Okay, still few few students to join waiting for all of them Waiting for everyone to join Wait will wait for a minute All right guys, so let us start now Others will join meanwhile probably So today's topic is the kinematics Don't get intimidated by this drawing in front of you kinematics is not as difficult as this Okay, this is like application of kinematics. It's a robotic arm So if you are good at kinematics, probably you can design a nice robot. All right, so Let us continue So as I said the last class Yes, yes, we We have finished with units first chapter is over All of you have completed the assignment which I had given Type in yes or no. Have you done the assignment? Okay, so So we are going to start this topic kinematics write down kinematics is Basically, it's a study of motion right down. It is the study of motion Without getting into the cause of the motion Okay, it is like I don't care about what is causing the motion. I'm just Studying the motion of it. All right. I will just see How much it has traveled with what velocity it is traveling and I will not care about the fact that from where the velocity is coming and And What is what is creating the acceleration? How much force is acting on it? All that will not be talking about Here Okay, so things will be more clear if we talk about few examples, right? So for example, this robotic arm if you see here, although this type of complicated things will not study All right, so you can see here. I don't need to know that Relative to this link that arm will rotate like this Okay for that, I don't need to know any cause. It is constrained to move like that. There's a constraint relation Fine and this part hand can rotate about this axis Fine even this can rotate About that axis So all these things I need to just know how they are constrained to move Fine and then I will also understand if link 2 is moving with certain speed Link 3 is moving with certain speed then with what speed link 5 and 6 should be moving There has to be some sort of relation Okay, so these kind of constraint movement will result in the concept of kinematics and We'll be just studying the motion of an object. Okay, so I'll give you another example. Have you ever heard of that? automatic cleaning machine which Nowadays it has come in India also So there are these small robots circular disk sort of rotate, you know robots Which can just trace the path on your floor and it will clean it up entire thing. Have you heard of them? right, so Right rhombus so these These will consume the power also. Okay, so what will happen is suppose might might You know Intention is to clean the entire floor. Okay floor will have a lot of obstacles Tables are there chairs are there so far is their center table dining table. Everything is there bed is also there So these robots if I just design it in such a way that every time it hits some obstacle It takes a turn and start moving in another direction Okay, so if I Just design it like that the moment it Hit something it rotates itself a little bit and starts moving All right, this is like the basic of the design wherein you're just making sure As soon as it hits it should it should it should avoid going in the direction where obstacle is there that is all Second step what you can do is you can put some sort of ultrasonic sensors When the sensors throw the ultrasonic wave forward and it comes back The robot will calculate how much time it has taken for the ultrasonic wave to reach back So using that the robot will calculate that thus The obstacle is let's say one second ahead Okay, so the robot will not move in that direction then Fine, so this is second step, but then in both first step and second step You are not making sure that Same place where robot is cleaning the robot doesn't come back again It might happen that robot is just confined to an area and keeps on coming back to the same place again and again Fine so an intelligent robot or an expensive one will be one which will keep track Of where all I have already been I will not go to the same place again With all the sensors in place Okay, so by just you know analyzing the motion and how far the obstacle is how fast the robot is moving I will be able to intelligently design all the functions of the robot Okay, so kinematics application is much wider And it is it can be it is also used in Organ replacement not all the organ the hands and limbs basically Okay, so all these things taken care of there Now I am not getting into debate of how it is why it is when it is Because it is not part of your curriculum Okay, I just told you so that you understand that these are the application of kinematics fine So let us get into the chapter actually what it is about fine So kinematics chapter can be divided broadly into two sections. By the way, have you heard of kinematics? Kinematics word have you heard of? Which chapter with chapter? Okay, so kinematics Name is not there as a chapter name. Okay. In fact kinematics is Is basically a Section which consists of two chapters in your syllabus The first chapter is motion in 1d Okay, so motion in 1d Chapter number one and second chapter is motion in 2d fine So you have already studied the basics of motion in 1d in class 9th Okay, you already heard of it in class 9 wherein you used to apply the Equations of motion v is equal to u plus 80 s equal to ut plus half 80 square all that Right all that we were making sure that It is moving in one straight line only and that is the motion in 1d Motion in 2d little bit. We have already done in the bridge program the projectile motion was an example of Motion in 2d Okay, now same motion 1d was there in class 9th again. It has come up But then now that it has come up our When it has come up in class 11th. There is no limit up to which the difficulty can go up to Fine, it is up to you what level you want to stop you want to prepare only for the school exam There is only one limit up to which you can solve problems Fine, if you want to prepare for CET another level need and J another level if you want to prepare for J Advanced there is no level you just keep on, you know doing various types of problems fine So now it is much more open-ended so we need to study it in a much more broader fashion and Understand all the finer nuances that are there in motion in 1d chapter Okay, so motion in 2d since it is chapter with two will not talk about it much right now so motion in 1d Can be classified into three parts What is kinematics analysis of motion or study of motion fine, so the first part is the Is the algebraic analysis? algebraic analysis We'll be analyzing the motion using algebra or using the equations that is algebraic analysis fine Then second part of the chapter will be graphical analysis Okay, using graphs and all will be analyzing the Motion of an object. Okay, third is basically calculus oriented so in class 9th, we never Worried about the fact that acceleration can also change So when acceleration also changes you cannot use equations of motion you learn that anyways So then you need to use the calculus approach By the way, have you done the basic calculus in the maths bridge program? Basics I think is covered already right. So yes, we'll be using that To study the motion in the calculus approach topic fine Yes limits derivatives and integral also a little bit So now that kinematics is a study of the motion itself So here, I'm not going to assume that you have studied anything in class 9th. We are going to start from the basics All right, so if you think that your class 9th was not so great Here is an opportunity to correct that it is up to you if you want to do that so If we have to study the motion by any means graphical calculus or algebraic We need to understand what are the variables which we will be using to analyze the motion Okay, so if we understand our variables then we'll be using the variables only to understand the You know what the object is doing fine So introduction of variable is the first topic of the chapter then we'll start studying it Algebraically graphically and calculus oriented Write down introduction of motion variables. Tell me a few examples motion variables Anyone okay velocity acceleration displacement Very good. What else velocity excision displacement time Time is very important variable. Okay, don't forget that Then is that it all right? So the first variable We are talking about motion in one day Aditya, not the circular motion So introduction of motion variable in motion in one day. Okay. All right.  Yes, why not? fine the first variable the most basic variable is Position location So initial location if I don't know How will I get the final location? Suppose I tell you the displacement is 100 meters. Tell me where the object is You'll not be able to because you don't know Which was the initial location if initial location was x equal to 10 and you tell me the displacement is plus 100 The answer is 110 If initial location was 100 and displacement is plus 100 the final location will be 100 plus 100 200 Fine. So initial location is the first variable Okay, most of the time we assume initial location or position to be zero That is our assumption, but you should know that you are assuming it fine, so Initial location is represented by variable x naught in motion 1d It is represented by variable x naught fine Now x naught can be greater than zero x naught can be less than zero or X naught can be equal to zero So it can have any value X naught belongs to a real number set. It can be anything. Okay Next variable is time what I can say about time Any constraint on value of T Time is represented by letter small t, which you know already What I can say about T Can I say T is less than zero? Is it possible T less than zero? T less than zero not possible everyone T less than zero. Okay. What is T equal to zero T equal to zero means what? T equal to zero means what? Starting time. What is a starting time start of the universe? Starting time or initial time is start of universe T equal to zero Start of your experiment start of the motion. You're saying T equal to zero is start of your motion So whatever is was happening before your experiment or before your motion That is what time that doesn't exist Whatever is happening before your experiment or your motion that is negative time only That is negative time only Before suppose I assume that right now T equal to zero So when I started the class sometime back that is negative time. So T can be less than zero, but It moves in only forward direction. So how can I write it mathematically? Right time will always increase but time can be less than zero also according to what you take as zero But it will always increase. So what does it mean mathematically? What should I write? correct It means that change in time delta T should be Always greater than zero. It can never be even grid equal to zero Delta T cannot be equal to zero also. It has to be greater than zero delta T T can be less than zero delta T cannot be Okay change in time Getting it right distance distance is represented I mean it's length basically so you can say L or any other letter you can use XYZ So displacement L or Yeah, you can say D also What is the constraint with it can distance be negative? Can the length be negative? It cannot be negative. So distance covered should always be greater than or Equal to zero it can be equal to zero also if I'm standing at a point I'm not moving anywhere Distance covered is zero. Okay Then comes Displacement what is displacement? All of you have learned this anyways Displacement is basically shortest distance shortest distance between two points It's a vector quantity. It's a vector quantity Isn't it? We have learned about vectors already. So vector quantity is Represented