconds, it starts going the other way. With 20 meter per second. All of this happens. In how much time? 20 seconds. Okay. Now, whatever we have written. Is. Without signs. So if I consider that direction positive. Then the initial velocity is. Plus 20. Final velocity is how much? If initial is plus 20 final is what? Correct. Minus 20. I can as well say initial is minus 20 finalist plus 20. Then. And the time taken is. 20 seconds. Right. So average acceleration as you all know. Final velocity minus initial lost T divided by the time taken. The final velocity is minus two minus 20. Initial is plus 20. Divided by 20. So you will get. Minus two meter per second square. Some of us. Are getting plus two meter per second square. That is perfectly fine. You have probably taken. The other direction. If you take the time taken. That is fine. You have probably taken the other direction. If you take that direction positive. You'll get positive acceleration. If you've taken that direction positive. You'll get a negative acceleration. That is fine. Not a problem. Is it clear to those who got a different answer? Some of you said zero. Is it clear? Do this. Delta T is time taken delta T is T only. Don't worry about it. Delta T you can say final time minus initial time. So that is amount of time taken. Definition of average velocity you remember or not. What is average velocity definition change in velocity minus time divided by time taken anyone close to the answer. Okay. I think some of you are guessing the answer without finding it out. You just think that it is the answer. So that is what you're telling me. Are you using any, let's say equation of motion or anything like that to get the answer? Those who have answered just guessed it. So whenever you guess, please write down on your notebook. Whenever you guess going forward in physics, you will get it wrong. You have to write it down somewhere and keep reminding yourself. That's the nature of physics now. It will be deceptive. Okay. Shall I discuss now? All of you have tried. Okay. Look at this. I'll tell you the way of solving any numerical. Okay. So let all of you focus. S is not given. You have to tell me the answer in terms of V1 and V2 only. That is that are the only thing that are given. Okay. Listen here everyone particle is going with. Okay. Vishnu got the correct answer. Focus here. Everyone particle is moving with uniform. So can I use equation of motion here? All of you. Yes or no. I can use equation of motion because equation is constant. It is uniform. So I can use it. So that is what they're trying to tell us. So at point A velocity is V1 at point B velocity is V2. C is a midpoint. You have to find velocity at point C. Okay. This distance is D by two. Total distance is D. So half and half. All right. Now D is not given. All of you remember D is not given, but doesn't matter. In order to solve numerical, we assume it, whatever it is, it is D. And then at the end, we need to remove. We need to. Remove D from the answer. Okay. So let's say they're asking velocity at point C. So let's say velocity is V. Okay. And uniform acceleration is also not given to us. It is just written efficient in uniform. So let's say this direction attrition is a. So can you tell me what should be V in terms of. V1 or V2? Can you tell me what should be V? How to get that? Is it correct to write V square equal to V1 square plus 2A into D by two? Do you all agree? Type in. Right. So what, what I'm trying to do here, whenever we solve physics numerical first, I will try to directly get the answer. So there's nothing wrong with it. This is the correct answer. V is equal to square root of V1 square plus AD. That's a correct answer. This answer is correct, but we cannot use A and D. All right. So although this is a correct answer, but we need to remove A and D from the answer. So what we do, we go back to the question and see, can we use something else to find out what is the value of A into D? And the answer is yes, you can write V2 square is equal to V1 square correct plus 2A into D. Right. So A into D is equal to V2 square minus V1 square divided by two. So that you have to substitute here to get the answer. So V is equal to V1 square plus V2 square minus V1 square by two. Square root of that. So you will get it as V1 square plus V2 square by two. Square root. So again, final answer, not important. Understand the thought process, how we are getting the answer. In the first step itself, you write down what they're asking in terms of whatever variable you can assume. And then look at what are the extra things that are there in your answer and try to remove those extra things from whatever else is given in the question. Is the method clear? The way you solve the numerical, is it clear? Everyone, yes, Omkar, you can unmute yourself. I mean, once in a while you can unmute yourselves. Don't be completely mute. Yes, Omkar. Sir, I don't understand the part with what? Sir, can you hear me? Yeah, tell me, tell me. After the part where you'll assume the distance and the first equation, the second one, the V is equal to V square plus 80. Can you explain that part? This is third equation of motion. Do you remember this? Yes, sir. I'm using this between this point and that point. The final velocity square is equal to initial velocity square plus 2A into SSD by 2. Okay, sir. Thank you. Same way I have used between A and B also. Between any two points I can use that equation. All right, let us move ahead. Mute yourself. Yes, yes, we are going to solve more and more questions. Don't worry. All right. So we will solve more questions. Things will become even more clear. Okay, so this thing we have done, right? I hope you remember that we have done this. Distance traveled in the nth second, U plus half A into 2N minus 1. Okay, do this. Sir, will derivation be as in exams? Yes, yes. In your school exams they will ask. Okay, but not in the community exams. Get the answer here. Every physics question, draw a diagram. Okay, always remember that. With subject you guys are finding most difficult till now out of the three. Don't tell me physics. Physics. Okay, somebody said maths functions in maths. Hold on before you. Mathematics is going to grow now. Physics also. Chris, you are attending offline as well as online, is it? Fine, Raj Venkat got some answer. Anybody else? Anybody else? Assume that accession due to gravity, G is 10 meter per second square. All right, so suppose total height is H. Okay. So, the time of travel is how much? Let's say the last second is nth second. That is n seconds. So time of travel, I'll say total n second it takes to reach the bottom. All of you agree or not? Total n second it takes. Last second is nth second. Like for example, first second it travels this much. Second second this much. Third second this much. Fourth this much and fifth it is here. So, one, two, three, four, five. Fifth second it has traveled 9 h by 25. But the total time of travel is total five seconds, which is as you meant to be n seconds. Clear? So, I'm going to use this value of n. H. In which equation? I'll use it in s nth is equal to u plus half a 2n minus one. So, what is s nth? What should I write here? Tell me. What should I write instead of sn? Good. So, 9 h by 25. When you write 9 h by 25, which direction you're assuming positive? Down or up? Down or up? You're assuming down. So, throughout the question, throughout the question, you will assume downward to be positive. Once you assume something. So, 9 h by 5, 9 h by 25 is downward displacement that you're assuming plus. So, everything down is positive. U is how much? Initial loss is what? Initial loss is zero. So, that's gone. Acceleration, what should I write? Plus 10 or minus 10. Plus 10 because the equation is down, you assume downward positive. So, 10 into 2n minus one. This is your first equation. Second equation will be what? Everyone. What will be my second equation? Any guesses? Correct. I'm going to use s equal to ut plus half a d square. So, if I write, instead of s, if I write h, u is zero, half 10 into n square. Right? Do all of you understand these two equations? Anybody has any doubt? You tell now itself. Otherwise, once I'm done completely about to go, you tell that, can you go back to the previous slide? I have doubts. Okay. Somebody is asking how t square equal to 2n minus one. Where t square is 2n minus one. This is the formula for the distance traveled in the nth second. And this is the substitution for that. And this is the substitution for s equal to ut plus half a d square. Okay. Clear? So, as n is nth. As nth is the distance traveled in nth second. Derivation we did, right? Last session, did we do the derivative? Some of you saying we haven't done the derivation? We haven't. Yeah, we have done it. We have done it. You're making me do the same thing twice. All right. Anyways, the last class, that is why you are asking that. Anyways, any other doubt? Okay. Some of you are asking G is 10. Is it actual value of G is 9.8 meter per second square. And even that keeps changing at different parts of the void. But roughly it is 9.8. But in order for you to solve the question easily, consider it to be 10 meter per second square. Okay. Now, can you tell me an easy way of solving this? 10 to n minus 1 doesn't become n square. This equation is different. That equation is different. Okay. You have two equations. How will you solve these? What is the easy way of solving it? Two equations, two variables. You have to find h. What is the next thing you will do? Quick. Substitute. Okay. That is one way. I like to divide. I like to divide here. When I divide it, when I divide it, h is gone. Okay. So all of this is gone. I get a quadratic equation. All right. When you