So currently we are in uh chapter two verse and and section 2.3 and the title is this one and the purpose of this section is that we are going to consider the motion of a particle under a constant acceleration and we are going to show you four kinematic equations. Last time we have the first one. Now let me repeat that that thing right that derivation there. So suppose let's look at a of a particle. Particle means uh just point right a point that is moving in on one dimensional space and this point suppose that it has a velocity v initial in the beginning and then has the the final velocity vf final here right here. And what we consider is as follows. We want first assume that acceleration is constant. Right? So acceleration is constant. We have this constant uh acceleration since it will be constant. Then that means this constant acceleration should just be equal to the average acceleration. Right? So now acceleration here of course we use say a subx here a subx. And so that average acceleration can be written as Vfal minus V initial over T final which is T final we just call it T T final right T final we just call it T and T initial we call it zero okay so in this section we use such a consideration and so that will be divided by t minus0 like this right and therefore we can assume that now you know that acceleration is constant so this ax right we assume that ax X equals constant. Then under this assumption we have this equation V final equals V initial and plus acceleration time. And then this last time we call it equation one here. Equation one here. And of course this is all under the assumption that acceleration equals constant. Right? And some people may write it in a different way. they can consider vfal to be like a v as a function of time. They don't want to say the like a definite final. They can just say that okay this vf final you can consider this as a function of time. Then that equation will become v as a function of time equals v initial plus the acceleration which is a constant times time. So what does this tell us? So that tell us how the velocity will change with with respect to time. And in this case the velocity in in this example here I show you here velocity will increase linearly as a function of time. And then then what does how would you consider the slope of this line here? What quantity will give you the slope of this line here? So in in this equation here and in this equ this is one right equation one and this is also equation one here and in this equation which which v which quantity would correspond to the slope of this line ax okay so slope of this line would just be acceleration and since acceleration here we assume it's constant that makes sense right this this slope is constant I mean over this line here so everything makes sense All right. So, so far we have shown you equation one. Now let's consider the following. We are in section 2.7 here. And now let's consider the displacement. Right? Let's consider try to consider displacement in this graph. How would you find displacement in this graph? Can you see displacement in this graph? Any idea? I mean some of you have taken high school physics already, right? But some of you probably didn't. But it's fine. Just think about this graph here. Velocity changes with time and acceleration is a slope. But you see displacement. Let's assume that initial time was zero, right? And final final time was t. Now I wonder what's the displacement of the particle from t equals 0 to the final time displacement from geometry of this graph. Where where can you see the displacement of the particle? Where's the displacement area? Yeah. Yes. Yes. Okay. Great. Great. So, do you want to say the same thing? Yes. Yes. Okay. So, that's great. So, some of you have taken calculus 2 already or you are just taking calculus 2, right? And so but if you are not familiar with the concepts about integration here at the bottom line is you do need to understand that if you consider the area underneath this line right of the VT graph this area here here I use a shaded area here this area area will be just the displacement during this time interval right what is displacement displacement is uh xfal minus x initial right so now let's consider what does this shape look like look like here there's a right angle here right angle here right so apparently this is a trapezoid right so there's a formula that for you to compute the area of a trapezoid right so so that's what we try to consider however suppose that I forget about the formula and some of you may remember the formula okay let's assume that I forget about the formula then what what do I do if I don't have a formula for computing the area under like a underneath the area under underneath this line look like a tri trapezoid so that means that's the area of the trapezoid so what I do is is as follows so since we know that velocity increases linearly I'm going to consider the mean point here right so it's like a mean point between v initial and vf final let me call it v star okay that me v me point v star. Now I'm going to define one quantity here. I define define that v star to be 1/2 of v initial plus vf final. Okay. And now the next step once I have the v star I'm going to see what this this v star can can give can give us. And this v star you can consider this as the arithmetic average of the initial and final velocities. So that means it's the mean point here. Now if I try to draw a hor horizontal dash line here right all the way to so that this this dash line would intersect with the with the line with the slope right this line here and then then I try to find out the corresponding time for this v star. I call that time t star. I call this time t star. And that means at time t star. If I call the initial time zero final time t then what is t star compared to t? How is t star compared to t? So now just use geometry. Don't think about formulas or anything. If I have this okay so acceleration is constant, slope is constant. And here in the this geometry I have a V initial and V final. Take the mean point. I call it V star. Right? And figure out the corresponding time called T star. And now let me ask you how