 As you remember, from Monday I hope, a lot of what we do here is essentially just sort of an advanced physics one, except that we take into account a lot more instances where the acceleration isn't constant. We won't do a whole lot with actually varying acceleration specifically with time, but we will have periods where there are different accelerations, however, within the periods they're constant. We'll do a couple of those from today too, but for the most part we have to be able to handle any possibility that comes our way. And very often, and John, you probably saw some of this with that daddy we were looking at earlier, the acceleration might not specifically be known as a function of time that can be approximated with a function of time, and then from there we can use a lot of the relationships we are establishing as we went through things. So we didn't add this first possibility that acceleration is a function of time or is known as a function of time. This is the more common one we work with. For obvious reasons, if you have some kind of accelerometer, it's most likely that when it's running and when it's recording, it's recording as a function of time rather than as a function of position or velocity. So we had this possibility that there would be some function of time, whether it's known as actual data points or if the function itself is known, it doesn't really matter. Either way, because the acceleration is the derivative of the velocity, then we know too that the change in velocity is the area under this curve between any two time periods. So that can be very useful to us if we want to go from one to the other that between any two particular time periods, we can just take the area. Now if you have the acceleration, not as a nice function like this that's easy to integrate, but you have it as discrete data points, you can still estimate the area under the curve by simply calculating the area, actually looking at the graph and counting the squares underneath the area and using that as an approximation of the integration. Not at all uncommon to do that sort of thing. In fact, there are software packages nowadays that will take discrete data points and approximate the area under their body, that very method as they estimate what this area would be that we could then use that to find the approximate change in velocity from that area. We have, of course, the special case that the functional relationship with acceleration in time is such that the area is a constant, then simply the integration of this then becomes that, or the derivative of that just becomes that a equals delta v over delta t. So this becomes one of our constant acceleration equations. So that's a very standard, very useful way to work through dynamics problems, especially if you have some accelerometer that's recording the acceleration as a function of time as you go through things. I think the second possibility we looked at was that acceleration was a function of position. I think that's the order we did to them. It's not important what the order is just as long as we get something out of it that we need. So if we have some known relationship between the acceleration and the position, now this is something that might be a lot more likely you could do, especially nowadays, since every smart phone works like a GPS and then most of them have accelerometers in them anyway. I know the iPhones have accelerometers in them. You can get a relationship between whatever acceleration you're seeing and where you happen to be instead and that may come from the very same data stream of the type of thing you were looking at today in general with your trip into school. But this is also the type of thing you'd see in a constrained track of some kind like a roller coaster ride where the accelerometer, the acceleration is very much a position function of where you happen to be on the roller coaster ride and what happens to be going on at the time. But then we looked to what the area under that represented and if I remember between any two particular time period times you need to look at the area under this was the change in the specific kinetic energy. Remember I used that term on Monday, the specific kinetic energy? That's a term you'll come across not so much in this class. In fact I think this will be the only time you'll come across this term where the prefix specific is put on the front of something. It's very common in a class you'll take called thermodynamics. Again a dynamic course we're interested in the change of state of things, in that case the thermal change of state. This prefix specific means nothing more than as per unit mass. If we take the kinetic energy that we remember and divide by the mass we then get this term that we have here would have the change in that specific kinetic energy. This just has to be the specific kinetic energy at one point. But this is a very common thing to do, to divide through by the mass to get the quantity you're discussing per unit mass. It's also fairly standard that when you do that you then go from upper case to lower case for whatever the quantity is so that it makes it easy to recognize. We won't worry about that in this class. This is the last time that I remember that I recall we'll see this type of thing. But when you get to thermodynamics you'll do that kind of thing a lot. Where you have a total quantity of some thermodynamic property and if you divide through by the mass to get a mass per unit mass basis then you go from upper case to lower case and then still talk about it a lot in that way. Alright so just for your future edification we had that kind of thing. Then we looked at the possibility again that the acceleration can be constant and we came up with our second constant acceleration equation of that form. So there we now have two constant acceleration