by an arrow like this this arrow connects initial and final point Fine This is a displacement vector now Displacement Actually gives us more information than distance. For example, if I tell you that I have traveled a distance of 100 kilometer You'll have no idea where I'll be Okay, if I just tell you 100 kilometer I've traveled a distance I could have been back to the same location But if I tell you the displacement is 100 kilometer in the east direction, then you will exactly know where I am Fine so the displacement carries more information than distance All right. No mother. No. No fine. So displacement can it be Can it have any value? Displacement by the way is represented as s can s belongs to all real numbers It can have any value. It can be negative depending on which direction you're taking positive The opposite of that direction will be negative. Isn't it? Correct This is displacement All right Now speed what is speed Rate at which Rate at which distance is covered. So can speed be negative Can it be negative s? speed No, right speed. Let's say I represent by you You will always be greater than or equal to zero fine There is a concept of average speed also average speed is basically equal to total distance divided by Total time Okay, so suppose you have traveled a distance of hundred kilometers in two hours Your average speed is 50 kilometer per hour What does it mean? If your average speed is 50 kilometer per hour, does it mean that you are traveling at 50 kilometer per hour? Does it mean that it doesn't mean that okay in between my Speed might be 20 kilometer per hour then 60 kilometer per hour Then 70 then again 10 like that my speed might be continuously fluctuating Okay, but on an average my speed is total distance by total time which is coming up to be 50 kilometer per hour So if I traveled at a constant speed of 50 Then same distance will be covered in the same time This is what it means This is the speed and now the velocity the velocity as you know already is rate at which Displacement happens Okay, so the velocity there is a concept of average of that also average velocity What will be average velocity? Total displacement Divided by total time. Okay, this is your average velocity Fine, so these are the concepts that are very basic concepts And we are going to use these variables to understand few types of motion. Okay, and Now when you're traveling in a car Okay, the velocity keeps on changing, but this is the average velocity So what is that? Velocity which you see every now and then when you look at this pedometer Pedometer the velocity which is showing there. What is is it average velocity which is showing there or what? Instantaneous velocity That is instantaneous velocity. What is instantaneous velocity? What is present velocity current velocity instantaneous velocity? What does it mean at that moment? At that moment, what is a velocity? What do you mean by at that moment velocity is rate at which Displacement is covered So at that moment if you just talk about at that moment The displacement will not won't happen at a particular time itself, right at a particular time Displacement won't happen. So in reality you have done the limits, right in bridge program. You've done the limits Okay, so what happens is that? instantaneous velocity Let me write like this instantaneous velocity V It will be equal to limit limit delta t tends to zero Delta s divided by delta t Delta s is a displacement delta t is a time when you're considering very small amount of time a very small amount of displacement Then take the average of average velocity for that is very very small interval of time That is your instantaneous velocity. This limit will actually result in the derivative of the displacement So this is the instantaneous velocity. I hope no doubts No doubts fine, so let us see the You know Graphically certain situations. How does it look like? We are right now assuming that Acceleration is zero. Okay. I'm talking about uniform motion, right? So for uniform motion I just need these variables. I do not need acceleration because it is zero Okay, so for uniform motion We need above variables only so graphically graphically if I if I plot a graph between let us say the The displacement and time Always remember I am talking about an object that moves in a straight line Okay, this is a chapter where motion is happening in a straight line like this Okay, this is how motion is happening. Okay, so at times in our book the X axis is on the Y coordinate X axis is nothing but X coordinate of the motion where it is happening Alright, and the X axis is actually time Time axis time also changes fine. So now tell me is What kind of motion this is is this a straight line motion or not? Is this straight line or not? Okay, many of you are saying no many of you saying no The answer is yes, it is a straight line motion if it happens the reason is you know the coordinate this is your X coordinate So all you have to find out okay T equal to zero X coordinate is where at this time This is X coordinate at that time what is the X coordinate then X coordinate decreases They are what is X coordinate and then what happens is the particle is Basically moving like this it goes straight like that comes back Along the same straight line then goes back. So all that happens along one straight line only Okay, now tell me what is the biggest problem in this graph this problem this? Graph in reality Can never exist. What is an issue anyone? What's the issue in the graph? Non-uniform is fine object need not have uniform velocity, isn't it? Yes, I think I got it Correct shasha shashat also More than one X at a particular T very good. Even that is also fine Time is going back. So you can see in this zone time is decreasing Time can never decrease It can't even remain constant Fine, so this graph mathematically may be fine, but physically doesn't make any sense Fine, so this graph is Not proper fine. This doesn't exist. All of you understand, right? Tell me what is happening here? Can you describe the motion little bit? Can you describe the motion in this case anyone? describe it properly Anyone It is accelerating. Achha retarding fine completely you have to Okay Velocity becomes constant after some time. Is that correct after some time? What happens to the velocity after some time? What happens to the velocity? Okay, listen here Listen here. So that's the problem. You know you have you should not lose the common sense and you should always You know Be ready to apply your thoughts. So here at When T equal to something value of X was zero So when X was zero T was not equal to zero that is fine Then suddenly it starts moving very fast. You can see that for in a very small amount of time The X value has gone from here to here very quickly it moved Then in the same amount of time You can see this. This is the amount of time it has taken in the same amount of time now the change in the value of X is lesser and It is decreasing Fine so for the same amount of time the value of displacement is keeps on decreasing So the average velocity keeps on decreasing because in every two second it is covering lesser distance lesser and lesser You can see when it reaches here No matter how much time is changing the value of X is not changing. So what does it mean in this zone? If X is not changing with time, what does it mean about the velocity? The velocity is zero. It has stopped Velocity is zero Okay, the last is zero not constant. I mean the zero is a constant But I know what do you mean by saying that it's constant. So it is zero So the biggest mistake the student does in Graphically analyzing the kinematics is that they look at the curve and Think that this is the path This is not the path This is not the path the particle is following this you should be very clear Because X axis is time It is time. It is not some spatial coordinates If it the graph is between X and Y then you can say that is the path But right now it is moving in a straight line and the graph or X versus T looks like that Getting it fine. So this is clear now. Let's see how velocity will look like in a graph See this chapter has very little theory getting it this chapter has just You know two three equations are given to you and straight away problem solving. So do not expect stories now All the physics chapter going to be like that We need to just see scenario one scenario two what will happen this and that that's all But then your varieties of problems have gone up very hugely This is X. This is time. Let us say, okay Now if I say that This is the graph Not that let us Is this graph fine? Is there any issue with the graph? Particle is moving a straight line Is this fine? No, it doesn't go backwards at the top No, no, no, it's like this Is it fine now? Right. This is fine. So can you describe the motion a bit little bit? Can you describe what is actually happening here? Initial velocity is what initially? When the particle is started from here Initially the velocity is zero Initially velocity is zero. So for some time the value of X is Constant and then suddenly it starts increasing and velocity keeps on increasing For equal interval of time. I'm covering more and more distance fine now No, you can't say linearly increasing velocity or not You can't say that okay You can only say that is increasing linear or not you cannot say but just by looking at the graph You need to know equation of the graph also Anyways, so suppose this is point a this is point b fine you need to Graphically show how the average velocity looks like Where is the average velocity if I give I've given you the location of a and location of B Graphically of the show how the average velocity looks like all of you Average velocity is total distance by total time all of you drawn see you're telling an algebraic this thing Graphically of the show graphically understand In the graph how it looks like all right? So all of you are done.  This is a displacement all of you understand that this will be the displacement and this is the amount of time that will be taken That is the amount of time from point a to point b this much time will be taken This is T2. This is T1. This is X1. This is X2 right Clear right Okay, so the average velocity is basically X2 minus X1 Total displacement divided by time that is T2 minus T1.  