does this T star compared to the final time T? How are they related? Any idea? Do you know how how would T star related to T be related to T? Initial time was zero. final time let's call that T and then how is T star related to T yeah yeah that that's great so T star will just be equal to half T if you don't believe in this you can try it yourself just use some high school geometry you can figure out that T star is just half T right so why did I do this so now I want to compute it let me say again I want to compute the area underneath this uh line And that's the area of like trapezoid. But now suppose I don't want to use that formula for tri area of the trapezoid. So what I do is as this so consider horizontal dash line here. Now look at a right triangle. So let's look at this right triangle here. This part right. So here I use a red cutter to encircle this right triangle here. Right? This part this area here. Now this area suppose that by imagination I try to shift this right triangle to here so that I can fill up the space here right here I have a empty space here right but now I try to shift this right triangle and make it filling up this space here and this will become a rectangle you try it by yourself it will work right because this is the me this is the halfway of the time so that means this is t star equals half t Right? So if you consider this right triangle red one and you shift it so that it will fill in this space. Now this trapezoid will become a rectangle. Now that means do you I think everyone knows how to compute the area of the rectangle. Okay. So now instead of calculating the area of the trapezoid I'm calculating the area underneath I mean of this rectangle here. Right? That would be much easier. How do I compute that? So time interval is just t minus 0. Right? That's the base. What's the height? Height is just V star, right? So the for this rectangle, the base base is t minus zero. That's obvious, right? But the height is just v star here. So So how would you compute the area? Let me say that again. Hey, so now I say that x final minus x initial equals the change in position. And we claim that it's going to be equal to the area. Oh, and that area now I claim that it should be the area underneath the sorry the area for this rectangle. And that apparently is what? It's just V star time minus 0. Very very simple. That's a area of a rectangle. And what does this give you? V what is V star? V star is defined as the this thing right. So now let me rewrite this will be 12 V initial plus V final times T. So what does this tell us? We have figure out an expression for displacement. This displacement right now let me come to here. So now I have vxfal x initial plus v initial oh sorry plus 12 v initial plus vfal times time. So that's our second equation. Okay. So this second equation will give you uh is rather to displacement actually right but if suppose you you know initial position right and you know the initial and final velocities then you'll be able to get the final position from this second equation here right so again we you you can practice those questions in the homework assignment right homework set one but but now we tell you the equations to use first equation this one right is about velocities these how would you obtain the final here? So this is our first equation and this is the some variation a different form of the first equation right you can call that this one you can call out one prime is fine but it's basically the same as equation one but just written in a different form right and this is called equation two here equation two here you can obtain the final position now let let's consider the following very often we may not know final velocity in the beginning in a question so now we are going to consider the following if we do not know vf final how would you obtain obtain the say xfal right so now let's consider the next equation so now we are going to do a substitution right so now we're going to prove the next equation very short proof so now we're going to substitute sub sub equation uh one into equation two right so after substitution let's see See here substitution you have xfal equals x initial plus 12 uh v initial and plus now vf final is repl by that so plus v another v initial plus a a time t and times another t here okay and now you have v initial right divide by two so eventually you will get this equation xfal equals X initial plus V initial * T then plus 12 * A T² right so this is our third equation pay attention to the difference between uh equation two and equation three here what's the difference equation two here you have V initial and V final right equation three you don't see V final anymore but you have you still have a V initial right and you have acceleration. So if that problem gives you acceleration right and give you time and want you to obtain xfal then you can for example you can use s such equation here right so that's the third kinematic equation we have obtained for you okay so any question no right so so I I hope that's that's clear so pay attention toes equation two and three and now let's let's there there there's one more equation we want to tell you And I think I need to erase some things. Hey, let me erase this uh this thing. So now what we are trying to consider is as follows. We still start from can equation two again, right? Start from equation two again. But now what we're going to do is as follows. In some problems you can look at equation two. Sometimes uh sometimes we may not know time explicitly, right? or sometimes or in some on some occasions we may not know time explicity in the question. So to for such a question if we want to use equation two we must figure out the time first and then we can use equation two right. So now we're going to derive a new equation from equation two and equation one again and see what we can do about this. So suppose we do not know time right suppose uh we we do not know time and so we try to rewrite equation one right so now let's just rewrite equation one suppose time is unknown but from equation one we can try to obtain expression for time which reads as follows you have vf final minus v initial