So this problem here homework problem 1.9 let's start with part C of this problem part C of this problem we consider the following the the the object the crate here is moving with a certain initial uh velocity and that initial velocity is negative right so say negative 5.5 m/s and this thing will move but apparently there will be a turning point right because eventually what what's the final final velocity? Final velocity is actually passive, right? Final velocity is a positive quantity here. I didn't draw it to scale, but f val equals 11, right? 11 11 m/s. So apparently there's a turning point somewhere. So you you should picture in your mind like um what happened? What what's really happening here? And now let's consider carefully this thing will move right toward the left but at some point it will come to a come to a stop right so come to a like a momentary stop at that point we can call that velocity let's say that code v star v star equals zero right v star equals zero at that at that particular location and then you will continue moving toward the right so here I just uh I didn't draw it very rigorously but you know what what I meant by such kind scenario, right? This thing moves toward the left but it comes to a momentary stop and right after this moment you the directional velocity would will be switched to the right and the velocity will increase then all the way to here right something like this now if you consider the VT graph let's try to consider a VT graph what what does the VT graph look like if you have doubts you come come back to the uh consider consideration here V as a function of time and V as a function of time is say that it the V initial equals 5.5 right m/s and what about acceleration currently um there's acceleration but we are not sure yet but let's consider the vf final right vf final I didn't draw it to scale but vf final let's say it's here and vf final here equals 11 right 11 m/s so that means V final is somewhere at some final time f actually time actually we do not know right we are not sure of the time but let's assume that the final time is somewhere here T final here and the V final is here so that's the final point here and how would you get the acceleration you have some V initial which was negative right vf final is positive then how would you get this acceleration is acceleration constant or not it's constant right and how would you interpret the acceleration we know that it's a constant acceleration so it's that means it's okay to connect the dots right so your VT graph should look like this and you connect the dots and the slope will give you acceleration right so this slope will just give you acceleration it's a motion with constant acceleration right and now consider the following uh first They want you to figure out the just the displacement for the overall trip. For the displacement, what what is the displacement? So suppose let's come back to the so this is a real space real space. Suppose this location is your x initial and this location is your xfal. Then where's your displacement here? Suppose this is x initial. Right? And this is your xfal. So then how would you figure out the displacement here? Displacement is defined as xfal minus x initial. Right? So delta x is defined as xfal minus x initial. So therefore the displacement is just this one. This is your displacement. Right? This is your delta x which is denoted by xal minus x initial. And does the turning point matter when you compute displacement in part C of the problem? Does the turning point would the turning point affect your answer? Okay. So, so this is something we can think about after we finish part D right. So now it may not be clear but here I just want to say that when you consider displacement it's relevant to a process right that process but that process is only relevant to the final point and the initial point of the process it doesn't matter I mean the I mean the inter the intermediate uh procedure does not really matter. So that means if this thing right so this thing here means let me use this ball to represent the car the sorry the crate here so this ball will move right with a initial velocity then you come to a momentary stop right then come back to here how would you compute displacement so pay attention to the difference between distance and displacement so here if I wonder what's the distance from here to here then this this is the distance right and what's the distance from this point to the final point. This is the distance. But here if you only care about displacement, it doesn't mean the intermediate those that distance or that distance does not matter. What matters is the where's the initial where's the final and if you use the final minus initial then you get the displacement. Right? So now let's look at part C of equation. How would you solve part C of the problem? Just use the one of the four kinematic equations. You use let's use the fourth one. V final square equals V initial square then plus 2 * 8 * times the displacement. Okay, this part is just the displacement. It doesn't matter the turning point will not be reflected with this equation. And then you solve the problem like this. 11 square right 11 let's say 11 m/s you square it equals -5.5 m/s you square it then multiply by 2 * 3.65 6 by m/s squared then multiply by the displacement which is unknown right and this displacement uh last time I call it delta x prime because I try to stress out the fact that it's um it's different from par a and par bh so