All right, so this value is Altitude divided by the base Which is nothing but tan of the angle the line connects From a to b. This is theta This should be equal to tan of theta Okay, theta is what theta is angle angle made by the line joining Angle made by line joining a and b Joining a and b with x axis Fine, so you can extend this line further Even this angle is theta only Okay, so all of you understand everyone Tan of theta will be the average velocity graphically All right Let's talk about how graphically the instantaneous velocity looks like Okay Same curve I will take This is X. This is T This is the curve Now at this point I want to find the instantaneous velocity at that instant Okay, then over here you need to make sure that you know if I consider a and b If we consider a and b tan of this Line is the average velocity between a and b. Okay. Now if I bring b Closer to that point a over here. This will be a1. This will be b1 Then tan of the average velocity between a1 and b1 is the average velocity between a and b1 Now if a1 and b1 comes Very close to that point Very very close Can I say that then average velocity between a1 b1 is actually the instantaneous velocity at that point? Can I say that? If a1 b1 Comes very close to this point capital a Then the average velocity between a1 b1 is the instantaneous velocity of that point All of you understand But I know what I'm asking you just answer that first Everybody understand Okay, so now if I Bring the a1 and b1 very close to a and then connect a1 b1. It'll become tangent It'll become tangent fine, so graphically Graphically instantaneous velocity Is tan theta of the tangent at that point Fine So instantaneous velocity Is the tangent is the off angle made by tangent At that point Fine So this is how graphically average velocity and instantaneous velocity looks like fine How do we know angle of the tangent that is a different issue? Okay that When you solve problems You'll understand later on Okay, and if you know already dy by dx Is what? Very small So not dy by dx in this case it will be dx by dt This is actually tan of theta Dx and dt are very small It is like you're bringing x2 and x1 very close to each other T2 and t1 very close to each other So if you know how to take derivative Then just differentiate You'll get the value of tan theta that way Understood nishant And there are other ways also The thing is that you have to take tan of angle And there are 100 ways to find out the angle Getting it all you have to know is Whether you are aware of those or not There can be using coordinate geometry Graphically you can find out the angle You can be very careful with pen, paper and scale You draw the tangent Find out the angle and then find the tan of that angle Okay, there are many ways All right, so let us take few questions on whatever we have done Just now. I hope it is visible clearly All right, so the answer Some of you already got the answer Others also solve Answer is Two times the radius So basically, you know Again keep your Thought process clear Let's say the person starts from here in a server path Of radius R One revolution it completes in 40 seconds One revolution So in two minutes, 20 seconds How many seconds are there? 120 Plus 20 That is 140 seconds 140 seconds can be written as 40 into 3 Plus 20 So basically it makes three full revolution Full revolution doesn't add anything You are back to the same point Three times And then after that there is one half revolution also So you are back to here point Three times And then you are here From point A you have gone to point B Displacement is basically shortest distance between the Starting and end point So this will be the displacement which is two times the radius Got it? All right, next one No, I'll put a poll Many people are giving different different answers All of you attempted Done all of you Okay Here is the poll, all of you please take it So here is the outcome The correct answer is C C for cat All right, so again here you need to analyze it properly A wheel of one meter rotates forward It rotates forward Half a revolution on a horizontal ground The magnitude of displacement of a point of wheel Initially in contact with the ground So this point Which was in a contact with the ground earlier After half the revolution Will reach where If it is full revolution it will be again back to the bottom Now it will reach the top Half the revolution Getting it? So this was the earlier point P And this is the final point P Fine Displacement is that Sorry Displacement is That distance Okay Now that distance how to find you drop a perpendicular here All right This distance base distance P2Q This distance is how much base one How much it is One full revolution it will travel two pi r distance Half revolution it will travel pi r distance So this is pi times r And P2Q distance is two r So this distance Let's say this displacement s Will be equal to root over pi r square Plus two r square Fine R is one so pi square plus Four So like this you do this question Okay Again the level of questions will not be your class ninth level Okay Level of question will be slightly difficult than What you have learned already so don't expect that Same chapter will have same level of questions So you need to learn You know these are like basic questions Whatever we have done just now These are you they don't even qualify for Basic competitive exams But still we are doing it because we are learning things right now All right, let's take up next few You can answer one by one Okay first one people have answered That's fine shashwat Yes the option B is correct for the first Particle covers half the circle of radius r Distance will be pi r And displacement will be two r it will be on the diametrically opposite end Right when it covers half the circle Second one quickly What is the answer you're getting People are answering different pick one option All right, so yes most of you are correct Answer is a Okay, how to solve this one person goes 10 meter north Let's say like this 20 meter to the east This is 10 this is 20 Displacement will be this much So that is root over This is not motion in 1d It's like combination of two motions both are in a straight line Actually this question should not be there but it's fine So this is 10 root 5 Okay, this is the answer Now you might be wondering whether 10 root 5 is 22.5 or 25.5 Just square it 25 square is 625 So if this is less than 625 square of that the answer is a Or you find out square root square root, you know how to find out Next question Okay, many of you have already answered Correct So the answer is a Over here How to solve this car moves half of its time At 80 kilometer per hour rest half at 40 kilometer per hour the distance covered is 60 what is the average speed now We have this habit of taking average right in these kind of questions Always stick to the formula formula for the average speed is total distance by total time Since it is going along a straight line only average speed and average velocity both are same Because everything is going along the same straight line in the same direction So stick to the formula use your imagination to solve the question not to invent a formula So total distance 60 divided by total time taken Total time taken is what? plus t So this will be 60 divided by 2t Fine Now we need to find out what is t How to get that We know that 80 into t Plus 40 into t Should be equal to 60 Fine So t will be equal to half 1 by 2 hours Just substitute here you get the answer In this case The average of 80 and 40 Is also 60 Fine But then it is not always Because the times are equal it comes out to be same But if I tell you 3 fourth of the time It travels 80 kroner per hour 1 fourth it travels with 40 kroner per hour Then the answer is not 80 plus 40 by 2 Okay, the answer will be 60 divided by Can you do it yourself? Let us see Let's say it travels 3 fourth of the time 3 fourth of its time like this And rest of the time it travels with 40 kroner per hour So what is the average velocity now? 70 Orificating 70 So now you can see that it's not the average So now you have to write something like this That total distance Is 60 60 divided by 3 fourth of the time Plus 1 fourth of the time which is t only 60 divided by t Here I have assumed total time to be t Here I had assumed total time to be 2t So I have taken t and t for half of times Okay, anyways So 60 divided by t And 3t by 4 Into 80 Plus t by 4 into 40 Into 40 Is equal to 60 So from here you get the value of t We just substitute here and you will get the answer Next Set of question last Before we start the next topic This one First one, what is the answer? All of you Everyone No charge word, just check once What you have done Use a formula Formula you have to use Then all of you The answer is D D for Delhi Okay Again it is Total distance by total time So first let's say Distance is this much Then that much So travels half the distance Let's say s by 2 And this is also s by 2 So total distance is s s divided by total time Time for the s by 2 is s by 2 divided by v1 For the second half the time is s by 2 divided by v2 So you will get it as 2v1v2 divided by v1 plus v2 Okay, harmonic mean Fine, do this next one Yes, that's correct Anurag, that's correct Which option, Arithra Okay, you need to just tell me once If you keep on typing multiple times, same thing You know it doesn't go down well Yes, Hariaran Siddhanth No Yes, Pranav There are two pranavs, is it? We have two pranavs Pranav, Dharmagudi and Pranav There is one more pranav I think he is absent today Anyways So yes, the option is b Most of you have got it correct Fine Ratio of numerical values of average velocity to the average speed Average velocity is what? Total displacement by total time Average speed is total distance by total time Okay, displacement will always be less than or equal to distance Because it is shortest distance between the two points So that's why average velocity will be less than or equal to the average speed When it will be equal to average speed What should happen when it will be equal to the Both the things will be equal The average velocity and average speed when they will be equal Straight line and in the same direction also Straight line and in the same direction Fine Then only Alright guys, so this is just a section, a small part of the chapter that talks about the uniform motion Now let's talk about the accelerated motion also And that is the bigger section of the chapter Right now Right now Accelerated motion with respect to motion in 1D We are talking about Okay So can the direction of the velocity change in the motion in 1D? Can it change? Can direction change in motion in 1D? It can change right? It can move forward and then backwards also Fine So the acceleration can be there In two different ways By the way, the definition of acceleration I always, you know I start assuming sometime that you have done in 9th But I'll do it again Acceleration, write down Acceleration is basically rate change of velocity Now when I say rate of change of velocity Do I mean only the rate at which direction is changed The magnitude is changing Or can I say rate at which direction also changes That also creates acceleration Pranav left kya? Pranav dharma gadi Stop typing message Pranav what do you think? So rate of change of velocity with time is the acceleration Fine Now mathematically how to write the acceleration Acceleration is basically v2 minus v1 Their vector difference Okay v2 minus v1 is you're adding it Sorry, you're subtracting it vectorally It is difference between two vectors Divided by delta t Okay This is called the average acceleration This is average acceleration All right Now in motion in 1D In motion in 1D What will happen is that You know everything will be along one straight line And direction you take care with sign This way it will be positive That way it will be negative So you need not include lot of concepts from the vectors You can handle it One dice and take positive other negative And then take the difference You'll get the average acceleration Fine There is a concept of instantaneous acceleration also Instantaneous acceleration Which can be delta v by delta t Where delta t tends to zero Okay This will be equal to dv by dt This is the acceleration All of you have understood probably Now let us see the graphical interpretation of it How does it look like graphically So on the y-axis you have let's say velocity On the x-axis you have time Fine So suppose this is the Motion of a particle Can you describe it a bit Anyone What is happening here It is decelerating There is a deceleration happening Okay You can see that initially the velocity is almost constant From here You need to you know think in the very open mind From here to here no change of velocity Okay I am talking about the motion in a straight line So there is no change of velocity Is there an acceleration initially No No acceleration Then velocity increases from that point It increases faster and faster Right So acceleration is growing Right from initial point to the final point Acceleration is continuously increasing Okay Can this line become steeper than vertical line This is what kind of