 All right, should be able to get the videos up there. Sorry, this weekend, I'm gonna take too long and get the software I need and all that. So, did you see if that's working out? What are they linked on Angel? Huh? What are they linked on Angel? It's like the activities. Yeah, it'll look tight. I do spot. Did you get a satisfactory answer to your question? I'm sure my answer wasn't satisfactory. Well, I don't understand your answer, but I think I actually got it right in the first place. You did it right, yeah. I just looked at the numbers and was like, oh, these are the most possible methods. Yeah, but this is the problem where it tells you to find some spot between the sun and the earth, where the gravitational attraction is the same either way, where the two gravitational attractions cancel each other. So, there's some spot here where though the earth is much smaller, if you're a lot closer to it, you'll have a gravitational attraction that's balanced by the gravitational attraction of the sun. And you solved the problem. And did everybody see his post on Angel? No? I explained why nobody jumped into hell, did they? He solved the problem and then said, but this can't be right because why aren't we floating then at that spot? Why aren't we at that spot where we're not attracted to the earth any more than we're attracted to the sun? Or why isn't at least something floating at that spot? The same as the numbers are really close to the distance to the earth, to the sun, but I forgot to go 10 times 10 to the 11th. There's actually like 260,000 conferences right there. Yeah, it's, it's, it's, it's, this isn't the scale drawing by the way. It's, it's somewhere off of the Earth's surface that that point is, but, but your, your main part of your question was, why aren't we there then? Why is, why don't we float at that spot? Or why couldn't we put something there and it just sits there? Or why isn't there a bunch of star junk there? Coffee cups and other stuff that the astronauts have been thrown out of the space station? So, is that? No, no, there, there is indeed a spot between the earth and the sun where there's zero gravitational attraction to either one. Do you have this Earth is in orbit around the sun? Earth is in orbit around the sun, don't forget. And what is required for any object to go on a circular, circular path? Centripetal acceleration. We have centripetal acceleration here as supplied by our gravitational attraction to the sun. So that's what keeps us in a circular orbit. If there was something at this point where there's no net force in either direction, much less towards the center of the circle, whatever's right there at that, that spot of zero gravitational, zero net gravitational attraction, whatever's right there wouldn't stay there because very shortly later, the earth would move to some other spot and there'd be no reason for that object to follow it and certainly wouldn't follow it. Maybe even if it has some velocity of its own, if there's no net force on it at that time, it's gonna go on a straight line, not in a circular path. It would not follow and stay at that same spot. However, there is a spot, a little bit closer into the sun than that spot is where there is some net gravitational attraction left over. So I'll draw it a little bit more, that's more attraction to the sun because we'll say it's a little bit closer than the original spot we were talking about, maybe that's back there, where the net gravitational attraction to the sun is sufficient for that circular orbit as centripetal force and something can be placed right there in orbit if it has an appropriate velocity for that orbit, remember that there's a balance necessary between the force, the velocity and the radius of the circle, F equals M times B squared over R. You gotta have that right balance. There is a place where the net force is balanced if you have the right velocity for that radius orbit and something between the earth and the sun will stay there. That's not where these satellites are because these satellites orbit the earth. But there is a satellite there that's very useful for being a solar observatory because there's never anything between it and the sun. It will always work. It's called Soho, Solar Observatory Helios and I don't know what. Go Google it. This spot is called a Lagrangian spot. This is known as Lagrange one, L one. And that's what I put in your note. I said that there is such a spot. That's one of the spots that I know. There's another one over here on the other side of the sun where the attraction to the earth, well, the attraction of the earth would be very tiny, it's so far away, but the attraction of the sun is just right for it to stay in an orbit and will always be the opposite of the earth. So it's a satellite we'd never see and never be able to communicate with. So no sense putting it. However, you need a place to hide. That'd be a great one. And there's also three other Lagrangian points. I think there's also one here. We got a little bit more gravitational attraction to the earth, little bit less to the sun, but it's just enough to be just right for an orbit that always stays, that satellite would never see the sun. So solar panels wouldn't be necessary. And then there's two more that are off to the side here. And I believe the gravitational attraction of the moon comes into that one as well. So there's five Lagrangian points where