get a quadratic equation, it becomes easy to solve then. Okay. So 9 by 25 is equal to 2 n minus 1 divided by n square. So 9 n square is equal to 50 n minus 25. Okay. So this will give you the value of n. This quadratic equation will give you the value of n. All right. Now, what if one of the roots comes out to be negative? If n comes out to be negative, then what you'll do from this equation? Let's say n comes out to be less than zero. One value is less than zero and other values more than zero. One value less than other values more than zero. Then what do you do? Take more than zero. Ignore this less than zero one. Time cannot be less than zero. Fine. Okay. So this is how you solve this particular question. Fine. So, you know, when we were in grade nine, then also we were using the same equations, but I hope you are realizing slowly and slowly that the difficulty level of the question has increased slightly. Okay. So that is why you need to do a lot of practice. Otherwise we will remain the same level, which was in grade nine. Just a second. Okay. This we have done, right? This is the graphical question. Do this question. Everyone. So what was the solution of the previous one? I have solved the entire thing. What do you want to know? Final answer is it? Yes, just for verification. You will get n equal to five. You'll get. Okay. Once you get n equal to five, now you can get the value of H. Is it clear? Yes. Okay. Do this question. Everyone. Somebody asking is today only practice problems. My dear friend is 95% of your preparation. Do you think you have learned any new theory? What do you learn in ninth? Have you learned anything different in theory? No. So if you are only interested in theory, you don't need to study in 11th, but then it's all about numericals. So focus on it. All right. But yes, we are going to do a little bit of theory also relative velocity. I'm going to introduce today after the break. We'll talk about theory. But right now problems. What does exclamation displacement curve represent? Area of that represents what? Do you remember that? Area of exclamation displacement represents? Look at your notes. Area of acceleration versus displacement represents what? Look at your notes. None of you answering it properly. What was it? No one. No one got it correct till now. I don't know why you are not looking at your notes. v square minus u square divided by 2. This is the area. Okay. Now do it. Get the answer. Sir, please. Hold on. I will launch a poll for this question. Okay. I'll give you one minute here. I did explain last class. You were not there when we were talking about the graphs. Who asked this? Yeah, Zayan. You are there. So why are you asking? Why divided by 2? You forgot. Just go through that previously. I remember now it was v square. We cannot go back again and again. We have lack of time. So better come prepared for the classes. And if you do your assignments, then you will not forget. All right. So should I launch the poll now? I launched the poll. Let me launch. By the way, has anybody got the answer? Unnecessary. I'm launching the poll. Otherwise. Okay. Couple of you have got it. Pick the correct option. All right. So I'm ending the poll now. It is all over the place. Actually, people have picked option A. Some of you be some of you see. So let me not show you the result of that poll right now. So the area is this. Area is this. Right. There is no doubt about it. So we will first find out the area. So let's say this area is a one. This area is a two. So total area would be what? Even plus a two or even minus a two. How do I have to consider? How will I have to consider? It'll be a one minus a two. Remember below the below the x axis. If you consider negative area. Okay. So a one is half of base, which is 10 into height minus half of this basis. Five into height again. Although that is not given. I'm assuming that is also five only minus five. This is what it is. All right. So you will get here. 25 by two. 25 by two is. The particle starts from rest. So because it starts from rest, the initial velocity is zero. So this is equal to final velocity squared divided by two. Right. So initial loss is zero. So final is square by two. So final lost is five meter per second. So answer is a. Is this clear? Type in quick. I think it was direct formula substitution. We are directly finding the area and equating what that should be. Some of us might be finding this also tricky. The reason is because you have not remembered that what area represent. All right. It is not tricky. It's a direct formula substitution kind of question. All right. Okay. How did you get V squared divided by two? So now you are there in the last session. In the last class where you're there. So there we discussed entire derivation, how it comes the area of area displacement curve. Do you not remember? I'll just quickly. Listen, listen, listen here. Let's say if a, this is expression. This is displacement. So efficient is constant right now. So area is what efficient into displacement. This is acceleration and this much displacement. This is efficient displacement. And if efficient is constant, I can say v square equal to use square plus two a s. So a into s, which is the area comes out to be v square minus u square by two. All right. So that we are doing only for the constant equation. But this I have proved in the last class that whether efficient is constant or not area is always v square minus u square by two. So this derivation I did for you when efficient is constant. Clear now. Okay. All right. Should I launch the poll for this question? Here is the poll and comes the poll. Use your common sense. Okay. There is no equation. Nothing like that will be used here. Common sense. Only two of you haven't taken the poll. Why? What happened? Now only one person is left. Okay. So I'm ending the poll now. This is what we have. Most of us have picked B. Some of us have picked A, C and D also. Let us see what. What is the problem with A? Can anybody tell what is the problem with A? You can unmute and tell. Distance can't be negative. So it can't go down. Distance cannot be negative. But where is the distance negative here? So it's reducing back to zero. You can't. The distance only increases. Okay. That's a problem with A. What is the problem with C? The same. Plus it can't be negative. In fact, it is becoming negative. Can distance be negative? No. The same issue with D. Distance cannot go down. Distance always increases. That is true only with D. Okay. Do this. Quickly do it. This is... These are the questions where everybody knows how to solve. But still people make silly errors. I will launch a poll after one and a half minute for this. Swarely, you joined just now. One hour late. What happened? Sir, can you help me? Yeah, tell me. You can type in. No, no, no, no. Type in, type in. Okay, fine. Okay. All right. Should I launch the poll? Poll is coming in your way. And here it is. The poll. Ending the poll now. Ending it. And here it ends. Share the results. Most of us have picked option C, but substantial amount have picked B as well. Let us see. Okay. So before I proceed, I'll again tell you that some of us are in a mode of, let's sir do everything. I will copy it down in my notebook. Then I will read it after the class gets over. That mode will not work for the comedy exams. This will not work. Okay. You should not wait for me to solve so that you copy it down. Otherwise you can just copy down from the, from the modules. You can copy down everything. It is written better than my notes. All right. But then we are here to discuss about the doubts, discuss the questions. And if you have any doubts, you ask. All right. So don't worry about, you know, getting any question wrong or thinking that what sir will think about me, what others will feel about me. So all that is nonsense. Don't bring that in your head. All right. And anyway, those who are answering most of the time, they're getting it wrong only. Right. So it is fine. Whatever you get it or not get it, you need to interact. Fine. Don't just sit without interaction. Okay. Anyways. Anyways, let us solve this question. Now, ball is released from rest from the roof of building of height 45 centimeter, 45 meter. Centimeter. So the height is 45. T equal to zero. Find the height of the ball from the ground at equal to two seconds. Now, some of us may be in a hurry. You might have found out height from the top. They're asking height from the ground. Right. So they will make you do the silly mistake. They will try to induce all sorts of mistakes from your side. Anyways, so at two seconds from T equal to zero dropped. So which equation you use to find how much it has moved UT plus half 80 square. Let us try to use GS 10 only. So S is equal to you, which is zero half into T square. So two into two. So as expected. 