acceleration this is If it becomes perfectly vertical How much is this acceleration From here to here What is the acceleration It is infinite Infinite acceleration Okay In almost no time Its velocity can suddenly increase Fine So There will be a constraint in the velocity time graph That it can't be steeper than the vertical line Fine No matter what you do So suppose I need to represent graphically the average acceleration Same thing No difference This is let's say A this is B Average acceleration Can be represented as Sorry Can be represented as tan of the angle This chord makes Tan of this angle Okay, how it is coming You can see here Again the right angle triangle is created This distance on the graph Is V2 minus V1 From here this distance is T2 minus T1 This angle is also theta Tan of theta is V2 minus V1 divided by T2 minus T1 So average acceleration graphically Is the tan theta of the angle made by the chord What do you think instantaneous acceleration will be graphically Everyone What it will be Angle made by the tangent Tan of that angle Right Correct So all of you Are now experts of the graphs So this is velocity time Suppose it is like this I want to find the Instantaneous acceleration at a point Then I have to draw a tangent And I have to find out what is this angle Tan of that angle is the acceleration Which is also equal to derivative Of velocity versus time At that point Okay So if velocity is given as Equal to 2T square dv by dt Is 4T So suppose at this moment T equal to 4 seconds So acceleration will be 4 into 8 38 meter per second square So like that you can Identify the Acceleration I hope everything is clear Crystal clear No doubts Anything Anything All right So let's proceed Further So you know We are only done with You can say 5 to 10% of The theory in this chapter And your knowledge of class 9th Was up till here So whatever was in class 9th Everything is covered More or less Okay We have to just introduce equations of motion also Then whatever you knew till now Will be covered And then will be going beyond it So again Coming back to the Accelerated motion I'll just graphically give you Few situations You tell me whether you understand The graphs properly So that analysis becomes easier This is x and this is t What is happening here You need to describe it Single word answers are useless Velocity starts positive Retards until some negative value Very good Correct So initially the velocity is Positive How do you know velocity is positive initially How do you know How do you know velocity is positive Because x is increasing Initially the value of x is increasing It is going in the positive x direction You can see When it is going from this point to that point The value of x has increased Isn't it So the value of x is increasing So we can say that The velocity is positive Till where it is positive Till where From here till the topmost point From a to the topmost point The peak of it The velocity is positive And then the value of x starts decreasing So from B to let's say point C Its velocity is negative Because x is decreasing It is going in the opposite direction So on a piece of paper You can draw it like this The velocity is like this And then it takes a turn And comes back Okay, this is how it is First velocity is positive And then velocity becomes negative I hope this situation is very clear to everyone now Can you identify a point where velocity is zero At the peak Don't confuse it with projectile motion It looks like a projectile So don't confuse it with that So the topmost point The velocity is zero What is the reason Topmost point The velocity is zero What is the reason What is velocity by the way Velocity is tan of Velocity is tan of angle The tangent makes with the x axis So how much How much is the angle This tangent is making with the x axis What is the angle that this tangent makes Zero degrees So tan of zero is what Tan of zero is zero Okay You need to also know that Tan of theta Will be greater than zero For theta between zero And 90 degrees Angle theta Will be less than zero If theta is more than 90 Less than 180 Okay, so these things Better if you remember it Okay Now here the angle made by the tangents With the x axis In this zone Angle is acute This is the angle Fine Whereas angle made By the tangent in that zone Is obtuse This is the angle Fine You always take this angle You don't take this one Okay So the tangent is going like this It changes the orientation slopes over And goes down like that So angle is zero at the top Okay Zero or you can also say that Zero or 180 And then it goes like that Fine So I hope this is clear to you And you can say that The velocity when it goes from A to B The magnitude of velocity is decreasing or increasing The velocity is increasing or decreasing Velocity increases This angle theta From A to B becomes zero It is decreasing No, you can't say constantly decreasing You never know constantly or non-constantly You're assuming it Just decreasing Fine So till point B Till point B The velocity actually decreases At point B velocity becomes zero And then velocity becomes More and more negative More and more negative Okay So can I say that This is a case of deceleration Continuous deceleration Between A and B Deceleration or not Between A and B deceleration Okay Between B and C also it is deceleration only The magnitude of velocity is increasing But deceleration is the acceleration Opposite of the initial velocity After sometime velocity will become in the direction of deceleration only Then it will keep on increasing Okay So downward Downward opening curve will always be represented as deceleration Write down somewhere that this is the negative acceleration Okay, this is how you represent deceleration in xt graph Now Same way This one Same way This one Is what? Oh that way you're saying that you're throwing some ball like that and then it comes Yeah, you can say that Correct This one A B C Now you can see here That right now velocity is negative The slope is negative Fine Then the velocity becomes zero at point B Just like in the earlier case And then velocity becomes positive over there Because angle is between 0 and 90 Okay, so you can see here that acceleration is always in the positive direction Because finally the positive direction the Positive direction the velocity has been reached So acceleration you can say is positive here If you say negative acceleration is deceleration Then you must say that positive acceleration is acceleration Okay, all right. So just one more graphical thing Now what we are doing is numerical only So this is not part of any theory as such But I know that there will be a struggle if you don't understand these things So let's say this is No, no, no, it need not be constant acceleration Do not confuse it It depends on what kind of curve it is Okay, it depends on what kind of curve it is All right, listen here everyone Suppose you have a situation like this Like that This is object one that is object two Okay, can you describe their motion compared to each other What is happening Will these two objects meet after some time These two objects will meet after some time or not If two objects meet at a point then what should be same What is the condition one object meets the other What should be the same Time should be same and what else My position should be same location Just because velocities are equal Does it mean that they will be meeting at a point Two objects are going somewhere One goes like this other is going 50 kilometers away Both of their velocities are equal Does it mean that they meet No, right Their location should be same at the same time Does that information this graph gives It doesn't give that information So you can't infer it You can't say that object two has overtaken object one You cannot Just by looking at this Okay, what you can say is Object one starts from rest Object two has some initial velocity What else you can say Anything else Object one Starts rest Object two has initial velocity What else Object one has greater acceleration How can you say that it has greater acceleration This angle is more than that angle So tan of greater angle will be greater So object one has greater acceleration Compared to object two Getting it clear to all of you And that's all we can infer that's all Shashva that is a wrong statement First acceleration is moving at faster pace When you say pace always means velocity Don't say pace with respect to acceleration Okay, can I say that object one is behind object two Can I say that I cannot say But if object one is behind object two Then object one will overtake object two after some time That time you cannot determine looking at the graph This is not the time when they will meet This is the time when their velocities are equal That's all Okay So quickly few conclusions then we'll move ahead If it is a constant velocity how the xt graph looks like xt graph constant velocity Straight line Straight line The good thing about the straight line is that You know The slope of the tangent at every point is that straight line only Okay, so straight line Is a constant velocity This one is constant velocity Constant acceleration In velocity time graph how it will be Straight line Right A straight line Like this Okay Now can I extend this line further What does it mean Can you describe the motion for this situation Object started going backwards Fine So you can see that Velocity is positive here and then it became negative It is going in this way Velocity became zero and then started going backwards But on graph you can see that it looks like one Continuous one straight line Fine So yeah Sometimes graph will be Not so intuitive also This is constant Acceleration Okay Now let us see how displacement Can be identified From vt curve Vnt curve Okay, there is always this confusion Acceleration or deceleration So it is acceleration or deceleration Okay, just throw this thing from your head That acceleration and deceleration are two different things Okay, deceleration is also an acceleration only But negative Okay, so always keep that in your head But sometime in question if they say deceleration Just take it as negative Take it as in opposite direction of initial velocity That's all It is also an acceleration but in opposite direction Don't confuse yourself Yeah, yeah That is fine In second case it is negative acceleration Because slope is negative But it is acceleration only Okay, so I was talking about displacement Write down the displacement from this Suppose You're going like this This is velocity This is time All right And you have reached till a point where time This time is t equal to t0 Okay So what is a total displacement? Mathematically, what should I write? Displacement in this case will be what? Consent velocity this is So V into t0 V into t0 So graphically how does it Represented Area of the graph Right V into t0 is the area Under VT graph Okay So area under VT graph is the displacement Now we have identified that is true for constant velocity What if velocity is not constant? Let us say It is like this It's going like this This is velocity And this is time Can you prove that The displacement in this case also Is the area under this curve Can you try proving that? Try proving it You don't need any integration or anything of that Done Write area run Fine So I think you have covered little bit in mathematics also If you take very very small width of time Let's say this width is very small Let's say delta t Okay So can I say that area of this is represented by the displacement in delta t This zone can be considered to be a straight line Horizontal straight line So I'll say that for delta t time The velocity is almost constant only So V into delta t is the displacement in delta t time