equations. If you have a constant acceleration problem, either of these two equations could apply to your problem depending upon what it is that you'll what you have in the problem to work with. If you have the two velocities and the acceleration and you're worried about distance traveled then that would certainly be the equation to use. And then we had the third possibility, acceleration as a function of velocity. This is very much the nature of fluid dynamic drag. That acceleration is some function of velocity. Very often it's linear with velocity squared, even velocity cubed depending upon the situation. So this is also a very common form for us to come across in the real life study of the physics behind what we're doing. And if you remember then that came to a relationship that looks something like that. And I requested that you remember that because there will be times when we come across it and it's just not an obvious one. It's not one that you're more familiar with. We didn't come across this particular relationship in physics one. So you have to be careful with that. But you don't forget it. Now that in particular led to these very same constant acceleration equations that we have anyway. So it's not going to be of any great use to add anything more to those. However we did have the, oh well that's not quite true. This one did lead to the constant acceleration equation. One half a t squared plus v i t if we left the delta t in there. Remember we got to this by eliminating delta t from the problem, d t from the problem. If we leave it in there then we can integrate the equation to get that. So that's our third constant acceleration equation. Then of course there was a fourth constant acceleration equation that did not come from any one of these standard starting points. You may remember what it was. The fourth constant acceleration equation. Those of you who had me for physics one, you remember I told you you get these four equations tattooed somewhere. So you could always refer to them. Now it would be a good time. So what? It's using the fact that the average velocity, which we know to be delta s over delta t, the instantaneous velocity is d s d t but the average velocity is delta s delta t and you can arithmetically calculate the average velocity by taking the start velocity and the finish velocity and dividing by two as we would any mean calculation that we would do between two numbers. So we have those four constant acceleration equations at our beck and call as we need them. Alright, so there's a moderately quick summary of where we've gotten to on Monday. Looks sort of familiar from physics one, I hope. Alright, so let's look at a problem. One of the less common ones, students aren't quite as used to, and that's this deal where we have an acceleration as a function of velocity. This is very common, as I said, in fluid dynamics problems. So it's very common when looking at something like a shock absorber or some kind of damped oscillator. They use this very fact to do the job they're supposed to do which is to decrease the acceleration of a problem by using the fact that the acceleration can be a function of velocity. So if we have a piston in a cylinder and that cylinder is filled with fluid, through the cylinder we have a couple small holes. Sorry, through the piston we have a couple small holes. So as the piston moves through the fluid, depending on which way it's moving, the fluid has to travel between those holes as the piston moves up and down the cylinder. Since the drag force can be a function of velocity, then the acceleration, too, is a function of velocity. In fact, it's typically linear to the velocity itself, so we might have a relationship of something like that. Why minus sign? Drag causes the speed to decrease, resulting in negative acceleration. Drag is a friction force. It's always in the opposite direction the object is moving. So if the piston as pictured happens to be moving to the right, the acceleration then is going to be to the left. It's going to cause the piston to slow down if it was moving at some speed. There's no force on it to maintain that speed. It's going to coast down to a stop because of the acceleration in the opposite direction it was moving. And that's exactly what a shock absorber is meant to do, or a dampener in an oscillating system like your shock absorbers are. All right, so we have that set up there, and we'd like to do a couple things with it. So let's first express the velocity, and we have the acceleration express the velocity as a function of time. We know the acceleration is a function of velocity. Let's figure it out as a function of time. Any recommendations, any ideas and what we can do with that to get to that point? We have acceleration as a function of velocity. We want to find the velocity as a function of time. If we use our acceleration, our definition of acceleration, then we can link those very things. We have the acceleration as a function of velocity. We want to find the velocity as a function of time. So we can use this very relationship that we have right there. So moving it around a little bit, then we can integrate between some initial velocity it might have, and then some later velocity it'll have over some initial time period. We'll take the initial time v0, just to make things simpler, to some later time t when it achieves whatever this velocity might be. If we can complete this integration, then we'll have velocity as a function of time, because we'll keep it sort of open-ended at the top limit there. So this side integrates very easily. I'll do it, so you can do the harder one. Joe's getting excited because he knows the integration is coming up. We know the acceleration is... Hang on, we need something else here. Oh, I'm sorry. Sorry about that. We did that the wrong way. Since the