All right. So welcome everyone. Good evening. Waiting for everybody else to join in. Okay. By the way, when you joined the meeting, did, did it ask for the registration or something? They asked you about your email ID is named required for attendance actually probably. So I'm waiting for others to join in quickly. So just a minute and then we'll start. Have you guys done the homework already? The workbook I've shared and the assignment that I had given anybody tried that? Not the PDF. So, you know, the trick is this only that you move with the class. You do a chapter in such a manner that you don't have to go back to that chapter again ever. Fine. So that way you'll be very confident moving forward and yeah, you'll not waste time reading the chapter again. So you without doing chapter number one properly, you move on to chapter number two. You'll see that you'll keep on coming back to chapter number one and then chapter number when you're in chapter number three, you'll keep on going to chapter number one and two. So don't do that to yourself. Work hard and at least at the start you should really, really work hard because that will give you a lot more confidence. And still few are just few more seconds will wait. Yes, what happened? I don't know. What do you want? You tell me what you want. Okay, then we'll tell. Sir, I was not getting many questions. So should I do all the questions I can do and then come back to the hard ones? If you only do the question which you can do, you're not learning anything. You already know how to do them, and you have done that. So what new you have learned? Okay, so if you are struggling with a question, you're not able to do it yourself, discuss with your friends, put it on the group. Do everything that is possible and then say that, okay, I'm not getting it. It should not be that, okay, first time you tried, you're not able to solve it. All right, let's move on to the next question. Then you're wasting your time. Okay, so struggle to the question if you're not able to get it. Understood? All right guys, so today we will continue with the kinematics as in the motion in one dimension. All right, so last session, last session we had started motion one day and motion one day was the first chapter of kinematics, wherein we have, you know, discussed the basic, you know, the variables which we should be using to track the motion and analyze the motion, fine. Like for example, the displacement, velocity, acceleration, speed, distance, all these things we have learned. And the most important variable was time. Okay, so we have learned these variables and then these variables can be put on a graph as well. So we can say x axis is time, y axis is displacement or x axis is time, y axis is acceleration, right? So all of that we have done. So we have analyzed the situation graphically. We have seen different kinds of scenario how it looks on a graph. All right, and then we have also understood that same situation we can analyze by using the kinematics equations, right? These kinematics equation that is v is equal to u plus 80, s equal to u2 plus half 80 square and other two equations. These equations are valid only for special scenario where acceleration is constant, right? Right, so we have learned, so we have learned about these constant acceleration equations of motion, fine. And we have solved a lot of questions on it and we have understood that the one very common example of constant acceleration is when object is moving under the influence of only gravity, right? So this is what we have done in the previous class. Now in today's session, we are going to discuss few more cases which can be analyzed graphically because graphic analysis is not yet over. Then we will learn one of the most important tool of kinematics which is the calculus, okay? So I assume that Akhil might have introduced to you a little bit of calculus and only that much is required in physics. You don't need to be a master of calculus, okay? The basis of calculus is enough. So we are going to discuss how calculus will help us to analyze a situation in which acceleration is not constant, right? And once I know that how to analyze a motion in which acceleration is not constant, I can analyze any situation, isn't it? Okay, because the equations of motion are restricting us to analyze only those motions which have constant acceleration. Now calculus will help you to analyze the motion in which acceleration is not a constant. It can be anything. It can vary. So then you can, you have been now given a free hand. Once you know the calculus, you cannot say that this question is out of my curriculum. Everything under the sun comes under your curriculum now, especially in community exams, fine? And then we'll be discussing about the relative motion, okay? How the motion will appear to a person who himself or herself is moving, right? So these are the agendas of today's class. So we'll be taking it up one by one, alright? By the way, what was the last thing we have done in the last class? Anybody remember last thing in the last class? What we have done? Elephant question we have done. Okay, distance traveled in the nth second. All right. Okay, so let's proceed. So there are few graphical scenarios that are pending from my side. So let us discuss them. The first one is graph between acceleration and time, right now, fine. So let us see how the graph, I mean, what does the acceleration time graph tells us, okay? So this is a simple graph. Let's say I'm taking a constant acceleration graph. The x-axis I have time on the y-axis I have the acceleration, okay? Now, when I'm plotting the graph of constant acceleration, how does it look like? Everyone, constant acceleration graph, how does it look like? Straight line parallel to time-axis, correct? Parallel to time-axis, which is the x-axis. So this is the constant acceleration getting it. Now, what I'm trying to do is I'm trying to understand what are the various things related to graph. Is there a physical meaning to it? And there are only two things in any graph in kinematics. What are these? First is slope. What is the second thing? Anyone? And area, only two things, okay? Slope is zero. Zero slope means that acceleration is constant, okay? So what does area represent? Area of this graph, what does it represent? All of you try doing it. It's a rectangle, right? This much distance is what? This much is the amount of time it is taking. And this much is A. So area is what? Area is 80, right? So area is small a into t. And it is a constant acceleration case. It's a constant acceleration case. So I know that v is equal to u plus 80, alright? From here, 80 comes out to be v minus u. So area of efficient time graph is change in velocity, okay? But then we have proved it only for the constant acceleration, right? What if the situation is like this? This is the complete graph, let's say, okay? This is, let's say, complete graph. This is, let's say, A1 area, this is A2 area, okay? So what I can say about A1 minus A2, difference in areas. One area is above the time axis taken as positive area. One area is below the time axis. I'll take it as negative area. So I'm adding the two areas. So what does it should, what it should be? Anyone? Let's say velocity over here is u. This is u velocity over here v1, over here it is v, okay? So A1 is what? A1 is v1 minus u, right? And A2 basically is a positive quantity, but there's a disillusion there. So actually A2 is negative there, right? So that is a minus A2 I'm writing here. Minus A2 is what? v minus v1. This area is finite velocity minus initial velocity before this expression started.  So A1 minus A2, if I add these two equations, I will get A1 minus A2 to be v minus u, okay? So even if there are multiple accelerations which are constant in a graph, then also the area under the graph should be equal to final velocity minus initial velocity, okay? Now what if, what if the acceleration is changing then what it will be? Let's say this is a situation, A and T, okay? Let's say this is how acceleration is changing, all right? Now I want to find out what does this area represents? Anybody knows what does area under velocity time graph is area under vT curve is what? Everyone, area under vT curve is displacement, right? It is displacement whether area, whether the velocity is constant or not, right? So we have proved it for constant velocity. Now if velocity is changing, then how did we prove that? How did we prove that even if velocity is variable, area will be the displacement. How did we do that? What does it prove? We have considered small strips, correct? So we are going to consider small strips like this, okay? So if I consider this small strip, this strip thickness is very, very less. This thickness, let me call it as dT, okay? Can I say that for this very small thickness, acceleration is constant? Can I say that for this small strip only? Everyone, right? So area of this strip is let's say dA, okay? dA is A into dT. Let's say from distance of from here to there is A. This is A, okay? So dA is ADT, all right? Now since the thickness is very, very small, this ADT is also equal to change in velocity between this point and that point, right? Because the acceleration is constant. We just proved that for constant acceleration, area is changing velocity between final and initial point. And my final and initial points are within a width of dT. So ADT is equal to dV and ADT is the small area also, okay? So when I integrate, so I'll get dA is equal to dV. So this is equal to that. Getting it, so when I integrate, my limits will go from when I am here, the area covered is zero. When I am there, total area is A, initial velocity U, final velocity V. So I'll get total area still to be equal to V minus U, okay? So whether the acceleration is constant or not, it does not matter. Area under acceleration time graph is final velocity minus initial velocity, okay? Which part? Did you guys not do the definite integral? You guys are seeing it for the first time? U and V when I am writing like this? Okay, you did only indefinite. So Pradhyan, hold on then. We are going to discuss the integral anyways. A little bit discussion we'll have. So how to do definite integral? You'll understand. There's nothing very complicated here. You'll understand when I will teach you a little bit about definite integral, okay? Alright guys, so let's quickly get into the, sorry, let's quickly get into the calculus in kinematics because that is more relevant. And test is on 14th. All of you know this, right? Test is on 14th, everybody? Okay, so guys, write down this thing. Calculus in kinematics. Now, calculus in kinematics