 We've set up the main type of coordinate systems that we're going to use. In fact, they overlap a little bit. As we're going to see here, we've got a couple problems coming up. It's very easy to look at the problem in terms of two of the coordinate systems we used. The normal tangential coordinate system is very, very useful for some kind of curved path, which is pretty common. It's certainly a circular path, and all kinds of different machines in the life. It's very, very common. But we tend to see things well in the Cartesian coordinate system. So we'll relate the two as we go through this business for the next couple weeks. What I want to look at now is this idea of relative motion. What we mean by that is how the motion of one object looks to another object, as if you were in a car looking at other cars going in other directions, other speeds with other accelerations. There's a motion that you perceive that car to have, as if you were in a stationary coordinate system not moving. Everybody's experienced this, I think, at some time or other. If you've ever been in a parking lot and the car next to you starts to move when you notice that it starts to move, but you think that you're the one moving, and you panic and reach for the brake ever gone through that? Yeah, especially after a few beers. I'm glad I'm taking it. He said it, I did. And this is one of the students over 21. Also, it's very interesting if you're ever on an airplane and you look at it and you see another airplane flying by. It looks like they're sort of skidding sideways when you know that most likely they're finding a street line. So that's the idea of relative motion. It's pretty easy to set up for us so that we can get right into it. Some object, maybe my car, moving with some velocity and some acceleration. It doesn't matter which direction these things are. I have to arbitrarily pick something just so I can draw it. And we're worried about how that looks to another car. In this case, I guess a pickup truck going some other velocity, maybe more, maybe less, maybe the same. It could be any possibility. And just for interest's sake, we'll say that the acceleration is in the other direction. So we have two cars traveling in the same direction. The car behind is picking up speed. The car in front is decreasing in speed. But it doesn't matter. It's all relative. So we can establish our business here first if we come up with some kind of position between the two. So from some arbitrary origin, we'll say that car A is at some position, some point. And of course that's varying with time since he has velocity. But it allows us to locate these two objects from some arbitrary reference. Now the relative motion that we're concerned with is going to start first with the relative position. So for example, if you're in car A and you look down at car B, it appears to be that distance away. That's the position of B relative to A. So that's going to be our subscript notation for relative motion now on. So this little bit here reads as B relative to A. How things look in A as you look out at B. What is B doing relative to what A is doing? That's very easy to set up from what we've got here. So A is simply going to be xB minus xA. And that's the relative position of B with concern to A. And drew it as one dimension, handled it fine with the outing vector sign, but it's just as true if we did this in two and three dimensions. This is a full vector equation. And the book presents it a little bit differently. What the book likes to do is say write it down and I'll show you why I don't like to use it myself. It will do something like the position of B is the position of A plus the relative position between them. If you notice the two equations are exactly the same thing. I prefer this and the choice for you is purely up to you. I prefer this just because it's so much easier to remember. B, A in that order, B, A in that order. If I do this, I always have to remember what's the order here on the B and A. It's just harder for me. My unlimited brain power just does an awful lot easier with that business there. So do as you wish, but that's what works for me. They're the same thing anyway. It doesn't really matter. But that will always be the case. We're always reading this as B relative to A and it's always the difference between the two in that order. Well, the next step, I guess, then becomes what's the relative velocity between the two. And that is nothing more than the time derivative of that equation. So the velocity of B relative to A. If you were in car A, how fast would and in what direction would B appear to be moving? For a case like this, we've all seen this on the highway. You can very easily imagine as you're catching up to a car in front of you, it can see as if that car is coming back to you and that's the relative motion we're talking on. And again, it's just the time derivative of the position equation and has exactly the same form as does the position equation itself. Without any great stretch, then, of course, the acceleration equation follows. And there's everything you need to know for relative motion. Except we'll find in a minute that there's a little bit more to the acceleration equation that just appears there, but we'll get to that when we get to a particular type of problem when that shows up. So that's it. That's all the preparation I can give you for relative motion. Oh, no, I can't. I can do something else. We actually have two types of relative motion. In this case, the two objects that we're talking about are independent of each other. There's nothing that A does that's going to affect what B does in terms of its velocity, acceleration, and position, other than you can throw in crashes and the like. That's an entirely different point. These are independent objects, and so we call this unconstrained relative motion. I mean, the two are free to move as they will, as they wear, as they what? They're completely independent of each other. So we've got everything we need. Maybe we'll put the position vector as something a little bit more generic than using X. It allows us just to be a little more general with two-dimensional motion, even three-dimensional. It doesn't matter. Bless you, it's all the same. So I'm just writing down what I've got there so we can do some other things on the board. And then the acceleration equation. That's it. That's all I can do for you