 All right, welcome everyone, waiting for others to join in. Good afternoon, everyone. I think a few more will join in. How was the test experience? How was the test? Challenging, OK. So anybody got, let's say, good experience but dissatisfactory scores. Yeah, I told you, right? You might have seen the scores of the top few kids but then don't go by their scores also because nobody is invigilating the test, right? So you never know who is taking it honestly or who is giving into the temptation of looking at the solution in the internet, right? So but then assuming that each one of you have done it honestly, which is a very tough assumption but anyways, they should not make any comments on that. So in this test, if you're close to 200, you're doing really, really well, OK? Don't worry about all that. But then if your score is near to 100, it clearly shows that you haven't done the assignments properly, 100 or below 100, OK? So your scores in these type of test will fluctuate very, very hard, all right? If you are honest with yourself, you have done the assignment properly each and every week, you're doing that. You just mute yourself, but now mute yourself. So each and every assignment, if you're doing it systematically, honestly, you'll be doing exceedingly well. But if you're like one of those kids who thinks that I can prepare, let's say, one or two days before the exam, these are not those kind of exam wherein you can prepare one or two days and hope to do well. So I hope you got a learning here that the preparation for these exams is continuous, OK? Continuous, you have to keep on working hard and keep on doing the assignments. And then only at the end, the result will be satisfactory, OK? And I'm sure many of you have learned many things during the exam, which is beyond the just concept. For example, if you're not getting continuously three, four questions right, then you will feel that, oh, timer is running, what to do, whether to go in physics or maths or to do chemistry first. So that kind of problem happens or not? Have you faced those kind of issues during the test? No one faced? OK, yeah, there are a few. If you have not faced those kind of issues, something wrong with you, OK? All of them face as such issue. Time is a big factor in the exam, all right? Which we ignore. When we do the assignment and when we solve problems sitting in a table, we have a smartphone. We chat with people. We don't care about the time running it. But then in exam, suddenly time comes in and starts ticking. And then the game begins, and it will lead to many problems. And I'm sure that you have, have you done any silly errors in this exam? Could not decide whether to attempt or leave question where I have doubts, OK? Yeah, you might have seen that you have done many silly errors also, OK? So apart from concepts, forget about whether you know something or not. Apart from that, I'm sure that you guys might have lost at least 20 to 50, 20 to 50 marks on just these kind of issues, like not managing the time properly, multiple silly errors, getting panicking, getting panicked in the middle of the exam. So all those things will contribute to 20 to 50 marks. Fine. So and they will go away only when you take a lot of test. OK, fine. So and there was one question. I think question number 18 in physics. There was I think two options have the same answer given. And that question is a bonus mark. And this happens in the actual exam also. There are some errors in the actual exam, wherein if it is found to be wrong, people are given. Bonus mark on that. Now, if you are like, you know, that how come I'm not getting the answer? You spend two, three minutes, ten minutes on that one question itself. You're losing too much. OK, that question itself may not be correct. So you have to learn to let it go and move forward. Fine. I didn't understand when to take risk and take a guess or to leave a question. OK, that is a common thing. And everybody has their own risk appetite and the way you should take the risk also, you know, nobody can teach you OK, that how to take risk. It is your own appetite. Like on that day, if you are like between option A and D, you want to pick one, but you're not sure. But you think that a could be correct. OK. So you have to reason out within yourself, which one to pick. I can't say that pick D or pick A if you want to take a risk. OK, I usually say that, you know, unless you are reasonably sure, don't take a risk. OK. I will not take risk. I will just go by whatever I know. OK. In chem, one of the molecule, OK, fine, sir. We have viewed the result again. Yes, you can view highest mark as per the result is 266. I guess you can check. You can click the results and it will be there. In this type, you should attempt everything because there is no negative mark, isn't it? So you should attempt all the questions and but then from zero to 10, sorry, zero to 100 or zero to 1000, you know, there are there can be any integer as a correct option. So that is why there is no negative marking. So if you don't know what is the answer, chances of you getting right in the integer type is very, very less. So guessing won't help actually in integer type. But anyways, you should guess because there is no negative marking. OK, any other doubt and any other thing that you want to discuss with respect to exam taking strategy, anything that you all right, guys. So what is a good score? I told you, right, near about 200 is decent. And I'm sure that those who have been doing the assignment and homework systematically would have got near 200 irrespective of whether you consider yourself very intelligent or not so intelligent. You would have got 200 easily if you would have followed the assignments and everything. With section to do first, second, third. It is totally up to you only. OK, if somebody tells you that there is some magic formula is fooling you and I don't want to fool you. So it is your own comfort level. Ideally, what people say is that I mean, there are multiple strategies. OK. One strategy is hold on. One strategy is to attempt the. You know, your favorite section first favorite as in which you are good at. And the surprising thing is the toppers who are toppers in the J exam top 100 top 200 top 500 also. Their favorite subject is chemistry. The reason is they can finish that section in 30 minutes and very accurately because that doesn't take a lot of time. So that gives you confidence that you have 30 minutes extra. And then in between you slice, I mean, in between, you put your, you know, mediocre subject or you can people say that you can take math in between because that takes relatively longer time and then you take physics. That is what people say. Majority say, but majority need not be correct here. I have seen many cases in which people are going sequentially and they are able to do very well. It is totally up to your comfort level. And one more strategy. Hold on, hold on. I'll answer. One more strategy, which I have never tried, but I have heard many times that top ranker, the single digit rank all in their rank one, two, three, all those guys, what they do is that they don't think that one section is physics, one is chemistry, one is math. They consider the entire paper as if it is one subject. OK, and they what they do is that, you know, they have multiple passes to the paper. They'll keep on looking for the easy questions to solve. So first pass, they'll scan entire paper and they just click question number one, two, three. They'll find, oh, question number three looks very simple. So let me solve question number three first. They solve question number three like that. They keep on finding the simple questions till the very end. So all the simple questions are exhausted. OK, then they again go through the entire paper. Look at the next difficult type of questions. They solve them and while doing multiple passes, they have the map in their head that, OK, fine, there was one question which I could solve, but that will take a little bit longer time. So they mark it also in the piece of paper that question number five, I have to go back and do it. OK, so like that, the top rankers do top 10, top 15 rankers. But that doesn't come just like that. OK, so that is another strategy. How many of you think that they have left some easy questions in the exam? No one. You always think after the exam, when you look at the solution, you'll feel, oh, my God, this one was very simple question and I should have tried it. OK, that should never happen. Even if you are, let's say, attempting, let's say physics first. So scan the physics paper, find the easy questions of physics and solve them. All right. First, you solve easy question. That is like thumb rule. OK, that should never be this thing. Whether the question is easy or difficult. They have the same marks. So what you should do, do the easy ones. Leave the difficult ones to the later time. Then do the easy ones of chemistry. Then do the easy ones of maths. Then you can pass it all over again. Fine. And the most difficult ones, just leave it. Why you have to solve it? Why you have to spend time in it? Anyways, you're not aiming for 100 percent marks. 100 percent is 300. Right. You write now your aim is to maximize your marks. Maximization happens when you get a lot of marks for lesser amount of time. And that will happen only when you attempt the easy ones first. OK, so all of these things you should keep in mind in the next test that will happen about a month from here. And it may happen that whatever we have done till that point in time from day one, everything will come in that exam. OK. All right. So all of you are there in the bridge program. Is there anyone from this batch who was not part of bridge program? Anyone? OK, so all right. OK, Namrata, Kaushik and Pooja, you you joined little late. OK, so but then have you watched the videos of the bridge program? I've circulated those. Have you watched the videos? OK, good. Pooja and Kaushik, you all right. So why I asked that because today we are going to do motion in a plane and the vectors is the integral part of it and vectors we have already done in great details. Fine. So I'm I'm assuming that you guys have you guys have gone through the vectors videos at least right. Kaushik, you've gone through the vectors, right? All right, great. So good to know that. And today we'll start this chapter. So let us continue. And yes, don't get disheartened by the marks you have got. You know, it is very easy to get disheartened by that because, you know, I can understand few students whom you think that would not have been so much ahead of you. Suddenly, you'll see their marks are at least one hundred point away. So there can be multiple reasons for that. Gap and that gap, the how fast it has created that fast. It can go away also if you are ready to do work systematically. And never prepare for these tests one or two days before. It will not help. OK, you have to continuously keep on working hard. Fine, enough of Gyan. Let's do that chapter right on motion in motion in a plane or 2D. Now motion in a plane is a logical extension to the previous chapter. So you will find that it's very, very simple way they have extended. Whatever we have learned in motion 1D, same thing they have extended to motion in 2D or motion in a plane. So there will not be a lot of theory in this chapter. This chapter will be full of numericals. So, for example, I'm not