equals a * t therefore time can be written as vf final minus v initial over a. Okay, suppose now we call so here is a new equation, right? So now in this new equation we substitute this new equation into equation two. So then equation two will become this thing xfal x initial plus 12 v initial plus vfal times time. But now since we assume that we do not know time, you just substitute this into the expression for time. So that will be val minus v initial then divided by a acceleration right like this right so now let's consider the following so we play with the algebra so first shift x initial to the other side of the equation so then you have this thing and then you multiply both sides of the equation by a factor of 2 a say see here you have things on the denominator It doesn't look good right because so now now we want to multiply both sides of equation by a factor of 2 a 2 a and then this left hand side will become 2 ax x x in x final minus x initial and on the right hand side equation let's see what we have here so you have let's consider algebra here right here you have look at this thing you have Vfal plus V initial multiply by VFAL minus V initial. Right here I just just flip these two terms. I call it V final plus V initial. Then multiply by V final minus V initial. From high school algebra do you know what this is? How would you simplify this expression? Expanded expanded. Then what what would you see in the end? This will give you right. So, so that's V final square minus V initial square after expansion you have this. So that's after simplification. So now we can substitute that into. So on the right side equation now you have V final square minus V initial square. Now you can see the effect is this equation no longer has has time no longer has any explicitly time dependence. So now finally we can write down one more equation. I I need some space. Anyway, so so let me erase this because I need a space here. Need a space here. So and also erase this and this you know that right we did the substitution. So let me also erase this as well. So after this this is what we have flip the I mean switch between the right hand side and left hand side equations. then you will be able to obtain this relation. V final squared equals V initial squared plus twice the acceleration times the displacement. Okay. So and that's that's what we call equation four. Right now we have the so-called four kinematic equations already four. In this last equation, you usually use it in a situation where you don't know time explicitly, right? Because in this equation here, you don't see time explicitly. So if in a question you don't see time explicitly, then then it's likely that you could use this. But still depend on what kind of questions you encounter. Right? So VFAL square equals V initial square plus 2 * the acceleration times the displacement. Remember this part correspond to delta X, right? This part would just be delta x displacement displacement but usually we write it in this way. I mean textbook they usually write in this way here equation four. So in this section we have told you the so-called four kinematic equations already. Right? So you can read the textbook and you will see the derivations and their derivation can be not exactly the same as the der derivation I gave you but you can refer to it. I mean the derivation I mean depend upon instructors and and so on. Right? So, so actually in the textbook you will see those equations but you just need to use the four that that I presented to you. So th those equations they they list it five because they treat that the the quantity I call it v star they call it something like v average here but uh there are some some details details I mean whether that's called v average or vstar that's something we can if we have time we discuss later right so that var let me say again that v star in the textbook they just call it v average right but I would say that I call it v star just to make it general to make it general but for detail if you are interested in what's the difference between V star and the V average early on then you can ask me I mean after class but but in general we we just treat it as some expression right this V star treat as some expression that's the second equation but basically we have the four kinematic equations right this the the one two three on a test I use these four right so you don't need to worry about the things in the textbook too much. But in the on the test, I start with this one, right? The first equation and the and the second one and the third one and the fourth one. Those four equations will be printed on the first page of the exam, right? So that means those four equations one, two, three, four will be printed on the Yes. Yes. Go ahead. Yeah. Are all these for constant acceleration? Yes. Only for constant acceleration, right? Yes. Yeah, you can also do do the integration. Yes. So if you have have taken F calculus 2, you have the skill to do integration, right? And al also you can use the formula for a trapezoid as well, right? But here I told you this method is because of the concept. If you use this method, you got a concept that why we what's the meaning of this vstar and then once you have the replacement var times the time interval will just give you the give you the displacement and you get this equation. So but the method is not unique. I mean there are there are different methods of proving those equations. You just need to use one proof. Yes. Yeah. Yes. That's that's why I explained there are four equations actually. the the second one in the I mean this is by the book that one is is just some additional formula I don't treat it as a kinematic equation right this one this one the at bottom we are just treated as a expression they call it average velocity but actually that's the arithmetic average of the two the can be I mean that equation in at the bottom there can be a bit confusing I mean by confusing is what's the difference between vstar and v average. So since you brought up the the question that second equation here I call it v star right this v star v star equals 12 v initial plus vf final. Okay but in the textbook I mean in