Displacement in delta t time Like that I can keep on adding all the displacements If all these displacements If I make smaller smaller rectangles like that I can keep on adding them And I'll get total displacement Yes or no So total displacement will become Equal to the area of graph here also Area of graph Here also is the Displacement Okay So remember these basic things Have you done this in class 10th sorry class 9th Whatever we have just done No problem So we are going to start the next section of the Chapter which is equations of motion Okay Write down So you know that we are doing the graphical part And the algebraic part together I'll again show you the classification which we have done We are doing algebraic and graphical analysis together Calculus approach will be doing at the end Fine and there's one more actually small topic Which is relative velocity that also will be doing At the end write down equation of motion Now equation of motion If acceleration is zero Is very straightforward Isn't it Velocity into time is distance Average velocity is total distance by total time Only these two equations are there And it is simply you can find out You don't need to worry much about it So that case we are all we have already done in a way So case number one Acceleration is zero That is manageable Case number two Acceleration is constant Constant acceleration Okay, let us see how we can deal with this scenario Okay, by the way These equations of motion you already know right So what these equations of motion Let you find out I mean why these are important Why these equations of motions All three of them Why are they important so important Anyone Fine So basically why the equations of motion All of you are correct only the way you are saying But the clarity is missing little bit So equations of motion Actually helps us Helps us find out future In future what will happen Based on present scenario Okay Based on what is happening right now You can predict what will happen in future That is very powerful tool Right Because future you haven't seen So you can just modify the present variable You will be knowing how much it will impact in future What is the impact of it later on So if you change the acceleration You will know what will be the final velocity If you change the initial velocity You will know after five seconds What will be the effect in the displacement So things like that So basically equations of motion Are always right down It always applied between between two points in time So before you even think of applying equations of motion You should first think Between which two points are applying the equation Because you always need two points Initial and final Okay So initial and final sometime will be given in the question itself That initially this happens and finally that happens But at times between initial and final You can have any number of initial and final What I'm trying to say is Suppose initial time is equal to zero Finally what happens is equal to 10 seconds Okay So you can say that my initial is zero Final is equal to one second My initial is equal to two seconds Final is equal to five seconds So between A and B You can any two points you take And you can connect them By using velocity, Acceleration, Time and displacement So four variables you can use To connect the two points in time So that is the importance of the Equation of motion Fine So now let us consider case number two graphically Where accretion is constant Is this a plot of constant acceleration Between A and B It is a plot of constant acceleration Right And yeah On the y axis velocity On the x axis like always time Usually time is always on the x axis Velocity time graph Yes Okay And and it has Initial velocity also At point A At point A At point A Let's say initial velocity Is U Fine At point B The initial velocity Is let's say V Rather than writing A and B Let's write one and two In clarity one And this is point two Okay You need to tell me If let's say from here to the time is T Okay Let's say T is the time You need to tell me the total displacement The object had Okay From point one to point two All of you Graphically it is represented by the area Right So find out that Area How will you find it It's a trapezium You can as well divide it into a rectangle and a triangle That is also fine Just find the area Area will be the displacement We have just learned about it This is let's say area one This is area two Displacement Will be Graphically equal to some of these two areas A1 plus A2 What is A1 equals to UT This is S A2 is what Half of base is T Altitude is what? This much is V minus U Right So V minus U A into T This is the equation All right This is the equation of motion itself No problem with it But sometimes The constant accession is given Let's say accession A is given to us So can I use accession In this equation So if accession is given How do I write accession to be equal to A is equal to what? Between point one and two Constant accession is A How to write A A is equal to what? A is Average accession is Changing velocity Divided by time Is what we have studied Right? So accession is V minus U Divided by T So from here The first equation of motion You'll get it as V is equal to U plus AT Anyone has any doubts till now? Anyone? So from here You can see that V minus U Is equal to A into T Right? So V minus U Is equal to A into T That you use here And you write S equal to UT plus V minus U is AT So it becomes AT into T Which is AT square So you have eliminated the final velocity From this equation From this equation You eliminated final velocity And you have only S Two equations of motion How these two equations of motions are different With each other In first equation There is no displacement term Okay? There is no displacement in the first equation In the second equation There is no velocity Getting it? I want you to derive the third equation of motion In which there is no T So find out that I mean you know what it is Derive it Find out Eliminate the value of T Substitute the value of T from the first one On the second one You'll get it Let me know once you're done Done? So just substitute the value of T To be V minus U by A In that equation You'll be able to find out That V square Equal to U square Plus 2 As Fine? This is the equation You have three equations of motion Remember that these three equations of motions Are vector equations Fine? So if they are vector equations You need to take care of the directions As well And how you take care of the directions Everyone? How we take care of the directions By using signs The component you have to use in motion in 2D Motion in 2D only Caused a sine theta will come But here everything is along one straight line Okay? So if it is in one straight line Only in one direction Or in opposite direction Two options are there So one option is positive The other option becomes negative So by using the signs You can take care of the directions In motion in 1D at least Okay? I hope it is clear So we'll take a break now Everyone? Or is there any doubt? Clear? So we'll take a break Right now it is 6, 0, 2 We will meet at 6, 17 Oh, oh sorry 4, 0, 2 Usually I have class from 4 to 7, 30 So I give them break at 6 So psychologically I have just saved 2 hours This is your break All right guys So I hope all of you are back Are you able to hear me? All right What happened Arunoff? What happened? Anyways All right So let us continue from where we have left We have just derived the equations of motion Three equations of motion Graphically Okay? So since we are talking about derivation Let us see How else can we derive them? Okay? So I am assuming the basics of derivatives And basics of integral You are aware of Okay? If not You will be slowly anyway When we solve questions You will get used to it So acceleration is constant Right? So acceleration can be written as dv by dt Okay? So from here dv is equal to a dt Have you done basics of integral? Everyone? Little bit of integral have you done? In bridge program you have done Bridge? Okay great Nice Yeah yeah little bit only I require Little bit only So Akhil sir has done it Very nice So let us continue So you can integrate this A is constant So left hand side I am going from initial velocity to the final velocity Time is from 0 to t So what will be the integral of this? How to integrate it further? Anyone? Are you able to see my screen clearly? Screen is visible Shashwat The problem is probably the internet speed at your end Okay? It's like YouTube If your speed is lesser The resolution goes down automatically So you can rejoin Or you can see that somebody else If they Else is using the same broadband connection Ask them not to watch videos on it Then bandwidth will be more available for you All right so now you can take A out of the integral Because A is constant Fine? So you will have integral of dv What it will be? V only? And then you substitute the limits V minus U Upper limit minus lower limit This is equal to A into t From here V is equal to U plus A t Got it? You have done indefinite But there is a definite integral also You have to just simply put the limits For example Over here integral of dv Is V But if you have limits like U and V What you have to do is First substitute the upper limit In whatever integral you got Minus the lower limit So this is how you substitute the limits When you integrate That is the only difference between The definite and indefinite As far as physics is going So all of you have understood this one All of you understood please tap in quickly Then I will go forward Okay so V is equal to U plus A t How? U is constant or it changes Everyone Is U a constant in each velocity? I don't know I can't make everybody wait for 2-3 minutes Now then you watch the recordings Let's move forward You can anyways watch the recordings Is U a constant or not? Initial velocity U is constant V is the velocity Which depends on the time Whichever time you are finding out Fine So I can write velocity V as Dx by dt Yes or no? Everyone Right? So dx will be equal to U dt Plus A t dt Now you integrate this U is constant comes out of integral Like this 0 to t 0 to t And let's say 0 to x All of you do this And let me know what you get Are you getting U t plus half A t square? All right so integral of dx When x goes from 0 to x Is x only Integral of U dt is U t And A t dt integral is half A t square Like this See t dt Integral is t square by 2 Then you put the limit 0 to t It will be t square by 2 minus 0 And then you know you just eliminate t from these two equations You will also get V square equal to U square plus 2 As Fine So we can derive the same expression By using the By using calculus or by Doing it graphically also All right Now is the Main part Knowing the equation is not enough Knowing the equation is just 2% of the work 98% is how to solve problem Okay So now we have known that there are three equations of motion Constant acceleration There are three equations of motion So we'll be taking the constant acceleration scenario And see how we can utilize these three equations So can you tell me the most common case of constant acceleration Common case of constant acceleration Gravity any object near to the earth surface Right any object near to the earth surface And it is experiencing only gravity force let us say Only gravity force is being exerted on it Know the force then it is a constant force So constant acceleration Fine So motion under gravity is a constant acceleration case And we have done it In projectile motion also That was much more complicated than this straight line motion So how will be the motion in straight line When it is we then when we are talking about the gravity Anyone How does it look like Give me one scenario in which motion in