going to tell you that V is equal to U plus 80 for, you know, I'm not going to derive that again that we have done in motion 1D. I'll just tell you how to use V is equal to U plus 80 in motion in 2D. That's all. OK, so this chapter can be divided into three chunks. All right, the first one is projectile motion. And I'm sure you guys have attended the bridge program and those who have not watched the videos, we have done projectile motion. You know, we have done up to the level of, you can say CT, or up to the level of starting of the mains level difficulty. So we have already done the school level projectile motion. So you'll not face any trouble if your aim is just to finish the school curriculum. OK, so when it comes to projectile motion, I'm going to again brief you up a little bit and then we'll directly jump into the problem solving of different types of situations. OK, then we have circular motion. Circular motion is another scenario that is motion in 2D and or motion in a plane. So we are going to study how to analyze the circular motion. There might be cases when the object is revolving with constant speed or let's say it's speed keeps on changing with time. OK, what is the acceleration? Whether I can use V is equal to U plus 80, all those things we are going to study in the circular motion. OK. And then we have relative motion in 2D. So relative motion in 2D is no different from relative motion in 1D. Whatever we have learned in the previous class about the relative motion in 1D, exact same way we are going to analyze relative motion in 2D. But just that here we'll be taking components. All right. So you'll see how we can analyze it. All right. Now, now that we have to start the chapter, I can take either of these three. OK. Now, projectile motion we have already covered little bit. So we will start from circular motion. Then we'll be doing relative motion in 2D. And after that, we'll get into the projectile motion. Fine. So write down the circular motion first. Circular motion. You can speak also whenever I ask you to speak or ask any doubt. Please speak. It's not that I'm, you know, I'm against you speaking in the class. You can talk. But not in between while I'm speaking. Don't start speaking. Fine. So tell me what is circular motion? Anyone? Motion about a fixed point? Motion about a pivot? OK. Tell me what is circle? What is circle? Circle is what? First define what is circle? Locals of all points that is fixed distance from the point. OK. All right. So basically, if you draw a path, if you draw, let's say, a line, a curved line or a path so that every time you are at a fixed distance, you are at a fixed distance from a point, then it will trace a circle. OK. Now, let me draw a circle. It's a circle. OK. This is a center. So whenever a circle is given, center should be given and radius should be given. So a particle, if it moves, if it moves along the circle, we call it as a circular motion. Now, tell me, does this point need to be fixed? Center needs to be fixed or not? Can center move? Can center move? Why it has to be fixed? When the earth is revolving around the sun, is the sun fixed? Sun is not fixed, right? Correct. The relative distance has to be fixed. Relatively, it has to be fixed. Fine. Relative to the center, the distance should be fixed. Center can be moving. OK. But then that will create a complicated scenario. We are not going to discuss that, but I'm just throwing the idea to you that the center need not be fixed. One more example I'll give you is that the fan is revolving around, let's say on top of your head, fan is revolving. So fans, fan is moving in a circle. Every point on the fin is moving on a circle. Let's say this is the fan. Each and every point, each and every point is moving in a circle. Yes or no? So whether you talk about this point, this point, that point, everything is moving in a circle with respect to this point. Now, tell me whether the center revolve around a circle with respect to that point. Is that correct? Is center moving in a circle with respect to the tip of the fan? Distance is fixed, right? It is moving in a circle. It is moving in a circle, but you are not able to see it. Why are you not able to see it? Because your position with respect to center is at rest. So you'll not feel as if center is moving. OK, but if you stand on the fin of the fan, then with respect to you, fin will be at rest and center starts moving in a circle with respect to you. Relative velocity will be circular motion. OK, fine. And these kinds of concepts will be clearer with time. Right now, I'm just throwing ideas to you and telling you that in a circular motion, center need not be fixed. OK, so now when we talk about the circular motion, we'll be talking about one more thing as in how to analyze the circular motion. So there will be some parameters in a circular motion. Tell me what are those parameters? If some object is moving in a circle, what are the parameters that you should know to describe its movement? Radius is no radius, center and speed. Is that it? No, circumference will be known by when I tell you that radius is there, circumference you can find out. OK. OK, somebody is saying that we should know whether it is going anti-clockwise or clockwise. That is also fine. You should know which direction it is moving. Right. What else? Time is a variable which has nothing to do with the motion as such. OK. Velocity. Right. Velocity you should know which direction the velocity is. When particle is moving in a circle, which direction is the velocity? Let's say over here, which direction? Where it is trying to move along the tangent? Everybody is clear, right? OK, so there will be. Tangential velocity like this. OK, now over here, which direction the velocity will be at this point, it will be upwards because tangent is upwards. Isn't it? So this is upwards. On top, it will be this way. Over here, it is like this. Now is the velocity constant, even if the magnitude is constant? If magnitude of velocity is constant, can I say velocity is fixed? Velocity is not fixed even though the speed is constant. Fine. The reason is it continuously changes its direction. Continuously changes direction, getting it? So. Even if, write down, even if speed is constant, then also velocity keep changing as direction. Is changing now? If velocity keeps changing, will there be an acceleration because of that? If velocity keeps changing, there will be an acceleration or not? There will be acceleration, right? What is the formula for acceleration? Everyone, if velocity keeps changing, acceleration is delta V by delta T. Now, this is average acceleration or instantaneous acceleration. This average, right? This is average acceleration. How to find the instantaneous acceleration? At a particular moment, what is acceleration? How to find that? Take delta T tends to zero. When we are talking about a very small time interval, then the average acceleration becomes instantaneous acceleration. Now, let us try to find out what is the value of that? And by the way, the magnitude of speed is given. Speed is let's say u. It is constant because speed is not a vector. Let's try to find out the acceleration from this formula, delta V by delta T. I will erase this. All of you have to draw these diagrams very neatly. You are doing it for the first time. Probably you will not do it ever again. So, do it properly the first time. Draw a nice circle like that. You will not be able to draw that nicely. But anyway, try it. Let's say these are the two points, point number one and point number two. It is going with constant speed. So, if I draw the velocities, it will be like that. These are the velocity directions. The magnitude of these vectors, which you can say the length vectors, the length of that, the magnitude of these vectors is u only. So, let's say the velocity over here is V2. This is vector. V2 vector is different from V1 vector. The magnitude is u only for both of them. Now, if you draw V2 and V1 together, let's say you draw here. This is V2 and this is V1. Just bring the vectors parallel to itself. Move it. Just make sure their tails are touching each other. This is V2 and this is V1. All of you have understood the vector diagram here. How we have drawn everyone. Everyone understood this. I have taken V2, moved it parallel. I have taken V1, moved it parallel and made it look like this so that I can find out V2 minus V1 because I am using triangle law. I have to create a triangle to find V2 minus V1. Everybody understood this, right? Okay. Now, tell me where is V2 minus V1? Where it is? Where is parallelogram? No need to draw parallelogram. In this triangle itself, can you find out where is V2 minus V1? This is V2. Where is minus V1? This one is minus V1 or not? Opposite of V1 is minus V1. This is minus V1 now. Change the direction of V1. You will get minus V1. V2 minus V1 is a sum of these two. It will be like that. So this is V2 minus V1. You can call it delta V. Everyone is very clear that this is delta V. You can ask doubts. Don't just sit if you are not able to understand. Delta V by delta T is the average acceleration. So the direction of acceleration is in which direction? V2, V1 or V2 minus V1? V2 minus V1 changing velocity which is what? It will be like that. It is like this. So now I have to write what is delta V in terms of speed V2 minus V1. If this is theta how much is that angle? Everyone What is the angle between V2 and V1? Extend this line. This is the angle which I am asking you what it is. No one. How much is this? Tell me how much is that? This angle is what? 180 minus theta. So this has to be 180 minus theta. So this angle should be theta. Right? So this is theta. Okay. Now tell me is the triangle this triangle O12 is it similar to the triangle PQR? Explain the last part. Last part was this. This angle is 180 minus theta. This angle is theta. Because the total angle is 180. From here to this is a straight line. Right? This is a straight line. So it is total 180 degree. That is 180 minus theta. So this is theta. And that is the angle between V2 and V1. So I am putting it theta here. Okay. Tell me not congruent not congruent. I am asking you whether they are similar. This triangle is it similar to that triangle? Both are isosceles triangle or not? Both are isosceles or not? O12 two sides are equal radius and radius. PQR two sides are equal. Magnitude is speed which is constant. PQ length is speed. PR length is speed. So both the triangles are isosceles. This angle is same. So even these two angles should be equal. So they are similar triangles. Now can I say that PQ I am using red this one. Can I say PQ by PR PQ is a length. PQ length divided by PR length is equal to O12 length divided by O1 length. Can I say that? No. Why no? The two triangles are similar. The ratio sides predict O2. O2 and O1 are both same. O2 and O1 both are radius only. It does not matter. Aditya is it fine? But 1 2 should not be there, right? 1 2 comes instead of PQ. Oh wait wait wait wait wait Sorry sorry. Yeah you guys are correct. See some of you have identified it. QR divided by PQ this is what I meant is equal to 1 2 divided by O2. Now is it fine? Corresponding radius ratios. Okay. Sorry for that confusion. QR Can I say is magnitude of the change in velocity? QR It is magnitude of change in velocity. QR is delta V. So magnitude of delta V is that length QR. So delta V magnitude divided by PQ is what? Length of PQ is what? Anyone? Speed U only. It is speed not the radius, speed. Now tell me if a speed is U and delta T is very small if delta T is very small what is the length 1 to 2? What is the length 1 to 2? 