an object put there at that radius orbit with the appropriate velocity. Remember for a circular orbit, you need a velocity that's always perpendicular to the force, the centripetal force. There's five spots where you could put something there and it would stay there. And then you could say, why aren't we floating there? Well, we could. You could be the first astronaut to go live there. All right. So watch Angel for homework questions and answers to help each other, especially since I understand that you're finding some of these challenging. Just a lot of them, is that it? So it's only this first chapter that had more than I think three on any one day. But that's because all the West are impossibly hard to do. What'd you say, Pat? I know Bobby would never say anything. All right. Any other questions? On Monday, Monday, this is where we were. We've established a couple things so far. The very first part was nothing new. We established that velocity is the time rate of change of the position. Remember, we're working in rectilinear motion. Freight line or one dimensional motion. These are, with what we're starting with, things can either go only left or right, or they can only go up or down or whatever direction, but there's no possibility of taking a corner of any kind, at least in what we're establishing to start with. And that's why I don't have any vector signs on here, like we would have had by a little bit later in the physics one term. All we need here to establish different direction is a plus or a minus sign, and some agreement to some convention, what the plus and what the minus means. Usually, we'll take something like plus to be in the right direction, minus being the left, plus to be out, minus to be down, and less we want to say something else, and it just has to be something we all agree on. But that's all we're looking at for right now. We won't be there for very long, in fact, if you look at the schedule. Monday, we started with curvilinear motion. We're done with rectilinear motion. So, we had this also decided that maybe that's just too much for us to write so we can do a shorthand business of just calling it s dot. Not very useful in rectilinear motion, but it will be in 2D and 3D motion when we have motion in different directions. And we can illustrate that with this notation and keep them separate. We can then talk about velocity in the x direction and keep it separate from the velocity in the y direction and even the z direction if we want to. So it'll be very useful once we get into other dimensions to use that dot notation. We also have that acceleration was the time where it changed the velocity so we can also use the dot notation on there. And we can even take it a step farther because the velocity itself was a time rate of change and time derivative which gives us other possibilities for the notation and the flexibility that those provide. There's no universal, I can't say uncategorically that one is better than the other at all times. There are times when it's a lot easier just to use v instead of s dot, but there are gonna be times when using the dot and the double dot notation. Can't have some advantage to that. Then we also discussed three possible situations, three possible problems that we could look at where either the acceleration is a function of time and then we discussed with that. We'll review those real quick here in a second. What do we do with that kind of problem find out what's going on with the velocity and the position. A could be a function of position itself which is pretty much how you live your lives whether or not you accelerate in your car is determined pretty much by where you are if you're near a stop sign, you're gonna be decelerating. Once you stop there, complete full wheel stop, fellas, then you're gonna accelerate from there. How hard and how long you accelerate may depend upon what else lays up the road is. Our corner coming up is in traffic being able to make a turn, whatever it is. And then the last possibility that we looked at was that acceleration was a function of velocity. So we looked at those three cases and we'll take them a step farther right now as we review what we got from them. So for this first case, we got it down to the point where delta V which remember is the change in the velocity. We could always get if we got the area under the AT curve, which is what presumably we have in some measure here. Maybe it's an approximate data taken curve or maybe we actually have a function and we can graph it and have the curve or just integrate it. Either way is the same to us, but we knew if we had the acceleration as a function of time, whatever that might look like, we could find the change in velocity between any two of the same time points. If we could either integrate the curve numerically, just do the integration if we know that function or an approximation of that function or you can always integrate by counting up the little squares underneath. The sort of a sort of a brute force way to integrate if you need to. And we knew that the area under that curve would represent the change in velocity between those two time periods. How do we handle the fact that in this illustration part of the integration is below the t-axis? What does that represent? What does that mean? Area tells us only something about delta V. So if we have a negative area, as we would if we did this integration, no matter