the pre previous sections we also define a so-cal v average or the var. What is that? What is v bar? That's delta x over delta t. and which is xfal minus x initial over tfal minus t initial. Okay. So actually under certain condition v star will will be equal to this. Okay. Which condition? Let me say again. So this v star is the arithmetic average of v initial and vf final. Right? And in the previous sections we talked about some v average or v average x. is defined as the displacement over time and that's x final minus x initial over t final minus t initial these two quantities under certain condition they are the same thing so that's why the test they call they call them by the same name they just call it v average right but I would say that there are some condition which may be difficult to capture when are they equal under which condition are they equal to each other it's not obvious right the textbook assume that it's obvious but it's actually not obvious. Hey I can any idea doesn't okay I can tell you a condition they are equal only when acceleration is constant when acceleration is not constant they are not equal right okay so that's why I say that this second equation don't worry too much about it I just treat it as expression so we only have four kinematic equations on the slides you will see five let let me say I can on the slides you see five but actually the second one from the textbook I don't treat it as a kinematic equation just an expression right and and that expression tell you that you have something called I call it v star I just try to be general but under the assumption that acceleration is constant this v star is just the average velocity right so so that's it so and another thing I can tell you I mean if you still wonder what's their difference I can tell you recall that we have talked about one example Yes. Oh, so you mean here? So, so this is t square a a subx. This x is a subscript and times t square. So, okay. Yes. So, just just t square just t square. Yes. So, any other question? Okay. So now now let's consider the following. Remember I talked to you about example of a person right walk from point A to point B right and then that person walks back from B to A again. And suppose the p the velocity of a person from A to B is called V1 right and velocity of person from B to A is V2. And remember how I compute the average speed right and how I compute the average velocity average velocity is actually zero. Now in this case if you consider this V1 and V2 here some people will think okay to compute the so-cal average speed can I take the arithmetic average of the two say if I take arithmetic average of the V_sub_1 and V2 can I get the average speed in this way right what does this mean and in this example here that V star is not equal to the average speed that we compute. You cannot use this. You cannot call this average speed. It's just because acceleration is not constant actually, right? Because you can see that there's a turning point here. At this turning point here, the velocity suddenly switch into this direction. That means there's some acceleration here. So that means acceleration is constant, not constant. So in this case when you try to define some quantity called v star is not actually equal to the it's different from average velocity or average acceleration right so this v star is something else so that's why I don't like that second equation that second equation the there's assumption there but not not everyone will sense the assumption so it's kind of risky to use that right so so that second one so use the four equations I present to you on the whiteboard right so so this is like a counter example where you can define V star but this V star is not what you want right if you take the simply take the average of the speed the forward speed and backward speed that arithmetic var average doesn't have any meaning because that v star in this sense is no longer is not equal is different from average velocity nor average speed of the for the whole tree okay so there's something something deep here and it's relevant to what's the difference between vstar and V V average right there's a difference but when acceleration is constant they are the same so that's why that that we only have four kinematic equations okay so so this one the one at the bottom here is a bit confusing so I don't want to use that as a kinematic I mean it's not a kinematic equation just an expression so so that's it okay so any question here no okay good so so now I think you you got a point here and move on okay so we finish the derivation. I think all the all the sections all the large sections we we have to finish those right so that's that's not not a surprise we we but some of you have learned those in high school just also keep in mind that all the four equations the condition is acceleration must be constant when acceleration is not constant you cannot use any of the four equations right all the four equations the condition is acceleration must be constant when acceleration is not constant don't use those four equations Right? Okay. So now let's move on to talk about other things. Yeah. So let's consider uh let's say some some let's say homework problems for example. Right? So let me just show you some structure here. Some of the questions we have finished early on already like the first question when you look at homework pay attention to the code on on the right right side. Okay. See the right upper right corner. You see a code right here. S E R means sur. That's the author of the book. PS means Yeah. Yes. It's fine. No. No. You don't go. Okay. Did you sign up for web design? We design web design. It's on web design. You did did you when when did you enroll in this class? When? Oh, June. But did you attend my first lecture? And no, web design talk to me after class, right? We design is is I think if you have sign up sign like enroll in web design, please raise your hand. Yes, it's free enrollment. So So okay, so let's move on. So um Surway P is the author of the author of the book PSE means physics for scientists and engineer engineers and 10 10 means 10th edition 2.1 means chapter 2 section one right