acceleration is a function of velocity, you need to collect that there. That's why things weren't going where they wanted them. Now this will integrate between time 0 and some time t, and now this between the velocities, v0 and some later velocity v, whether it's increased or decreased, will depend upon a lot of the things going on in the system. All right, so this we can integrate a little more... a little more... a little more tightly. Sorry, this will be okay. Yes, because I brought the a down. Yep, actually let's just put that k over there where it's easy to slip it in, because I brought the dt up, took the a down, and then integrated both sides. So this side integrates k delta... actually, since the bottom limit is 0, which is to minus kt, and then that side integrates to, remember it's off hand. Yeah, the natural log between the two limits, and since they're subtracting the natural log and then we put them back together, they'd be v0. Or in a more complete form to answer what we were looking for, the velocity is a function of time, that if we solve for the velocity as a function of time, we get v0 e to the minus kt. And what's that look like on a vt graph? Starting at some initial velocity at time 0, then what does the speed do? And remember, this is the piston being unforced, undriven, just simply coasting down at this acceleration where the acceleration is a function of velocity. Well, it'd be a decaying exponential curve. Looks something like that. From the same starting point, let's express now the velocity as a function of time. No, wait, sorry, we did that. What am I looking for? I think the position is a function of time. That makes more sense. Doesn't make any sense to do the same thing over again. Express the position as a function of time. All right, how are we going to manage that beast? And then I'd also like to see what it looks like to an xt curve, where x is measured from some initial position to some later position as a function of time. We do have the velocity as a function of time, and we know then that that's the derivative of the position with respect to time. So we can move it around a little bit. Reintegrate, we'll integrate from 0 to x, and from 0 to t, taking the initial time as 0 itself. That's how we can integrate this. v0 e to the minus kt dt between the semi-open and limits of 0 to t. Who's got that integral? Negative v0. David? What? Negative v0. Integrates to negative v0 over k between those limits. So it just becomes then t. So our position, our change in position, but if we do it from t equals 0, we have then the change in position as a function of time. We saw the e to the negative kt. Oh, you didn't give me the whole integral. Yeah, so the inner, yeah, it's like e to the u du. So we need then, we have the t, and then what part goes with it? Not, it's just e to the minus kt when I bring the x to the minus 1. v0 out in front, e to the minus kt, and then since it'll be the zero power, just that. Divide by negative. What part? The whole thing. Oh, the whole thing over negative k. Let's squeeze that in. v0. We would have caught that looking at the inner. Is that a little better? Yeah, that looks better. Okay, and so that with respect to time looks like assuming it starts at zero. Then what does that look like? It'll reach some or approach some maximum position something like that. Alright, the last possibility is to put the two together where we have the express the velocities of function of the position x. The idea isn't where to start on this one. Acceleration is a function of velocity. We're looking for the velocity as a function of position. What? You didn't forget it? I told you to forget it. I told you you would forget. Okay, we know several of the pieces of this. I'm starting using the x for position here. No, it's arbitrary. This is k v dx equals v dv. If we collect things a little bit so we can do the integration we'll leave the k over with the dx and then put the v dv over v. Well, that's certainly convenient. Then cancel that. And so then we can integrate that minus k delta x equals delta v. And then that graph looks like remember if x equals zero we have some initial velocity and that velocity decreases as the piston moves because of the fluid drag. So it's going to be a decreasing function that looks like a straight line. It's a constant, minus k. Minus gives us the decrease. Something like that. We could have also gotten that from the first two the one I already raised and the second when we got over there if we just put those together and eliminated dt. We would have gotten the same thing. Alright, any questions before I clear up? We'll stop the tape here. Any chance to reset my taper? Now we're going to go to sort of quasi constant acceleration problems those where there's approximately one acceleration and one time period and a different acceleration and a different time period and we're going to go through these to try to figure out what the different parts are that we don't have that need to go with that very same problem. So a couple things we need to remember that I can't imagine you possibly forgotten because we've been doing it in the physical one but the velocity at any point is the slope of the position curve and because of that as well then the change in position is the area under the velocity time graph if you happen to have it that can be very useful to us. It's also true that at any point the acceleration is the velocity graph and of course then the area under the velocity graph between any two points of interest is the area under the acceleration graph if you have that acceleration as a function of time. So keep those in mind plus keep in mind the constant acceleration equation this is always true if acceleration is constant then we have a couple other things that are true that one just says if the acceleration