what method you do an integration below the axis, you get a negative area. If you get a negative area, you have a negative delta V. In other words, the object for that portion would be what you say, slowing down. Area above the axis is a positive delta V. Area below the axis is a negative delta V. Bless you. If we're numerically integrating, that just gets handled. You might not even see that. But if you're graphically integrating, actually measuring the areas, you have to remember that we do have such a thing as negative areas. So graphically that looks like, I'm sorry, numerically that looks like that mathematically. We have the possibility that A is a constant. In that case, we can do this integration, do it almost instantly, and we get that. Friends, the constant acceleration equations. Frank, you're at a disadvantage. Everybody here, well, DJ, I don't think you did either. Had me for physics one. You didn't have me. So you and Frank and DJ are at a disadvantage because everybody here, Pat didn't either. God, all right. So anyway, in physics one, as we got all the physics one, all the constant acceleration equations, every one of my students every year has always gone out and gotten those tattooed somewhere. They're that useful. So they might share them with you sometime. What tattoo artists you used in town, they got to be pretty good at those equations by now. So you can go get the appropriate tattoo of these constant acceleration equations. What do we? We'll get you some free if you feel bad for it. I'll do it for you. We were thinking again, I'm glad we didn't invest in the tattoo stuff, now that Andrew's gone. I don't know, I'd have been a waste of time. Then we looked at case two. See if it's okay if I'm gonna be able to slip it in here in this narrative. Yeah. Case two, if you remember, starting with the assumption that whatever we know about acceleration is dependent upon position. We got that down to VDV equals a ds in differential form. If we want to, we can write it as s dot ds dot as double dot ds. This is one, students tend to forget that this is available a lot because this is not something that we've necessarily seen. Certainly haven't seen it very often. So we integrated that to see if we do indeed have the acceleration as a function of position. What it might tell us. So whatever that function might look like, if we integrate this equation between two points, then the area under there, area under that function gave us what? Remember, gave us what? This is the change in position right here. Just right off that axis. Gave us what? Not directly, it could lead to it. But what did it give us directly? If we had that area, what was it we had? Change in the quantity V squared. From that we can get the change in velocity. So if we had the area under this particular curve, that's what we'd get directly from that area. Don't forget delta V squared quantity is not the same as delta V quantity squared. So be careful with that. What it looked like, I guess, was let's say V two squared equals V one squared plus one half integral, okay, same thing. There is mathematically, there is graphically the same business. Then again, we have the special case possibility that A could be a constant. If it is, it's an easy integral to do. It comes right out of the integral. We integrate dS, that's delta S. We have V two squared equals V one squared plus one half, sorry, something's wrong. It's not one half, I don't know where I got that. Oh yeah, I do know where I got it. Cause this one half will come over, so this is actually two, two A delta S. And there's the second of our constant acceleration equations, same from case two. We can do something else with case two just to give us another tool in all of this. Remember what we're trying to be able to do is solve any of these accelerated motion equations we can come across, whether the acceleration is constant or whether it's not. Most of our time in physics one was constant acceleration. We'll do some of that, but we won't do exclusive by that. All right, so there's another way to look at this case two that we have there. Let's rearrange things a little bit. I'm gonna take the same differential form we have there and just do it like this, which is no big deal. Remember what we're looking at is what happens if A is a constant. Let me return that here. Actually we can do this even if it isn't, but what we're gonna do next requires that it be a constant. So we can integrate this S one to S two. Integrate across V, V one to V two. Of course, this left hand side is nothing but delta S. The right hand side integrates. Remember here A is a constant. It can come out of the integral. We're then integrating V dV, which we already did over here anyway. It's just we had to leave the A where it was in that instance. Now we're doing the A equals a constant instance. And so that integrates to one half V squared between the limits of V one and V two. Well, that's no more than we did there. There's nothing really new yet. Well, other than A is a constant. But to get something new, if I took this any farther, we just get that and there's no use to us. We don't have anything new. So I'm gonna do something a little bit different. I'm going to, let's see. Let's let V two, V squared, V two. Let's let V two equal V one plus the amount that the velocity changes in a time delta T plus A delta T. Essentially what we've got here is this is V