and op is just some some problem operational problems and so on right but anyway so that means when you see 2.1 means that question is rather than 2.1 which 2.1 of the section 2.1 of book which we have already covered then you should finish that question as soon as you can. Is that clear? Hey, so the code is on the upper right corner when you see the corresponding section we have already finished. Just just finish that question of the homework as soon as you can. Right? So like this question you are supposed to finish it. This question apparently we have talked about this right this question here again this question here that I just talked about don't just calculate the arithmetic average that's var and that's not it's not equal to average velocity nor average speed because in this case there's acceleration here at a turning point so so it's not constant acceleration right so v star doesn't have a meaning in this problem here right so now let's Move on to here. This question is also relevant to section one. So section one. So again look at the code. Look at the code. And this question like the code says section five. We also have already covered. And now starting from question seven is about this section here. Some questions are rather uh simple. This one I will just leave it for your own exercise, right? Because now you can see that just look at this very very quickly. You know the displacement, you know the time and you will slow down to a final speed and you want to figure out the original speed, right? You know the displacement, you know time and you want to get the original speed provided that you know the final speed. Then which equation to use? We have equations one, two, three, four. Let me say that again. One uh two three and four. would you use to solve part A of the problem? Don't worry about details. Just pick a equation. I mean find out the known and unknowns, right? You know displacement, right? Displacement means that 32 is the Xfal minus X initial, right? Now 8.8 seconds is time. Final speed is VFAL. And to find V initial, which one would you use to to be efficient? get it efficiently. Which equation would you use? First one. First one. Uh here. So in the first one, do you know acceleration? I mean in the problem, do you see acceleration here? Do they give do they give you acceleration? H no. Okay. So we don't know acceleration yet. So to get the or initial speed any idea any other idea it's good thing second equation hey so for this problem now let's look at the second equation here we know displacement final minus x initial is known we know time we we know the final speed now they want you to obtain initial speed so use part a equation use the second equation to figure out initial three. Yes. Oh, okay. So that's also a good question. What you can do is you don't need to assume X initial nor Xfinal. Just shift this to here X then you become XAL minus X initial and just call this 32. So it's not clear when you are using this equation when they just give you a displacement just shift X initial to the RSI equation and call this this would be called displacement right and just you 32 is this quantity here right so good question so it's always good to think right you don't expect yourself to to just know the answer immediately unless you you try to search for solution online right but we don't we don't recommend you to search for solution online you need to think right so thinking process is very important so part B of the question find the acceleration now assume that you know initial speed final speed and you know time you know displacement which which equation would help you figure out acceleration then which one just tell me equation number one two three four which one can you figure out acceleration equation one hey so that's great now you can use equation one acceleration So is that clear? So this one is very straightforward. So as I would recommend you practice practice this thing here. This question here this question. Okay this question I will let you work on it. This one this question is also not not really difficult. It's kind of straightforward. And there's a watch it right. There's a watch it. When you click on watch it button that means you will see a video that provides nearly the full solution. Right? So this one you will work on it on your own. Next one. Okay, let's work on this one. This one can be a bit confusing in some parts of it. So, read this question first. See if you can let me the light scheme so that Sorry. Yeah. Try to find my printed hand out. I don't any. But anyway, let's let's just look at the screen here. So, first um you have a crate, right? And you know the acceleration which is constant and you know the time interval which is also uh over a certain time interval but however time interval is actually is it known or unknown? Not really given right time interval is not really given. Final velocity is given. Right? So now part a of a question if final velocity is given sorry if initial velocity is given then what is the displacement right so now part question which you use the fourth one right fourth one would be good so now I need to erase the white board I'm Okay. All right. So, let's see. Uh, equation four and let me write down equation one somewhere. Yeah, I have equation two. Equation one is this. Equation one is V final equals V initial plus AX * T. This is equation one. And equation four. Equation four, let me write it down here. Right? V final squares equals V initial square plus A the displace. So now we have the four equations. Now we want to use this equation to to solve uh let's say part a of a problem. Hey. So this is a homework problem. Which home? Question nine. So homework set. Homework set one. Question nine. Right. Part A of a question. Let's try to use this to solve it. Solve a problem. We know what is the final. Final is 11. So you have 11 m/s. You square it and v initial let's say is 5.5 m/s you square it. Then plus 2 times the acceleration two times the acceleration 3.65 m/ second squared. Then multiply by displacement right. So this where can you find the displacement? That's what we want. So displacement let's call it say delta x. Okay