20 meters is there in the option, but answer is what? What is the answer? Is 20 meters the answer? Answer is 25, which is 45 minus 20. They're asking from the ground. Look, look at the thing here. They are making you do the silly mistake. Yes or no. This option is sitting here. What does it mean? What does it mean? This means that they want you to make silly mistake. All right. And if you are not careful enough, you are anyway making silly mistake and they want you to make the silly mistake. Then higher chances that you will make. Fine. So I think most of us have picked option B. Right. Am I correct? See. Okay. Good. But some of us have made silly mistake. All right. Here comes the next objective question after one and a half minute, I'll open up the poll. So should I launch the poll? Some of you are saying they are done. Yeah. Mentally done will be wrong. Every physics question make a habit of drawing little bit of diagram, writing the equations, getting the answer. Okay. Maybe CT level questions you can, you need not draw, but if you don't have practice of drawing the diagrams and stuff like that, you'll never be able to do some tricky ones. Shall I launch the poll? Everyone. Okay. Here is the poll. My voice is breaking. Is it, is it true? Somebody's saying my voice is breaking. No, no, no. It's not. So check your internet connection. Check your internet connection. I'll just show you my internet. Speed. Okay. So that. So the thing is the voice may break because of your internet connection also. Okay. So I have only this much, but still that's good enough 50 Mbps I have. And probably just check in your house itself. Maybe some of maybe your brothers or sisters, they might be watching YouTube or something like that. Just tell them wait for some time. Okay. So bandwidth you need to release if internet is, it is less at your home. But yeah, sometime it happens on my side also, but not today. Anyways, so let us solve this question. There is a parachutist who bails out and falls from five meter when his parachute opens. For after five meters after dropping five meters, the parachute opens. So before opening the parachute, it will gain of initial lost or not. It'll gain the initial loss or not. Oh, do I need to launch the poll? Sorry. I forgot that. Take the poll quick. So you're telling your internet speed 240 nano Mbps or 240 Mbps. All right. So I'll end this poll now ending it and it ends. People have mostly picked up option B, but option A is also there. Let us see. All right. So we have a situation here falls for five meters. So this is a free fall. Okay. The deceleration produced is one meter per second. So after this, after this free fall, there's a deceleration. They are saying deceleration. But remember, deceleration just means acceleration in opposite direction of velocity. That's all. Okay. Reaches the ground reaches the ground at zero velocity. You need to find what is the height from where he or she falls. Let's say total height is H. Okay. After five meters. Parachute opens. Okay. Now it will by this time, this person have gained some initial velocity before parachute opens. So after this point, till this point, efficient is vertically downward efficient was equal to G. After parachute opens, what direction will be the acceleration is 123 after point to two. Acceleration is upwards. Acceleration is one meter per second. Squire in upward direction. And when it reaches the ground, the final velocity is zero. Clear. This is the situation. Now you need to split the question into two parts. Yes or no. One to two and then two to three. Why? I had to split the question into two parts because although efficient is constant in one chunk of the flow, like one to two efficient is constant. Two to three is also constant. But throughout one to three, it is not constant. Are you getting it? So between one and three directly, I cannot use equation of motion because efficient is changing between one and three. I can use equation of motion between one and two. And then between two and three. Understood. All of you type in. All right. Now, now you remember how we solve the physics question. We will try to directly get the answer. H is asked. So from two to three, this distance would be what? What do you think this distance is? Two to three is H minus five. Right. H minus five. We have final velocity zero. We don't know initial velocity. Equation is minus one meter per second square between two and three. So which equation should I use to find H between two and three? I'll be using V square equal to U square plus two AS. Can anybody tell me why I have chosen to use this equation? Why not the other two? We don't know anything about the time. And we know about the displacement. So displacement should be there in the otherwise what do you will find? Right. The final is zero initial velocity. We do not know. Plus or minus. Minus it will be minus two. Displacement is one or accession is one in a position of velocity. So if downward is positive, upward is negative. Two into one into displacement is what? S is positive or negative. Positive negative. I'm taking downward positive. So this placement is positive. So two accession one and S is H minus five. So this is my first equation. And this should give me the value of H. But what is the problem? Problem is I do not know. We don't know. What is you? So how do you find you? You do find you between one to two. It is a free fall. So for a free fall, you can use this same equation. You can use this same equation for the free fall. Right. So you have final velocity is actually the initial velocity of two to three. Final velocity of one to two is initial velocity of two to three. Two square is actually V square zero. Okay, plus two into G is 10 and displacement is five. So from here, you come out to be 10 meter per second. Okay. You come out to be 10 meter per second at point two. Now what? Just substitute the values. So 10 square is 100 minus two H plus 10. So from here, H comes out to be 110 divided by two, which is 55 meters. Is it clear to all of you type in? This is how J main questions would be. Everyone type it. Just spend couple of seconds here. Go through and type in. Is it clear? Yeah, this is J main. J main level you can say. Some of you are saying that do we need to subtract five from the height? No, they're asking from which height the bailout. So it is written right. It is bills. It bails out. Fine. Yeah. Let us move forward. Do this, everyone. This is a numeric value type. You get X as integer. After this question, probably we can start something new. Then once that theory is done, we can again come back to the problem solver. You can directly type in your answers because no point putting a poll for this. But they got something. Others sake got something. I hope you have converted kilometer per hour into meter per second. Any other answer? Gayatri got something. Okay. All right. Different different answers. People are getting. Let us solve it now. Let us solve it now. Vishnu also got something. Narayan got it. Chris got it. Fine. Everyone. Truck is moving with this velocity. If you convert into meter per second, how much this is 20 meter per second. So the driver can produce the deceleration. Now, you know, they, they say deceleration, but you need to keep your head clear about it. Deceleration means opposite direction on the velocity. So if you say initial lost is plus 20 meter per second. Expression would be how much? Minus two. Fine. The stopping distance of the truck is something. If reaction time of the driver is 0.2 second. So for. For 0.2 second. What will happen for 0.2 seconds? Acceleration is zero. All right. For 0.2 second. Acceleration is zero. So they are asking about how much distance it will move. So there'll be two distances. For 0.2 second. Acceleration is zero. For rest of the time. Acceleration is minus two. So distance traveled for 0.2 second. Is the initial velocity into 0.2 second. So that is four meters. Then I have to find out distance traveled. When initial is this. And it stopped with because of the deceleration, which equation should I use? Ford. Ford. That is V square equal to U square plus two a s. I know that final loss is zero. I know my initial lost e 20 square. It is then exhibition is minus two. So two into two into s. So what it comes out to be as comes up to be 400 divided by four. comes out to be s comes up to be 400 divided by 4 that is 100 so total distance is total is 104 104 it is now what is given stopping distance is 13x so 104 if it is equal to 13x is equal to what 8 and 8 it is this is an actual question of j main few years back somebody asking how did you get 4 meters look at what they are writing they are writing reaction time is 0.2 second reaction times means what reaction times means what reaction times mean that before you apply the break it takes 0.2 seconds okay you will react you will not react with in 0 seconds right so but then before you apply the break deceleration won't happen so for 0.2 second the truck will move with the constant velocity so it will travel a little bit of distance which is the loss into time which is 4 meters okay and then deceleration will kick in clear Rajvan cut so total displacement is 4 meter which is because of his reaction time plus deceleration time the deceleration distance all of you type in is it clear all right okay okay so should we continue doing numericals or should we do something new we can come back to the numericals later also right so let us go do something new I have just too many questions prepared for you but we will do something new write down relative velocity now has it ever happened to you I hope all of you have traveled once in a while in a train okay has it ever happened that you are standing in a platform and you you think that your train has started moving and then you later realize that it is not your train it is a train next to you that is moving has it ever happened right it happens to almost everyone and then then what you do you look at the platform and see that okay I am moving relative to the platform or not forget about the train you will then stop looking at the train you look at the platform and then you realize that you are stationary right so that kind of study is what we are going to do now fine that is called relative motion or relative velocity okay and unfortunately it is removed from your NCIT curriculum okay so in your school they will not teach you but then it is a very much an important part of any commodity exams you write fine so let us proceed now why you see something moving relative to yourself first of all when we say an object is