one plus delta V. That little bit there, that little tail there, that's nothing, oh, what I had down here and we just raised a second. That's just A equals delta V over delta T. So I'm gonna put that in there and do this integration. So I get one over two A between the limits of V one and V one plus A delta T. Go ahead and put that in. We got one over two A. Put in the top limit. So I have this quantity squared as the first limit. So I'll work that out as we go. That's V one squared. Remember that we're squaring this business here. So we'll be foil on it. But we'll do it a little bit faster than that. Plus two V one A delta T. That's the middle term. So I have that squared plus those times itself twice plus that squared plus A squared delta T squared. Did I do that right? That's the top limit of the integration put into our result of the integration. And then we subtract from it the bottom limit put in, which is just minus V one squared. As we notice one, we have V one squared minus V one squared. So that cancels with what we're left over. We have A on the bottom and an A in each term. So that A cancels that one, cancels the squared. Remember we couldn't have done that if we still had V one left over in there. Those cancel each other. So we're left with, let's say delta S equals one half delta T squared plus this one half cancels that two. We have V one delta T. You will also recognize that as one of the constant acceleration equations. If you don't, check your tattoo. If you don't have one, check somebody else's. Then go get one. First exam without your constant acceleration tattoo. Right, Colin? Probably even refer to it a couple times over Christmas. Check. I think that space shouldn't make it mandatory when you're driving a car that they project on the windshield with some light shining up there so you'll always see it with constant acceleration, but you'll always keep them in mind. And when do you do more accelerating than when you're in the car? All right, so that's three constant acceleration equations. Guys and gals, how many are there? There's four. So the fourth one, fourth one isn't really anything we derive, but it is something we have to keep in mind. And that's that anytime we have A equals a constant V over delta T is also constant because A equals delta V over delta T. So of course that's true. We then look at what's happening when something is accelerating with constant acceleration. Well, of course it's going to be a straight line. It could be negative, could be flat, could be down, or positive like I drew it, but I have to draw something so it's just arbitrary what I drew. Then remember, whatever velocity we start with, V one and whatever velocity we finish with, each velocity is simply the arithmetical average of those two numbers because it's a straight line in between them. Because it's a straight line in between those two, the midpoint is halfway in between. How eloquent is that? And don't forget though that the average velocity is delta S over delta T. And so there is our fourth constant acceleration equation. Anytime you have a constant acceleration problem, those four equations are at your disposal. Every time. Be careful though. Students very often try to apply these when we don't have a constant acceleration problem and they don't work. But is it a much brighter group than that group? Any questions there before I clear the board and we do some other stuff such as just for reminders, we'll do a constant acceleration problem. A simple one here. We got, here's the town limit. Trying to establish something about a legal jurisdiction over this automobile problem. So we need to know exactly where everything was happening in this problem. So we're gonna need to find where object, some object, some, well, we'll put a car here. Let's use my car. I like it when we use my car for something. There we go, that's my car. Look out the window, you'll see my car in the parking lot area right there. Initial velocity, 15 meters per second. Going to constant acceleration problem here to start with and that acceleration is four meters per second squared. All right, so the first question we wanna find is find the distance traveled in two seconds. So two seconds beyond that initial spot, 500 meters, find how much distance was traveled in that time. DJ and Pat, we boiled this down in physics one to a very straightforward way to solve all these problems. The trouble is I just gave you four constant acceleration equations. One of them works, three of them don't. And the problem is how do you figure out which one of the three, which one of the four you keep and use? I always thought it was kind of like, almost like magic when I saw the problem solved in the book, they always said use this equation, they never said how they figured out to use that equation. So in physics one, we were able to boil it down to a way that we could instantly and accurately decide which of the four equations to use at all times. Is that right, boys and girl? We did just that. Goes like this, I'll give it to you real quick for your digestion and then I'll even give you a sheet of the constant acceleration equations that they all had. That's what they took down to the tattoo parlor and had put, it was up to them where. Every constant acceleration equation, a problem has only five possible variables in it. There's only five things that can come up in any one of these problems. Five