I mean the the the concept is actually different from par a and par b here but actually you will get the same answer if you call this delta x prime in this case you that you would you would this will give you such answer which which look like uh let's see final is the positive one right so then would be a positive displacement which look like 11 squared and minus here 5.5 squared so will disappear right because you if you have negative 1 square it become one so you can but you just keep in mind right 5.5 squared but this negative sign will disappear but there's another negative sign here right and divide by 2 * 3.65 then you figure out the answer. So that will be your displacement and you figure out once you figure out the answer you will see that this answer is the same as the answer in part A of the question right just the same answer. So actually we don't really need to call it delta s prime is basically you will get the same answer here but anyway so so that's the idea we have here. So so now that's part C. Now let's look at part D. What's the difference between part C and part D? Part D they want you to obtain a total distance. How would you obtain a total distance? Okay. So think about the following displacement. If you compute displacement it's a bit but if I want you to compute the distance what does distance mean? Remember that that is traveling toward the left from here to here and that displacement we can call that data X1. So we are working on part D right now. Right. So part C already finished. This is was part C, right? Let's look at part D. Part D, you should look at two two different displacement like one displacement was from here to here. So this one you can call that delta X1. Okay. And once you get this, last time I told you that take the absolute value, you get the distance, right? So that's the first displacement. The second displacement you consider this one from from this momentary stop that location all the way toward the end. That's your delta x2 here. Okay, you have delta x1 delta x2 and you take absolute value of both so that you get the total distance traveled. Okay, so part of the question you wonder what's the total distance traveled. Let me erase this. Okay, because I need some space here. So this is very important right? So although not many of you probably have learned about high school physics but this question is rather if you if you wasn't pay attention if you weren't pay attention to the the physics concepts. So now let's look at the part D of a question. Part of your question I told you the how would you get the total distance. So you grab the first displacement then take the absolute value then grab the second displacement then take the absolute value and get the sum then you get a total distance traveled right but why did I need to do that later on I will elaborate the meaning but now thing is how would you get delta x1 okay so consider the the trip okay the travel toward the left that displacement is called delta x1 so you call You consider this as V initial but you consider V star as the final right. So there therefore to obtain delta X1 how would you do that? You have V star squared plus sorry equals V initial square plus 2 * the acceleration time delta X1. That's how you obtain delta X1. Right? But and you will see that delta X1 is actually negative. How would you see that it's negative? Sub do the substitution, right? Once you do the substitution, you will see that you have 0 squar equ= uh -5.5 squared then plus plus okay plus 2 * 3.65 then times the displacement delta x1 right then once you figure out you'll see that delta x1 equals what? delta x1 will be equal to uh this thing. So you have a negative one squared. So that negative sign is gone, right? But there's another negative sign when you shift the terms. So it will look like this. You would like negative of negative of 5.5 squared over 2 * 3.65 like this. Okay? Remember there's a additional negative sign here. So that's that's why delta xy is negative. Does that make sense? So come back to look at this graph here. Okay, look at this graph here. Where's your delta x1? This graph here. Where's your delta x1? Delta x1. If you have some concept about calculus, you know displacement from the very beginning the turning point here is a turning point. Right? This time let's call that t star and that v star equals zero. Okay, this is the turning point velocity equals zero. So where's your delta x1 actually here the area under this line here is called delta x1. So delta x1 is it positive or negative? Negative area. Okay. So that makes sense. Okay. So this is a geometrical meaning. So doing the calculation itself is good but you also need to know the geometrical meaning on the VT graph otherwise the concept would not be very clear to you right so you are computing this it's a negative displacement and you take absolute value so that it becomes distance right oh so now let's let's figure out next how would you get delta x2 so this is delta x1 right later on you can take absolute value and put it here and then how about delta x2 There actually is a second process. So here you have V star equals zero. Treat this as an initial right and then this will become here right. If you treat that as initial then how would you write down the next equation? Next equation would look like the following. Next equation would look like uh val² v star 3 square plus 2 * the acceleration time. Okay, so that's how you figure out delta x2. Now let's substitute in the