a straight line happens When it is under gravity When you drop a ball When you throw a ball up When you're throwing the ball down Okay in all the cases The acceleration will be constant And it is motion in a straight line So let us solve questions now Because that is what the mechanics is all about Mechanics is not about theory So let us take few questions I'll be giving you questions now Write down Train is traveling right now with 20 meter per second Okay, this is the velocity of the train It takes a U turn U turn in 20 seconds Without changing its speed Without changing speed So in this process of taking a U turn What is the average acceleration Find out What is the average acceleration in this process of taking the turn Sorry Straight line Everything is straight line Whenever I give a question here It is straight line only Okay And train will be traveling in a straight line usually right Anyways Usually I mean it can take a curved path also Okay No U turn need not be straight line It can take a turn like that And take a curve bend and then finally it is back All right So I can see some of you have got 0 also as answer Is velocity changing or not Has velocity changed Speed is not changing Has velocity changed Velocity has changed So there should be acceleration If velocity is changed there should be acceleration I am talking about average acceleration What is the definition of average acceleration Final velocity minus initial velocity divided by time It takes full 20 seconds for completing the U turn Final velocity If you take the initial velocity V1 To be equal to 20 meter per second What do you think final velocity is V2 is what 20 only Minus 20 It changes the direction It changes the direction Direction is accounted by the sign of the value Okay So V2 is minus 20 minus of V1 again minus 20 Divided by T which is 20 So you are getting it as minus 2 meter per second Square Okay This is the answer I hope it is clear to everyone it is not zero Acceleration is not change in speed by time Change in velocity by time Direction taken into account All right next one is this Particle is moving with uniform acceleration Uniform acceleration Okay From A to B Along a straight line It has a velocity of V1 And V2 At A and B respectively Fine C is the midpoint Midpoint between A and B You need to find out velocity V3 at point C How much it would be No that's not correct Anything you do Anything They there might be multiple ways you solve the problem It is not like your class 9 10 10 Fine There are multiple ways So you have to keep your mind open about it And see how will you approach Okay I can see if you have answered Multiple answers people are giving Fine I'll solve it now I'll solve it Let's say You know you do not know the distance between A and B But what you do know point C is midway So you need to use that information Okay so let's say If you have to use the information You need to assume the distance is x And then C is x by 2 from A A to B is x And this is x by 2 And this is point C Fine So I can say here that V2 square Is equal to V1 square Plus 2Ax This is my first equation So excretion remains constant throughout A to C, C to B, A to B, everywhere And I can say that V3 square Is equal to V1 square Plus 2Ax by 2 Okay so from equation number 2 I can write down V3 square Is equal to V1 square Plus A into x Plus A into x Now if I multiply by 2 Throughout I'll get this And the first equation was Anyway V2 square equal to V1 square Plus 2Ax And then I subtract these two equations 2Ax is gone Which is not known Getting it? So you'll get the answer of V3 To be equal to root over V1 square Plus V2 square divided by 2 Anyone has any doubts here? All right next question This one Object moving with constant accretion Constant accretion is A Initial velocity is U To find out the distance traveled Distance traveled In nth second of the journey Nth second of journey So you need to find distance traveled In one second only But that one second Is the nth second Fine So how much distance This object travels in the nth second Okay at T equal to 0 Its velocity is U Find out Not asking formula here Formula I know I am asking you to derive it in a way No no no that is Sorry Anurad Yeah yeah yeah That is That is what When you say that distance Is equal to Un Plus half A n square All of you tell me What does imply What does it mean What is this S Right now Distance Traveled This is the distance From T equal to 0 To T equal to n seconds Total distance it is I am asking you to find out at Nth second of journey What is the distance Nth second is what Between which time and which time Anyone Suppose I want you to find out the Distance traveled in The fifth second of the journey Fifth second What it will be Distance traveled till Distance traveled till five seconds Minus distance traveled till Four seconds Is that The distance traveled in fifth second Is that the distance traveled in fifth second Distance traveled This is nth At nth second Till n second this is a distance Okay till n second Till n minus one second What is the distance U into n minus one Plus half A n minus one Whole square So can I say that Sn is This is a distance traveled Till n second So distance traveled in the nth second Can I say it is Sn minus Sn minus one Is this correct All of you should answer Can you find out how much it is Hold on Wait Hold on a second Tell me That Fifth second Starts T equal to what T equal to five Till Sorry T equal to Fifth second Fifth second of journey Starts from T equal to four To T equal to five right It's the fifth second starts After fourth second is over So Then it will continue till Fifth second gets over Okay So this is one second gap So that is why Snx will be equal to this All of you quickly tell me what is the answer Rearrange the term and tell me what it is All of you Are you getting this U plus half A Two n minus one All of you are you getting this So if you're getting this This is what it is So some people consider this itself As the fourth equation of motion Fine so you can treat it as if it is Another equation of motion Apart from the three equations which you already know Fine so for example this particular question which I am giving you You can solve only by using this particular formula Okay No doubts right All of you have understood this one Everyone Here is the question right The question is A body Covers a distance of 20 meters 20 meters is covered In seventh second Okay 24 meters Is covered in ninth second You need to find out how much distance it will cover Distance covered In 15th second All are just one second time periods But which one second that is different Do it Okay some of you already answered Answer is correct Those who have answered Till now are correct No Sridhan Check the silly errors you have done probably Direct application of formula S nth is U plus half A Two n minus one Use that A is not known U is not known So from first two conditions Get the value of U and A And then use it in the third So like see the answer is 36 Okay No shortcuts nothing No nonsense Okay You can apply A P G P E H P But there should be proper logic A solid logic Do not Do not look at the answer and then say that Oh I can take geometrical progression Okay Even if that may result in the correct answer But if there is no logic then that is useless Okay It will destroy your logics When you write in the exam All the exam when you write If you start assuming a lot of such things It will never be clear in your thoughts So you will not I mean forget about J advance then J advance then out of picture Forget about it J image you can do little bit here and there If you start assuming Why I am telling you all of you right up front Because once you start developing this habit Starting started to assume Creating your own laws and rules You're you'll be filled up with Lot of nonsense in your head Then what will become more important Is throwing these nonsense from your head away Rather than putting something new in your head Okay So these nonsense will start destroying your chances Keep your head clear SNH is U plus half A 2N minus 1 Right So using the first condition 20 is equal to U plus half A 14 minus 1 13 Second condition 24 should be equal to U plus half A 2 times 18 minus 1 17 Okay What I want is S which should be equal to U plus half A 15 to 29 Right 29 I want this So just solve this equation Get the value of U or A And then substitute here You'll get the answer Or if you have a better way of solving the equation Directly getting the value of this expression itself Then that is different Okay So if you rearrange U and A such that it becomes this Then left hand side will be the answer Okay, but if that is not possible Get the value of U and A And substitute here All right So let's move forward Now next thing is Motion under gravity Motion under gravity is another example Of constant acceleration only So we'll solve Four or five questions on it All right So here is the thing Once you have attended the bridge program You'll find out this chapter very very simple So let's say you are throwing an object from here This distance The height of the building is 100 meters Okay You're throwing an object with 10 meter per second upwards You need to find out when it goes up And then when it comes down It reaches the ground How much time it will take to reach the ground All of you Take GS10 No no it is not projectile motion Projectile motion will be curved It's motion in a straight line It goes up in a straight line comes down But yeah the way you solve it Very similar to the projectile One answer I've got others Which equation will you be using s equal to u t plus half a t square you'll use Right because displacement is given to you Initial velocity is given to you And you're asked what is the value of t What is the value of s What is s In this case What is your initial point and final point Which two point you're taking initial and final This is your initial point And this is your final point Between these two points what is the displacement Which direction you're taking positive Up or down So displacement is minus 100 You all forgot projectile Displacement is minus 100 between i and f It can go up to any height But it has when it comes down displacement is Minus 100 only Okay U is what What's U U is plus 10 10 t Plus a is what Minus 10 Into t square Okay So this will give you the answer for t This will create a quadratic equation You'll get it as 5 t square Minus 10 t Minus 100 is equal to 0 And then simplify it further t square minus 2 t Minus 20 is equal to 0 So this is the quadratic equation you'll get Solve this you'll get the answer Okay And yes Sometimes we have this tendency that Oh it goes up Then comes down So I'm sure some of you have done like this Then like okay fine I'll take first between this point and that point Let's call it as g I'll first write down i to g How much time it takes Then g to f What is the time Total time will be t1 plus t2 Some of you might have tried like this You can do like that also You'll get the correct answer But why you're complicating it Right s equal to u2 plus half t square You can directly apply between Initial and the final point Getting it? Final answer you can solve this quadratic equation Okay that it will be All right next up Any doubts anyone? Anyone has any doubts? No doubts doubtless We'll proceed Next question You are in a balloon It is This balloon is moving up with acceleration of 2 meter per second square Initially the balloon was at the ground Okay It is continuously accelerating 2 meter per second square upwards fine After 5 seconds After 5 seconds You have dropped a ball You need to find out how much time the ball will take to hit the ground Yes Shash was got something Others All right so some of