you guys already know a little bit, okay? So for example, just give me a second. Isn't this a calculus equation that velocity is equal to ds by dt? What is this? What is this? A differential equation? Have you guys seen this equation before? V is equal to ds by dt. V is equal to ds by dt. Now, tell me when this equation is valid for constant acceleration or any type of motion, V is equal to ds by dt. What do you think? V equal to ds by dt is valid for every type of motion or only constant acceleration. It is, yeah, I don't know, once only, okay? You guys reply only once. So this equation is valid for every or for any type of motion, okay? And what else you guys know? Anything else? Any other equation you guys know? Correct. A is equal to dv by dt. Even this is valid for any type of motion, okay? Yes, okay? So these are there. Now, these are the founding stones of the calculus in the kinematics, all right? Now, this will give you velocity. What about speed? How do you get speed? Let's say I want to find speed. What is the equation for speed? Distance by time is average speed. Then velocity is displacement by time, according to that definition. Give me a differential equation, like we are given velocity ds by dt. So u is what? You have to differentiate the magnitude of the displacement, okay? Magnitude. You don't care about the change in direction with time. You only care about the change in the magnitude of the displacement. So that's why you need to differentiate the magnitude of the vector, not the total vector, getting it? We'll be taking a lot of numerical examples. You'll understand how to use all of this, okay? Right now, just write it like a theory, okay? Now, there is one more very important differential equation, which is like this, okay? So acceleration is written as dv by dt. So right now, you are writing acceleration as a function of velocity and time, okay? Now, if you multiply, if you multiply and divide by the ds vector, like this, okay? You'll see what you'll get. You'll get ds by dt. I'm rearranging the term into dv by ds. Now, what is ds by dt? ds by dt is velocity. So acceleration can be written as v dv by ds, okay? Now, acceleration is written as a function of velocity and displacement s, all right? How is a vector? a is acceleration, right? Ansh, a is acceleration. It's a vector quantity, okay? Now, this is the differential equation. These are the differential equations, which are always true. Now, but then these differential equations will tell you what is ds. What is dt? What is ds by dt? What is dv by dt? This equation, for example, a is equal to dv by dt. This will not tell you what is v. This will only tell you what is dv by dt. So how to find v from here? To find v, you have to integrate. You have to integrate, all right? So now you'll see how integration will be used here. So, for example, is audio lagging? Everyone? Is my audio lagging? No, no, it is not lagging. Yeah, I'll take a poll. Is my audio lagging? Is there a lag in the audio? So you can see there is no lag in the audio. Those who are experiencing the lag, log out and immediately log in, okay? Fine. So what I was saying that you need to integrate to get the value of velocity from this equation. So how to integrate it? So dv can be written as a dt, like this, okay? And then you have to integrate this, all right? What is the integral of dv? Anyone? Integral dv is what? Integral of dv. You guys know integral of dx? There's how much? Okay, remember, integral of anything like this is a variable itself, okay? So you guys remember this? Integral of this is x. So integral of dv is v, okay? Integral of dv is v. Fine. Now what is the integral of a dt? a into t. Yes, there will be a constant also, correct? Plus c will be there, but I'm not putting it right now. I'll tell you why later on. So integral a dt, will you be able to integrate a dt integral? Just like that. a need not be constant. Acceleration a need not be constant. Will you be able to integrate a dt? Please write down, we can't integrate two variables. What are the two variables? a and t, okay? So if a would have been constant, what will happen to a? If it would have been constant, what will happen to a? a will come out of integral. a will come out of integral. You can write a outside the integral and then integral dt will be t, okay? Else if a is not constant, if a is not a constant, you write a in terms of t, then only you can integrate it. Otherwise, you cannot integrate integral of a dt, okay? So I'll repeat myself again with an example. All of you listen here. There is no hurry, okay? I'm not going to rush. This is the first time you are seeing the use of calculus in physics. Everyone. So what I'm trying to say is that integral a dt, you'll not be able to integrate unless you know what is a, okay? So if a is constant, a will come out of integral and this integral will become integral of a dt and it will become a t. So plus constant c, okay? This is what you guys have learned, right? And if a is not a constant, for example, a is given like this, a is t minus 2t square. This is how a is changing with time. Then you need to put the value of a there and instead of a, you'll write t minus 2t square into dt, okay? Then you'll open the brackets. It'll become integral t dt minus 2 will come out because 2 is constant integral of t square dt. So what it will be equal to, everyone? Indicate t dt is what? You guys know the formulas? You don't know the formulas? Okay, a few of you know it. I'll just write down the list of formulas that you must know at least for today, okay? And majority of the time in physics you're going to use these formulas only, everyone? Integral of x and dx. I think it is done, isn't it? Anyways, all of you write down. There's a constant also plus c. I'm not writing plus c, okay? So these are the few formulas that you should know and whatever other than this formulas are required, I will tell you as the time progress, okay? Sir, if we only got a list, but we don't know any questions. I mean, you're doing the questions now, right? This is the list. You have to use this now. Integrate t dt. Which formula should you use? Which one? First one or second one, third one? Which one? The first one, what is the value of n there? Instead of x, t is there and n is equal to 1. So it will become t to the power 1 plus 1, which is t square divided by 1 plus 1, which is 2. So t square by 2, integral t dt, okay? So this is t square by 2. What about this? How much this will become? Everyone, minus of 2t cube by 3, alright? 2t cube by 3 and plus constant because you're dealing with the indefinite integral, okay? So now you understand, you will not be able to integrate anything under the sun. So for example, x dy, you cannot integrate it. Unless we write, unless you write x in terms of y, you can't integrate this. You cannot integrate, fine? So remember this, okay? Now I'll tell you what is definite integral, okay? Definite integral. These are indefinite integral, wherein a constant will come in the expression, fine? So definite integral is simple straight forward extension of indefinite integral. It is not, it is not completely new, okay? So what it says is that if integral of fx dx is gx, okay? Integral of fx gx, dx is gx. Example is integral of x dx is x square by 2. So fx is what? If you compare these two, fx is what? x gx is x square by 2, alright? If this is true, the limits, if you write x1 to x2, so it will be x1 to x2. Once you get gx, you have to write it as x1, x2 like this, x1 and x2. What are x1 and x2? x1 and x2 are the points between which you are integrating. They are not integrating till infinity. I am only interested to integrate between initial point and final point. So I'll write what is the value of x initially and what is the value of x finally. So these are x1 and x2, okay? So this will be equal to gx2 minus gx1. So how to put the limits? It will be x2 square by 2 minus x1 square by 2.  Constant will not be there if you put limits. So is it clear to everyone? Everyone is clear? Quickly type out. We'll be taking lot of questions anyways. Is it 60s to 70% clear? We'll be taking up questions, don't worry. Okay. So let's take a question quickly, alright? So what I am saying is the velocity is given as 2t minus 3t square meter per second. This is the velocity given to you, okay? You need to find displacement. How much is the displacement? How much is the displacement between t equal to 1 seconds and t equal to 2 seconds? Between these two time periods, what is the amount of displacement we are having, okay? And whenever I am writing these equations, I am assuming it is motion 1d because chapter is motion 1d, everything is under that. Object is going in a straight line, okay? So don't worry about some angles and components. Do it, everyone. Two people have answered already. Everybody else? Okay, people are getting different answers. Okay, how will you proceed here, everyone? How will you proceed? Is accession constant or not? Everyone. Can you tell me, accession is constant here? Is accession constant? How to find acceleration? How to find acceleration? Velocity is given to you, accession is dv by dt. How much it comes out to be? How much is accession? 2 minus 60. So you can see accession is not constant. You cannot use equations of motion. Then how to get the displacement? What is the relation between velocity and displacement? Velocity is dx by dt. Why I am using x? Because everything is happening in one dimension. So I am calling that displacement in terms of x coordinate, okay? So dx will be equal to v dt, okay? What I have to find out? Displacement, right? So I will integrate what? X because I want to find what is the value of x. This I will integrate and that I will integrate, okay? Limits will be for x, what is the limit? Initial location is what? Initial x location is 0 and time is what? t is equal to what? At x equal to 0. At x equal to 0, what is time? 1 seconds. Then at x equal to x, t is 2. So I will get the value of x between 1 and 2 if I put the limits like this. That is the only difference between definite and indefinite integral, okay? Everything else remains same, formulas and everything. So from here integral dx is x. Limit goes from 0 to x. v is 2t minus 3t square dt integral 1 to 2, okay? So the value of x is equal to 2t dt integral is t square 1 to 2. 3t square dt is t cube 1 to 2, right? So it will be 2 square minus 1 square for the first, this is t square, limits 1 and 2, minus 2 cube minus 1 cube, right? So the value of x will come out to be 2 square minus 2 cube, which is 4 minus 8. That is minus 4 meters. Now go through this question and let me know whether you have understood it or not. You can ask me any doubts. Everyone, I didn't get the last step. What? What is the issue there? This is 2 integral t dt minus 3 integral t square dt. So what is t square dt integral? t square dt is what? t cube by 3, right? So 3 will go away. Then t dt integral is t square by 2. 