1 to 2 length D theta That's fine. 1 to 2 distance is covered with what speed? What is the speed with which 1 to 2 distance will be covered? U only? See if delta T is very less arc length will be like a straight line. Isn't it? Theta is very less. Arc length is very less as delta T is tending to 0. So the arc length 1 to 2 is like a straight line. So if the speed is U and time is delta T So what is the distance 1 to 2? U delta T Isn't it? This distance 1 to 2 is U into delta T U into delta T If delta T is very small All of you able to understand this Everyone please type in yes or no No Okay Listen Fine So if delta T is very less the particle is moving in a circle from 1 to 2 like that This is arc in which it is moving. So this arc length The length of the arc Can I say it is speed into delta T? That you understand? The length of the arc In delta T time If the speed is U It will be U into delta T With that speed it is covering constant speed Okay? So U into delta T is arc length And if delta T is very small The arc length will be very small Right? So if arc length is very small Then The straight line And the arc length distance will be equal That is why 1 to 2 distance Is same as the length of the arc Which is U delta T Now is it clear? Where is it clear now? This is very important People ignore it They just mug it up for the Derivation in the school They don't understand it But I am explaining it slowly so that It comes U delta T is 1 to 2 That divided by O2 is what? O2 distances Radius R Okay? So from here You will get magnitude of Delta V Divided by delta T Equal to what? U square by R Okay? All of you have understood probably This is the magnitude of the acceleration Now if delta T is very small Which direction the acceleration will be? What do you think? Direction of acceleration will be Towards the center Towards the center You can see here Direction will be towards the center Okay? Direction of Centripetal acceleration Will be toward the center Why are we taking Mod of delta V Because we are taking length Length cannot be vector So I am taking magnitude of that Okay? QR cannot be a vector It is length Mod of a vector is a length Of the vector Anyone has any doubts in this one? Whatever you have done? Till now R is radius R is radius This is the Coming to that, this is the centripetal acceleration This is centripetal acceleration Okay? Yes, instantaneous At every instant this acceleration will be there Okay? So now that we have got The value of acceleration Let's look at the scenario In Little bit more detail So till now we only considered The velocities This is the direction of velocity Here This is the velocity Okay? That's the velocity These are the directions of the velocity Now can you draw the direction of accelerations? Everywhere Speed is constant Previous page you want to go to Alright so All of you draw the acceleration Acceleration at these Points Done everyone This is the center Where is the acceleration at this point? At the bottom point Acceleration will be Upwards towards the center Here towards the left This one Bottom This one this side Okay? This is the acceleration Now One thing is very clear That if an object moves In a circular path There must be an acceleration Towards the center Okay? And what creates the acceleration? What you need to create acceleration? From where acceleration comes? Force There has to be a force Okay? If it is moving in a circle What kind of force should be there? Which direction the force should be there? Everyone There should be a force In the direction Towards the center Okay? Towards the center There has to be a force To create the acceleration Towards the center Fine? So That is essential But a force towards the center Is a must Because there must be an acceleration towards the center So now you think of any Circular motion example Any example Earth revolving around the sun Sun is trying to pull towards itself Continuously Then You tie a Tie a stone with a string And rotate it String applies a tension force towards the center Circle So every example that you take There will be a force To create this centripetal acceleration And that force Is called centripetal force To create this centripetal acceleration Okay? But that will take care of In laws of motion chapter Right now I am just telling you for your knowledge Okay? Now tell me Is acceleration constant? Because what do you mean? Point mass can't rotate on its own It has to take a bigger circle Is acceleration constant? Everyone, magnitude of acceleration is constant But is direction is constant? Is direction constant? Direction keeps on changing So if direction keeps on changing How can you say the acceleration is constant? So when we talk about vectors Vectors have two things Direction and magnitude Both you have to see The circular motion In the simplest of the circular motion In which speed you are taking constant Even there also The acceleration Is not constant And velocity is also not constant So can you use v is equal to u plus at here? I cannot use directly Okay? Because acceleration is not a constant Right? So One thing is very clear That Magnitude Magnitude Of the velocity Magnitude Of the acceleration You don't need to track You already know It is fixed Speed is u And acceleration is v square by r So you only need to track what? If you want to analyze the motion You only need to track Where it is Okay? Where exactly the point is If you find the point is here Okay? If you know that After some time the object will be here You will know that acceleration will be in this direction Velocity will be in that direction Okay? So you only need to track You only need to track the direction Where it is And what is the best variable To track The direction or the location Where it is in a circle What do you do? Can you think of any variable To track The location in a circle motion Angle Angle, right? It is angle only So if I know at what angle The object is From its initial point I know everything about it in a circle motion Fine? So that is the reason why Circular motions are best analyzed In terms of Angle Okay? Not in terms of position Positions Or rate of change of position Which is velocity Or rate of change of velocity Which is acceleration Works best In a linear kind of motion In a circular motion Magnitude is already known to you You want to track the angle Where it is right now So you should use angle to Find out You should analyze angle You should track angle not the position Now let's see how we can track the angle Okay? So Suppose there is a circular motion Okay? The object Goes from Into that point With speed u This angle is Let's say delta theta Radices r Okay? This is the arc length Arc length Is let's say Delta s Okay? So delta s is how much? In terms of r and theta Theta need not be small What is the definition of Definition of angle you know right? Arc length divided by the radius is theta Okay? So delta s is r Delta theta So this is The first relation If you are tracking angle theta R into theta will be the distance Covered by the particle in the circular motion Okay? Fine? Then if you divide delta s by delta t This is equal to r Let's say delta theta by delta t Okay? Now if you put limit Delta t tends to 0 What do you think left hand side is? What is left hand side? Distance covered per unit time Where delta t is very less Instantaneous speed it is? Is it speed or not? Distance by time it is Right? It's a speed Okay? Let's say speed is u So u is equal to r Into d theta This delta theta by delta t Can be written as d theta by dt And d theta by dt Is referred as angular velocity The symbol is omega Okay? So we have these two relations If you tell me what is angular velocity And radius I can tell you what is the speed Okay? Because angular velocity Into radius should be equal to Speed Fine? Everybody understood right? If you differentiate speed What you should get? Why change in theta? Theta will be changing right? It will Keep on moving like that Theta will keep on changing What is the issue? Quickly type in Or you speak Siddhan Oh no no no Theta is the Initially The particle is here let's say As it moves theta changes First this is theta then that is theta This is theta Theta keeps on changing From zero it has become 90 degree If it is 90 degree particle Will reach there So you are tracking how much angle It has made from its initial position Okay? Now what is dU by dt? What do you think this should be? Will it be centipital acceleration? dU by dt Rate at which speed is changing This is not rate at which velocity is changing by the way This is rate at which magnitude of velocity changes Which is speed what it is Omega is angular velocity yes Omega is angular velocity What is rate at which speed changes? Everyone Rate at which speed changes So basically you are changing the Changing this Right now it is this much After some time the velocity will increase this much After some time velocity will further increase this much And then you know velocity keeps on increasing But it remains tangential Magnitude of acceleration No? Ajay tell me Tell me one thing If speed would have been constant All of you listen If speed would have been constant Okay? Then if I differentiate speed What will I get? I will get 0 I will not get acceleration Speed is magnitude of the velocity Magnitude of velocity will be fixed If speed is fixed Fine? But suppose speed is not fixed Speed is changing with time Then if you differentiate du by dt You will not get 0 Then what it is What it is represented as It is the It is the It is the acceleration Which is in tangential direction Radial acceleration will be always u square by r This will always be true Okay If tangential acceleration is u At any moment Tangential acceleration is u At that moment Radial acceleration will be u square by r That will not change But if u is changing Apart from Radial acceleration There will be tangential acceleration as well Okay? This will be equal to r times d omega by dt This is at And as at This is equal to r times Alpha Alpha is a symbol for the Angular acceleration Okay? Alpha write down is Angular acceleration So, you know You should not be surprised at These values here Because When we discussed about the linear variables Okay? When we discover the linear variables We have found out The variables Position Displacement Rate of change of displacement Rate of change of velocity Now I am tracking with theta So I should find out what is delta theta What is rate of change of Theta which is omega And what is rate of change of omega Which is alpha The thing is that you have been dealing With the velocity Acceleration and displacement Since class 7th or 8th Okay? That's why you are familiar to them But this is no different from that scenario Instead of displacement You have delta theta Instead of velocity You have omega Instead of acceleration You have angular acceleration All of you clear about It Coming to a direction Hold on Imanchu Is it clear whatever we have discussed just now Is it clear Is valid only if speed is changing If speed is not changing Alpha will be 0 80 will be 0 But radial acceleration will be there Radial acceleration will be u square by r Whether speed is changing or not Now tell me what is the SI unit of Delta theta Everyone What is the SI