is constant then the acceleration always equals the average acceleration and we have our other constant acceleration equations just rewriting what we had a couple minutes ago in a less condensed form than in the last one in every buddy's paper we have those four constant acceleration equations what? we would have caught that looking at the units the units wouldn't have worked out thank you David so we have those four constant acceleration equations if you remember from physics one if you took it with me I know we went over it this way there are possible variables in any constant acceleration equation sorry, problem and it's any of these five that show delta t delta s v1 v2 and hey those are the only five possible variables in the constant acceleration problem notice that only four of them have any one of these equations whatever four constant acceleration whatever four variables are part of the problem you're working on three of them you'll know one of them you won't that tells you which one of these equations which one of these equations to use whatever four variables you've got in the problem one of these equations has those four variables that's the equation you use I remember taking this course and well physics one and this course and just not understanding in the sample problems in the book how they knew which equation to use they just started using one of the equations and I finally realized if you lay it out this way there's always four in any one problem three of them are known one isn't and those four tell you which one of the equations to use alright so let's let's do a sample problem here imagine the position of a bicycle is known it's very the same type of thing that John would have gotten off his GPS software as a function of time and we'll use good standard American units of feet and let's say it went something like this after about ten seconds the bicycle would have covered approximately a hundred feet and then at some time later say thirty seconds not about to scale it had covered another portion of the distance such that position was varying in that regard and then was linearly changing thereafter actually I need to fix this this is why you should take nose and chalk just need to bring my hundred down a little bit so it looks something like that so we have a curved section linear section and it reaches a distance of about 500 feet after 30 seconds for that first section maybe we'll call it section A the position is proportional to T squared and then of course for the second section it's linear which means it fits the form S equals M T plus B and B is the Y intercept so let's find for each of these graphs find the graphs we've already got the position now so let's find velocity as a function of time and acceleration as a function of time for these graphs alright well we know the velocity is ds dt we're not going to be able to find that until we have this functional form of the two sections for the the data that we have for section A how can we figure out what K is if we figure out what K is then it's a fairly straightforward manner of differentiating it once and then differentiating it twice to get the acceleration how can we figure out what K is Joe you look like you knew for a second you need to figure out what K is you also need to figure out what M and B are we need to do this but we need to do this separately because it's a different functional form for each section any ideas to find K we can put any one of the data points in any data point that lies on the curve we can put in here and solve for K so we have a point which is at 100 feet and it took 10 seconds to get there and we can solve for K do it in your head 10 squared in your head any units on it well if we have seconds in here and feet over here this will be second squared this would better be feet per second squared for acceleration it might be certainly got the units for it is that the acceleration very very simple for me to write down equals A and I'm done with this part for section 2 sorry for section A well is this section a constant acceleration section the position is a function of we take the first derivative to get velocity that will be a function of T the second derivative will be a function of T squared well it will just be a the constant the second derivative will get rid of the function of independent amount of time so it is constant acceleration in this section is that this acceleration we don't have to guess let's take it a little farther now that we know this let's do this derivative we don't need to guess the equations will tell us what to do we've got d dT of one foot per second squared T squared I don't like writing it down without the units even if it's a one out in front by one is that right? two comes down reduce the power by one we get two feet squared T is that the acceleration which one of these is the acceleration is it this is it that or is it neither delivering again A equals dvT that will solve this problem once and for all for us we've got dv per second squared T that's an easy one to derivative we get dv per second squared as the acceleration why wasn't that the acceleration because the acceleration as defined by one equation is twice as okay well we knew that when we got down it here but if we're just looking at this there's some way to tell that was not the acceleration that was only half the acceleration or we have a constant acceleration equation that looks like this normally has dvT there but our initial velocity is zero so I just left it off so whatever's out in front of the T squared which is our K is one half A and so by the time we got to the bottom it all agreed so now we know the velocity and the acceleration for section A velocity is 2T squared with all the positions and all the velocities being in English units and the acceleration in section A is