2 will go away, okay? You have to do it yourself. If you just copy it, you will not get anything. All right. So any other doubt? Anyone? Anyone has any other doubt? How did t square becomes t? This is what we have done. This is this one. See this gx2 minus gx1. Whatever you get after integral, you have to put up a limit that is 2 minus 2. Instead of t, you have to first put t and then minus, then put 1. So 2 square minus 1 square. Here it will be 2 cube minus 1 cube. Upper limit, lower limit. That is what we have done here. Clear? Any other doubt? Anyone else? All of you have understood? Please type in yes or no. Then I'll proceed. Okay. Fine. So looks like everybody understood. There are a few more formulas, which not sure that you guys know it. It need not be just sine x. It can be sine kx. What is the integral of this? Anybody know this already? Minus 1 by k cos kx plus constant. The constant will come in the denominator. Integral of cos kx dx. This will be equal to 1 by k sine kx plus c. Okay. So these are the few formulas, which could be handy. All right. And you will anyway be remembering it. You will be solving so many questions that you'll tend to remember all of this. All right. Now take another question. All of you, since we are learning it for the first time, we should take a lot of questions on it. What is given is acceleration is T minus T square. Okay. Initial velocity of an object is 2 m per second. You need to find what is the final velocity. Okay. In the T equal to 3 seconds. In 3 seconds, if initial velocity was 2 m per second, what is the final velocity? Everyone. Okay. Anybody else? I've got two answers. All of you, quick. Try it out. I mean, you may not get the correct answer. It is all fine. It doesn't matter how many of you are able to solve it, how many are not able to solve it. Okay. These are just for the practice. All right. So let's do it now. Everyone. Now here clearly, you know, expression is a function of time. All right. So it is not constant. Again, I cannot use equations of motion here. So what I'll do is that I'll write expression as DV by dt. And this is equal to T minus T square. So DV can be written as TDT minus T square dt. Okay. So when I integrate this, I'll write like that. Limits. What will be the lower limit for velocity? What should I write here? Lower limit. At T equal to zero, what was the velocity? Velocity was two. Then at T equal to three seconds, let's say velocity is V. T is what? Lower limit T is what? Zero, right? It starts from T equal to zero, then goes to three. Zero to three. Like that. Okay. Integral of DV is V. Then you have to put a line like this. Write down lower limit, upper limit. This is equal to integral of TDT is what? T square by two, right? Limits are from zero to three. Minus integral T square dt is T cube by three. Limits are from zero to three. Like this. Now, what should I write here on the left-hand side? What should I write? Aditi, what should I write here? Correct. Upper limit minus lower limit. Instead of V, I'll write upper limit and then lower limit. T square by two. What should I write here? Upper limit three square by two minus lower limit zero square by two. Instead of T, you have to put the upper limit.  So it is three square by two. Minus three cube by three minus zero cube by three. So zero is zero, right? So zero cube doesn't change anything. The velocity, I'll just tell you the final answer now. Velocity will be minus of five by two meter per second. Okay. So when you simplify this, you'll get that. Don't make any silly errors here. Fine. So this is the answer for the velocity. Okay. One more question is this. Acceleration is given as sine of pi T by two. This is the acceleration. Clearly it is not constant. It fluctuates as a sine function. Okay. Initial velocity. Initial velocity is zero. You need to find final velocity in one second. In one second, how much will be the final velocity? Look at the formula above. You don't need to convert in degrees. As long as you know what is sine of pi by two, what is sine of, you know, pi and all those you already know, why you need to convert in degrees? No. Sine of pi by two is what? Everyone? One, right? Sine of pi by two is one. And cost of pi by two is zero. Okay. So you know what it is. You don't need to convert in degrees. In fact, the radiance is the SI unit. You convert degrees into radiance in physics, not otherwise. But then till now you have been learning degrees. So that's why you're comfortable with the degrees. Everyone, how should I solve it? What should I write as? Okay. I'm waiting. All of you tried. What should I write as? A should be written what? dv by dt is A. This you write it as sine of pi t by two. Okay. So dv is equal to sine of pi by two t dt. Now you can integrate. Integration can be done only in one variable. You can see left hand side, the variable is V. Right hand side, variable is T. So now you can integrate. Okay. So velocity initial is zero. Finally, let's say V time is zero to one. Okay. So you'll get V is equal to integral sine pi by two t is what? Minus of two by pi cos of pi t by two zero to one. Okay. So minus of two by pi. First I'll put T is one. So cos of pi by two minus cos of ten put T equal to zero. This one cos of pi by two is zero. So you'll get velocity as two by pi meter per second. Okay. Everybody understood this question. All of you getting it. Let's take one last question. Minus two by pi. Why? See here there. This is zero. Minus of cos zero is minus one. So minus one into minus two by pi becomes plus two by pi. But if you make such silly error, there will be one option which says minus two by pi. And you'll mark it. You'll not get zero. You'll get minus one. Okay. So be very careful. All right. Now let's say acceleration is two x. Acceleration is two times x. What is given here is that at x equal to two meters, at x equal to two meter, velocity is given to you. Velocity is two meter per second. Okay. Then at x equal to 10 meters, final velocity will be what? Okay. This is the last question on calculus as illustration. Solve it. What do you write A as? Should I write A as dv by dt? What should I write A as? If you write dv by dt as A, what is the problem? The problem is your unnecessary introducing time variable. So dv is equal to two x dt. You'll not be able to integrate the right hand side first. And you don't know the limits for the time t. Limits for x, you know. So this won't help you. Okay. So you should write A as v dv by dx, which we have learned v dv by dx. Right. So you can see here. So we are using this one. Function of velocity and displacement. Acceleration. Okay. So v dv by dx will be equal to two x. Right. Okay. So from here v dv is equal to two x dx. Then you integrate this lower limit for velocities. What? What should I write? Limits for the velocity 2 to v. Right. And for x, x will go from where to where 2 to 10. It is written. It goes from 2 to 10 where velocity goes from 2 to final velocity. Okay. So I will get it as v square by 2. The limits are 2 and v. This will be 2 into x square by 2. This one. So from here v square by 2 minus 2 square by 2 is equal to 10 square minus 2 square. Clear to all of you. So v square is equal to two times of 10 square minus 2 square plus 2 square. So velocity is root over same thing. I will not be calibrating it. So you simplify it. You will get the answer for the velocity. Now when you take a square root, you get plus and minus. Which one will you take? Or both are correct. You will get plus minus when you take square root. Will you keep positive or negative or both are correct? You keep both. Why? As I tell me, initial velocity is in positive direction or negative direction. Initial is positive or initial is positive. Acceleration is positive throughout 2 to 10. Acceleration is positive throughout or not. Acceleration is positive throughout. So how can velocity become negative? Right? Velocity will keep on increasing. Isn't it? You know, just one more question because this is important. Acceleration is given as 6 times T. Okay. It is given as 6 times T. You need to find out the displacement X between 0 and 1 seconds. A is given and I am asking you to find the displacement X. What you should do here? What you should do here? Acceleration is a function of time is given. But what do you want? What do you want? You want velocity as a function of time. Isn't it? All of you, you need velocity as a function of time. Then only you will get displacement. Correct? So how to get velocity as a function of time? You have to integrate it 2 times. Okay. Acceleration is given as dv by dt. This is equal to 6T. Okay. From here dv is equal to 6T dt. So when I integrate this, my initial velocity should be given here. I am assuming initial velocity to be 0. It need not be 0. Okay. I am assuming it to be 0 for sake of simplicity. So whatever the initial velocity come here. Initial velocity is 0. Okay. The final velocity let's say is v. v is the velocity at any time t. So that is why upper limit is v. Lower limit is 0 because initial velocity at t equal to 0. I am assuming it to be 0. And it need not be 0. So t equal to 0 and t equal to t. So from here you will get v is equal to 6 into t square by 2. Or you will get 3T square. This is the velocity as a function of time. v is equal to 3T square. Now I will write v as dx by dt. This is equal to 3T square. Right. From here dx comes out to be 3T square dt. Integrate this. Limit goes from 0 to 1. x initially is 0 and then whatever it is x. x will come out to be t square dt is t cube by 3. When you put limit you will get x equal to 1. Okay. So this is how you solve this type of question. Fine. So I hope you guys are now reasonably sure. I mean you cannot be 100% sure just by solving few questions that you are done with it. No. Okay. But then at least if you guys now think that if you do enough practice you will get a hang of it. Then that is what is required as of now. Okay. So let's proceed forward. There is one more graph. Graph between expression and displacement. Write down graph between a and x. So let's first take a constant expression case. This is x and this is a. Okay. There is a constant expression case. Right. You need to tell me what does this area represent? Slope is 0. Right. 