unit? Radiance Radiance What is the SI unit of omega How should you write? Radiance for angle Radiance per second For angular velocity And what about angular acceleration Radiance per second Square This is angular acceleration Fine these are the SI units Right? There is no doubts here I will just Ask you a simple question here Then we will proceed Suppose There is Alpha and omega are the vectors Yes Alpha and omega are the vectors And they are the direction Is along the axis Along the axis Okay If it is clockwise you can take it as negative Anti-clockwise you can take it as positive Hold on Previous slide Omega Alpha and angular velocity Omega, theta and Alpha Take anti-clockwise positive Clockwise negative Okay that way you can take care Of their directions Now suppose the speed is Suppose there is a circular motion going on With angular velocity Omega Angular acceleration is alpha At any moment At a particular moment Angular velocity was omega And angular acceleration was alpha The radius Of the circular motion is r You need to tell me The direction of the acceleration At this point At the bottom point What is the direction of total acceleration All of you do this Let me know once you are done You have to show the direction of total acceleration What is the direction of total acceleration In this case No one Tell me the direction of the centripetal acceleration Which direction Centripetal acceleration will be in this way What is the magnitude of centripetal acceleration Anyone Centripetal acceleration Centripetal acceleration Is u square by r What is u in terms of omega What I can write u in terms of omega Look at your notes u is what r omega So r omega whole Square Divided by r So this is omega square r Okay And what about tangential acceleration There will be tangential acceleration also How much This is omega square r This is alpha into r alpha So total acceleration will be what This is 90 degree Right Total acceleration will be Vector sum of these two accelerations Okay The direction of total acceleration will be this All of you clear about it Anyone has any other doubt Is it clear If there is a Tangential acceleration You need to account for that as well Anyone has any doubts I am waiting still Quickly ask Shashwat is it clear Direction of alpha will also change With theta right No no alpha Alpha and omega and theta They are either clockwise or anticlockwise So alpha will remain anticlockwise Omega will remain anticlockwise How Expression is along the tangent That is tangential zeggen Tangential acceleration is along the tangent Radial acceleration is towards the center Alpha is not Along the tension Alpha is not Alpha into r Tangential expression is along the tangent Alpha is The anti clockwise direction The actual direction is when you curl your fingers like that term is the direction of alpha which is coming towards you from the axis that is the direction of alpha. Tangential acceleration has to increase the speed so tangential action should be in the direction of speed only otherwise how to change the speed fine so till now we have introduced theta omega and alpha. It is like introducing what is displacement, what is velocity and what is acceleration. But until as you get a relation for example previously we had a relation v is equal to u plus a t s equal to u2 plus half it is square until as you get a relation among theta omega and alpha. All these definitions are useless every time I have to tell you what is theta what is omega and what is alpha you'll not be able to analyze it. Okay so now let us try to find out what is a relation among these variables alpha theta and omega. Okay so write down angular variables linear variables your school started right ypr school has started 8am to 1pm. From tomorrow sir. Tomorrow and what about the other HSR? HSR starts tomorrow ypr also tomorrow. Okay but in HSR it is 8 8 30 to 3pm right. In ypr it is 8 to 1pm only. So angular variables delta theta correspond to displacement s. Omega corresponds to velocity corresponding variables okay alpha corresponds to acceleration. They are similar variables okay and by definition by definition you can write down omega as d theta by dt here you can write down velocity is ds by dt. Then you can write down alpha is d omega by dt. Here you can write down acceleration as rate of change of velocity. Okay then for constant acceleration. For constant acceleration mathematically we have solved these equations these differential equation if you solve it you will be getting v is equal to u plus at s equal to ut plus half at square. And v square equal to u square plus 2 as for constant equation you get these three equations. Now do mathematics know I mean does mathematics care whether you're writing omega instead of u or s instead of theta alpha instead of a u instead of omega mathematics is blind mathematics doesn't care. So if for these equation you get this solution so for constant angular acceleration what should you get write down all of you your own instead of u omega is there instead of s theta is there. So the first equation we want omega is equal to initial angular velocity plus what are you able to understand alpha into t. What about a second equation what should I write instead of s what should I write delta theta right amount of angle covered is equal to omega not t plus half alpha t square. What about the last equation omega square equal to omega not square plus 2 alpha delta theta but these equations are valid only if. Angular equation is a constant okay. Now what if angular equation is not constant then what you will do what you have done when a is not constant we use calculus right when a was not constant we have used differential equation and calculus to solve it. Similarly here if if alpha is not constant you cannot use these three different these three equations of motion rather you have to use it directly integrated omega is d theta by dt. Okay then you have alpha equal to d omega by dt any other differential equation you can think of alpha equal to any other differential equation that you can think of just like acceleration can be written as v dv by ds remember that. Alpha can also be written as omega d omega by d theta. Fine so do not hesitate to use these differential equations sometime alpha will be given in terms of time then you have to use this one. If alpha is given in terms of theta use that one easy to integrate. Okay, and you will be understanding it more when you solve questions only just knowing it is not enough. Fine so this is the basic concept these are the basic concepts on how to analyze the circular motion getting it how do we differentiate vectors all these vectors are either in clockwise or anticlockwise direction upwards or downwards. So you just take care of its science okay make it a scalar details will be clearer when you solve questions it is not a theoretical concept as such fine let's take up questions only like everyone write down question number one. So an object is moving with angular velocity of two revolutions I can write like this two rpm two revolutions per minute it will be very slow right let's keep it as 120 revolutions per minute. Okay this is the angular velocity. The what happens it it starts decelerating with angular iteration of deceleration of 10 radians per second square. Sorry not 10 pi radians pi by two radians per second square it starts decelerating. Okay, now you need to find out how much angle. It will rotate before coming to rest. Okay, how much angle delta theta it will rotate before coming to complete rest. This 120 revolutions per minute. Okay hurry up and got something see you need to convert it into SI units rpm is not SI unit. What is the angular velocity everyone in SI units. One revolution is how many radians omega is given as 120 rpm. Okay, so one revolution is 2 pi radians this much radians per minute now to find out how many radians per second so that divided by 60 this much radians per second so now you have converted it into the SI units that is 4 pi radians per second. Okay, and hence now you can use the equations of motion constant deceleration constant. Which equation should I use to find delta theta before it comes to rest omega square equal to omega not a square plus two alpha delta theta. Correct. You can see this is very similar to motion in one day. Okay, I will just solve one or two questions. Then I think you will understand it is very similar omega not omega is final velocity it comes to rest omega becomes zero omega not a square is four pi square. Alpha is what here what should I write instead of alpha minus pi by two so minus of two. Into pi by two delta theta fine so angle it will rotate is how much 16 pi 16 pi radians how many revolutions it will have before coming to rest. Before coming to rest how many times it will revolve this divided by 2 pi right this divided by 2 pi will be the number of revolution because one revolution is 2 pi all of you understand right. Raghuram a Kumar have you understood. Yes, sir. Understood. Sir, I'm having a lot of power issues I have I don't know sir. Okay, it is getting recorded. Don't worry. Adwik have you understood. Adwik. Chalo are now. Is it clear. Rehan Rohit. Shashwat. Okay, okay. Simple right. If you keep things simple in your head, it will turn out to be a very simple problem. Okay, don't complicate things in your head. I'm again again telling it is exactly same as motion in one day that you have just. Okay, fine. Next up. Let's say angular velocity. Omega is. Um, let's say. 10. Minus T squared. You need to find out. How much time it will take for the particle to stop. How much time the particle will take. Before it stops. Yes, and how much time. This particle will take before it stops for the first time. Okay. Himanshu got it. It was a straightforward question. What is the condition for it to stop? How many of you started integrating and differentiating? It's not required. Omega should be zero. Just put 10 minus T squared to be zero. That is the condition. Omega should be zero. Isn't it? 10 minus T squared. When it becomes zero, the particle will stop revolving. So T will come out to be under root of 10. Okay. Fine. Now tell me how much angle it will rotate before coming to rest. How much angle it will rotate before coming to rest in terms of radiance. Can we use v is equal to u plus 80 or s equal to ut plus half of T squared? Everyone. Can I use that? Can I use delta theta equal to omega naught T plus half alpha T squared? Can I use it? The answer to that is no. I cannot use. The reason alpha is not constant. What is alpha differentiate omega? You get d omega by dt, which is alpha minus of 2t. It is function of time. Okay. You cannot use that equation. So you have to do something else that something else is calculus. My dear friends do it. Okay. Should I solve it? Fine. So all of you pay attention here. Omega can be written as d theta by dt because I cannot use the equation. I have to use the differential form 10 minus d square. So d theta is equal to 10 minus T squared dt. Okay. So from here, you'll get 10 dt minus T squared dt. So when you integrate this d theta, now T will be from zero to root 10. T you already found out before it stops. Zero to root 10. Zero to root 10. You find out from zero to whatever angle it has rotated delta theta. So delta theta is 10 into root 10 minus one third of T cube. Okay. Which is 10 to the power 3 by 2. All right. So it will be two third of 10 to the power 3 by 2 radiance. Everybody understood this. Any doubts? Please ask quickly. What is the doubt? Integrate delta theta. D theta integral is theta only. Last step integral is 10 dt. 10 dt is 10 into T. Then putting upper limit and lower limit root 10. So 10 root 10 minus T squared dt integral is T cube by 3. So 1 by 3 comes then limits will be zero to root 10. Okay. When to use integration, it's up to I mean it is your knowledge. Okay. Or it is it comes to the practice. Nobody can teach you that. Okay. Fine. Now you should use integration and this is how you have to use. You can still understand when to use integration. But how to use it? You will understand with the practice only. Okay. You solve questions. Assignments and then only you will understand how to. All right. Okay. Let's see whether I have any more question on it. Try this one. This one is a simple problem. Okay. Everyone should be able to solve it your own. I'll tell you the answer. I'm waiting waiting. You guys continue solving. I think you need to put it as a number. What is that? Yeah, that is fine. Everyone solve this question. All of you. It is going with constant speed. It is given that is most steadily. Okay. Seven revolutions in 100 seconds. So how many radians? Seven into two pi radians in 100 seconds. 