relative motion other than let's try a couple. Most of them are very easily put into things like travel type equation problems and like. So let's imagine a train traveling at constant speed along a track. So we're looking at it from above. It's got a velocity of 60 miles per hour. And it passes over a freeway upon which a car is driving. And we want to find out the relative position of the two, or relative velocity of the two. So the velocity of the car is 45 miles per hour. And the two travel paths are about 45 degrees apart. So let's find the velocity of the train relative to the car. So first thing you need to ask, given that question, find train relative to car, which is B and which is A? Because if you get them wrong, then you have the opposite of the answer we're looking for. Because rA relative to B is just the opposite of B relative to A. So we have to have the letters straight to even use the proper equation. So which is B, which is A? And the velocity of something relative to something, whichever letter comes first determines what vector algebra, vector arithmetic we do next. What did you say, John? Well, no, I need two letters here. I think the train and the car. Is it T relative to C, or C relative to T? TC. TC is John's wild guess. I thank John for coming today. Bill? It's T relative to C. Yeah? Relative to C. It's a shorthand notation for those words, and we'll always take it to read that way. We might not specifically have it in those words, so we have to be careful as we get a little better at these. But it's then nothing more than the difference between those two vectors in that order. So I guess VT, we can just leave it in miles per hour, no reason to move it to anything else. This will be 60 miles per hour in the X direction. VC is 45 miles per hour, 45 miles per hour in the cosine 45, plus J direction, and all of that, that miles per hour. Does that look great? Since the sine and the cosine are the same, that's out, and all you have left with that is just figuring out what the numbers are. 28.2i minus, let me see, 31.8j miles per hour. So if you're sitting in the car, you look at the train, it appears as if it's traveling with a velocity in that direction. We can sketch that out if you're in the car, and you look at the train, it's going to appear to have a velocity of about the same in the two directions. So just give it that length minus J. It's going to appear, I guess it would make more sense to put that up here above because that's where the train is. That would be the velocity of the train relative to the car, something like that. So just give it a little bit in the I direction and about the same, give it a little bit in the J direction. And that's just what the train, the trouble we have really imagining this stuff is anytime we look at a train in a situation like this, we also see all the other stuff. We see the train track itself that's not moving. We see the trees and other cars there that may or may not, most of the trees don't move, but the cars do. It's a little tough to just look purely at one object to get this true sense of what it appears to be doing because that's the idea we're looking for. Is that the relative velocity? It's a little easier to imagine in planes because there's no other reference thing. The clouds are there, but they're usually so far away that there's no reference as you look at it in another plane. How about the acceleration of the train relative to the car? What's the acceleration of the train relative to the car? The car was constant. Yup. Zero. There's nothing there. Neither one of these have any component to them and you can see that this is a constant as well so even if we took the time derivative of that, time rate of change of that, that would also be zero. So it appears to the car the train's coming near and near to it but at a steady speed. So don't panic, there's an underpass there anyway. Which we obviously neglected to change the difference in the height between the two. Okay, questions on that before we try a couple others? Very straightforward. You need to just work with the definition of the relative motion equation and very carefully take each of the two vectors and you'll be able to put the rest of them together. We'll be using this in some detail later. Oh, there was something else I wanted to do with that so let me just put it back up real quick. Trains here, going 60. Park here. There's another way we can look at it too if you write out the equation and then actually make a vector drawing and there's another possible way to solve it if you prefer. If we draw those vectors, the train vector looks something like that maybe, however long it takes to represent 60 miles per hour. The car's going 45 relative to that and we want vt minus vc so that's 60, 45, that 45 degrees might be something about like that and then this is vt relative to vc and if you prefer then you can use the law of sines and cosines to figure out the other sides based on your own. So if you're more comfortable with that you can certainly do that. You just have to remember the law of sines and cosines which I never can plus this seems straightforward enough to me for most of the problems but it is a way to either double check things or maybe find things in other ways that you wouldn't have normally. Okay, that was a real quick aside. Very few students I've noticed over the years actually like to draw the vector triangle so that's why I almost forgot to bring it up. Okay, so another problem then. Imagine a car passing under an overpass from which it's dripping a little bit of water. I want you to find the velocity of the drops relative to the car. What would it look like the drops are doing the instant they hit that windshield? A couple of little details I'll include. The drops are coming from an overpass six meters over the windshield and the car's velocity is 100 kilometers per hour. So I want you to find the velocity of the drops relative to the car and actually sketch what that looks like. So maybe a D for drops and a C for car if you wish. What are we looking for? D slash C or C slash D? Which one are we looking for? D slash C, C slash D. D slash C, yeah. So I need the velocity of the drops relative to the subtracting of the velocity of the car. Velocity of the car is easy. That's just in the I direction, 27.1 meters per second. You just have to make the change there because it's not easy to get