0 slope means expression is constant. What does this area represent? Think of it is a naught x naught. It's a constant expression. Right. So which equation of motion you see the multiplication of expression with displacement v square equal to u square plus 2x. That is valid here because it is a constant expression. Right. So if you see here v square equal to u square plus 2ax a naught x naught should be equal to final velocity square minus initial velocity square divided by 2. So graphically this is what the area under the ax graph implies. Okay. And if it is like this, let's say this is a1 area. This is a2 area. Then a1 minus a2 even this is equal to final velocity square minus initial velocity square divided by 2. It is valid. You know graphically things are much more stronger. You can directly get it at get it as this. If you try to do it using equation into first write down the equation for the case number one when excision is constant then second case when excision is constant and then solve the two equation but graphically it is very strong. Just find the area. Total area is final velocity square minus initial velocity square divided by 2. Okay. What if excision is changing with X then what will happen? Whatever happens with the constant equation, same thing used to be true for the variable equation also. Right. Will it be true this time also or not? This is a this is X. Let's say. All right. This is how excision is changing. So what does the area represent? Think over it. Yeah. Make a rectangular strips. The method will remain the same. Every time we'll be doing the same thing. So if I take a strip like this, a small strip, then I can say that since the strip thickness is very less. A excision is constant. Okay. The thickness is DX excision is let's say A. So area is what area is a DX. This is a small area. Okay. This should be equal to this should be equal to change in velocity across this square by 2. This will be equal to the V square by 2. Okay. So total area when you integrate DA will be equal to integral of the V square by 2. Okay. So over here also area is final velocity square minus initial velocity squared divided by 2. So whether area is sorry whether excision is changing or not total area is final velocity square minus initial velocity squared divided by 2. Okay. Any doubts till now? So we have completed the graphical analysis as well as the calculus based analysis. Anything related to these two things? Anyone has any doubt or any issue? Any problem? Tell me. How is X V square by 2? Where is X? What do you mean? See the first line. Right. This is area of the strip area of the strip. Okay. This is the area of the strip. Now in this area thickness is so less. I'll assume that excision is constant. Okay. If excision is constant. We have already derived that the area will be final velocity square minus initial velocity square by 2 or you can say delta of change in V square by 2. Delta means final minus initial. Final velocity square by 2 minus initial velocity square by 2. So if thickness is very less delta is replaced by D. Okay. D is very small change in V square by 2. And integral of anything of D is that thing only. So DX integral is X. D of V square by 2 integral is V square by 2. But then yes, I can understand this is not very intuitive. Counter intuitive it is. So this will come with the practice only. Okay. Anyone has any other doubt? Has any other doubt? Okay. No doubts. Doubtless. Fine. So let's take a few questions and then we'll start the what is that? Relative motion. Okay. This is a good opportunity for you to learn here itself. We are doing a lot of questions on calculus. Don't keep it for the later. And all of this is NCRT level. We haven't yet gone to the mains level. Remember that V is velocity. Okay. V is velocity. First one. If you are able to solve it, type it out. Ideally, you would have wanted X in terms of T. But problem is T is given in terms of X. Right? So how will you solve it? If T is given in terms of X, how will you get the acceleration? How to get velocity first? If T is given in terms of X? How to get velocity if T is given in terms of X? If X is given in terms of T, you get dx by dt. Take a derivative of X with respect to time. dx by dt is velocity. But T is given in terms of X. How to get velocity? Yes, Abya? No, no, no. There is no other formula. Same formula you have to use. Should I solve? Nobody else? Okay. All right. No one. How to get velocity if T is given in terms of X? All right. No one is able to do this. T is equal to AX square plus... See, I am not interested in final answer. Okay? You tell me final answer is this. That is useless. Learn how to solve it. That is more important. T is equal to AX square plus dx. I will not be able to get dx by dt. But I will be able to get dt by dx. This is what? 2AX plus B. Fine. So, I will take inverse of this. dx by dt... Will it be the velocity? Right? dx by dt is 1 by 2AX plus B. What? How? Enough? Inverted, right? Second step. Are you differentiated? dt by dx... dt by dx is equal to A times... How do you differentiate? X square? Then, how do you integrate X? What do you get? 2AX plus B, you will get. All right. So, this is the velocity. Now, what I have to do? If this is the velocity, I will have to differentiate it again. So, dV by dt. Everyone, what it is? dV by dt. Minus of 2A divided by 2AX plus B whole square. Okay? Into dx by dt. I know 1 by 2AX plus B is velocity. This is minus of 2AV square. And dS by dt is also velocity only. So, this will give me minus of 2AV cube. Okay? So, that's what d... Everybody understood? See, all questions will not be same, right? We have to do different questions. And there will be... There will be millions of the varieties of questions. You can't mug up every type of question. You didn't get dV by dt step. Derivative differentiation of 1 by 2AX plus B. What it is? How you differentiate 1 by 2AX? If this is, let's say, Y, then dY by dt is what? 2X plus B, sorry. It is a chain rule. Okay? You differentiate it. What you do is that, assume... This I thought you would have done it. Okay? Assume this to be t. So, dY by dt, first you do. This will be 1 by t then. Right? So, dY by dt will be minus 1 by t square. So, this will be t square is this. This is dY by dt. Okay? Now, but I want to find out dY by dx. dY by dx is dY by dt into dt by dx. dY by dt is this. dt by dx is 2A. Have you done chain rule in maths? In maths, you have done the chain rule or not? Same thing here. Looks like you haven't done practice of calculus. You haven't done the assignments? Anyways, do the next question. All of you. Next question is lot simpler. Everyone try it. It's very simple question. What's the formula for average acceleration? What is the formula? What is the formula for average acceleration? Gurman, what is the formula? Final velocity minus initial velocity divided by the time taken. Right? Now, do it. You don't need to have any integration differentiation here. Final velocity is what? When t is 5. Okay? Velocity at 5 seconds is v5. This will be 5 minus 5 square. That is minus 20. Velocity at 3 seconds, 5 minus 3 square. Is it how much? 5 minus 9 minus 4 meter per second. So average acceleration. Final velocity is minus 20. Minus of initial velocity minus 4. Divide by time is what? 3 to 5. Time is 2 seconds. Okay? So average acceleration is minus 16 by 2. Minus of 8 meter per second square. Yeah. Calculus is just like a tool. You don't need to use all the tools every time, right? Wherever screwdriver is acquired, you don't need to use plier. Right? So you need to decide when to use calculus. Nobody can tell you that, okay, fine. In this question, use calculus. In this question, don't use calculus. Right? It is your own judgment that you have to practice and evolve. But over here, I think it is direct formula. So always you should first think directly how to get the answer. And then if something is changing, if acceleration is changing, right? And they are asking you something to, you know, evaluate as such. Then you have to, suppose I, in the same question I asked you, what is the displacement? Well, acceleration is not constant. But I'm asking you what is average acceleration? Average acceleration is that there is a formula, final velocity minus initial velocity divided by 2. And that formula is valid for whether acceleration is constant or not. So I'm directly getting it. Why will I use calculus? Isn't it? Okay. I hope no doubts. Solve this one. Okay. Gourman, Schittich, others. It's a simple question. Try doing it. And it's till now nobody got it correct. And now you got hint. You got minus one. Remember that. Next time also get minus one if you are over confident. Okay. I can see everyone is making that error, which is started by, I'll not take names. So look at it. What they're asking? They're asking you to find speed. Can speed be negative? No, right? So minus 40 is not correct. 40 is correct even though you get minus 40. So area under 80 graph is change in velocity. It is not change in speed. You're getting velocity, not the speed. Okay. So this is, let's say area A1. This is area A2. So A1 minus A2 is final velocity minus initial velocity. A1 is 4 into 10. A2 is 4 into 20. Particle starts from rest. So U is zero. So this is V. V comes out to be minus 40 meter per second. And we are in a hurry to answer quickly. So we will say C is the answer. Okay. Save that extra one second to again check what is asked. So answer is D. Okay. Graphical question, simple 80 graph. Area is change in velocity. Next up. Now I'm giving you questions on everything we have done till now before even starting the relative velocity. I'm giving you questions on everything that we have done. Okay. These are the simple straightforward questions. All of you will be quickly answering it. Okay. C to definitely not correct because speed cannot be negative. What about D? D for Delhi. Time is going back in, right? Time has increased till here. Till this point, time has increased. And then, you know, it is going back in time. Time cannot even be stationary. It can't even be 90 degree. It has to tilt a little bit. It has to travel forward. So both D and C, they are not correct. This one. Oh, she has answered quickly. Others, what are you doing? I will put a poll for this. I'm giving you one more minute. Okay. Because there will be some difference between class 9th motion and class 11th motion, right? So don't expect that the difficulty level will be similar to what you have done in class 9. You are in 11th now. So get rid of that mode, that thinking that, oh my God, it is difficult, different. It has to be. It's all, I'll launch a poll now. Here are poll. Tell me, I'm ending the poll now. This is what it is. And yes, answer is A. How it is A. Let me just tell you