14 pi by 100 radians per second is the angular velocity. Okay. What is linear velocity? How to find linear velocity? You got omega. So how to get linear velocity? Omega into R. Right. You have learned that omega into R. You have learned that with radius 14 pi by 100 into 0.12. Which is equal to how much? Velocity will come out to be 5.3 into 10 raise to power minus 2 meter per second. Okay. Now what is the acceleration? How to get acceleration? Is there a tangential acceleration here? Everyone. Is there tangential acceleration? There is only radial acceleration. How much is that? U square by R. So use it 5.3 into 10 raise to power minus 2 whole square divided by the radius which is 0.12. How much it is? Have you calculated? How much it is? 2.3 10 raise to power minus 2 meter per second square. You are able to do this one? Are you finding circle motion difficult? Everyone. Is it difficult? Do you want to solve one more question? Is it difficult or simple? Everyone. Difficult. Okay. More questions. This one. Okay. I can see many answers people are getting. Should I tell you the answer? Pranav, what you got? Pranav and Arav. What are you doing? What did you get? Answer is 3 is to 1. I can see many of you have got it. 6 is to 2 is 3 is to 1 only. Fine. So a wheel is subjected to a uniform angular expression about its axis. Initially angular velocity is 0. So omega naught is 0. In first two seconds it rotates by an angle of theta 1. So naturally you will use this delta theta is equal to omega naught t plus half alpha t square. Fine. So first two seconds. I am assuming that it starts from. Yeah. It is given actually. It starts from rest. So delta theta in first two seconds is half alpha into 2 square. Okay. In next two seconds it rotates at digital angle delta theta 2. So let's call it as theta 1. Now how to get next two seconds? How much it rotates in total four seconds? It will rotate theta 1 plus theta 2. Right. Next two seconds is this. This one will be half alpha into 4 square. All of you understand these are the two equations. Everyone it rotates additional delta theta 2 in additional two seconds. So total angle will be theta 1 plus theta 2 in four seconds. Is it not clear? So theta 2 will be equal to half. Alpha 4 square minus half alpha 2 square. Fine. So this will be half alpha 4 plus 2 into 4 minus 2. Okay. So you will get it as 6 alpha theta 2 theta 1 is equal to 2 alpha. So theta 2 by theta 1 is 3 is 2 1. Right. Should we take one more? Should we solve more questions or move ahead? Till the break let's continue with the circuit motion. Okay. Anyway 430 will take a break. These two I hope you can read it. Is it readable? You're able to read it right? Yes sir. Okay. Do it. Okay. I can see two people have answered already. Others. Direction of angular velocity will be constant. Final answer. Correct. Charon got it. Charon is very quiet today. What happened? Suddenly Charon becomes very sincere student. Right. So yes the answer. Let me take a poll since I have the poll with me. All of you take the poll for the two questions. All right. For the first one answer is C as majority is saying and the second one answer is A. Again the majority. Let's see. Particle is moving in server path with an expression of A. If velocity gets doubled find the ratio of expression after and before the change. So we are assuming velocity is constant. Acceleration is what? Velocity is square by R. Radius is fixed. So A1 is proportional to V1 is square. A2 is proportional to V2 square. So A1 by A2 is 1 is to 4. But then here after and before ratio is asked. After is A2 before is A1. Okay. So that is a 4 is to 1. See 1 is to 4 is also there in option. So you if you are in a hurry will pick one is to 4. A particle is next one. A particle is moving along a circular path with uniform speed. Through what angle does the angular velocity change when it completes the half the circular path. How the circular path is this from here. This is the initial velocity. This is the final velocity, right? So through what angle does it angular velocity or its angular velocity not the velocity angular velocity direction remains fixed, which is towards you this anti-clockwise. So it remains fixed. So there is no change in direction in the angular velocity. Now, can you tell me what is a change in the direction of the linear velocity? What is the angle change in linear velocity? 180 degrees correct. Last two questions on circular motion. Then we'll start something new. Hope it is readable. Are we done with? Yeah, yeah, we had that. Didn't understand the first one. What two bodies concentric radius are smaller time periods are same the ratio between the center penetration. What is the doubt here? What exactly is the doubt? So what masses are given why are you getting confused with it? You know the formula, right? Don't worry about additional information given. This is how JEE exams are. They will confuse you with unnecessary information also at times. I can tell you these two bodies are in the moon or in Jupiter. Then you'll start wondering why they are in Jupiter. Okay, I'll put a poll. Charan is requesting for the poll. Okay, here is the poll. People are getting impatient. Here is the poll. Yellow. Okay, take the poll. All right. So I'll end the poll now. Only one person hasn't taken the poll. Charan hasn't taken. All right. So this is the poll result. It is completely mixed. You know, the first question is a confusion between A and B. A and B. Okay. So now here is the thing, guys, when they say that ratio between Centipede Recreation, you don't know whether it is the first one divided by the second one or the second one divided by the first one. Then you have to always take first one divided by the second one. Okay. By default, it is first by second. Second question. There is a big confusion. I think all ABCD people have been giving equal weightage. So let's see what are the answers now. First one, radius are capital R and small r. Their periods are same. Is it written their velocities are same? Their velocities need not be same. The time period are same. Time period is what? Time period is 2 pi r. Divided by the velocity. Okay. So velocity is what? 2 pi r by T. Acceleration is V square by R. So this will be 2 pi r. The whole square divided by T square into R. So you can see that acceleration comes out to be proportional to the radius. Yes. All right. Fine. This is the first acceleration. Second acceleration will be proportional to small r similarly. So A1 by A2 is capital R divided by the small r. So I hope you have understood. There was a twist in the question. Some of you might have assumed that velocity is constant, but it is written time period is constant. So you need to get the velocity in terms of time period if you are comparing. Okay. Clear to everyone the first question. Second one I am doing it now. Car is moving along a server path of radius 500 meters with a speed of 30 meter per second. Speed increases at a rate of 2 meter per second. So what is this? 2 meter per second square is what? Tangential acceleration. Tangential acceleration is given to you guys. 2 meter per second square. Okay. Then at that instant magnitude of resultant acceleration they are asking you to find out. So let's say at that instant the particle is here. It will have a centimeter acceleration. How much? U square by R. And tangential acceleration is 2 meter per second square. So anybody got U square by R? How much it is? 30 square by 500. This is 9 by 5, right? 9 by 5 meter per second square. So how to get the resultant now? This 90 degree resultant of these two vectors will be? Minitude will be what? 9 by 5 is 1.8 meter per second square. Sum of the squares of the whole root. Okay. This will be equal to root over 2 square plus 1.8 whole square. Fine. This is what? This is 4 plus 1.8 whole square is what? 1.8 whole square. 3.24. 3.24. So this is roughly under root of roughly exactly 7.24. Hold on. Hold on. Don't message root 7.24. So this cannot be 3 because 3 square is 9 cannot be this cannot be that. So this will be equal to D. Fine. So you don't need to find out what is square root of 7.24. After getting it just evaluate with the option which one could be root of 7.24. Okay. So that's it with respect to circular motion. All of you have understood. Is there any doubt still? If not, we'll take a break clear to everyone. All right. So circular motion won't leave you. When we'll be doing a rigid body motion circular motion will come again with full flow. And that is considered to be the most difficult chapter. There the circular motion will be like a big object rotating. So each and every point will be rotating in a circle. That will take care later on. Okay. Not now. Fine. So right now let's take a break now. We will meet after 15 minutes. So currently it is 440. We will meet at 455. PM. This is the break. Yes. Resultant means some vector summation. Just give me one more minute guys. Okay. So all of you there. We can start. I hope you guys have understood how to deal with the circular motion. You have to just practice a few more questions and then it will be perfect. Okay. You can hear me right? Am I audible clearly? Fine. So next topic that we are going to take up is relative motion in 2D. Write down. Okay. So relative motion in 2D as the name suggests it is the motion in a plane only but there are two objects that are moving here. Fine. What happened? What do you want for now? We are doing something else. Projectile motion will do later on not now. Okay. Fine. So all of you focus here. So relative motion in 2D. It is a concept very similar to relative motion in 1D. But here it is not like motion 1D where both observer and the object both are along one straight line only right in motion 1D relative motion. But in relative motion in 2D observer can be moving like this and object can be moving like that. Fine. So both the velocities need not be in a single straight line. So this is what we are going to analyze here. Fine. But the basic concept remains the same which we will write. The observer. I don't observe. Observer observes the world after subtracting own velocity from objects. Okay. So when we subtract