you are getting root 5 as the answer Let us see whether that is correct or not Okay someone is getting root 2 also Most of you are getting root 5 Let us see We'll solve it now So tell me Tell me when I am When I am when I am dropping the ball while I am in a balloon What will be the situation for the ball? Will it start from rest? The velocity initial velocity of ball will be zero What what will be the situation? Initial velocity of the ball will be zero Everyone type in yes or no Initial velocity of the ball will be zero right? Initial velocity of the ball is zero yes or no Remember inertia? Inertia of motion Okay so here is the pole result Initial velocity of the ball will not be zero Okay I mean with respect to the person who is dropping it That is zero with this Suppose I am dropping it I will feel as if I am dropping it from the rest But in reality it will move with my velocity Which velocity I am going up with that velocity only With that velocity only the ball will move up Fine the situation is like this Suppose there is a plane that is moving forward Right you drop something It will not have zero velocity If it has zero velocity it will drop vertically down It has a velocity of aircraft That is why it goes like that Same thing here also Okay so After 5 seconds when I drop the ball The velocity of the ball is the velocity of balloon And velocity of the balloon is what? Use v is equal to u plus 80 for the balloon Acceleration is constant 2 meter per second square So velocity is 2 into 5 10 meter per second This is the velocity of balloon When I drop the ball The ball has initial velocity of 10 meter per second upwards Okay I need to find out what is the height balloon has achieved Also because this is what the ball will travel Okay that you will get from s equal to ut plus half 80 square for balloon So h will be equal to ut zero plus half into 2 Into t square 5 square which is 25 meters Okay so h is equal to 25 meters Understood all of you Okay so this is related to what balloon is doing Because there is a relation between what balloon does And what the stone will do So that is why we first found out what the balloon will do Before you drop the stone Now stone is dropped What is the acceleration of stone? Will it be same as balloon? Acceleration of stone will be how much? First tell me acceleration of balloon Sorry not the balloon Acceleration of the stone how much? Acceleration of stone 2 plus 10 minus 10 to what it is When I have dropped the stone after dropping what will be its acceleration Okay so here is the thing guys The inertia happens only for the velocity It doesn't happen for the acceleration As soon as I drop something The acceleration will not be same Okay, acceleration will be because the force acting on that mass So when I have dropped the object it is under the gravity only And if only gravity force is applied Acceleration is always g G is the acceleration downwards So I'll say Acceleration is minus 10 meter per second square downwards Velocity will be velocity of balloon But the acceleration of balloon will not impact anything on the acceleration of this stone Because there is no such law You are creating laws yourself The law is only for the velocity An object that is moving with constant velocity Will continue to do so until there is an external force applied on it Fine So as soon as you drop it Immediately after it tries to move with the same velocity at the balloon There is no such law for the acceleration All of you clear that Acceleration will be minus 10 meter per second square For the stone as soon as you have dropped it Everyone please type in yes or no Is it clear? Okay, so somebody was asking me why the velocity of stone is 10 So I think I have repeated multiple times I'll tell again This is because of inertia When you drop something from moving Station when you yourself are moving and you drop something Immediately after dropping that object will try to move With the same velocity with which you yourself are moving So that is why the velocity In each of the velocity of the Stone which you're dropping Is the velocity of the balloon Which is 10 meter per second upwards Getting it? All right, now we need to find T So what I will do? I have the displacement 25 meters Right, so I'll use this s equal to ut plus half at square Right So displacement is This is my initial point Point number one Final point point number two So between these two points I will be using the Equation as you do ut plus half at square s is minus 25 U is 10, so 10 into t This will be minus half into 10 into t square Okay, so this is 5 t square minus 10 t minus 25 Is equal to zero Or Further change it t square minus 2t Minus 5 Is equal to zero So this is a quadratic equation And solution of this quadratic equation is the answer Everyone has understood Please type in yes or no Let me know if there is any doubt quickly Very important question this was Okay, I'm going to the next question Let's take one graphical question Suppose There's a plot between x and t Okay And This is how the graph is So Not like this exactly Sorry about it Okay, take care whatever it is Now suppose this is a point one Point one this is Okay You need to find out Find out another Point p on the graph another point two on the graph So that The average velocity Average velocity between one and two Is equal to instantaneous velocity at two No mathematics nothing Just think you'll get the answer Think it in a straightforward manner Done Anyone Anyone got it anyone Should I do it now I think couple of you have got it already probably Okay, I'll do it So if I take the second point in such a manner That at that point the chord is tangent So the slope of this line Is what instantaneous velocity at point number two Because it is tangent at two There's also the average velocity between One and two because this is called Between one and two Fine So you need to draw a tangent from point one To whichever point the tangent could be drawn So if it is tangent It is slope of that line Is the instantaneous velocity at point two And slope of that line is also The average velocity between one and two Clear I hope clear Clear and is it clear So let's go to the next graphical question You know there are any guesses how many varieties of questions can be there More than a million Okay You can spend your entire life Just mastering one chapter Still when you are 90 year old I'll give you a question you'll not be able to solve So there is no limit up to Where the difficulty can go So how will you know That a chapter is Over from your side How will you ever know that you are done You can spend your entire life Learning it Right So it will not be like your class 9, 10, 10 that You should be able to get all the questions Then only the chapter is over If that is a mode the chapter will never get over So you should be able to solve No, no, no That is not the mode for now So you should have It is a more of it is sense Whether you have a sense of you whenever you are solving Let's say if you solve 10 difficult question You're able to solve 8 out of 10 I'm talking about higher order questions If you're solving 8 out of 10 correctly You should move ahead to the next chapter Okay So that should be the mode Let's go to the next Draw this On the y-axis we have velocity On time axis we have t Okay This is t is 0, 2t This is 20 seconds The this end is 20 seconds All right What else Velocity time graph is given as shown Rate of acceleration and deceleration is same And it is equal to 15 meter per second square Rate of acceleration and deceleration is same Which is this Okay, if average velocity during the motion Average velocity is given Average velocity is How much 15 meter per second Okay from t equal to 0 to 20 seconds If the average velocity is 15 meter per second You need to find these things The value of t Then maximum velocity How much the distance traveled Distance traveled with uniform velocity With uniform velocity Once you are able to get all the values Then only tell me the answer Permissions and maximum velocity Everyone Enough time is given now Get the answer You know what something As I tell me one by one What is the answer for the first part Everyone You know why you're repeating yourself What is the answer for the first part Did not get Okay That is strange Nobody is getting it So average velocity is given as 15 meter per second Average velocity you have to use The answer for the first part is 5 seconds Okay Average velocity Total distance by total time Total distance is what The area of this The area of that is the total distance So how to get that You will have half of Half of See what is this distance This one This is what 20 minus 2t 20 minus 2t it is Isn't it So sum of the parallel side 20 plus 20 minus 2t Into distance Into the distance between the two parallel side What is this distance Everyone How to get that distance This one Pranav Do it yourself That I have changed the value of acceleration It's not the same as given in the books Anyways What is the distance between the parallel sides How to get that Over here What is the value of velocity Is that the distance between the parallel sides or not What is the velocity here How to get that V is equal to what We'll be using v is equal to u plus a t A is given as 15 Right So v is equal to 15 into t Okay 15 into t Fine So Pranav in the book It was Given as a equal to 4 meter per second square If you use a equal to 4 meter per second square Yes, you'll get t equal to 5 seconds But over here That is not the answer Okay Sir This in Hold on Hold on This into 15 into t This is the area All of you understand This is the area Right This is the area And this is the distance covered That divided by the total time Total time is 20 This is equal to 15 meter per second So anybody solve this What do you get t equals to Arnau you're saying something So So what is the distance On the first parallel then B2t minus t which is t So You wrote 20 minus 2t 20 minus 2t right So from here to here it is t From there to there it is again t only Both are t Because they have to maintain symmetry It stops finally here So this is 20 minus 2t So and anybody solve it to get the value of t Anybody So what I did was I took the average velocity in a multiple way Into total time And then I got total distance covered In doing that time And then I found the area under the graph and I Solved No If you use this acceleration You'll never get t equal to 5 Okay So what is the value of t Anybody solve it to get the value of t Try solving it and tell me what is the answer Pranav now tell me what is the answer Solve it Sir I have a doubt Yeah tell me Isn't 20 seconds the same as 3t because t then t Then another t I think he meant 4t Because t then it's t to 2t So it's B2t Yeah t to 2t is still 1 T to 2t Let's 2t plus t it's got 20 Alright hold on Sorry this is not correct 2t is a time from here to here Okay t is a time from here to here It is marked Okay from here to here to t So total distance is 20 So 20 minus 2t Um Ah sir then 20 is 4t Then you get t is equal to 4 seconds You just divided by I'm sorry 5 seconds Divided by the Anybody what is the answer No even this should be t okay They have you have to maintain symmetry Even this should be t Accurations are same both sides Amount of time it will take to go from here to here It should be same as to go from there to there This is t this is also t So so you get That is 2t so Equal to t so you get 20 is equal to 3t Yeah that's a yeah Yeah so you get 20 by 3 Hariar and hold on So Hariar and listen here From here to here 20 Yes sir This much is t this much is t So remaining is what They have given no the whole thing is 2t they have given right So you want to get 2t plus t equals 20 Yes sir that's all Want to get what 