numbers, right? What's vinyl? So you 11 m/s squared equals star which is 0 squared and plus acceleration 3.65 uh m/s squared time delta x2 then you can obtain delta x2 right what does delta x2 look like? that x2 will look like 11 squared over uh 2 * 2 sorry three sorry three sorry three sorry three sorry three sorry three sorry three sorry three sorry three sorry three sorry three two times 3.65 65 something like this right and you that will be in terms of meters finally will be in terms of meters right delta x2 will it be positive or negative positive right how you now let's look at the graph here where where would you see delta x2 where do you see delta x2 from this graph here starting from where I mean we have computed delta x2 like this right but now I Wonder what's the geometrical interpretation of data x2 on the graph on the vt graph where is that well from you start from t star right start from t star and end at the t final this is your tal and this area here is your delta x2 right this red one this is your delta x2 here okay so that's great right so now you take absolute value that means you only care about the area you don't care about whether it's positive or negative. Then once you have computed okay the sum delta x2 is here right take the absolute value then take the absolute value of each then take then take the sum then you will get the total distance traveled in terms of meters then you are done okay so now do you get the meaning of this now now let me also ask you another question right so in part c of problem we have obtained delta x now in part d you obtain delta x1 and delta x2 respectively then compute the total distance in this way. But how is the answer part C that delta how is that delta x related to delta x1 and delta x2 right? Do you see what I meant in part in the previous part of the problem called that delta x or delta x prime depend on what what how do I call it right? Call it delta x or delta x prime here and that delta x the total displacement right. How is the total displacement related to delta x1 and delta x2? How are they related? So if you know calculus well you know that how they are related basically that they are related by delta x or delta x prime right. So par c the answer to par c should be equal to uh delta x1 plus delta x2 without the absolute values. So that means if you simply try to get the sum of delta x1 and delta s2 without taking the absolute value then you get the displacement. So what that means is as follows geometrically you have a negative area here you have a positive area right. So now if you look at this region of the positive area. So this negative area is smaller right positive area is bigger. But if you are what you are doing is as follows. and try to in that positive rate right red right triangle here you look at you you can try to single out a certain area and this area is equal to the absolute say so this is this is a positive area that is equal to this area in magnitude try to single out a certain equal in man with delta x1 and then this part okay this part once Because here you have a negative area right and when you do the sum without absolute value this part will be cancel this area which is positive which will be cancelled by this negative entirely since they are equal in magnitude. So therefore the remaining area here the remaining area here is delta x. So that means geometrically your delta x represent this remaining area here. Right? So do you see what I meant? Let me say that again. Delta x1 is negative. Delta x2 is positive but is larger magnitude. When you do the sum without absolute value, this ne negative delta x1 will cancel out partially part of delta x2. So this part is got it's cancel out by the negative area. So the remaining area here of this trapezoid represent your delta x or delta x prime if you would like to call it delta x prime here. Hey here. So that's that's the geometrical meaning here. So now do you know what you are computing right now here? Do you understand what you are doing at this point? Okay. So so hopefully hopefully you understand what you are doing with this problem here. This problem the meaning is is very I mean it's deep. It's not I mean it's not as simple it appears. If you under this understand this problem completely then you understand the whole whole story here. I mean this section seven here. Let's move on. Hey any question up to this point? No. Okay. So let's move on. Yes. Find a point. Okay. So basically this is not real space. Okay. This is a velocity versus t graph. What I did was to conceptualize what we meant by displacement. Displacement can either be positive or negative. When the displacement is negative that means you have an area which is underneath uh v equals z right underneath this horizontal v=0 line. this area is negative. But when you have a positive displacement, that means that area is above the the x-axxis, right? Above the x-axis. So now when you do the sum of the two displacement, that would just mean that the negative area will cancel partially this large pos larger positive area. So partially cancelled. So then the thing would that you give you obtain in the end is the remaining area that remaining area is actually the to the total displacement. So this total displacement is the geometrically means this remaining area here right. So this the thing right by definition is x final minus x initial. So that means geometrically if people ask you where's how can I interpret the total displacement then you can tell them that