whether we should subtract their magnitude or we should subtract vectorily. What do you think? Any guesses? We need to subtract like a vector. Vector subtraction should be there. Okay. Vector subtraction. Fine. So let's take example. A simple example we'll take. Let's say a person is going like this. This is velocity of the observer. Okay. There is an object that coming down like that. This is let's say V2. So can you tell me how this person will see this object falling down? In what direction it will see how it will be? Everyone. The observer will see the relative velocity which is V2 minus V1 towards the center. Where is the center? The observer which has velocity V1. Okay. Won't be able to see V2. Observer which is moving with V1 will be seeing relative velocity which is what the objects velocity minus observer's velocity. So observer will subtract everything with its own velocity and then look at the world. Okay. So like this it will be. So V1 is like this V2 is like that. So I need to add minus V1 to V2. So what should I do? This is V2 minus V1 is this opposite direction. This is minus V1. So V2 minus V1 will be I'll be completing the triangle. Like that. Fine. So this is the way observer will feel the object is coming. Now I'll tell you the real life example here. So when you guys have you ever ran with the umbrella when it is raining or you sat in the bike while it is raining? Although that is very dangerous. Don't try doing that sitting in a bike in a rain with umbrella. Okay. So when you run, when you run with umbrella and rain is coming down, you keep umbrella like this or you slightly tilt it forward. What do you do? You tilt it forward. Is the rain coming forward or rain is coming down? Rain is coming down. But why you tilt it forward? There's no wind. Forget about wind. There is no wind. Why you tilt it when you're running? And faster you run, more you have to tilt or not? Faster you run, more you have to tilt. Okay. The reason is when you are running, you're not seeing the true velocity of the rain. It will appear to you as if rain is coming like that towards you like this. You have to tilt in that direction. So you have to tilt in the relative velocity direction. Okay. Sometime rain coming slanted, not vertical. No, no, I'm not talking about that. I'm talking about a normal scenario. Why observer won't be able to see the object? Observer is able to see the object. But observer himself is moving now. So velocity will be different. Velocity will be different. It is like this. Suppose you are in a train. Train is moving. So you'll feel as if trees are going back, right? These are going back against your motion. Similarly here also. But that was motion 1D. Everything was happening in one straight line. Here it is motion in 2D. Okay. Is it clear, Himanshu? Observer, if it is running with V1, observer will not feel as if the rain is falling vertically down. Observer will feel that rain is coming towards it in a slanted direction. This is what it means. It will see that rain is coming in this direction, which is the blue line. So the observer won't see the object how it is actually moving. Yes, that's what I'm saying. The actual motion of the object won't be able to... Observer won't be able to see it. Because observer is running while looking at it. Any other doubt, anyone? Please ask all your doubts. When we cross the river, when there is a current, that is a different thing. We'll talk about it. River boat questions. Any other doubt? If I am running through rising steam, the direction will be opposite. What are you saying? Not able to get you. Acceleration, it will be separate, right? Yeah. Acceleration will be dealt like this only, separately. If observer is accelerating like this, you have to draw the diagram for acceleration also. Acceleration has nothing to do with velocity. All right. So I think let's discuss the river current situation now itself. Why to wait? All of you. So there will be one very common scenario, which has to do with the river boat problems. So there will be situations in which it will be like this. A river is going like this. This is velocity of river. Okay. And there will be a lot of things happening here. Observer is going forward, then coming backward. Swimmer is there. So we need to understand how to analyze this kind of scenario also. Okay. So if you are a swimmer, let's say, okay, if you are a swimmer, you are trying to go like this. You're trying to go like this. What will happen to you? You have no nonsense. Otherwise you leave it. Go. What will happen? You are trying to go like this. You'll also move with the current, right? But is the swimmer making any effort to move with the current? Swimmer is making any effort or not? Swimmer is not making any effort. Okay. So the total velocity will be what? The effort what swimmer is making, plus effort what the river is making. So let's say velocity of river is VR. Total velocity will be what? Total velocity will be some of these two velocity. It will be like this. So observer will feel that as if observer is going straight, but then in reality what happens is the drift velocity, this is the velocity of the river gets added to it. Observer goes slant like that. Okay. This is a true velocity. This is what observer tries to do. This is what river tries to do. So both efforts get added up. Fine. What would that total velocity root over whatever is a velocity of the river square plus velocity of the observer square? Observer of the swimmer. Now tell me what is the velocity of the swimmer with respect to the river? With respect to river, what is the relative velocity of the swimmer? Everyone. V resultant minus, No, no, no. You are finding out with respect to observer. I am asking you with respect to river. With respect to river it will be what? Total velocity minus velocity of river or velocity of observer minus velocity of river. Which one? A or B. Which one do you think? Look at what is written. The first line in this page. What is written? Observer sees the world after subtracting his or her own velocity from the actual velocity. What is the actual velocity? V0 or Vt? Which one is actual velocity? Vt is the actual velocity. Right? Total velocity of the swimmer is Vt. That is the actual velocity. From that you need to subtract the river's velocity. Which is Vr. So what do you get? Sir, is it actual velocity or the object's velocity? Because for the river the object is the person going. Let me finish first. So like I said, it is actual velocity. Look at the first statement. It is always the observer looks at the world after subtracting his or her own velocity. Here who is the observer? River or the swimmer? Who is the observer? Tell me. Swimmer is the observer? Pranav, you are getting it? Who is the observer here? River is the observer. And swimmer's velocity is what? V0 or Vt? Now you speak Pranav. What do you want to speak? Who was speaking? Arna was speaking. Arna, what was the doubt? Tell me. Sir, I got it. I just got confused. That is why I said first let me complete my statement. So again, total velocity is Vt. Vt minus Vr is the relative velocity. Which is what? V0 only? Which is V0 only? So what the swimmer is capable of? That is a relative velocity. The capability of the swimmer is a relative velocity. Plus the capability of river is getting added on it to get the total velocity. I guess little bit of confusion is there. We will take up questions and it will be lot more clearer. Let's take up questions. We will take the river questions only first. Suppose all of you draw a river like this. How do we know if swimmer is travelling faster or something? Hold on. Solve questions. Then many of the doubts go away when you take up the question. This is anyway not a theory. Swimmer, boat, do you think it's a part of theory? No. Theory is just what is displacement, what is velocity, what is acceleration. Everything else is numerical only. So we are discussing numerical without even taking one numerical. So now take a question and realize what is happening. Suppose the width of the river is 100 meter. Width of the river is 100 meter. Now swimmer can swim with a velocity of 2 meters per second. This is hypothetically 2 meters per second. 2 meters per second is the capability of the swimmer. Then here the velocity of river is 1 meter per second. The velocity of river is in this direction. This way. And swimmer is trying to move across as shown. This is how swimmer is trying to move. Fine. You need to tell me these things. The total velocity of the swimmer B time to cross the river C. How much deviation or what is A to B? What is A and B? A is a point right across from where it started. B is a point where he will land up. So what is the distance between A and B? Try it out. It will be clear when we try this question itself. Others, I can see many of you have answered. The true velocity will be what? Velocity of the person plus velocity of the river. So his actual velocity will be like this. So if I make a full triangle, this is the capability of the person. What the river is doing? This velocity plus that velocity is this one. The true velocity. Good to see many of you have got the answer. So the true velocity is root over this 90 degree. So resultant will be root over this velocity square plus that velocity square. It will be 2 square plus 1 square which is root 5 meter per second. Now time to cross the river. Time to cross the river. How to get that? How to get it? This is a true velocity right? Do I know displacement? Displacement is 100 in what direction? Displacement is given in this direction. That direction displacement is given right? Do I know velocity in that direction? If this is theta, the component of velocity in this direction is what? Vr cos theta which is this one only. That is 2 meter per second. So along that direction along the direction of displacement the magnitude of the velocity is 2. So time will be 100 divided by 2 which is 50 seconds. Now I need to find AB. AB is in which direction? AB is in that direction. In that direction what is the velocity of the person? Which is Vr sin theta that is 1 meter per second. So AB is 1 meter per second into 50 time. 1 into 50 seconds that is 50 meters. Now tell me is there any doubt in this particular question? Anyone has any doubts? Feel free to ask doubts. Don't worry about. Didn't get the time. See in order to find time what we do? If it is moving with constant velocity we take up distance by velocity. Distance by velocity is time isn't it? So distance is 100 but 100 is not in the direction of velocity. So I have to find the velocity in the direction of displacement which is 2 meter per second. So displacement is 100, velocity is 2. So time is 100 by 2 but suppose you know the displacement in that direction. Suppose you know what is this distance. If you know that distance then you can as well do that distance divided by root 5. That will again come out to be 50 only. By the time this person goes from here to here displacement in that direction should be 100. Velocity in that direction is 2. So time is 100 by 2. How is AB coming in? AB is what? Displacement in that direction? What is the velocity in that direction Anurag? Velocity in this direction 1 meter per second and in what time? How much time you need that distance? By the time 50 second happens it will be reaching there. So that velocity into 50 is 50 meters. Any other doubt? No doubts. Let's take up one more question