just the two feet per second squared you get sloppy with units I don't have to do it too okay so we get that then for section A what about section B oh yeah the position is T squared again we would have caught that if we put numbers in with our units alright so that's section A where are we going to fit in section B we've got the position well we at least know it's linear what we don't have is the slope and the intercept how are we going to find those oh man see this is rough because you can't think back to calc for this you've got to go back to pre-pre-calc for this which for me was 40 years ago so I have an excuse you don't well we could do the same thing we did before put in how many data points will we need to use in this case we'll need two we have two we have two unknowns we need two data points two known points well we've got two so we can put those in not to belabor the point because that's not the situation here you can pretty quickly come up with those the slope you can get right off here in 20 seconds it goes 400 feet so that's what 400 feet over 20 seconds or 20 feet per second t and then I believe the intercept comes out to be minus 100 it's there about minus 100 and then real quickly we can do what we need to do again derivorate those units are in feet what it's just that we're too excited during derivoration too much and so the acceleration is 0 alright let's do this let's plot the velocity as a function of time and then we can do the same thing for the acceleration as a function of time for section A what's the velocity look like for section A it's linear with a slope of 2 and an intercept of 0 how far, how fast is it moving after that first 10 seconds it'll be moving 20 feet per second so we put the two sections together should look something like that is that jive what's the acceleration graph look like then it's 2 at between 0 and section A it's just the slope of the velocity curve which is a constant is what do we have 2 that's feet per second squared and then 0 then 0 for the rest let me ask you this what's that area represent and can we double check it upon any of the others to make sure it all agrees what's the area under a velocity time graph equal to distance distance traveled between those same times so this area is 10 times 20 times a half is 100 and that's how far it went for section A this area see another 20 seconds at 20 feet per second 400 feet and it went another 400 feet in section B so these graphs agree going up and down derivative coming down and the area going up what's this area cool it David take a break kick back a little bit Joe was saying darn he was doing great what is that area equal to could be nothing just because we can take the area on their graph doesn't necessarily mean it means something does this area here represent anything Bill you're not yet change of velocity change in velocity between the same two times 10 seconds and 2 feet per second that area that change of velocity is 20 and that's just what happens here change of velocity between there at the same point what about section B here's a tough one what's the area under the graph for section B negative negative area how about the zero area meaning the change of velocity was zero and that's exactly what we saw there do these do these three graphs taken together look like something else you've studied in your recent past doesn't this exactly the way the shear moment diagrams go the load the shear and the moment diagrams have exactly the same relationships as these ones do alright I'll let you do one a little bit see what you can put together for it this time I'll start with the acceleration graph you come up with the two other graphs the velocity and the position graph it looks like something like that so here's the 10 seconds we have an acceleration like that 10 meters per second any time we get numbers down we better get the units down as well and then after 10 seconds we have negative acceleration well let's say these are test results for a car in fact we'll make it my car very my car better you're a bit of a realist sometimes David so test results for a car what does the velocity time graph look like what does the position time graph look like just like in the shear moment diagrams things change in the three graphs at the same time for each of them we'll just run this one out to some some end time t maybe we'll call it t prime some end time t there and I don't know what that is what is your acceleration when t is about 10 seconds again at 10 seconds it changes abruptly from that to that I don't see it labeled your lower line there oh the value for it alright minus two that's what you needed too many things to think about way too many things to think about up here go under all kinds of pressure alright and I'll give you that it starts from a standing stop we'll measure things from there and that was its initial velocity trial for you guys works the same way our shear moment diagrams worked and put numbers to these different important spots here through the problem and then things like total distance covered final velocity intermediate velocities if you can come up with them those kind of things this goes well we'll have a get out of class question we can start early go to 35 remember I tried to go to 20 after on Monday what do you got Tom remember the acceleration is the slope of the velocity curve and the pressure started on these pieces remember one thing that's unknown is what this T prime is see if that piece of it can come into it filled and okay do we need our work oh meter where should I set it there's our work oh meter we're we're right about there maybe Travis I think is already done so his work oh meter is a little bit lower can that help some got a little bit there alright the slope of the velocity curve is acceleration and acceleration is 10 here then we