it's here it's the area of this trapezoid provided you the you know the fact that this delta x1 is smaller delta x2 is bigger here so that's total displacement is here right but but that's just for this problem for some problems things can change right delta x prime in this problem delta x prime is is a positive quantity right so That's for this reason. But but generally if you are interested in more interpretation you can let me know after class. Right? But but this is what we cover here in class here. Right? All right. So hopefully this is not boring to you because uh this is so important right this concept is so important and and so so that I would consider this as a great problem. I mean this problem is great. Now let's look at another problem which may does may not be simple to you. This one right question 10 also. Usually I put this on a test very often, right? This question it's like a chasing problem here, right? So you have two objects who which are moving. So sorry, two persons which are moving. Now let's read this question here and and we I'm going to erase the whiteboard in the meanwhile. So after this problem we we will finish this section at this I mean we'll finish explaining about this section and we'll move on to the next section after this problem right we'll move on to moving on to section eight but so make sure that you keep working on the home so so here we are in question 10 of the homework right it's due um sometime in September Right. So make sure that you you the design is specify on web design. Right. So I think some of you know know may know how to solve this problem but I think many of you probably did not know how to solve such problem here. If you have seen problem in high school presentation in high school some of you yes so that's great so I think then if that's the case we we can try try to explain it more more quickly about this kind of problem here right in this problem it's a more involved because you are going to solve a quadratic equation first right but of course first difficult part is we should set up the problem first set up a problem First so when I tell you about the skills of problem solvings usually I stress on the uh importance of so-called senses of geometry. So for solving each problem usually I don't provide procedures right procedures is something you can develop I mean procedure of problems but what we consider important will be the so-cal developing a sense of geometry so that's why each time I show you some graphs or show you some some pictures so that you can develop some sense senses of geometry that's important for problem solving very important so now here you have two persons so one this hockey player let's call it person A right and the other person we can call out person B here and the situation is uh as follows so how person A originally was standing still somewhere right and person B passed by it at the first time then after certain period of time say 2.4 4 seconds then the first player say person A makes up his mind to chase the second player right then that means after 2.4 4 seconds the first person person A starts moving and now person A moves with a constant acceleration right person A moves with constant acceleration but what's the initial velocity for person A did you see I mean from reading the statement for the for that hockey player let me call out person A what's his initial velocity zero okay that's great it starts from rest right but person B was starting with a uniform speed. We assume that it starts with the uniform speed. So now the importance of this problem you consider that the two persons they can share the same clock right. So like the same clock is ticking which can measure the the things motion of the two persons. So this problem can be a bit difficult to address. I mean the physical meaning is a bit difficult can be a bit difficult to address. So it will be helpful if you start with uh some some diagrams or plots or something or a picture right so you usually it's good to have some pictures in mind. So let's look at this thing here. Suppose this is a location where the two persons they met at the first time right location here and let's say that person A is here person B is here they happen to be at the same location that means you you can say that X A and XB they are equal right at this location here right so two persons are at the same location but after this person A is not moving but person B is moving with a constant speed over certain time is that time given yes right and constant speed is also given. So that means person B will travel with a certain distance and that distance equals the the speed of person B. Okay, person B is moving with a constant speed. So we don't need to worry about acceleration of person B. Person B doesn't have any acceleration. And here this is not a this is not a diagram, right? So sorry this is not a di a graph. This is not a graph but this is just some schematic view like a compare the two motion of the two persons. It's like a motion diagram basically right but this is not really a graph like a I for this problem I didn't really draw the VT graph or AT graph but this is just a a a diagram for real space here. So here you have VB the velocity of person B times the that per that time that interval that it moves. I call that time interval t star right that means means this is a real space but here I call this t equals z for person b and for person a is the same t equals z here okay so let me clarify so let's assume that person