on river and boat. One last question. So suppose the distance between the two banks of the river is 100 meter. There is this swimmer which can swim at a speed of 2 meter per second. The river has a velocity which is also 2 meter per second. The swimmer wants to move in a shortest path. In a shortest possible path to reach the other side. Find out how much time it will take. How much time it will take to swim? Velocity of river like this. Which direction that total velocity should be? Tell me. The net velocity should be in what direction? Papandibidu flow. Papandibidu flow that is what the swimmer wants. Total velocity should be Papandibidu flow. It is not like earlier case. So the actual velocity should be in this direction. But I know the actual velocity that is total velocity is the what swimmer can do plus what river is doing. So this vector sum should be valid. So how to complete the triangle? This is the velocity of the river. This way. So which direction the swimmer should try to move? What do you think? Which direction? Everyone try it. It should try to move like this. It tries to move like that. The river will add its velocity and it will become straight. So this is V0. So what is this velocity then? Oh sorry. This particular case cannot exist with these. Otherwise the vertical velocity come out to be 0. Let's take it as 1. Take it as 1. So now tell me what is the true velocity? Vt will be what? Use Pythagoras theorem. If you use Pythagoras theorem V0 square is equal to Vr square plus Vt square. Isn't it? So V0 square is 4. Vr square is 1 plus Vt square. So Vt is root 3 meter per second. Its total velocity is along the direction of the displacement which is 100 meters. So time will be what? Everyone T will be what? 100 displacement divided by the velocity 100 by root 3. So this is the time the swimmer will take. Swimmer wants to go against a straight line across the bank in 90 degrees. Yeah, I'll repeat it. But please ask doubt. What is your doubt? Don't tell me repeat. What does it mean? Tell me what is your doubt? Everyone, what is your doubt? Shouldn't V0 be towards the other side? Why? V0 plus Vr should be Vt. So if V0 is on that side, then if you add Vr, it will become like this, further slant. This will be Vtotal. And you don't want that. Why did you do it like the previous? Every case is different to above. And here also we have done that only. Look at the previous case what we have done. Previous case, swimmer is trying to move. Swimmer is not actually moving in the 90 degree. Swimmer thinks that he is moving. Okay. And while it is moving, the river will drift along with it. And the total velocity becomes slanted. But over here in this case, swimmer is smart. Swimmer knows that if he has to go in a straight line, he has to go in this way so that when drift of the river happens, its velocity will become straight. Okay. Sir, I doubt. Sir, for this question, will the deviation be 0 or the length over here? Because I didn't understand that. Okay. Tell me, will there be any deviation? Well, the swimmer will land up here only or somewhere over here or there, where it will be? So, directly opposite. So, what is the doubt? Sir, is the deviation 0 or is the deviation the length of AC? I am not getting what is the doubt. Sir, my doubt is that in this case is the deviation 0 What is the definition of deviation? Sir, how much it has deviated from its original path? So, how much it is here? AC. So, is it 0 or not? Sir, how is it 0? Because the swimmer is going straight. His actual velocity is straight. Swimmer thinks that he is going slant. Swimmer is trying to go slant so that the river takes him along in a straight line. Okay. He is supposed to go to C. No, Hariharan. That is not... If suppose the river is not flowing, the river has zero velocity, then the swimmer will go to C. Like that. Okay. Because that is what the swimmer is trying to do. But if the river is flowing along, then river velocity will get added to the velocity what swimmer is trying to do. So, swimmer is trying to do like this. Velocity of the river will get added up. River will push it this way. Swimmer goes that way. So, the velocity becomes straight then. Getting it? If river is at rest, it will go like that. River has no effect then. The swimmer is swimming on the water. Okay. The swimmer effort is apart from what water does. So, swimmer effort plus the effort of the water is the total effect. Deviation is not ac... It's velocity. This is the total velocity. This is what if a person is standing on the bank like this, this person will feel as if swimmer is going straight like this. Swimmer is trying to go like that. The river will push it this way and swimmer velocity will become straight. So, this person will feel as if swimmer is going straight along this path OA. Okay. So, more than drift, we can call it X code. Anything you call... But you have to define what do you mean by that. Initial path was towards C. Yeah, you can say that. He has started swimming towards C. Started swimming. But then river velocity is also there. River will push it in this way. While he is swimming, river is also creating an effect. So, velocity of river will get added up. Everybody understood? If river is flowing in the other direction, should the river swim to be? Yes. That is very correct. If river is flowing like this, then swimmer has to go like that so that when velocity of the river gets added up, then it becomes straight. If river is going in other way. Okay. Remember one very simple thing. The actual velocity is what swimmer is capable of plus what river is doing to the swimmer. This is the total velocity. All of you understood? Please type in yes or no. I will not move forward. These are the river questions. There can be river boat questions also. River swimmer question also. There can be swimmer and boat both in the river. But then I don't want to further complicate it. First you need to do basic level of questions. Get used to it. And then only I can complicate the situation further. But then I enabled you so that you can try out any question by the way. Now let's talk about the scenario. I mean it's a very common question which is like this. All of you. A person is there with an umbrella. This person is running with let's say five meter per second, which is higher than the, no it is lesser than the Usain Bolt's speed. Usain Bolt is nine meter per second I guess. Anyways, who cares. So the person is running with five meter per second. There is rain coming vertically down. There is no wind. There is a rain that comes vertically down with a speed of three meter per second. Three meter per second. You need to tell me which direction the person should tilt his camera so that, I mean whatever you have to find out which direction the person should tilt the umbrella. For obvious reasons. Do we count accession due to gravity for the rain? No, no, no. Assume it is coming with a straight constant velocity. Opposite to the relative direction of rain. That's not how you answer in physics. It should be quantitative. You should tell that angle should be tan inverse of this like that. You have to basically find out which direction the person will see the rain coming down. Will the person see the actual velocity of rain or velocity of rain minus velocity of himself. What the person will see velocity of person is let's say VP and this is VR. Person will see, see VR. The person will see VR minus VP. Which one? Which one this person will feel? Second one, right? VR minus VP because the person is looking at the rain while running. So person is the observer who is observing the rain. So person or the observer will look at the world each and everything by subtracting his or her own velocity. So he has to subtract his own velocity from the velocity of the rain. So this is velocity of the person. 5 m per second. This is velocity of the rain. 3 m per second. So you need to invert it so that it become minus velocity of the person. And this is what is VR minus VP. So this is how it will feel the rain coming towards now. What is the magnitude of this velocity? Everyone with what magnitude the velocity will come towards the person root over 5 square plus 3 square. It is hypotenuse. So magnitude is this length. This length is square plus that length is square under root. That is root 34. Okay. And this is the angle with the vertical. That's a theta. So tan of theta should be equal to 5 by 3. So angle with the vertical should be 5 by 3. Is it clear to everyone? The person will feel the rain is coming with a velocity of root 34 at an angle of tan inverse 5 by 3 from the vertical. Everyone is able to understand this. All of you, shouldn't theta be the angle? Theta is angle only. What is the doubt? Theta is the angle so that tan theta is 5 by 3. How do you find theta? Once you learn inverse technometric, you will understand. Theta is tan inverse of 5 by 3. Okay. Any other doubt? Should we solve more questions on these topics? Frankly speaking, this is not, this should not be part of theory. I'm just taking numericals and making you understand how to solve numericals. Okay. There is no theory as such. So let's take few more questions. Solve this. Make your judgment. This is not a good habit to ask about the question. Okay. Once you attempt it, your own judgment should be there. Otherwise, every time when you write exam, it'll be like, oh, somebody should explain me the question. Okay. Once you try it based on your judgment, then we'll discuss it. Okay. So I think many of you have tried already. All of you attempted. Nikhil, have you attempted? Okay, fine. Let us do this. First thing, again I'm repeating, I think I've repeated that many times. First thing to solve any physics question, draw the diagram. Okay. Doesn't matter how simple or difficult the question is. It's a matter of habit. 400 meter wide, the river is flowing at 2 meter per second. It is flowing with 2 meter per second. It is flowing at a velocity 10 meter per second with respect to water. Now it is given with respect to water. So it is like capability of the swimmer is given to you. This is the capability of the boat. This is what boat can do if there is no velocity of the river. Right. This is the velocity. When I say relative velocity, that is also capability. So on top of it, river can add its velocity. Fine. So capability of the boat, like that you should read. Capability of the boat is 10 meter per second. Okay. And it is moving in the direction perpendicular to river. And this direction is the direction of the relative velocity or direction of the actual velocity. It is given that it is sailing at a velocity of 10 meter per second with respect to the water in a direction perpendicular to river. So it is a relative velocity direction or actual velocity direction. Read it again. There is no hurry. Okay. There is no hurry. Read it again. I'll read with you. A boat is sailing at a velocity of 10 meter per second with respect to water in a direction perpendicular to the river. So this direction is with respect to water or the actual velocity direction. This is the relative velocities direction because they are talking about velocity with respect to the river only. Isn't it? So don't get confused in the words. Okay. This is the relative velocity and this velocity is given to me. Relative velocity is the capability of the boat. The boat is trying to move like this with 10 meter per second. Okay. But will boat be able to do that? Will boat be able to move like that? Anyone? What will happen when boat is trying to move like this? The river will take it along. The river will add its own velocity onto what boat is trying to do. Total velocity will be in this way. This will become the total velocity. Fine. So what is the magnitude of total velocity? Root over 10 square plus 2 square, which is root over 104 meter per second. Time taken by the boat to reach the opposite bank. What is the component of... Okay. All of you, this is the actual velocity. This red line is the true velocity. If you want to find the component, you have to find component of the true velocity. So component of this red line velocity, which is the true velocity in the direction of 400 meter displacement is what? What is the component of the true velocity in the direction of displacement? What is the direction of 400 meter displacement? What is the component of velocity? Let's say this is theta. Total velocity times cos theta, which is 10 only. Hypotenuse times cos theta is base, which is 10. Okay. So time is what? Displacement divided by 10, which is 40 seconds. How far this one? B part. I think something like this we did. B part, how to do? 40 seconds is moved with what velocity in this way? 2 meter per second? 