possible variables. No constant acceleration problem will have any other variables but these five. Well there might be a slight form difference but basically these are the five variables possible. Of course, acceleration is a constant acceleration problem and might be part of the problem. As it is in this one, there are A equals. Also, the initial velocity, like we've got here, it could be part of the problem. The final velocity might be part of the problem. We don't have it here and it wasn't asked for but it could have been. It's also possible we're talking about how much distance was traveled in these problems. We are asked to find that. A certain amount of time goes by in all these problems. No constant acceleration problem can ask you anything other than one of those five, or ask, that can deal with anything more than those five things. Maybe you're given x one and you're supposed to find x two but we just convinced that down to delta x, it's always the same. Five possible variables. Then what, old students of mine? When you look for it, it's only gonna involve four of those. One of them is left out just because of the way the problem is written. We don't have in this problem any given or asked for v two. So only four variables are important. In our case, A v one delta x delta t. Didn't ask anything about v two and it wasn't given. So it's not part of the problem. Go back now to the four constant acceleration equations I just gave you. I don't have them on the board. I have one on the board but I don't have the others. Go back to those four. Only one of them has these four variables in it. All of the others have a v two in it. There's the equation you use. Works every time, doesn't it? Only one of the equations has those four, none of them. Three are given, one is not. It's unknown. And that's the one you're supposed to find and you just use that equation to find it. Works every time. What equation is that? I think it's the one we had just before that. Delta x equals one half A delta t squared delta t. In fact, we don't even have to do anything with it. It's all right there and it's proper algebraic form and we just plug each other and finish up. Make sure your units are okay. Meters, meters, seconds, seconds. Yeah, everything's fine. You can just plug it in and solve for it. No sense even bothering with any of the other equations because they have v two in them. When you plug that in, what do you get? Get who? 38 meters and 40 meters extra. That's 30, that could be 240. Oh well, it's better than what I had written down first which was 118,000 meters, 118 kilometers. That's you guys driving. You could go that car in two seconds, I could. So it's what, 38 meters? All right, so then the police can say, well, that's out of our jurisdiction. We'll have to share a few for down the road and handle this one. As if, sure, if anybody can catch that car, can have it. All right, just one more step of practice. Find the position x when the speed is 25 meters per second. Once again, four things are involved. Three things given, one thing not. Four things involved, decide which equation to use. The other three equations won't work. Just in this case, the problems change slightly. But still, three things given, one thing not given. Is that a question, Jay? Yeah, we don't know the time. Well, I don't know, let's see. We still have A, same acceleration. We still know that it was going a certain velocity, V1. We're given, what is this? Yeah, this is, in this case, we could call V2. Maybe you want to call it V3, just to separate it from whatever V2 was in here, but it never came up, so I will not belabor it. And then, what's the fourth variable? Well, we don't have this in there yet. We want to find out how far it went. You have to pay attention as this x is measured from the town line, or is this x additional or what? But those are details. Those are just details, that's how it's called. So, three are given, one is not. It's only one equation is going to work. Isn't that a beautiful method, make it simple? Yeah. Aren't there ways you had an E for physics one? Yeah. Do you remember who you had? That was at Clarkson, right? No. No? Oh. Was that, oh, that was high school physics. You took college physics, because that wasn't, was that calculus-based? Yeah. Yeah. So, did that work as college credit? Yeah. But how come I had you again, for a minute? Oh, the counselors did that. Yeah, I see. The counselors, you tell them physics and calculus, and their eyes roll back in their head, and things start spinning around, and they start getting like this, and they have to go lie down in the back, and then during that time, they're having a vision of what class you should be in, and it has no reality to it, so. We try to tell them, please don't counsel the engineering students, please send them down to us and they won't do it. So, we're gonna have to edit that part out. All right, so only one equation works. Which one is it? U squared equals B1 squared plus 2A. In this case, we're using delta X, not delta S. That's the only equation that has those four things in it. The time is not a factor in this problem, in the second part of the problem. It's a different time, maybe it's asked later, but it's not asked now. And so, you get what for delta X? 50 meters? Let's see, please God, let us agree on that one. Yeah, 50 meters. Now, if in any of these problems, you are asked for the one variable