how you should arrive at the answer, okay? The final answer is not important. I'll tell you how to get to the answer. Now here you can see, ball is bouncing. After hitting the ground, so direction of velocity is changing or not? Directional velocity is changing. All of you agree that the velocity direction is changing and it changes immediately, okay? So if initially it is positive, after collision it should be negative or vice versa. So direction of velocity should be changing. And you can see in option B and D, the direction of velocity is not changing. It remains positive or it remains negative. So B and D you should eliminate, okay? Now it is two options, okay? You have two options, all right? So based on these two options, you need to pick which one is correct. The A and C, you can see here, the only difference is there is a linear graph between velocity and height in C and in A there is not a linear graph, okay? So you know that velocity, how it changes with the height, V square equal to, it is dropped, right? It is dropped. So initial velocity is zero. This is 2G into H. So you can see that it's not a linear relation. V square is proportional to its height, okay? So that's how option A is correct. So you have to do only this much to arrive at the answer, okay? Don't overdo things. But then you can analyze it further. For example, can you tell me when ball is dropped from the top, this is the graph for that or that is the graph for that? One or two, which one? When ball is dropped from the top. It'll be one. You can see here, the height is D. Height is D, velocity is zero. Then it goes like this. When height becomes zero, velocity is maximum. And then it bounce off. So from this point, where it goes immediately? From this point, it goes to that point, which is A or that point, which is B. Where does it go? From here. A or B. It goes to A. Because height will remain zero only. Immediately height will not become D by two. And then it moves up by a distance of D by two. So like this, you can fully analyze also. But then if it is just about getting the answer, you can get it just by doing one step. Yes Oshik, what happened? Quickly. Okay, next up, graph one, velocity is reducing. Velocity is reducing. How can you say that? Velocity is not reducing. How can you say that? Velocity is zero here. Velocity is on the y-axis. It goes like this. And it is maximum when it is about to hit. You're dropping something. When it is about to hit, velocity will be maximum. So let us take up this one. I hope all of you are able to see it. Anyone? Maximum acceleration. Graphically, how will you find where the acceleration is maximum? Velocity versus time. The slope is steepest. Wherever the velocity increases, fastest. That is the maximum acceleration. Okay. So what is the acceleration between zero and A? How much it is? I assume this is 20. It will be 20 minus zero divided by 20 minus zero. So this is one meter per second square. Acceleration of AB, how much it is? It is zero. Acceleration of B to C. Velocity as C is 60 minus velocity at B is 20 divided by time. So 40 minus this is let's say 30. This is how much? Four meter per second square. Acceleration of CD how much it is? Final velocity is zero. Initial velocity is 60. How much time? 80 minus 40. So this will be minus 60 by 40. So that is minus of 3 by 2 meter per second square. Acceleration at 40 second is not defined. Fine or shake. Left hand limit is different. Right hand limit is different. You cannot differentiate velocity there. If you are slightly this side Acceleration is something. Slightly the other side Acceleration is something else. Okay. We are assuming this to be 30. Okay. So clearly option D is correct. B to C. By the way, one point I wanted to tell here. Suppose C to D comes out to be minus 5 meter per second square. Then what is the maximum acceleration? Four or this? Which one is maximum acceleration? Minus sign means what? Minus sign signifies what? It signifies direction only. Don't judge the magnitude by the sign. Nothing to do with the magnitude. So when you are saying which one is higher compare the magnitude. Okay. Compare the magnitude. So 5 is the maximum acceleration even though you are getting minus 5. Right. All right. So we'll do a few more questions. Questions the nth second formula. Just look at it nth second formula. No one. I'll just write down the nth second formula. Should I solve also? Or should I wait? Should I solve? Okay. Should I solve everyone? All right. So here the last second it travels this much distance where H is the height. Okay. So it is dropped. Initial loss is 0. So nth is 9H by 25. This is equal to half of G times G is the acceleration due to gravity. Now you can claim that acceleration is negative. So will be the 9H by 25 even this will be negative. So I am keeping both as positive. This will be your first equation. What could be the other equation? Because there are two variables H and N both are the variables. So what will be the other equation? You can use S equal to UT plus half AT square. If I put T equal to N what will be S? If I put T equal to N what will be S? S will be H. So H will be equal to U is 0. So half G N square Okay. Now they are asking you to find H. It is better if you find N first. So if you divide first equation from second you will get 9 by 25 is equal to 2N minus 1 divided by N square. So from here if you solve this quadratic equation you will get N equal to 5 seconds. Okay. So once you get N equal to 5 seconds H will come out to be half into take G as 10 I am saying so. Right here G is 10 into N square that is 25 that will be 125 meters. Everybody understood this? Everyone? Type it any doubts? Anything you want to discuss or ask? No doubts? Okay. We will take one more question then we will take a break. After the break we will take up the relative velocity then. This one. Rectilinearly means it is straight line. Okay. Rectilinearly means straight line. What is the formula for average velocity? Everyone? Formula for average velocity.  That is the average velocity. Gourman got something Oshik, Shittij Everybody should answer. Okay. It is a simple question. Gourman, Oshik and Shittij get the value of average speed. Average velocity you got? Get the value of average speed now. Everybody else? What is the answer? When are you school starting? And are you no are enough? That is not. I think in NAPL and in NPS Yashwanpur some of the schools they have started. So the answer is 0. Answer is A. How it is A? Because total displacement is 0. How to get total displacement from velocity? It is not a constant acceleration is not constant. So how to get the displacement? We will use calculus because I know V should be equal to dx by dt whether acceleration is constant or not V should be equal to dx by dt. You know you can use calculus even when acceleration is constant forget about equations of motion. Okay. Once you are in that mode you can use calculus approach everywhere but then that will take some time just that I am telling you that calculus approach is a generic approach it is valid everywhere. So this is how you will integrate now X is 0 initially T is what? 0 0 then T is 2 seconds 2 seconds X. So X will be equal to 4 times T square by 2 0 to 2 minus 3 times 3 times T cube by 3 0 to 2 Okay. So this is 2 T square you will get 2 times 2 square minus 0 square minus of 3 and 3 are gone. So 2 cube minus 0 cube So this will be 2 into 2 square minus 2 cube that is 0. So displacement is 0 displacement is 0 so average velocity will be 0 total displacement divided by total time 2 0 meter per second now if somebody ask you find out the average speed then what you have to do any idea anyone this is the average velocity find the displacement by time is the average velocity average speed is distance by time so for that you need to get the value of distance how to get distance average is you know so speed should always be positive okay speed should always be positive but 40 minus T square can become negative so basically I need to integrate this 40 minus T square I have to take mod and then integrate this so I need to check when it will become negative so when it will become negative when 40 minus 3T square is less than 0 or 4 minus 3T less than 0 or T is greater than 4 by 3 seconds okay so this integral I will not solve it completely those who are interested I will just I am just extending this question so 0 to 4 by 3 you have to integrate 40 minus 3T square DT then next up you have to integrate between 4 by 3 to 2 what you should integrate everyone what you have to integrate 40 minus T square or what you have to take mod of it 40 minus T square T square is negative between 4 by 3 and 2 so multiply that with minus 1 to make it positive so now you have to integrate 3T square minus 40 TT okay once you integrate this you will get the value of distance covered that distance divided by 2 is the average speed okay answer I don't know I don't care I am just telling you how to solve it so this is till now we have covered the graphical approach calculus approach algebraic approach all kinds of approaches we have covered to analyze the motion in 1D okay after the break we will be taking up a new kind of scenario wherein 2 or 3 particles are moving together and there is a relative motion between them it is like for example I am running behind the bus bus is also moving so how much time I will take to catch the bus bus might be accelerating I might be accelerating so things like that will be taking care of after the break fine so right now it is 605 we will meet after 15 minutes 620 don't miss the last part come on time it is very important section we will do it after the break okay alright bye for now alright so assuming all of you are there someone left during the break no auro is there anyways we will start okay so let's continue hmm relative motion aushik somehow auro and aushik sound similar to me yeah but somehow initially I was confused aushik and auro they are same but then yeah of course they are not no no no they are not same of course alright so write down relative motion here we will be talking about how you know objects are observed by an observer that himself or herself is moving what does it mean it means that if observer moves while looking or observing will the observer be able to see the true velocity of the object or not the actual velocity it won't be able to see the true velocity then it able to observe adding it for the observer of course it's a living being but then it can be an object also a sensor so won't be able to observe the absolute velocity has it ever happened that you are on a train in a platform and then you realize that