know the slope is 10 down here since the acceleration is constant then the slope is constant it's a straight line so let's see what's that area represent this area here represent change in velocity and we're starting from v to 0 so it'll be the final velocity after that time period that area is 10 seconds by 10 meters per second so we're going to end up at 100 meters per second at the end of that time period then what then what happens after that section decelerates as we use in the common terminology has negative acceleration slope is only minus 2 until when yeah until t prime but what do you think that is t prime is 0 yeah it's a test car if we don't bring it back to a stop somewhere we just let it keep going we've lost the valuable piece of test equipment so we can pretty well assume that it's going to come back to a stop and the driver can get out of it so that slope is 10 meters per second squared this slope is minus 2 meters per second squared what is that area what is this area across here maybe it helps if we call them section a and b again like we did before what's this area in section b represent still represents change in velocity the area under the acceleration curve between any two times is the change in velocity how do these two areas compare maybe this is delta a delta b a delta b b for the two different sections how do those two areas compare they've got to be equal whatever speed it picked up here it's got to lose here to come to a stop you can use that then to find out what a t prime is what is t prime we have a 100 an area of 100 meters per second here we need to have a minus 100 meters per second here it's a minus 2 so we need another delta t of 50 seconds so we know that t prime then is 60 seconds because we started that time period in t what about this second one we're going to have to go up for that part what about this second graph velocity starts at 0 what's that tell us on the position graph no the velocity is 0 here what does that mean for the position it means the slope for the position graph so we start at 0 slope and then steadily increase the slope to here then what happens the slope steadily decreases back to 0 the only way we're going to be able to draw something like that is a curve that starts at 0 slope until it reaches 0 then we're going to have something like that remember the second derivative of a curve tells you it's curvature the acceleration is the second derivative so it tells us the curvature of the position graph acceleration is positive curvature is up acceleration is negative curvature is down so what is the final distance then travel how can we figure that out David I told you to take a break my problem I'm going to do the principle how can we fairly easily figure out what this position is the total area under the velocity curve is delta s and so that's 1 half of 10 that's 500 and 1 half of 15 times 100 that's what 25 500 500 plus 25 1000 total distance of about 3000 meters there's other ways to find that that's probably as easy as anything just calculate the area and you're trying to do the shapes alright last little thing you get out a class question and the weekend starts what you don't have a weekend alright here's a city limit sign 500 meters from that sign r why don't we make it my car traveling on the ground thank you David caught that the velocity of 15 meters per second if the acceleration is 4 meters per second squared find and you get 4 minutes I'll only make you find one thing otherwise you'll never get out early find the position at 2 seconds assuming that acceleration is constant and that's your get out of class question you get it, you get a door note you don't get it, you stay here for the weekend I want to know how far from town it is you got it Chris? you want to go earlier? alright let's say that's not what I had I never do anything wrong as we know this on David double check that remember I'm asking you how far from town it is so your number doesn't make any sense Chris started 500 meters from town from there with that velocity at that point that acceleration at that point where does it end up 2 seconds later does that help, now that I drew it up there you can just measure it just do something it's going 2 seconds so it's just not too tense 2 seconds no, look where it's pointed do you drive backwards when you drive? look at the car, the town's here oh my goodness 500 plus if we were in Arizona you'd be lost that's wide open spaces out there I guess not you would have been so if this was a tombstone they would have shot you full of holes now you have another number that's fine that's not what I had is that what you had? that's not at all what I had maybe we're going to be here for the weekend anybody have something else? David let's see yeah that's what I had oh wait no that's not the right thing nope no I looked at the wrong number alright we'll do it together want to find where it is at 2 seconds so we can call that S sub F we have 4 things in this problem well we're looking for S but that's the same as delta S so we'll represent that that way we've got velocity acceleration and time which equation has those 4 things in it change of position velocity acceleration and time t squared plus b1t so the initial position is 500 meters 1 half the acceleration 4 meters per second squared times 2 seconds squared what's that? half that's 8 times 4 is 32 and you guys had 38 38 that's 2 squared and then the initial velocity oh I've had more on here same as 0 there plus 15 meters per second plus 2 seconds that's 538 you sure? 0.5 when you multiply 0.5 by 0.5 1 acceleration by 0.5 4 meters per second squared here? oh in my head see I have like 740 meters so yeah that's the nature of a little red sports car so I don't know I don't know what that is alright here's what we'll do next week I'll let you go 2 minutes early