2t plus t equals 20 seconds Like the difference the individual Correct correct 2t plus t equal to 20 so I mean that's nice You directly get the value of t So t plus 2t is equal to 20 So t will come out to be 20 by 3 That is fine So tell me how will you get maximum velocity Where the maximum velocity will be Which point The top most where the velocity is constant So this is this is the maximum velocity on the graph It is clearly visible That this is the maximum possible velocity isn't it Distance travelled by the travelled in uniform velocity Will be simply equal to whatever is the velocity Modified by 20 minus 2t Okay fine so how much time we have We have five minutes we can take another question Okay this one all of you Car accelerates from rest at a constant rate alpha Okay it accelerates From rest rate of alpha Okay for some time for Some time and then after acceleration After which it decelerates then deceleration happens Okay decelerates at a constant At a constant rate of Beta to come to rest getting it The total time is given as total time consumed is t seconds Fine you need to find out these things Maximum velocity the total distance covered Total distance Do it What is the point of maximum velocity where it will be First tell me that where it will be maximum velocity Just before deceleration Just before deceleration Where the acceleration has stopped Acceleration will keep on increasing its velocity As soon as acceleration is gone Velocity will be maximum there You have to find that You need to first understand what all things are given to you Initial velocity is given Final velocity is given In between acceleration is changing its magnitude So it will be better if you draw the graph Velocity time graph This is how the graph will be right So from from zero to let's say t1 It accelerates from t1 to t2 It decelerates Any doubts in the graph This is the maximum velocity Now do it total time is given t We have to use that somehow Sir is maximum velocity alpha t plus alpha square t by beta minus alpha Why are you announcing the answer You can just type it out Everybody else is also doing it Type it out whatever answer you get Put it on the message Don't disturb others Everybody See between point A and B constant acceleration Between B and C Again constant acceleration So you can use the equation of motion between A and B And then between B and C That is a hint Total time is given to you So if the time from here to here From there to there Is t2 Oh wait This is t Okay So if this is t2 from here to here It will be t minus t2 This is t2 Okay Now can you do it So here let us say This is the initial velocity So here the velocity is 0 Okay So let us try to see How we will get the relation Between A and B The relation will be velocity of point B And let's say v Okay so v square equal to u square No not u square v is equal to u Plus at this I'll be using between A and B So v is equal to u is 0 And accession between A and B is what Alpha So alpha into t2 This is my first equation Now between B and C I'm using the same relation This one only Now final velocity is how much Everyone What is the final velocity Final velocity is 0 Right So I'll be saying v is equal to 0 u is equal to v only v Minus now Because there is deceleration Beta times what What should I write here t minus t2 t minus t2 Okay see again there are few Whenever I ask something Okay Until is there is a discussion Or there is a doubt from your side Please do not speak Then others Why others are attending classes If you tell the answer to everyone You should type it out Arunov understood Arunov Have you understood Yes sir I'll type it out Thank you for that Just few seconds back only I told And again you You're doing the same thing Right so Okay so now u is equal to Now v is what v is alpha into t2 Right so alpha into t2 Minus beta t minus t2 Now from here you get the value of t2 Isn't it So t2 will come out to be equal to what t2 will come out to be beta t divided by alpha plus beta Yes or no This t2 So how will you get maximum velocity Just substitute over here Alpha into t2 Is the maximum velocity So v is equal to alpha beta divided by alpha plus beta Into t I hope all of you have understood how we are Solving it Everyone understood Type in quickly Right guys so you know The thing is that total distance You can just find the area under this Area of that will be the total distance Which will be half b part It will be half Into base which is t Into maximum velocity which is alpha beta divided by alpha plus beta into t So this is your Distance Altitude Which is maximum velocity which you already found out Half of that Into the base So you get the maximum total distance also Fine Okay guys so we are Done with the first part of the chapter And you know There are still few things that are left from this chapter Those few things I'll just tell you Relative velocity Is one which is Which we haven't done yet And the calculus we haven't done yet Fine And yes we haven't done the Variable acceleration cases also But these are the small topics It will get over in an hour or so Theory part of it And next class We will be doing at least two hours of problem practice Only problems So do you want Higher order questions like Advanced level questions Or basic levels What what Because today we haven't done Any advanced level frankly We have been limited to Mains neat or below that level also What do you want Next class after we are done Okay so what I'll do is that I'll mix it up 50% will be 50% Logical reasoning I mean frankly speaking Those are useless Fine Logical reasoning should be developed From Solving questions If you solve questions your logical reasoning should develop If you try to read it like a theoretical questions You're wasting your time Okay you're wasting your time That is how frankly someone can tell you Okay if someone is telling you that You can develop your logical reasoning By reading the books theory again and again That person is fooling you Okay if somebody tells you there is a shortcut Again you are getting fooled No shortcuts No learning of theory All that is useless What you should do is that Immediately after the class You should start solving questions Don't try to just read the theory again Whatever we have done Okay start again reading the theory And from different books you keep on reading theory And just remember all the garbage in your head All that is going to be useless You need to develop Problem solving aptitude How to think openly You can see in this question itself Do you think that your knowledge of equation of motion Has to do anything with solving this question No, it has nothing to do with we are using v is equal to u plus 80 only Right But someone will be able to solve this Someone will not be able to solve this The only difference is that That someone who is able to solve this Has done lot of problem practice Without looking at the solutions directly So he has undergone lot of struggle Frustration and able to develop the problem solving aptitude Okay, so I am not answering only to Shashwat I am telling to everyone Okay, so make sure that You solve problems Mark my words 95% of preparation is solving problem Only 5% is theory Okay, so guys that's it from my side We will meet again next week And complete this chapter And yes, there will be assignment given to you That is the minimum possible work that you have to do As I told last class Bye for now Here's Hariharan Correct Minus b Minus beta Did I write plus beta If yes then correct it


Christians can be attacked, deceived, tempted, and influenced by demonic beings, so Christians often do need deliverance from strongholds, addiction, and torment. But what if Satan's greatest lies is that Christians can be demonized or full-on possessed? This lie keeps the believer distracted, trying to fight against demonic ownership when they should be focused on overcoming demonic deception. This is why many believers get stuck in the constant cycle of spiritual defeat. The Scripture declares in 1 John 5.18, We know that God's children do not make a practice of sinning, for God's Son holds them securely and the evil one cannot touch them. Yes, Christians can be attacked by the enemy but never owned, for the believer is God's possession and the dwelling place of the Holy Spirit. For a more detailed and biblical look at this topic, subscribe to my channel and watch my teaching. Can Christians be demon possessed?


The Joint Forces Air Component Commander, or JFAC, has tasked your combat rescue helicopter squadron to stage out a pop-up airstrips, which are isolated, accessible, and hundreds or possibly thousands of miles from the main base. You can leverage this location to launch your aircraft even further into contested regions to support combat operations, and because of its very temporary status, lodging and communication infrastructure don't exist. Or the JFAC has ordered you to support an area roughly the same size as Boston, west to Albany, and south down to Miami, which is around 246,000 square miles. To solve these geographical problems, you pre-position your helicopters, maintainers, and extraction force all around the area of responsibility to more effectively support the service members most at risk, which changes every three to five days. In both examples, you have relatively safe places to stage on or near landing strips, but you don't have food, shelter, or connectivity to receive updates or intelligence. You do have C-130 support, and they've got a culture of if it fits, it ships, but you don't have any support once you get there. How do you most effectively and repeatedly make the mission happen? You adopt the concepts and techniques of the overlanding community. The overlanding community has spurned the development and production of the lightest, most durable, and most versatile solutions on the market. They've integrated these products on various versatile off-road platforms. The key to success is ruggedness, durability, versatility, quality, efficiency, and most importantly, value. These systems are self-contained, self-propelled, and ready for anything. The only modifications we'd need to make is the integration of compliant, compatible, robust and resilient communications and information technology systems, which are all modular and in the Air Force inventory now. By adopting the systems and solutions of the overlanding community today, we achieve the agility necessary to win now and tomorrow. The overlanding community says, when you take your overlanding system, there's no real destination. The goal is to become reliant on your vehicle for transportation and the gear you pack for shelter. Everything you need should be available within your system. You should be equipped to tackle all types of weather and any type of terrain you face and any challenge that comes your way. What overlanders do is eliminate the logistical barriers to their adventure. Airmen will use the same technology and techniques to eliminate the logistical barriers to projecting combat power forward to truly austere locations for short duration basing with the intent to project air power even further. What you're about to see is an example of an overlanding system. Jim McDonnell is a 13-year veteran of the United States Marine Corps. He's used to being out in very austere locations. Here's what his rig looks like. Here I've got a 6 by 8 foot awning, I've got a 2 person tent up top. I use solar to power up everything. For my comms, I have a CB, it's four wheel drive, lockable, differentials, but on this thing I can survive for a week for two people. Perfect for overlanding, if you had a four door Jeep, it would be four doors plus the trunk space and the top. That's going to be my next upgrade, the Jeep Gladiator.