you have started moving alright but then immediately after few seconds you know you come to know that it is not you it is the train next to you that is moving has it ever happened so this is one thing which you can observe I mean I have observed many a times so this thing you might have observed everyone might have observed suppose you are on a moving train okay you know that the trees are at rest okay there is no mystery in that trees are at rest okay suppose you don't know that trees will be at rest suppose you don't know so how will you feel the trees are moving in which direction you will feel as if trees are running behind you trees are going against your motion right so that is what you are observing forget about the truth truth nobody knows right absolute velocity doesn't even exist fine so what we do in kinematics is whatever we observe we analyze it right but over here if observer is at rest on the ground the observer is observing something trees are at rest and when observer starts to move observer feels that trees are moving so there are two different velocities for the same object the trees velocity was 0 when observer is at rest trees velocity is something when observer is moving okay so what is going on here observer is not observing the true velocity observer is observing the relative velocity so basically what happens is that observer observer observes observes the world by subtracting or her motion so basically what happens is if if so if velocity of observer is vo this is let's say velocity of observer okay and u is the velocity of object then what when I say velocity of object I mean true velocity true velocity of object is u true velocity of observer is v0 so what observer will observe anyone what observer will observe minus v observer will subtract observer will subtract his or her own velocity from the object the observer is observing so when observer is observing trees u is what velocity of tree was 0 right so for trees u is 0 so what observer will observe observer will observe 0 minus v is negative of your own velocity right so you will observe trees are going with minus v velocity so if you are going like this trees will be going backwards this is what it is okay and this is valid not only for motion in 1d motion in 2d also will be talking about relative motion this is true there also but you need to make sure that you are having vector subtraction here this is not like you find a magnitude and subtract it is vector subtraction getting it all of you let's take one quick example here suppose an object is going like this 2 meter per second there is object and this is observer observer is moving like this 2 meter per second so what is the relative velocity of the object with respect to observer what observer will observe what velocity observer will feel as if object is at rest because distance between them won't change observer will feel that object is moving when object is either coming towards the object observer or going away from the observer but over here the distance between object and observer will not change because both are going together like this so you will feel as if velocity is zero the best example is the earth with what velocity earth is revolving around the sun 12 kilometers every second 12 kilometer per second earth is revolving around the sun but you are not feeling anything two reasons one is earth doesn't vibrate while revolving around the sun second you are moving with the earth so you don't feel with respect to you velocity of earth is zero fine so let's take another example suppose this is the velocity of object let's say 2 meter per second and observer is going with 5 meter per second so how the observer will feel object is going in which direction and with what velocity minus 3 2 minus sorry 2 minus 5 which is minus 3 so observer will feel as if object is going like this 3 meter per second you are assuming forward direction to be positive so velocity coming out to be negative meaning that it is going in the opposite direction everybody is clear right no doubts as such no doubts just one more question then we will move ahead let's say the object is going like this 2 meter per second and the observer is going like that 5 meter per second so what is the velocity of the object which observer sees and in which direction assume right hand side to be positive assume right hand side to be positive tell me what is the answer that side is positive as in 2 meter per second is positive everyone so velocity of velocity of object is plus 2 velocity of observer is what velocity of observer is what minus 5 right so observer will see the world by subtracting his or her own velocity from everything it observes so it is observing 2 meter per second then observer has to subtract its velocity which is minus 5 2 minus of minus 5 you will get 7 meter per second plus 7 it is coming so you will feel as if the object is going in the positive direction this way with 7 meter per second even though its actual velocity is 2 meter per second so this is a relative velocity concept simply explained for motion in 1D only everybody is clear about these basic observations everyone please type in yes or no or if you have any doubts please type in I am waiting for your doubts everyone no dbts alright now you will see that it becomes little complicated that you subtract the velocity take care of the sign convention and things like that so there is one very simple way you can analyze the relative velocity you will not find it any textbook so this is based on our own experience I thought we could discuss that so because you know right now we are only dealing with velocity what if there is relative acceleration also what if velocity and acceleration both are there and in different different directions you will go crazy right so there has to be a very simple way to deal with relative velocity and this is what we are going to discuss next but before even discussing the simpler way you should understand the basics this is what the basics are anyways so write down velocity so the heading you know it is not part of any textbook as such so we can have our own heading so basically I am talking about here velocity or acceleration of approach separation so you will notice one very peculiar thing about motion in for relative motion that either object will be you know trying to separate from the observer or object will be trying to come close to the observer only these two things can happen fine there is no other thing that is possible so since only these two things are possible that either they are approaching or going away so we will use this to analyze the relative velocity okay concept is I mean the way we should the way we understand this very simple so we will keep it simple only think in a simple way okay all of you suppose these are the two points by the way you know it is not written on the observer's hat that observer is observer anything can be observer even object can be observer alright so don't start finding okay which one observer which one is object not required so suppose these are the two objects a and b okay a is going like this with velocity v1 b is going like this with velocity v2 the distance between them is L so use your common sense tell me what is the velocity of approach with what velocity they are approaching each other I am not asking the relative velocity of one with respect to other I am asking you how fast they are approaching each other common sense you can be wrong also a ja v1 and v2 both are pausing I mean both directions are shown that I am only talking about magnitudes here I will not consider science here velocity of approach is some of the velocity right velocity of approach is equal to v1 plus v2 simply it is some of the velocities because a is trying to approach b with v1 b is trying to approach a with v2 so both are trying to approach each other and total velocity with which the distance between them is decreasing is v1 plus v2 both are adding each other distance of approach is how much how much distance they will be covering in total L so time taken by them to cover the distance L is L divided by v1 plus v2 okay so very simple right now let's take another example now we will have expression also this is a that is b distance between them is L okay this velocity is v1 this velocity is v2 okay expression of a is this way a1 expression of b is this way a2 getting it now tell me what is velocity of approach how much it is common sense velocity of approach everyone a is trying to approach b or trying to go away from b a is trying to approach b or trying to go away from b I am asking a simple question a is trying to approach b or trying to go away from b a is trying to go away you are looking at acceleration or velocity talking about velocity with respect to velocity I am talking about velocity of approach right so they is trying to go away or towards B initially it's trying to move towards B if I just talk about velocity it is trying to move towards B isn't it what about B B is trying to move towards a or away from a look at velocity velocity I'm talking about forget about acceleration things are not clear are now what is the velocity of approach suppose efficient is not given now tell me actually is not there fine now tell me is trying to come closer to be what about B B is trying to come closer to a or going away from a it is trying to go away so net velocity of approach is what net velocity of approach is what V1 minus V2 V1 is trying to decrease a distance V2 is trying to increase a distance the total net net with what velocity their distances are decreasing V1 minus V2 so this is the velocity of approach okay now assume that they have acceleration let's say this is a1 and this is a2 tell me a1 is trying to decrease a distance or increase a distance look at a1 a1 only decrease right what about a2 a2 also same right so acceleration of approach is what what is acceleration of approach a1 plus a2 okay so you have velocity of approach initial velocity of approach and acceleration of approach and distance of approach is what L initial distance between them right now here is the power of this particular concept the thing is that you can use s equal to ut plus half 80 square for relative variables also the distance of approach is L velocity of approach is V1 minus V2 that into t plus half a1 plus a2 times t square so using this equation you can get directly the value of t for which these two particles will meet these two particles may not meet if let's say V2 is very no they will meet because acceleration is favoring it fine so you solve this differential equation sorry you solve this question you'll get the answer