moving okay then that object is always moving with respect to you only all right because if you actually look at it then you can say that earth is moving whatever is on the earth that also moving earth revolves around the sun sun is also moving galaxy is also moving in the universe is expanding so will you be ever able to say that something is at rest absolutely will you ever be able to say or find out that something is at rest you know absolute rest is it possible to say never never in fact there is no experiment in this world that can tell you whether you are at rest or moving with uniform velocity you just cannot experience the difference between being at rest or uniform velocity okay so but then yes when you are in a train as soon as the speed increases little bit the vibration starts your while you are on the rails so then you start feeling that you are moving but suppose vibration is completely absent okay then you will not feel that you are moving if the velocity of the train is constant okay so this is this has to do with while you are observing something what are you doing let's say I'm standing at the place and looking at something okay then if that object is moving it means that distance between me and the object is changing or at least the location of the other object relative to me it is changing then only I'll say that the person or that object is moving because if I start let's say if I start running and that object or a person also starts running then distance between me and the person is constant so I will not feel that that person is moving fine although I know for the fact that I'm running so other person is also running but if for a small for a split second if you forget that you can feel whether you are running or not then if the distance between you and the other person doesn't change then for you that person is at rest okay now the picture in front of you that lady is sitting inside the train that lady is looking outside so relative to this lady the train the trees are moving backwards the poles and trees everything is moving backwards although the common sense will be there and the lady will know that why it is happening okay but we are not here to talk about why it is happening it is kinematics it is happening or not that trees are moving backward all of you agree or not although you know why it is moving backward that is fine but with respect to you are you feeling that trees moving backward when you are in a moving train yes or no right so that is what we are going to study and this is a very very important study this is the basis of a general theory of relativity also which Einstein has founded in fact Einstein told us that not only the velocity distance even the time is relative okay so that becomes very complex then right anyways so moving forward moving forward right down let's not discuss relativity the different principles apply over there okay anyways so when I say relative velocity or relative motion okay I can say relative acceleration also so whenever I say velocity I mean the something similar for the acceleration also okay so if I write velocity u a b it means that I'm talking about velocity of a relative to b okay if I write x a b it means displacement of a relative to b if I write acceleration a b it means acceleration of a relative to b clear right so let us see how these things come out fine look at this although this this looks a little you can say mathematical don't worry only for the explanation purpose we are using mathematics here uh while doing the problem solving none of these things will be there don't worry okay now look at this everyone this is object a object b is located there object a is located here this is this is you can say a person standing on the ground like this okay now this person is observing a so this person will say that a is this much distance away clear person b will say that a is that much distance away all right now if you see that you can use triangle of addition here you can say that the x coordinate between b and g plus x coordinate between b and a if you add it up you will get the x coordinate between g and a does it make sense everyone let me say this is 2 comma 1 okay this is 0 comma 0 okay this would be let's say 4 comma this 2 comma 3 4 comma 2 this is what it is all right now the x coordinate g to b the x coordinate of b minus x coordinate of g it is 2 this value is 2 from b to a the change in the x coordinate is again 2 from 2 it has become 4 so 2 plus 2 you can directly measure from g to here that is also 4 only so 2 plus 2 4 nothing special is written here same thing okay now you can see here you can write x a b measurement of a's distance from b is equal to measurement of of a's distance from g minus measurement of b from g okay and if you differentiate it I think some of you will find it this thing little messed up lot of mathematics is there but don't worry soon everything will start making sense so just hold on for some time okay now someone is asking what is a point of understanding that graph the point of understanding that graph is this so x coordinate of a relative to b is x coordinate of a minus x coordinate of b all right so similar way you can write velocity velocity of a with respect to b is equal to velocity of a minus velocity of b b is the observer that can move all right same way you can write for the acceleration acceleration of a relative to b is addition of a minus x system b now let me quickly summarize it because I can sense people are getting confused here listen here everyone so let me go back and do this again all right so there is this object a there is object a all right which is observed by two people one is standing here and another person is standing here this person is at rest and this person is moving okay so both of them will see the same thing or both of them will see two different things how a is moving all of you if b is moving and a is at rest both of them both of them will see different things right now what I am trying to do I am trying to what I am trying to find out is what b will see in terms of what a is seen that is what I am trying to find out okay for that is what I am trying to find out so I am trying to find out velocity of a relative to b got it I know velocity of a relative to someone at rest till now we always dealt with velocity of an object relative to someone at rest okay so when we write velocity is 5 m per second it means that it means that someone who is at rest observing that velocity okay let's say some object is moving with 5 m per second someone standing here will observe it to be 5 m per second but what if this person also start moving with 5 m per second then what this person will see will it see 5 m per second what that person will see everyone if both of them start moving with the same velocity 5 and 5 a will observe what velocity of b 0 a will feel as if the velocity of b is 0 right so this is what I am trying to find out what I am trying to find out what I am trying to find out that if I move with certain velocity then what exactly I will see clear clear everyone is it clear or not okay fine now let us not get into too much of mathematics of it but this is what I am trying to achieve here fine so when that is understood then let us use little bit of common sense and then you will see that these equations make very easy sense all right for example let's say there is object a going this way 2 m per second this is a and b is going this way 2 m per second so what b will observe velocity of a to b what b will observe velocity of a to b 4 right all of you agree you can say right 4 that is what it will see fine so if you substitute here values if you consider let's say this equation is correct then if velocity of a is positive so you write plus 2 minus of velocity of b if a is positive b is negative so minus of minus 2 you will get a 4 only you get a 4 only that is what we have been getting it all right but if b also goes the same way as a is moving both are the same direction both are positive so 2 minus 2 should give you 0 all right now all of you understand why we are subtracting the two velocities to get velocity of a relative to b or should I simplify it more should I simplify it even more so that all of us are able to okay let me simplify it I will completely simplify the situation here and let us talk about exactly how we solve numericals then it become very very simple okay forget about all of these things right now for us for some time let's talk about look at this example for for instance for instance look at this example okay so here here you can see that the the blue t-shirt girl was running but from behind somebody else came and won the race okay why because that other person was traveling faster now does it matter how fast that red t-shirt girl is running or what matters is how fast relative to the blue the red girl is running what matters whether the red t-shirt girl will win or not will depend on how fast the red girl is traveling compared to the blue all of you agree right so if you just know how fast red girl is running you'll not able to find out whether she will win or not but if you know that the blue is running with 2 meter per second the red is running with 5 meter per second and there's a certain distance to cover you will find out who will reach first fine so in order to understand the relative velocity I can introduce a very simple concept here write down velocity of approach everyone write down velocity of approach now you'll understand very very easily so suppose this kid is running with velocity v1 and that kid is also running with velocity let's say v2 so with what velocity they are approaching each other tell me with what velocity they are approaching