a and person b they have the same clock or the same watch right their watch watches are synchronized that's that's why I say that they have the same clock so then their their watch or clock will start ticking at is mean when person B is at this location and when person A is at this location but they have the same clock that clock starts ticking when person A is here when person B is here so do with this kind of setting here again if you solve this problem in the past your instructor may solve it in a different way this that's why this question is a bit difficult because uh because different persons may try to solve it in a slightly different way but it's fine as long as your fe consideration as as long as your concepts are correct. It's the meth methodology doesn't really uh matter as long as your procedure is correct. Right? So let's assume that they have the same clock and that clock start ticking when B is at this location A is at this location then now both persons they start to move. A person B continue moving with a the same constant speed and person A is moving with a certain constant acceleration then at some other point here at a later time they they met again right so at this location here they met again so this final location I call this XA this final location I call it XB and final X and XB they met again so that X A and final X AA and final XB they are equal to each other Right. So now let's consider how far they travel after the right after the clock start ticking. So right after the clock start ticking it travel with another distance which is called VB times the time. Okay. So this T final is called that uh t equals z and this is called that tf final here. Right? So the t final. So VB * the T final is the is the F is the distance that additional distance that sec person B travels after his clock start ticking right so so that's the now let's look at person person A person A travels from here and at this location his clock start ticking and then person A ends at this location what's the total distance traveled so then this distance travel since the initial is zero Right. So, so therefore the distance travel would just be equal to I'm sorry 12 times the acceleration of person a time t final square right say say the final time is called tf final here right initial time t = z so 12 t acceleration of person a * t final the same tfal square then this will give you the distance traveled by person a so now since this is a chasing problem eventually person A catches person B. So then what's the first equation that you write down? Yes. Ael. Oh okay. So now from this equation so remember we have the kinematic equations right. We have the third equation which should look like XAL= X initial plus V initial * T + 12 A T² right? Then I can shift if you if you set this initial to be zero right and then your your your final final location for person A will be equal to is zero then that will be equal to 12 a t final square right that's where how you come the third kinematic equation do you recall that in the previous lecture we derive four katic equations right and use the third one v initial was zero you agree that person A start from from zero velocity, right? Person A starts from rest. So So X A equals this. Is it okay? Yes. Do you have any doubts or disagreement on this with this one? No. Okay. So now let's come back to the goal, right? What's the goal? So the unknown time first, right? We want to solve for that T final. How much time does it take to catch his opponent? Right? So that start with this equation X X a= XB their final should be the same right. So then then how do we write down the rest? You have 12 A a time T final squared that's on the left equation. How about XB? So do you know how I got XA? X a I got it from here right from the third equation. Then I can write down X a here. Now what is XB? Look at the look at the diagram I show you here. XB has two parts, right? First part was this one. Do you know this part? This part was given, right? You know VB, you know T star. So this part is given. And this part yet because you do not know T final. You want to solve for T final, right? So this plus this this plus this will give you XB. So what is that? So XB will be equal to say VB T star which is constant. Then plus VB T final something like this. Right? So now let me erase this equation because I need some space. Let me erase this. So now do you have any doubts on this equation here? Now I'm going to shift the terms. So now space unknown because here quantities are known for this problem. What are the known quantities here? For example, T star do we know 2.4 seconds that's known right? VB do we know about VB? VB is 2 meter per second which is also known right and do we know about acceleration for A is also known right so right so means what is the only unknown in this equation t final right tal is what we want right so clear now I'm going to solve this symbolically but of course you can solve it numerically But this is quadratic equation. How to solve it? You you sol. So now let's try to shift the terms. 