that's missing as a second part to it, then what equation do you use? I just have a question. Hang on, let's answer the question. If for a second part, we weren't asked to find something entirely different, what if we were asked to find what's the velocity after two seconds? Then, which equation do you use? Well, a more general answer from my Physics One students. Because now, all five variables are part of the problem. Because we just found one of the missing ones, in this case, that had to be delta X. Then, if it asks for what's the velocity at that point, now, all five variables are part of the problem. That means all four equations could, well, actually the other three could work, because they're the ones that have B2 in them. Which one of the three do you use? Whichever one you want. Whichever one you want, because all three of them are gonna get the same answer. Use the easiest one. You don't want to, for example, if you're looking for the time, solve the quadratic. In this form, if you don't have to do one of the other forms, because there's hazards in doing the quadratic equation on the fly. I think it's the one where you didn't find, just find the answer. I used only the one that you have, but you didn't solve already. That way, I can make a mistake from your problem here. When you found delta X, if you made a mistake, and you used that mistake, and you felt the X, and you don't want another equation to be solved twice. Well, if I understood your question, if we solve this, and then as a second part, I had to find the velocity at the same time, then find V2, you can't use the same equation. Not the same one. I mean, you use a different equation, but only with the nodes that you have. Oh, I see. Not the found one. Oh, okay. Yeah, I guess so. That or do it right in the first place. Or do it twice. Do it twice and check it. Always check your answers if you can. All right, that should be just a bit of a reminder or a refresher from Physics One and an introduction to my amazing method on how to solve these problems. So let's then do a non-constant acceleration problem. More likely the case in physics and engineering. Sometimes we can approximate it as constant, but can't always do it so. So imagine we have two charged plates. Already in Physics Three now. A couple hard, but we'll be soon. And that's when you do electrostatics and electrodynamics. So we have a charged plate which sets up a magnetic field between those plates. We're not gonna belabor the physics of that. We have a charge, two charged plates, 200 millimeters apart. And suspended between them is a little metal particle right in the center at 100 millimeters. That's its location at the start of the problem. Now as most of you will learn, the particle will then have a position-dependent acceleration will say that it's down. It depends on the charged plates. Again, we're not worrying about the details of the electromagnetic. And in this case, the acceleration is a function of position goes something like four s per second squared where s is the position measured from the top. So starting from rest, right then in the center, we then release it. It's subject to that acceleration. I'd like you to find the velocity at the bottom plate. With what speed will it hit the bottom plate? Absolutely. Oh, that's not suspended. You're reaching in there and holding it. Right. So I'm just trying to figure out what the initial velocity is zero. Yeah, initial velocity is zero. So we'll write that down. Remember, from that you need to get the equations, the concepts, the mathematics all put together and solve it. No, it starts in the center, right there, point one. We want to find with what speed it hits the bottom. We go to 1235, I think. So it's too early for a get out of class question. Yes, you gotta be careful with what you say just as much as I do. Well, the administration may see what I say. I can always show your parents what you say or maybe your past. All right, we talked about what we do with certain problems that are where the acceleration is positioned dependent. A short relation to that type of problem that had four variables in it. Do we have those four variables here? What did I tell you to do for position dependent acceleration problems? Of course, you don't remember. That's why you take notes, look in your notes. Didn't remember to write it down. Is the equation V2 squared equals V1 squared plus two? We're broke, we're broke. Something. Yeah, I think that would work because that came from our basic relation that we had, then we didn't take that and engraved it. V2 squared, well that's what we're looking for. V1 squared equals V1 squared plus, what was it? ADS. So V1's given, in fact, with zero. All you need to do is add interval. You've got the equation, the functional relation of the acceleration on position. You guys agree? We'll keep a general understanding that we'll use three or four significant figures on answers. I bet you got it because you did it. Did you get the same answer? No. That's with anybody? That's bad for you, because Bob's always right. I don't even care what his number is, he's right. You're wrong. Ruby, how are you doing? DJ, check, oh, we're both calculators today. Huh? Do you have a real one? That's one. That's awesome. 