each other all of you agree v1 plus v2 type in v1 plus v2 right and suppose a distance between these two is d so can you tell me how much time it takes for them to meet with this velocity they are approaching this is approach velocity so time taken is d divided by v1 plus v2 as simple as that as simple as that okay all right now v1 plus v2 what is v1 plus v2 v1 plus v2 is velocity of approach it is also the relative velocity of first with respect to second or vice versa also it is two this is one and this is two so this person will feel as if the other person is approaching with v1 plus v2 got it the first person will feel as if the second one is approaching with v1 plus v2 so that is what is velocity of approach got it is this clear type in very clear right these are very very important thing that is why i'm not proceeding because i think that some of you are not clear so i'm standing here itself okay so let's let them meet okay now one thing that will help you a lot is this statement all of you write down an observer sees the world after subtracting velocity of approach is relative velocity only but looking at it in a different manner easier manner when you solve the problems use velocity of approach and separation don't use relative velocity approach of solving questions okay this thing probably you'll not get in many textbooks so an observer sees the world after subtracting his or her own velocity or acceleration what does it mean it means that this is let's say this this kid is observer first kid is observer so this first kid will observe the velocity of the second kid after subtracting his own velocity let's say his actual velocity of the second kid is v2 the actual velocity of first kid was v1 v1 is negative v1 direction is opposite of v2 right so what the first person will see is v2 minus of minus v1 it has subtracted its own velocity fine that is what is velocity of approach so relative velocity and the velocity of approach one in the same thing fine but when you look at it as if it is velocity of approach then things will make more sense because velocity of approach is something which is easier to understand now after velocity of approach what do you think the next logical thing should be after approach there should be correct velocity of separation as well fine so tell me tell me if this person was going with v1 velocity that person was going with v2 what is the velocity of separation here velocity of separation everyone with what velocity they are separating v1 plus v2 v1 plus v2 both are going away from each other okay now here what is velocity of approach by the way how much is the velocity of approach velocity of separation is v1 plus v2 velocity of approach is what it is negative of velocity of approach sorry velocity of separation velocity of approach is this way velocity of separation is that way so one is negative of other all right just like if velocity is this way positive opposite direction negative so separation is opposite of approach or not so one is negative of the other okay is it still confusing should i give more examples more examples should i give okay let me give few more in fact send the velocity of approach you said was v1 plus v2 yes v1 plus v2 in that example it was v1 plus v2 both were coming together right both were trying to come close to each other so you need to add them up both are going away from each other so that is separation this is approach okay so you need to see what is happening then only you can say what is the answer for example let me give you certain situations here everyone let's say this distance is 10 meters two objects are there fine this object is going this way with let's say two meters per second okay that object is going this way with five meters per second okay you need to tell me you need to tell me first of all velocity of separation then time for the separation to become 20 meters these two things do it velocity separation is what you can type in first that correct most of us have got the correct answer velocity of separation is five plus two anybody has any doubt in this type in anybody has any doubts velocity separation is seven meter per second no doubts separation is seven meter per second means what with this velocity with this velocity they are going away from each other now tell me here what is the answer time for separation to become 20 meters initial separation is 10 so how much extra separation was done how much extra separation was done how much extra separation 10 more so with 7 meter per second extra 10 meter was covered so the time is 10 divided by 7 clear type in okay somebody is asking if same was velocity of approach it would be minus 7 yes minus 7 but they are not approaching each other they are going away you know it so shall I take another something similar to make concept even more better let us do that no still it is not clear for some I will keep on taking example till it is clear to everyone now you know remember the previous question suppose the previous question you have to do the previous question by let's say by usual method without using velocity separation suppose you have to do then what you have to do find out the distance travelled by two meters two meter per second velocity find out distance travelled by this velocity then add these two and equate it to 10 right so it was a longer way of getting the answer but if you use the losses separation which is indirectly using the relative velocity you will directly get the answer okay so that is why we are studying relative velocity we are studying relative velocity so that when more than one object is moving we can account for both the motions together clear when we use relative velocity we are factoring the effect of both the movements in one go itself otherwise you have to first look at what the first object is doing then look at what second object is doing then combine them get the answer answer is with that way also okay so let's take this one again this distance is 10 meters this velocity is let's say two meter per second this way and this velocity is that way which is seven meter per second okay you have to tell me time when both a and b meet and also location of a and b where are they i'm trying to make relative velocity very very simple just use your common sense you get the answer many of you got the answer already what is the answer of the first part okay so i'm trying to find out when a meets b when a meets me b so what is the total distance of approach total distance of approach is what distance approach is 10 meters right now distance between them is 10 it should reduce to zero or not distance between them which was 10 should reduce to zero so how much they have approached each other forget about that a might have moved lot more b might have moved lot more that doesn't matter what matters is how much was a distance before and how much is the distance now so total approach is 10 meters all of you agree this approach is 10 meters the loss of approach is what with what velocity are approaching each other seven minus two all of you agree with seven a is trying to go close to b but b is trying to go away with two so actually seven minus two which is five meter per second so the time is distance divided by speed which is two seconds many of us get the answer in this case now how much distance a would have traveled a would have traveled seven into two 14 meters b would have traveled two into two which is four meters so they will meet somewhere here where this distance is two and that distance is 14 this is 14 this is four sorry how do you get two seconds distance of approach is 10 and velocity of approach is five so distance is equal to velocity into time velocity is five so t is two what is the doubt is this clear does this make it clear no still there will be little bit of hesitation so what you can you know you can solve the same question by what you can say that in time t a will travel seven into t distance okay in time t b will travel two into t distance if they have to meet distance traveled by a minus distance traveled by b should be equal to 10 and you get the same answer okay but you can directly get the answer like this so start looking things like this so after the break we are going to discuss more examples okay things are not yet over I know still there is a confusion I will keep on taking more and more examples here make sure you get it very very properly in your head okay then you know you will be able to solve many tricky questions in a very very simple manner fine so let us take a break now and we will meet on the other side of it okay so can you all hear me type in am i audible am i audible all right so let us take few more examples here try doing this try doing this let us see everyone I mean I am asking you something to do here where I have not considered acceleration while explaining to you but I want to see if you can extrapolate and solve this question based on whatever we have learned till now let us see what is the velocity of separation here velocity separation is how much I am not solving I am just asking you the what is given here all of you agree 50 meter per second everyone right now not only velocity there is acceleration also what is the acceleration or separation tell me acceleration of separation is what good to see some of the correct answers good this action is separating or approaching one and one total two is approaching or separating both both are approaching one and one two is approaching so separating would be minus two okay so that is a hint now do the complete question get the answer aditya got something others never answer anybody else naren got something okay now tell me displacement separation how much away they have went from initial