12 a tal² minus vb tf final minus minus vb t star equals zero. Remember t star is not a variable right? You can circle the variables here. You have tal square which is a variable. Tal here is variable. Right? But t star is not a variable it's a constant right t star is given. So how would you figure out t final right? So use some formula right. So let's let's assume that consider the equation consider equation called alpha t final squar plus beta tf final plus gamma equals z. Right? So this is a for I want I want to use a formula for solving a quadratic equation. So what does the formula do? The formula do look like what tal if you want to solve that quadratic equation t final should be equal to 2 * alpha. Then here you have negative beta right plus or minus square root of beta squar minus 4 alpha gamma. Okay. Usually use abc but here I don't want to use abc again. So that's why I use alpha beta gamma. Right? Do you have any doubts on this equation here? Quadratic equation high school algebra, right? High school algebra. So now you can solve for that problem. But I need to consider what are the the coefficients. Right? Here we assume that this is alpha and beta equals negative VB and gamma equals negative negative uh VBT star. Right? So did I make any mistake? Hopefully not. Right? Okay. So now let's try to write down the expression. Write down try to write down this expression here. Right? So therefore 2 alpha should be equal to since 2 * 12 a 1/2 the acceleration. So that would just be equal to the acceleration. Right? So the thing on the denominator two alpha would just be equal to the acceleration a here. Right? Now let's look negative beta. You know what it means right? negative beta must be equal to the the positive right so negative beta will just be equal to positive VB here which is simple so now let's look at this term here what is beta squar minus 4 alpha gamma just let's compute this right beta is negative VB so beta square is just VB square okay negative VB square then minus 4 * alpha what is alpha four times alpha. So basically alpha is 12 a the acceleration right then times gamma gamma is uh negative of vbt star right so here gamma is negative of uh vbt star here right but vbt star is just some some constant and then you know you can compute that right so something like this so what does this thing look like in the square root So that will be that thing will do VB square then plus because here negative sign of another negative sign so it will become plus right so plus uh 2 * the acceleration time VB then times T star like this right so that's that's a thing called that's that's this part is inside the square root okay inside the square root you have this thing here then you compute tal right so tal will look like what here I'm not telling you a formula to memorize here I just tell you how to compute but actually you can just substitute in the numerical values and because here there is no further mystery inside such a calculation so basically what you do is obtain the tal final should look like this here you have the acceleration and here I have VB right plus or minus minus square t of VB² then plus 2 A VB T star and then you get two roots right two answers but should both answers be correct here you have plus or minus right how would you get rid of the unwanted answer time can never be negative so that means when you get the answer which correspond to a negative time you should drop it right time can never be negative so then you solve this then you solve for time right so this is a numerical problem and you will have different numerical inputs so just solve it yourself you get the time right and then you are done so but don't treat this as a formula right because if I change questions the you cannot use the same formula so don't treat as a a formula you should you should start from the very beginning such a consideration here Right. So where so the key point is they share a common clock and that clock start ticking. We assume that it start ticking when A is here and when B was here right this moment both clocks start ticking and they have the same initial position the same final position. So that's why we say to be and then you can solve for the once you have a equation here you have a quadratic equation in terms of t final right and t star is not a variable okay t star is a constant which is known so then then you solve for this quadratic equation then you'll be solved for the unknown time now part we finished part a right then what is part b how you figure out how far has person a traveled during that time suppose now I I have obtained t final right then next next question how far has person a traveled during that time how would you what kind of equation would you use what does it mean by how far what's a unit meters right so it's a distance and here since there's no turning point that distance is just displacement right you have finished the previous problem if there's no turning point Distance is just displacement. But if there's a turning point, distance can be different from displacement. So be careful. Okay. Here is there any turning point? No. Okay. So what which equation can you use to get the distance traveled by by which you you should try to obtain displacement? Displacement of person A. Which which expression? The answer is on the whiteboard. Which expression would you use to get the the dis the total distance traveled or the displacement of person A? Which equation? If you know, just raise your hand and speak up. Yes. Yeah. X A. Yeah. So, what's the expression here? Right. Okay. So, that's great. So, basically the answer is here. Part B of the question. The answer is here. Right. Once you get the final time then the displacement just just X a right X a X initial assume initial zero. Okay. So the final location X a X a minus the initial location zero will just be equal to this thing. Okay half a T final square. Then you are so this is part