0.2, what's in my car this time? It's 10 of me, is that all single of me? I don't know, I thought that was 6,200 of us. It's all over the years. Colin, do you agree with anybody? I don't think I've got the last answer. We have a comparison among the three, all of a sudden. Maybe you're right. Did you add, did you add 1.35? That's what I mean. Yeah, that's the first time I've ever got that. Oh, we're already wrong. Yeah, 1.25, you think it was percent? Huh? That might be right then. Is it 0.25? No, that's 0.35. I got 0.25. How are we doing? I like your way. Check with another group. AJ, what did you get? We got two answers. Ah-ha-ha. I'll check with the DJ. 0.25, 0.26. We don't need two answers. Certainly, they didn't get two answers from the square root, did you, because that minus wouldn't make any sense here. Long term, calling down positive. That's all I get. I think you got like square and then that's all I should get. All right. Let's say, was there any trouble integrating 4s? Integrates to what? 1.5s squared, is that right? This is not an indefinite integral. This is a definite integral. What are the limits on the integration? What is s1? It's not 0 from the top. The acceleration is a function of s. So we have a position dependent acceleration. We do not have a 0 on the lower limit. Upper limit, then, is the bottom plate, 0.2 meters. What, Bob? You can leave anytime you want. It sounded nice when you brought it up. Good idea. I'm out of here. Tim left his book, his calculator, everything. He said, that's a great idea. I'm out of here. All right. Here's the next part. How much time do you have to get in there and be ready to catch it? Should we make this a get out of class question? I don't want to just start something, barely get into it. So far, was it important? You have an old lady? Is that your mom old lady? Your mother-in-law old lady? No. Oh, your old lady old lady? Your wife? Are you married? You're not married? No. Girlfriend. He was just an old lady. Didn't he say it was my old lady? No. I said mildly important. Oh. My little one. My little one. It's taller than I am. No, it doesn't start from the top plate. It starts from here. Oh, so from halfway by the bottom? Yeah. So this is a class question. You're here for the weekend with me. Happened a couple times, but I got a class question. No one answered it. Ordered pizza, sleeping bags. It was cold. They turned the heat off from the door here now. So we could watch all the seasons of Dexter. Probovo on this weekend. How did I miss that? We can watch the University of Oregon play in the BCS for a certain amount of hours. Got it already? If you check with me and you're right, then you can know. See, that's what a get out of class question is. Well, you don't shout out in the answer question. You don't say, okay, that's good. You can know instantly everybody's got the answer. It's amazing how that happens. Luckily, luckily though here, you're wrong. Why'd you get that too, Booby? What I'll do is I'll put them up on Angel. And just print out another one. Frank, DJ, Pat, I'll put the concept celebration sheet. That's what I gave all my physics ones to. I got the answer. I put each out of that. Well, nobody called me back. We have .832 seconds. Possibly three seconds. No, you didn't tell me. Look, that's seconds. That's not what I got. Here. That's not what we have. We don't have A as a function of velocity. That's where that came from. We have A as a function of position. So you can't use any other relation but that. We are attempting something like, I see somebody doing it. Remember, whatever the integrand is, must agree with whatever the variable of integration is. We have an awkward one. One more privacy. You get it, Pat? No. Balan? Are you thinking, what am I going to do? Drag it down. Drag it back into the classroom. Mine, would you try? See, we have distance involved. We know the acceleration. We're trying to find the time. Do we have any kind of relationships that involve those? We can also throw in, of course, any velocity. For instance, W. You don't have to if you want to do any of the integration that we have. This is the one I ended up integrating but you've got to come up with velocity as a function of, then you will get delta T there. But you're going to need to know velocity as a function of position. You can get that if you integrate this not up to the definite limit of that, just to a generic limit of s. You can do this integral and that will give you v2 as a function of position. Then that you can integrate here and will give you delta T directly. v2 will be the square root of ds from point one where the problem starts up to any position s of the square root of that because that's v2. That will give you the velocity as a function of position and that you can put into here in any group. And that should give you six, five, eight seconds. Now that y'all have an answer, now you can go. To put this solution online? No, no, no. You should go try it first. Understand how to do this. You just, you do exactly what we did before instead of putting in point two for the upper limit. Put in s. Which means you can use any final position s because it'll make the whole thing a function of position. You'll have velocity as a function of position. Just complete this integral with that as the other upper limit instead of the point two. Once you have that, then put it in here, integrate between the same limits on s only this time we are taking it to the bottom. Give it a try out. You're pretty good at this. Put your head to it a little bit. You'll get it.