to the final position initial distance was 20 now it is 60 so how much is the movement away separation is how much 40 meters 40 meters is there they are asking t for it t is what so which equation will you use which equation will you use you can use s equal to ut plus half 80 square that is a surprise everyone the same equation you can use it for approach or separation all the displacement velocity and acceleration should be of approach or all of them should be of separation you can still use this okay now just substitute here s separation is okay I have written approach same thing is valid for the separation also separation is 40 initial velocity 50 into t plus half into minus two into t square so you have t square minus 50 t plus 40 is equal to zero sum of the one of the students asking how is separation 40 can anybody explain why separation is 40 why it is 40 so yeah so so initial separation is 20 meters so in order to get 60 meters of separation we need nothing but 60 minus 20 which is equal to 40 meters so 40 meters more correct so you don't need to you don't need to do anything to be separated by 20 right final separation is 60 so how much you have to work for 60 minus 20 which is 40 clear all right so what do you get from here has anybody solved it this quality equation no one but anyways when you solve this quality equation you'll get the value of t is this is this question very clear to everyone type in very clear do you want to solve more questions on the velocity separation and approach okay let us solve few more so I hope all of you understand by now that physics is not about theory that is very much clear or not no longer about theory do this when two objects are moving better to use velocity of separation concept okay somebody got the answer here no Arthur probably probably you may get minus 10 and plus 40 just check it again but meanwhile do this question but even 10 and 40 maybe correct also after this question we'll discuss that okay but first complete this yeah just answer once otherwise I'll feel as if if you answer five times I'll feel as if five students have answered all right shall I discuss now how much weight Aditya how much so close all right so let us discuss now b catches c you need to find separation between a and c when b catches c so do do you think that we need to first find out what is the time taken for b to catch c yes or no right so what is the velocity of the approach between b and c what it is with what velocity b and c are approaching each other everyone 10 minus you are you minus 10 approach with what velocity they are approaching 10 minus you are you minus 10 10 is going away you is coming close so it is you minus 10 okay so time taken for them to meet is total distance covered d with velocity approach you minus 10 this is a time taken okay now in this time separation between a and c is 3d right now what is the separation 2d from 2d it has become 3d so total how much movement away from 2d it is going to 3d so total total is d distance d distance right and it is separating the velocity separation is what between a and c what is the velocity separation plus five or minus five you guys are some of you are making a lot of silly mistakes use common sense common sense don't do it mathematically like this common sense apply they are separating or approaching first of all this is five that is 10 they're going away they're going away right clearly so separation velocity should be positive 10 minus 5 10 with 10 going away with five approaching so net net is five clear to everyone so d distance of separation with five meter per second 5 meter per second is the velocity separation so 5 into time taken d divided by u minus 10 this should be equal to d itself okay so from here you will get u as 15 go through it for 10 seconds go through it let me know if you have any doubts there is no acceleration here right then also a tricky question like this can be made can you please repeat the last step in the last step see the question has two parts right question has two parts the first part is you need to look at b and c okay when you look at b and c they're approaching with a velocity of u minus 10 the time taken for the approach is total distance approach which is d divided by the velocity of approach which is u minus 10 okay then you have to look at a and c forget about b then by the time b approaches c distance between a and c become 3d so from 2d it is going to 3d so total distance of separation is d the velocity of separation is 5 so 5 into time should be equal to d is it clear type in okay this is also j main level let's look at few more you want to solve more questions on it right I can sense I can sense still it is not very clear I'll keep on doing questions till all of you say that it is clear can you tell me what is the condition for this to happen what will be the condition so that the man in the car just reaches the friend the car just reaches the bike what will be the condition for that to happen forget about the solving part of the question tell me what is the condition for which the car just reaches the bike everyone what is the condition for that correct you can see that the car is going with a constant velocity bike is accelerating bike is accelerating right so if let's say if car reaches bike and before it reaches bikes velocity becomes more than car then will car be able to catch the bike bike will accelerate keep on increasing its velocity more and more right so the condition for maximum acceleration is that when the car reaches the bike both of their velocities match okay because if bikes velocity becomes more than car before car even reaches the bike then bike will run away okay so is this clear or not this condition is clear this has nothing to do with the equations this condition is clear or not again the condition is that the velocity of the car should be equal to the velocity of the bike when car reaches bike Arthur already got the answer anybody else wants to tell the answer I can see many of you getting the answer so tell me few things here everyone it is approaching or not we are talking about an approach here car is approaching the bike right so initial velocity of approach is what everyone the bike is at rest initial loss is zero and this is going with constant velocity of 20 meter per second so everyone understand the initial loss approaches 20 meter per second right okay now tell me the final velocity of approach is what finally when car reaches near the bike final velocity approach is what use common sense and keep it very very simple in your head keep it very very simple in your head final velocity of approach is what when the car reaches near the bike both the velocity will be equal or not just now we discussed this at the top it is written both the velocities are equal velocities of approach to become zero I don't care what is the velocity of the bike right when the car reaches here car velocity is constant 20 so bikes velocity should also reach 20 only it should not be more than that otherwise bike will just run away so 20 this side 20 same side this is bike this is car so approach is zero all of you clear those who are telling something else is it clear or not Pratik Aditya you can unmute yourself let us discuss unmute yourself sir I didn't understand how the final velocity becomes zero final velocity is not zero final velocity of the car is 20 only and the bike is also 20 so both of them are moving 2020 now tell me what is the velocities approach both are going this way only 20 so velocity approaches so it will become 20 minus 20 so 20 minus 20 is zero right Aditya is it clear okay now I want to find what is the value of a so accession of approach should be what in terms of a in terms of a what is accession of approach is it plus a or minus a everyone plus a or minus a is this way so it is going away right so everyone approaches minus a okay and distance of approaches what distance approaches what how much distance they're approaching each other parameters okay fine now you have to simply use v square equal to u square plus 2 a s over here final velocity approach is zero initial velocity of approach is 20 minus 2 a into 100 so that accession has two meter per second square all of you type in makes sense why accession is negative because acceleration is in this direction accession is trying to separate you're trying to find what is approach acceleration separation accession is positive okay but approach is negative not a very simple question but then yes we need to get used to these kind of questions then only you will get ahead of others otherwise you'll get the same thing which everybody else get lakhs and lakhs of students also get the same much you also get the same much only okay should we continue doing velocity of approach concept i do this how to know how to start with our numerical see there is no answer to that kind of question you know you're asking like how to play football kind of thing i can tell you theoretically this is the way to play football but till you practice a lot will you become good at it tell me Raj Venkat till you practice football will you be able to become good at it by just learning how to play football on piece of paper no right so you need to practice a lot of numericals and then only things will become you know streamline for you but yeah broad steps we have already discussed this is a last question on velocity of approach and separation some of the students are asking how do you know which formula to use when how many formulas you have three only three question motions you have or not only three you have so why there's so much confusion you can try doing one make using one formula doesn't work out then the use other one okay and