delta x and then you will be able to find delta x from from this expression. Right? So delta x will just be equal to uh let's see. So pay attention to this. There's a constant acceleration over. So initial velocity and final this is smaller. This is bigger. So acceleration is positive. That makes sense. So delta x will be equal to uh 11 squar minus 5.5 squar then over 2 * 3 you will be able to get delta x in meters. Right? But in your actual homework system you will see different numerical values. For our homework system all the numerical values are randomized. So different persons will see different values. So you just com work according to your I mean the in input values from your homework system. Right here I just give you the the strategy the methodology right here part A. Well done. Right. So what is the distance that it travels during the interval? So this question here we need to pay attention whether there is any turning point here because here what is delta x delta x means x final minus x initial right. So if there's no turning point you are fine the displacement is just equal is just equal to the distance traveled. So in this case in this case since there's part B of the question part B of question no turning point if there's turning point we need to be very careful okay no toing point so so displace uh the distance traveled is just equal to the absolute value of the displacement so just take the answer from part part a and then take the absolute value then then you are done so that will be in meters again right that's part B of version if no turning point now look at part C here suppose the initial velocity is negative and if it's negative then I uh what is the displacement during the time right you figure out the displacement again so now what you can do is use this equation again if you try to find displacement right so part C of part C have a question how you find find the displacement in in this problem here here uh we want to consider a different displacement let's call that okay so you can see code this in this way so suppose final is like 11 again but the initial velocity is negative 5.5 and plus the acceler uh plus two times the ex acceleration which is positive and times the displacement. Now, now this new displacement we call that delta x prime, right? It's like a different displacement. This okay. But then when you the algebra, you'll see that you'll see that you actually get get this by by coincidence you'll get the same value actually. But now see see how you have you have 11 squared but minus okay - 5.5 squared then divide by 2 * uh 3.65 something like this but since you have a negative 5.5 here right but you square it once since once you have square it then the expression is just the same as that expression there. So basically the answer you obtain is actually the same as that answer you obtained earlier. Right? So part C by coincidence the the answer is just uh just just the same as part A. But now the tricky part is part D. Okay. Let's consider what's how you solve part D of the problem. Part D of let me erase this. Okay. Because uh I think you are familiar with the four equations. But now we are So part D of a question we should be careful. So think about the real space. Let's consider the real space here. This is my positive x direction, right? The real space is here. And now let's assume that that that particle starts from somewhere but initial velocity was like negative. So let's say that it starts on at t equals z it starts from here and initial velocity was negative. And now in the end of the process the final velocity is actually positive. So but now let's see how it will work. But in veity is negative. So that means this thing is moving all the way to left but some point it will come to a momentary stop and then will change its direction. Okay. So that means this thing will move and come to a momentary stop. So at this point let's call that V star as some spatial you'll be zero. Okay. And then after this this is my turning point here. So this is point and then after this turning point you will go all the way to the right here will go to the right and then in the end so this t equals z was here right and then then at some final time this is t equals the t fi uh t final let's say that some final time here t equals tf final the final time here the final and vinal here is 11 m/s. Can you sense this? What's the meaning of this problem here? Here you have negative 5.5 m/s, right? That was an initial and then you the the first displacement was toward the left. That's your delta x1. Now when put in this way. Okay. First displacement is this one and to this point and then after this point it would be displaced. So the first displacement was toward the left. Then after it reaches the turning point it will turn and the new displacement second displacement is actually toward the right or that's your delta x2. And pay attention delta x1 is negative. Delta x2 is positive. So how would you solve this problem? So that means this problem actually to get part D. Okay. Now let's consider part D of a question. Part D of question we should be careful because we know that the answer we obtain from part C of the question that that delta X prime we obtain from C of a question actually correspond to the final initial. Okay, the final X final, right? And this is the X initial. So final minus initial your overall displacement. In this case, we call that delta X prime, right? Equals final minus X initial. But this is part D of equation. This is not what you want because from from this expression, you only get this displacement. But now you want the total distance. The total difference for part D of equation should be equal to absolute value of delta x1 plus absolute value of delta x2. That's what you want. Okay? So think about this. How would you obtain absolute value of delta x1? How would you obtain absolute value of delta x2 and then take the absolute value first and then get some you get a total distance traveled. Okay? So try to finish this by yourself, but if you have trouble, we talk about this next Wednesday, right? So thank you all and hope you all have a great weekend and see you again next