for the time for which they will meet so we'll take up another scenario then we'll see what to do next okay alright everyone draw this there are two objects a and b distance is L this is V1 V2 a1 a2 okay you need to just do these things find out velocity of approach find out acceleration of approach and distance of approach and then s equal to ut plus half 80 square you have to use here everyone do this velocity of approach is what minus half V1 plus V2 both velocities are separating accession of approach how much it is a approach a1 is trying to decrease the distance minus a2 distance of approach is what s approach is L so I can write here that L is equal to minus of V1 plus V2 into t plus half a1 minus a2 times t square okay now if I ask you how much time how much time it will take for separation to become I mean just write down the equation don't solve it same conditions how much time will it take for separation to become 12 now you have to find velocity of separation and accession of separation try doing it velocity of separation is what everyone this was velocity of approach you're talking about separation right from L it is becoming 12 velocity of separation is what V1 plus V2 both velocities are trying to separate acceleration of separation a2 minus a1 how much distance they have separated in time t what is the distance of separation from L it has become 12 distance of separation is L only 12 minus L okay so I'll be using s equal to ut plus half 80 square so L is equal to V1 plus V2 times t plus half of a2 minus a1 times t square now you might be wondering how come if they are approaching how come they can be separated they can I mean they can be anything depending on what is the value of a1 a2 probably when you solve this you will not get real values of t getting it so you can solve it either ways depending on question of course but have you understood how we use the relative motion concept here velocity of separation accession of separation everyone has understood type in yes or no quickly then we can take up questions ask doubts whatever doubts you have don't hold it will be taking questions of course on whatever we have discussed just now it's not very simple to understand just like that okay okay so let's take questions now on the relative motion concept so that elephant question you remember this one okay try doing it with velocity of approach concept all of you try doing it with velocity of approach concept there is no acceleration right no worries you can take time I don't remember the final answer I will solve it step-by-step so that you understand the approach and separation concept can you find out what is the time B will take to catch C how much time B will take to catch C that is step number one right so between B and C what is the velocity of approach anyone between B and C if velocity of B is you C is 10 what is the velocity of approach everyone answer it B is running behind C with velocity you C is going away with 10 so with what velocity B and C are approaching each other everyone answer it I'm waiting for everybody's answer I'll take your names 10 Gurman what it is? Oshik, Arabi, Aditi, Anjh, Priyam, Anjh has answered I'll not leave everyone must answer at least those who I've taken the names elephant B is going behind C with velocity you C is going away with 10 with what velocity B is approaching C that is relative velocity of approach velocity of approach is how much? Arabi, Aditi, Arya I haven't got reply from you everybody else replied waiting for your reply quickly Aditi answered Arabi and Arya what it is? Arabi now you're only one Audible Arabi not answer not looking at answer I am asking you velocity of approach is what? Between B and C between elephant B and elephant C what is the velocity of approach? Arabi, velocity of approach I am asking you're telling me option A, B, C what does it mean? I'm not asking you to solve it what is the B is running with velocity you C is running with velocity 10 what is the velocity of approach between B and C between B and C Arabi what it is so you can see that B is going towards C with velocity U and C is going away from B with velocity 10 so approach velocity is U minus 10 okay with this velocity B is trying to approach C what is the distance of approach everyone distance approaches D so how much time will it take for them to cover the distance when B meets C T is what D divided by U minus 10 I want to now analyze between A and C I can see that between A and C distance is increasing from D it has become 3d so distance of separation is how much from D it has gone to 3d so how much they have separated total D separation the separation has increased habits oh sorry between A and C is it okay from 2d between A and C the distance was 2d now it has become 3d the separate separation is D right by D units they have separated further the initial distance between them was 2d now it is 3d so the separation has increased by D so distance of separation is D between A and C in same time in this time only T is D by U minus 10 what is the velocity of separation everyone between A and C what is the velocity of separation everyone what is the velocity of separation between A and C C is trying to go away with 10 is approaching with 5 so with what velocity they are separating from each other C is going away is following it up so with what velocity they are going away from each other 10 minus 5 right 10 minus 5 so with 5 meter per second they are moving away from each other so 5 into T 5 into T should be equal to distance of separation which is D okay so from here you'll get U to be equal to 15 meters per second now you know you have learned something which is completely new you have never seen relative velocity concept so you'll feel a little bit of uneasiness that what is this new thing completely let me do it in an old-fashioned way but then don't you know get scared away by this concept you know there are many many complicated questions that can be solved in a very simple manner by using velocity of approach and velocity separation concept fine so keep on solving questions on relative velocity and try to learn struggle it through okay few of the students might be able to answer it quickly probably they have done it before that is the only reason so if you also spend let's say some time to struggle through it and understand then you'll also start getting it when exam happens it doesn't matter how much one knows how much one doesn't know it is all about whether you are able to analyze the situation or not right now probably some people might have mugged up a lot of concepts or they might remember many things they're able to solve it quickly but when exam comes after two years the computer exam knowledge doesn't matter your analytical mind will matter so do but then of course little bit of concepts you need to put it in your brain don't get scared away by this okay this is very important concept alright we'll take up few more relative velocity concepts question this one what is the condition for maximum value of acceleration a before you even start solving it you need to first understand what is the condition when the acceleration will be maximum what should happen for that what is the condition no no no forget about answer I don't I'm not interested in checking the answer I want to know in what situation a will be maximum what is that situation car should be able to catch up with the bike by even if even if acceleration is less if a is 0 also the bike is stationary then also car will be able to catch the bike right isn't it so what is the condition for the maximum acceleration again you understand even if expression is zero car will be able to catch it then what is the condition for acceleration to be maximum not mathematical condition physically what should happen time should be least okay time should be least don't you think when expression is zero time will be lesser to catch time should be maximum maximum time it should take to catch but what is the condition for it to catch condition for the car to catch the bike you're not understanding what I'm asking basically so so let me twist the question a bit here it is suppose see car is velocity of bike must be lesser than the car at one approach to it okay understand listen here all of you listen suppose see the car is going with constant velocity whereas bike is accelerating velocity of bike is increasing after every second it is kept on in it kept on increasing fine so tell me one thing if the bikes velocity becomes equal to the cars velocity before even car reaches the bike will they ever meet they ever meet no because accession of bike will keep on increasing once accession of bike is equal to the accession of car bike will just run away because its velocity will keep on increasing beyond cars velocity also so now tell me what is the condition for maximum acceleration so that car can you know meet the bike tell me you're telling me inequality you're telling me bike must be lesser than car it tell me equality equal to what when s is zero velocities are equal okay so when when car reaches the bike their velocities are equal so car has been just been able to reach the bike that is a condition for the maximum acceleration so again I'll write here for maximum acceleration reaches bike bikes velocity should be should be equal equal to the cars velocity all of you understand this forget about solving 100 questions one question you should you know do it systematically this is equivalent to solving 10 questions all of you understand this type in yes or no quickly for max if your accession is more than this acceleration before even car reaches the bike bike will run away if your accession is more than this conditions acceleration okay all right now I have got the condition for maximum acceleration now tell me initial velocity of approach how much it is everyone initial velocity of approach velocity of bike was zero initial velocity was zero so initial velocity of approach is what v everyone it is v final velocity of approach is what when they meet for maximum acceleration both the velocities are equal finally so velocity approach is what finally zero it is zero this is the strength of relate you know the velocity of approach concept you have both initial and final velocities and distance of approach is what 100 and by the way acceleration of approach how much it is with what acceleration they are approaching each other what is a approach car doesn't have any acceleration bike has acceleration what it is minus a very good very nice so which equation should I use looking at these variables which equation should I use to get the value of a quickly v