many times is very obvious you just write down what is given what you do find out and then you automatically find out that okay this formula i should be using here every physics question all of you listen carefully every physics question draw a diagram that is your that is the step which you should never skip even if you get a very very simple question then also draw a diagram then on that diagram write down whatever is given to you and then you can see that suddenly things becomes very simple to you okay and in that process you may see that some of the students are very fast in getting the answer but you don't need to worry about it okay just don't worry about who is getting the answer fast because you never know right people might be googling who knows so you just worry about yourself let others answer doesn't answer you care about yourself do it systematically and i'm telling you if you follow all the steps while practicing when exam comes you'll be the fastest amongst all then you'll be mentally able to draw the diagram okay so follow all the steps doesn't matter whether you're taking three times four times five times the amount of time required to solve a question doesn't matter you're just learning right now but first step is draw a diagram show all the values there should i show you how to do it in a step by step manner drawing the diagram and everything or should i wait okay do it yourself see i'll do it step by step manner but if you don't do it everything is useless step by step manner of solving question is there in the book also just copy it from the book right so first do it yourself whatever you can understand do it yourself even if you get the wrong answer doesn't matter what do you understand do it first i'll wait for another minute here all right so let us discuss this is there anyone who is about to get the answer anybody else who is about to get the answer okay sake got it let us see everyone focus here we are drawing a diagram here a train starts from a station with a constant iteration of 0.4 meter per second square so there is this train it starts with like this you have to draw okay look at it carefully acceleration is 0.4 meter per second square you don't need to read the entire question and then draw while reading you can draw a passenger arrive at station six second after the end of the train left the very same point so this is let's say the end of the train t so this t will reach here by the time the passenger is here when passenger is over there the train has moved forward right it has already left and there is a distance between passenger and the train okay so this distance is for the six seconds what is the least speed at which passenger can run and catch the train feel to find out with what speed passenger should run so that it is still able to catch the train okay this diagram is clear to everyone this is the way at least this much diagram you have to draw is there any confusion how you start drawing the diagram is clear or not type in yes for the train initial velocity is 0 it says it starts okay now now don't you think that it is better to find this distance between passenger and the train right so how will you find that which formula comes in your mind t is given initial loss is given efficient is given so in order to find distance which formula comes in your mind you have to find distance so now do i need to tell you that you have to use this or does it come automatically what do you think it should come automatically right because u v t are given you to find s this should come automatically all right so s is equal to u is 0 half acceleration is 0.4 costo what are you writing why that is for this into six square right so this is 0.2 36 into 2 7.2 meters all of you getting 7.2 now till now passenger was not there only one thing was moving so for that you don't need to consider separation and approach so this is one part of the problem okay only the train was moving now when passenger comes passenger is running so now train is also moving passenger is also moving so now you have to use velocity of approach and excision of approach and things like that when only one particle is moving you don't need to worry about it fine so now the problem is this now you have a situation like this in which the distance is 7.2 you have to find what should be the velocity here okay let's say this is u 1 passengers velocity is u 1 and the train now the train will have velocity or not when it has moved 7.2 meters it will have velocity so u 2 is the velocity it has acceleration also 0.4 meter per second square is acceleration okay so the velocity of the train becomes what after traveling for six seconds you can find out its velocity which is u 2 is equal to initial loss is 0 plus acceleration into t that is 2.4 meter per second this velocity is 2.4 meter per second till now all of us on the same page clear this is how it would be in j advanced so we are solving an advanced level numerical it's fine that you haven't got it okay don't worry about so till now is it clear or not type in answer what you want explain the last part last part is this see this is this is this situation is when only train is moving I want to find out what the train is doing when passenger has started running so the train although it actually started from rest but when passengers started to run the train had some velocity because it already has traveled for six seconds so I found out velocity when the passenger has started running from here by using v is equal to u plus at so this is a situation for which I have to use velocity of approach concept okay fine now tell me now tell me what should be the condition so that the passenger just reaches the train what should be the condition for it so do you agree that when passenger reaches when passenger reaches the train the velocity of the train final as your train should become equal to the velocity of the passenger if let's say before the passenger reaches the train if train velocity becomes more will the passenger be able to catch the train no so the final velocity of approach should be what finally velocity approach should be when passenger has just caught the train should be zero initial velocity of approaches how much u1 minus 2.4 u1 minus 2.4 and initial accession of approaches what sorry accession of approaches what plus 0.4 or minus 0.4 accession of approach minus of 0.4 right the distance of approaches what distance approach 7.2 7.2 distance they have to approach so which equation comes in your mind in order to find the value of u1 what should you do everyone do I need to tell you what one you have to do which equation you have to use you have to use v square equal to u square plus 2 as here right final velocity of approach is zero this is u1 minus 2.4 square minus 2 into 0.4 into 7.2 okay so u1 minus 2.4 whole square okay this will become equal don't type the same thing multiple times okay Zayan so this will be equal to what this will become equal to what 7.2 into 0.8 56 5.76 okay so from here are you getting u1 to be equal to 4.8 so this is what you will get I mean I'm not sure whether this is what you'll get I mean if you do it proper calculation you get the correct answer from here fine this is how it will be in j advanced okay so I'm not expecting that you should be able to get the answer yourself but I have to show you how it will be in j advanced or not should I have not shown you how j advanced questions are how many of you think that should I not have shown you so it is good that you know what it takes to solve a j advanced question right otherwise we will unnecessarily assume will unnecessarily sometime assume that okay they will be very simple and this and that right but then yes some of us are some of us may think that we don't know anything right some of us even I was when I was at your age and I used to go to some coaching and that one of the teachers there had shown some tricky questions and I started hitting that guy because he was giving the tough problems and then later I realized that he was helping me to do well and helping me to solve the difficult questions right so the trick here is that what you can do is I'm not expecting that you should get the answer here right if you get it that is good but if you don't get it then you know you have to struggle through and arrive somewhere and then ask doubts and learn to solve this question so you can either say that okay I don't want to do j advanced that is also fine there's nothing wrong with it you can say I don't want to do j advanced I'll not do the tricky questions fine but if you want to do j advanced you need to go through these questions and these questions do you think that you'll read theory and someday you'll wake up and suddenly you'll start getting these questions no you need to learn slowly and gradually it will happen and one day you'll see that you'll start getting it because you have practiced enough questions fine so let me go back to the neat ct and j man level questions in fact let us go and suck it you can unmute and speak so why do we have to assume that the person needs to run to the front of the train to catch it if the train's end is at the platform when he enters it won't it be enough that he runs as fast as the end of the train to catch it but if you read the question it says that the end of the train left six seconds okay it is not the train front is leaving end of the train is clear yes let us take up enough of velocity approach thing let's go back back and back more and more back do this everyone you can type in your answers as and when you get it so we assume gravity to be the same by 10 itself yes assume g is 10 assume efficient due to gravity is 10 vertically downwards okay arthur got