B of the question. So is that clear? Okay. So this question you should know it very well because each time each semester I put this kind of question on the test on a test here. So this is like a chasing problem here. All right. So so any question at this point? No. Okay. So if if not let's let's move on. Okay. So in principle we need to move on to the next section. Okay. So starting from question 12 there will be problems about the next section. Right. from question 12 all the way toward the end. So that means up to question 11 you have full ability to finish already. Right? So so from question one all the way to question 11 finish that as soon as you can. Then from question 12 we are going to start a new context here which is section eight of the book. Right? Section eight. So now let's look at section eight in the following section seven. If you are interested in examples in a textbook you can read by yourself. Right? I mean there are some examples in the text but basically we have already finished a very important problems already. Section A is called 3D falling magic. Right? So this new section here. So in this new section here let's consider a new scenario. So so far we have the skills about uh solving problems regarding an object moving with constant acceleration. Right? Constant acceleration. And we have derived the four kinematic equations for you in the previous lecture. If you miss it, just make sure you watch the video like very early this morning. I I I I send the video video to you all. If you miss that, make sure that you I derive the four equations for you. So those are useful. But now we are considering a new scenario here. So consider I have a basketball, right? Suppose now I try to toast this ball vertically upward. So that means once the ball leaves my hand, you'll be given a initial velocity, right? So once the ball leaves my hand, you'll be given an initial velocity. Now let's observe this motion here. Initial velocity. Once it leaves my hand, there's no other force except for one force. What is that force called? Force of gravity. Yeah. So you have I think all of you have some common sense about this. There's a force of gravity and we claim that near the surface of the earth force of gravity is held the value of the magnitude of the force of gravity is held as a constant. So that value we will tell you very soon. Okay. Now magnitude of the force of gravity is held as constant. So that force of gravity will produce some acceleration. Now let's observe the motion again. Okay. You are being so this ball is being given with some initial velocity. Then at the highest point. Okay. At the highest point what is the velocity? Zero. Okay. You also know that. So that actually that point is the turning point. So remember the question we solve early on. Right. a an object which is being given an initial velocity. So remember that that kind of question is not here on the whiteboard but we have solved problem like an object is being given a negative initial velocity and you will move all the way to here right and it was it will come to a momentary stop then you will move into the different direction and all the way to the final point. If you rotate that problem by 90 degrees in the say say clockwise direction. If you rotate that problem and into the vertical direction then you you encounter such a problem here. Let me say that again. Hey you you p you push this object so that you this object is being given with the initial velocity and this object is then it will reach a at a highest point that velocity we call that v star. theta v star equals zero. That's a turning point. And after a turning point, it the velocity changes its direction and now this object moves downward here. Right? Okay. So now you have a new setting of a problem here. So we don't really need to derive the new equations. But we just need to uh start with the four kinetic equations and rewrite them in in a different dimension. Okay? Rewrite them in a different dimension. So that means now we are focusing on vertical dimension. Okay? Okay, we focus on the vertical dimension and then try to rewrite four kinematic equations. So basically the consideration is here now this is your code is positive y-axis right and there's some some origin here right and this object suppose is being toast from some initial position here okay so here this location we call it y initial right and suppose this object is being toasted vertically upward here and it's being given a certain initial velocity v initial and now we wonder how this object moves right now we Assume that there is a constant acceleration that this object is being subjected to and that constant acceleration during the course of motion I we call that a y and h that ay a y actually points downward and we call that a y we know the value is like negative of 9 square the constant acceleration which takes a negative value just because positive y is upward but that constant acceleration points Right? So then you can imagine that this object is being slowed down. Then at the top it will come to a vstar equals zero then it will drop again and then then you will so that means after that the direction of velocity you will change to the negative direction. So we have such a setting here and next lecture we are going to write down the four command equations for this new setting here which is called 3D folding motion. Right. So now we will end the lecture here and thank you all and hope you all have a great day. See you again next time.