2 meter per second, right? This component of velocity is 104 sin theta, which is perpendicular only, and this is 2 meter per second. This is 2 meter per second. That's the best thing because there is a right angle triangle. You can take components. So 2 into 40. That is 80 meters. Everybody understood this? Type in whatever doubts you have. Quickly ask doubts. No doubts? Okay, we'll go to the next question now. All right. So let's take up this one. Equalty triangle. This is an equalty triangle. Fine. Vertex 1, vertex 2, and 3. Okay. So particles 1, 2, and 3 starts to move with constant speed in such a manner that 1 always towards 2, 2 always tries to move towards 3, and 3 always tries to move towards 1. Okay. So what will happen is this kind of motion will appear. They will basically converge. Okay. They will converge, and every time equalty triangle will be there, and with time, they will converge and meet at the centroid. Like that. So you need to find out how much time it will take for them to meet. How much time it will take for these particles to meet. Okay. Side length is, sorry, side length is A. Side length of the equalty triangle is A. Constant speed U. It's not a straightforward question. Okay. It's very tricky, but let's see whether we can solve this. All of you try it. Everyone, should I solve it now? Okay. Listen here. Take only 2 points, 1 and 2. So by the time 1 meets 2, 2 will meet 3. Isn't it? So now tell me initially 1 is going towards 2. Isn't it? This is the velocity U, and 2 is going towards 1. So 2 is going towards 3. Like that. This is U. Okay. Now here is a trick. Suppose 2 is the observer. Suppose 2 is the observer. How it will see 1? 1 is continuously coming towards 2. Isn't it? So relative velocity of approach. Velocity of approach for 1. Can you write down what it is? Velocity of approach. This is 60 degrees. U plus U cos 60. All of you agree? Component of velocity along the line joining is U cos 60. Like that. Fine? Everyone understand or not? See, you need to find out the velocity of approach along the line joining 1 and 2. Along the line joining. So you have to take velocity along the line joining 1 and 2. Fine? So the 1's velocity is along the line joining only. That is U. The velocity of 2 is like that. So you have to take its component along the line joining, which is U cos 60. So U cos 60 from here and U from there. So total velocity of approach is what? U plus U cos 60. Are you getting it? Distance of approach is what? Distance of approach is a that divided by the velocity of approach is the time. So the answer is 2a by 3u. So again you have seen the power of the velocity of approach concept here. Fine? The whole of 3. You can take any 2 points. You can take 3 and 1 also. Same thing will come. It is a symmetrical case. Whatever happens between 1 and 2. Same thing happens between 2 and 3. Same thing happens between 3 and 1. Why? Angle remains 60 degree because of the symmetry. The equilateral angle is there. It will rotate while rotating. It will become smaller and smaller and then it coincides. Why do you take cos? Because I have to take velocity of approach along the line joining 1 and 2. So I have to take component of the velocity of 2 because total velocity is not along 1 and 2. Total velocity makes angle theta. So when you take the component, you have found out the velocity along the line joining. Again, this is not a very easy question to understand just like that. It takes a lot of practice. Ideally, I do not give this question when I introduce a concept because you guys have not done the basic practice on this chapter. So once you do the basic practice, then you will start understanding these kind of questions also. A lot of questions. I will send you the homework and if you want more, you just text me. I will send you more. Do this. No, Anurag, that you cannot do. See when what Anurag is saying is we know that point number 1 goes to a centroid only. You find the distance from its centroid and divide by its velocity. But Anurag, it is not going in a straight line to the centroid. It is taking a curved path. You don't know the length of that path. Fine. That is why taking a relative velocity concept because relatively it appears that distance of approach is only a distance of approach is a but in reality they are traveling in curved paths. Displacement will be in that direction, but velocity keeps on changing. Velocity is not constant. Direction of velocity keeps on changing. Anurag do this. Once we are done with circular and relative motion, you will see in projectile motion also we can mix up all of this. Okay, Imanji got something. Others? Imanji, you did not get in terms of theta. Independent of theta. Answering party or party. Different answers people are giving. Still looks like concepts are not very clear. So it's okay. Just keep on trying. We'll take up questions. And these are not simple questions. So it is okay if you are not able to solve in the first attempt itself. Should I attempt or should I wait? Okay, I am attempting it. This distance. Are you finding this chapter difficult? Everyone finding it difficult? Not much. Okay, so see the previous chapter you already did in class ninth. So it was familiar to you. This chapter you have never, you know, done. So that is why you're finding little bit uneasiness. But then the truth is this chapter is simpler than the previous chapter. That is the truth. So if you just practice little bit more, it will be a lot comfortable with these concepts. Fine. Swimmer wishes to cross 500 meter wide river flowing at 5 km per hour. His speed with respect to water. What is given? Capability of the swimmer is given. That is a relative velocity. That is what swimmer can do. Okay. If he heads in the direction. If he heads in the direction making a little theta with the flow. Then the time takes to cross the river. Now in part a is angle theta given for its relative velocity or for its true velocity actual other relative velocity. Suppose it is not written. It is not written whether it is actual velocity or relative velocity. Then what you should assume. If it is given velocity is two meter per second forget about this question. If it is given velocity is two meter per second. Should you assume that to be real velocity or relative velocity? What is a normal practice? The real velocity actual velocity, right? If I just say that velocity is this you have to assume it is a real velocity. If I do not mention that it is relative velocity. Then you have to assume it is the actual velocity. But it so actual velocity makes an angle theta with the flow flow velocity is this. The person is let's say flowing going like that. This is the actual velocity. Fine and it makes angle theta with the flow direction. This is the direction of flow. So where is the angle you have to extend it? This is the angle theta. Angle between the two vectors. You have to connect them till today. Fine. Find the time it takes to cross the river. Fine. Now in order to do that I know the displacement is 500 meter. But do I know the direction of do I sorry. Displacement is 500 meter. But which direction my velocity is velocity direction is like that. Okay. Fine. So I need to find out its velocity. In that direction in this direction I need to find its velocity. So how much is that this is 180 minus theta. This is 90. This one is 180 minus theta. So 180 minus theta plus 90 plus 5 should be equal to 180. So 5 is 90 minus theta. So this is 90 minus theta. Okay. Okay. Yeah. It need not be 90. I'm asking what is the velocity in that direction? What's the velocity in that direction? Because the displacement is known in that direction. If actual velocity is given to you that's a V. V actual is given to you is VA. So we want VA cos of 90 minus theta. Right. This one will be VA sin theta. This is the velocity in the direction of displacement. So time it should take is 500 divided by VA sin theta. What is VA given to us? No. It is not given to us. Should be given to us. Either it should be given to us. VA sin theta need not be equal to 3. Right. It need not be equal to 3. Because the vector. No, no, no. VA is 3 you are saying. His speed with respect to what? Now then you are assuming in part A. They are saying that it is angle with the relative velocity. If you're saying that VA is 3. You're assuming in part A. They mean to say that the angle theta is with the relative velocity. All right, which is not true. Or they should have clearly mentioned that a, you know, relatively he is moving or at least they should write that. The person feels that he's going at an angle theta, which is not actually the case, but he thinks that he's going at that angle. Even that is not very clear. Then just for the solving this question, let's say we write here that if he feels that he is moving in a direction of angle theta, then I can say that this is the relative velocity. Okay, this is not the actual velocity. Just for the sake of completion, we have modified the question a bit. In part A. We are saying that the person feels that he is going at an angle theta with the flow. So this relative velocity is given to us. A swimmer is going with 3 kilometer per hour. So this is 3. Okay, so now we a is 3 and this makes an angle theta. So this direction the velocity is 3 sin theta. So like this, you can 500 meter into converting the kilometers also. Some issue with the question here, not clearly written. We'll just take up another question. Okay, basics of projectile, you know, right? Basics of because there is a question which includes projectile motion also. So you have anyway taken a test today till when the class will be. Class is till 6.30. I have message, no? It was 3 p.m. 2. No, wait. What is the timing of class? 2.30 to 6. Oh, God. Sorry, guys. Sorry, sorry, sorry. So we'll stop now. No, you never told me that class is over. Time is over. All right, guys, so. All right, sorry for extending the class. You guys can take up the questions as homework. Fine. Oh, you thought that I will scold. I don't scold like that. All right, guys, so take this as homework. I will, you know, send this thing across. So we'll see you next week. Bye for now. Take care.