something sake it got something should i share one write up on the velocity of separation and approach i have written one document on it with certain examples will that help okay just go through that document which i have written i have there was this article which i had published i'll try to find it out okay everyone let us solve it now you have a balloon this balloon is moving up five meter per second in upward direction t equal to zero balloon is at the ground at t equal to 10 second ball is left gently from the balloon so let's say balloon was at the ground and t equal to 10 second it will be some certain height or not now when you drop the ball the ball will be at rest what do you think when you when you are standing here drop the ball will the ball be at rest everyone okay some of you are saying yes some of you are saying no now tell me here if you have let's say if you are traveling with there is a fighter plane okay you drop let's say bomb to the enemy okay now this bomb will go vertically down or it will go like this one or two how it will travel why why travels like two why what is the reason answer is correct correct initially it will try to move it'll try to move with the same velocity as it was moving earlier that is the property of the mass it is inertia all right so initially it'll try to move with the same velocity as it was earlier itself so earlier the ball which was there in the balloon was moving five meter per second only or not so it will continue to move up five meter per second but what will happen soon because of the gravity it will lose the momentum and then come down is it clear type in all of you is it clear whatever I've told here is it clear so we need to find the speed of the ball at the ground right so first of all I need to find out what is this distance s isn't it so it was going up with five meter per second this distance is 10 into 5 which is 50 meters how many of you got 50 meters the distance balloon will go up to is 15 meters how many if you got it good so 50 meters is this distance so there's 50 meters and when you drop the ball when you drop the ball the initial velocity is what initial losses upward let us assume upward to be positive upward is positive so initial lost is plus five meter per second okay so displacement is what plus 50 or minus 50 when it goes down and hits here right minus 50 meters okay what about the acceleration plus 10 or minus 10 minus 10 minus it will be it is down right all right so up is positive right so acceleration is down so acceleration is negative or not right it is negative so we need to find final velocity now so which formula do we use what do you think correct so now do do we need to do I need to tell you which formula do you have to use once you write like this isn't it become very obvious like this when you write isn't become very obvious which formula to use yes or no those who were answering who were asking so u square is five square all right u is five plus two into minus 10 into minus 50 so v square is equal to 25 plus 200 fine so v is root over 225 okay which is how much it's root over thousand so two into 10 oh sorry sorry sorry sorry sorry this is thousand is 100 into 10 so v is root over 1025 which is roughly what 32 mute yourself mute yourself just type so you know we are getting more than 32 the reason is because you have taken gs 10 all right if you take gs 9.8 you will get much closer to the 32 fine so that is how you solve this question clear let's look at one more question okay sakheth got something others again they're asking the distance from the ground okay make sure you take care of it sakheth change the answer quickly okay I hope all of you have received the material is there anyone who did not receive the center module message me those who haven't received please message me with your complete address all right with pin code so shall we discuss this everyone two minutes you want okay let us discuss this already we have spent five six minutes on this question let us see a ball is released on the top of the tower of height h meters it takes t seconds to reach the ground so I can use s equal to ut plus half at square okay so s is h u is 0 so it becomes half into g is acceleration into t square so this is h you need to find out what is the position of the ball in t by three seconds so this is h so in t by three seconds s is half g into t by three square so that is one ninth of half gt square which is h by nine h by nine from top or bottom h by nine from top or the bottom from the top so from ground it is what h minus h by nine from the ground that is eight h by nine clear everyone everyone clear type in oh this is this is j main level the easy question of j means okay now don't ask every time we solve a question whether it is j main j advance or whatever it is okay just a question could you explain the last step again see instead of t we are substituting t by three all right so s becomes half of g into t by three square so one by nine of half gt square so that is h by nine all right maybe we can take one more do this probably the last question for today you get you want more time for this take three minutes then we start discussing nareen is the no not nareen patik is the first one to answer then nareen has answered anybody else gt aditya what is the full form of gt aditya it's aditya gt zayan samya what is your answer see most of the question that i solve with you guys i'm solving it for the first time and i do not know the final answer i do not know the solution why i do that because i want to think like a student so that when i solve it you will feel that how to start the struggle or how to approach a question it is completely i am looking at it for the first time so you will understand exactly how to approach a question all right so so that you can relate more to me when i'm solving a question fine so i don't know the final answer if you're asking is this correct or not all right aditya has answered shall we discuss now i can see some of your answer d some of your answer c let us see whether c is correct a ball is dropped from the roof ball is dropped from the roof of a building can reach the ground in five seconds so this is height h a ball is dropped it reaches in five seconds so like this you can draw a figure like that right i just need to do gravity g is vertically down so this is the first thing then if a ball is stopped after three seconds of its fall and fall again then what is the time taken by the ball okay so all of you agree first of all that h is equal to half g into five square all of you agree with this or not 25 by two times g is h right now after three seconds it stopped let's say it is stopped here so in three seconds what is the distance traveled let's take this distance travel is s1 so that will be h minus s1 total distance is h so s1 would be equal to half g into three square right so that is nine by two times g yes or no everyone so the remaining distance is h minus s1 which is 25 minus nine is how much 16 16 by two eight so eight g how many of you got till here this distance is eight g okay good so this distance is equal to half g into t2 square right because again the velocity here is zero so now i'll consider this point and that point between these two point i'm using s equal to ut plus half a day square u is zero right so eight g is equal to g by two t2 square so t2 is how much root of 16 which is four seconds option is d all right okay now let me summarize today we have solved many tricky questions okay so i hope you guys are not discouraged because of that fine the purpose of solving tricky questions the understand the purpose the purpose of solving tricky questions is so that when you do your assignments or when you practice yourself then you will be able to relate what we have done if we keep on doing the same type of problem very very simple ones so many of those questions then when you do the actual exam you'll see a lot of tricky questions then you'll not be able to score well now we have got exposed to many different like today we have taken so many varieties of the questions yes or no right so in in a short span of time we have done many varieties so what you can do is i would suggest that you please watch the recording once again all these questions if i were at your age i would not have got more than couple of questions in the today's class but what i would have done i would have sat and watched the videos so that i get a like when you watch the video you can pause you can watch it again and all that thing we can do don't worry about don't worry about that oh my time is gone i have i'm able to you know i have only four hours i'm able to watch the recording only i will not be able to solve 100 questions 200 questions no that doesn't matter at all it does not matter how many questions you solve the number of marks you get at the end is not proportional to number of questions you are solving it is proportional to how much struggle you are having while solving these questions okay so just do that hard work where you you can be as slow as you want but if you're understanding it you can watch it again if you understand it that is the best thing fine now somebody's asking where the recording will be uploaded all the recordings are getting uploaded to learn is where you do your assignments also i will share the link once again you don't need to ask for the class notes you don't need to ask for the assignment everything is there fine all right so that's it from my side it was a tricky session for all of you but i hope you have learned and got exposed to many new things today so thanks for coming in bye for now so you so bye bye so message me whatever is there message me i look like it thank you sir bye sir