square equal to u square plus 2 a s this I will use v is zero velocity of approach is zero initial velocity approach is v square this is minus 2 a into 100 right so a will be equal to v square by 200 all right and v is based v is given as 20 20 square by 200 that is 2 meter per second square so you can see that in one step in one step you get the answer by using velocity of approach concept otherwise if you try to solve it using step by step approach okay fine bike will move ahead by x so car will have to travel 100 plus x and then their velocity should be equal finally get the value of time distance it'll become a mess everyone appreciates I mean I have written here multiple steps but then many steps are redundant at the end you can solve it in a single steps everybody understand a cadoo drawing of car is not good this it's a bus okay look like a bus no enough now tell me is it clear to everyone so as approach is a starting distance between the two of the yes distance of approach is see distance of approach is how much distance they have come closer to each other whatever was the initial distance between them that much distance they have completed to meet right so whatever was a separation initially that much is the approach if they are meeting if suppose from hundred it has become distance become 30 so what is the distance of approach from 100 the distance between them become 30 distance of approach is what 100 minus 30 they have covered 70 meters if final distance is 30 between them they have covered 70 meters to approach each other then distance approach is 70 okay so let us see what other this one these are you know little tricky questions understand and I don't I'm not expecting that you guys will be able to solve it okay frankly because if I just do simple question you'll be able to solve it you'll not learn anything so I've taken tricky questions so that you struggle it struggle through it and then you learn something new out of it okay try attempting this okay anyone alright no worries okay got final answer let me check again final answer I don't have we'll discuss the solution don't worry by the way the first chapter is I mean the kinematics chapter one motion 1d is over all right so you can solve any question from anywhere now as long as it is motion 1d chapters question should I do it now wait so let us first analyze the question forget about solving it and getting the answer okay that's not important train starts from the station with constant equation so what is given is that it starts starts means initial velocity is zero and excision is given to us about the train passenger arrives the station six seconds after the end of the train has left after six second the passenger comes by the time train will move ahead and will gain some speed also right now what is the least constant speed at which passenger can run and catch the train fine so situation is like this this is let's say trains rear end this is the passenger okay this is train so let's say this is t and this is passenger p okay the the excavation of the train is given as 0.4 meter per second square fine the velocity of the train is not given as in we need to find out because this is not the initial velocity of train this is the velocity when passenger has arrived at the platform that is six second after the train has started so velocity of passenger is not given to us that is what we need to find out actually so I'll keep that as v and this is velocity of train is you so the distance of approach is let's say l l is what l is a distance developed by the train in the six seconds so l is equal to half into 0.4 into six square ut plus half it is square for the train how much it is 7.2 meters so l is 7.2 okay and what is shittage while I'm speaking don't type it distracts and I think I've told you several times okay then I can't keep on checking messages and then keep on writing on the screen also got it take care of that all right so now what is this velocity this velocity is after six seconds because I'm analyzing the situation after six seconds fine so v is equal to u plus atl use so this velocity is 0.4 into six so that is 2.4 meter per second all right this velocity is 2.4 meter per second there are two situations here situation number one train started and then six second has gone that is situation number one and then in situation number two passenger arrives and then running behind the train so this is a situation number two in front of you everybody is clear about it everyone everyone is clear the distance between passenger and train will be 7.2 the velocity of train at this situation is 2.4 acceleration is constant throughout that is 0.4 meter per second square everybody is clear now tell me what is the condition for least speed v with which passenger should run to catch the train what is the condition for that to happen when passenger is able to catch the train their velocities become equal if before even passenger catches the train if trains velocity becomes higher train is accelerating it will just move forward passenger won't be able to catch it fine so the minimum speed should be that when the passenger reaches the train their velocity become equal and because of that you know the final velocity of approach how much final velocity of approach when passenger meets the train zero right their velocities are equal initial velocity of approach is what initial velocity of approach everyone what is the initial velocity with which train and the passenger are trying to approach each other waiting for everybody's answer now okay passenger is moving passenger is moving with v train is moving with 2.4 what is the velocity of approach it is v minus 2.4 see it is very simple apply your common sense it will come out passenger is trying to approach train is trying to go away so with what velocity they are trying to approach v minus 2.4 okay keep it that simple don't complicate in your head it is straightforward fine so this is the situation distance of approach how much 7.2 meters and acceleration of approach how much a approach minus of 0.4 meter per second square this guy passenger has no acceleration train is accelerating but going away so approach acceleration is minus of 0.4 now i'll use v square equal to u square plus 2 as v square is 0 square u square is v minus 2.4 whole square okay minus 2 as in minus come because of 0.4 2 as 7.2 from here you directly get the value of v all right don't you think it's a simple method if you understand it you'll be able to get the answer in a single step otherwise the question can be lot more i mean this will make you write many steps i hope this is clear to everyone okay in case i mean it is perfectly right to feel that it is difficult if i would have been in your place i would not have understood the first time okay so what i would have done i would have gone back and watch the recording of the relative velocity concept again sit with paper and pen and i'm sure little bit of effort you'll make you'll be able to get it it's not something which is out of the world okay fine one last question we'll take one last question we will take this one solve it last question for today just write down the equation that is enough as of now step number one is draw the diagram to visualize what is happening straight line motion is straight line it is what is this velocity you this velocity is just sufficient for it to reach the top of the tower okay so top of the tower if it can just reach final velocity at that point should be 0 so v 0 u square minus 2 g into 90 okay so u is equal to root over 180 into g let's say this is 10 okay so i'll get root over 1800 don't simplify it keep it like this so now velocity is known to you okay two second later a stone is dropped so you need to find out where the this first stone is after two seconds after two seconds suppose it reaches here so how much is this distance how to get that this distance let's say h okay so i will be using s equal to ut plus half 80 square so s equal to u into 2 minus half into g let's say 10 into 2 square so this will give me the value of h so now the stone is dropped from here and this is another stone this distance is 90 minus h i need to find out how much time they will approach towards each other they will meet basically for that i need to know what is this velocity initial velocity of approach will come from its velocity at a height h that you will get from v is equal to u plus 80 so v is equal to root of 1800 minus g is 10 into 2 okay this is your velocity v over here okay so velocity of approach initial velocity of approach is what everyone in terms of v and u and everything tell me what is velocity of approach initially this one is dropped it is just dropped initial velocity of the top stone is zero velocity of approach is what velocity approaches v see this v is going towards the first stone got it execution of approach how much it is everyone execution of approach execution of this stone is g downwards execution of that stone is g downwards so execution of approaches other stone has zero velocity initial velocity after two seconds it is dropped it is not thrown execution of approach is zero both actions are same same direction it is zero execution of approach is zero so this is something amazing it will look as if it is going with constant velocity because execution is zero relative acceleration so basically v into t should be equal to 90 minus h from here you'll get time when they will meet 90 minus h divided by v constant velocity relative velocity is constant all right guys so that's it from my side today it got extended a little bit so many new things we have done in case you are feeling that some concept was very difficult and i can understand it is not easy to understand in one sitting it's not like your class 9th and 10th go back watch the recordings solve questions okay solve a lot of questions make sure that you do this these chapters while i'm teaching if you finish it off from your side you'll never have to go back and revisit them again okay finish it off school level so you'll take care of i don't have any problem with it you'll be able to do pretty well in school the main issue will be computer exams so that's why we're doing higher order questions okay so you when you see the school level question you'll see them that they are very easy and you'll be able to solve them easily so make sure you do a lot of practice on the computer exam front and do it in such a way that you don't have to do it again later on otherwise it will just pile up it will create a lot of trouble later on all right so that's it from my side we will meet again next week take care bye for now