the velocity of the drops in kilometers per hour. How do you find the velocity of the drops with your constant acceleration motion equations? Does anybody happen to have that sheet from Physics One? Oh, good. The rest of you have your tattoos. You have to make the assumption that the drops are in free fall so you know their acceleration, you know the distance, you know the initial velocity, you find the velocity at the base of the six meters and you know the velocity of the drops. If we sit down as negative, it would be minus j direction. So for the velocity of the drops, assuming constant acceleration, we have the acceleration of the drops is minus g. That minus, because I've already chosen down as negative, the initial velocity is zero and the distance of the drops all is minus six. So then you can find the velocity of the drops after they fall in six meters and we can finish the problem directly. That sheet, how? Sold again. The first one was free. If anybody doesn't have those constant acceleration equations I do have extra sheets in the office. Have a velocity for that. The speed with which the drops are hitting the car. Sound about right? 0.9 meters per second. We make it a vector minus j. Now we can put the two together. Minus 10.9 j minus the velocity of the car 27.1 and all meters per second. Make a sketch of that. What that looks like at the windshield to the car. And imagine you're in the car and you see these drops. You see it leave the overpass. What's it appear the drop is doing and you have to take everything else out of the picture, the bridge, the trees, that thing you have hanging from your mirror and what's it look like the drops doing as if you were not moving. It's easier to feel like you're not moving when you're at constant velocity. It's harder when you're accelerating. There's the sensation of fictitious forces when accelerating and accelerating your reference frame. We've got the vector here but double check what it looks like with the other parts to it. The velocity of the car or something like that with the velocity of the speed of the drop being about half that. I don't know, closer to a third that. The two vectors look something like that and the difference between them, something like that. Which is just what you see. It look like the drop is coming at you from an angle slightly elevated and coming towards you. What about the relative acceleration? Relative acceleration always has the same form. Do we know the acceleration of the drops? Assuming free fall we do and do we know the acceleration of the car? I don't know if I actually said it but I don't say anything about the acceleration. We have to assume the velocity is constant meaning that it's 0 and then the relative acceleration of the drops is just minus g. Don't forget acceleration is much less intuitive than its velocity and that holds true here with these acceleration problems of these relative motion problems as well. They're much less intuitive in terms of the acceleration and velocity. This I think, I don't know about you, this I can imagine. It's harder to imagine what the acceleration is just by casual observation. Questions before we make things a little bit more complex? Alright. So still all we need to do, all we need to remember is right there but now we'll give us a problem that's a little bit more involved. So imagine a car is coming onto a freeway so it's coming onto the ramp and then that ramp joins onto the freeway and we have one car here and one car here. So the details, this car, car B and this one car A each have a velocity in those directions and each is also accelerating. Of course B is trying to pick up some speed. A is not at all confident that he's going to pay any attention while merging so A is actually slowing down and then some of the other details will need a 60 degree angle like that and the arm ramp has 100 meter radius. So I have to draw it so that the acceleration velocity vectors are vertical just to make things convenient. If they weren't, we can easily put the coordinate directions so that they line up with one velocity or the other, it doesn't matter which. I just have to draw it that way. So we'll use that as our coordinate directions as we write out these vectors. Still need some values for those. So let's say VA has a speed of 18 meters per second and acceleration, absolute value of acceleration of magnitude 2 meters per second squared. B is not quite as fast trying to come up to its freeway speed is at that moment at 12 meters per second not a square. But acceleration 3 meters per second squared. So find the velocity of B relative to A. So you do that part, you've got more than adequate tools to do it. Don't forget this relative. These relative equations are full vector equations. Full vector equations. Don't forget it's not going to be sufficient just to add and subtract the magnitudes. There's occasional problems where that's true but this is not one of the full vector equations. This problem is very similar to the last one. You just have to come up with the two vectors and subtract them. I'll make your drawings too small on these. Lots of stuff to put in. If you didn't quite get the velocity of B straight down then just tilt your x-y axis a little bit. It's an arbitrary reference. Velocity of B is pretty easy. A just has a little bit of trigonometry in it. A lot of times you can just do these sort of on the fly. Minus 12j. Meters per second is the velocity of B. Velocity of A is minus in both directions. We already have a minus there. Pull out the magnitude. Cosine 60 minus 60j. All of this is then meters per second. Let's double check. We've got a minus from our relative motion equation. That's this first minus. The second minus is simply because A is moving in the minus direction for both coordinate directions as I drew x and y. Cosine 60i and sine 60j. Okay, that looks good. All right, everybody, Travis, you got that too? Okay, so you can just check those numbers. And we're going to minus the minus on the i. Minus nine in the i direction. Sorry, plus nine in the i direction. And 3.6j. Don't want to look like them. And that's all meters per second. Don't forget the units on these. So B relative to A. So A, it appears to A that car B is coming, well, pretty much straight towards them. Not quite moving away a little bit just because there's a component of B's velocity that's away from A. But it's all taken care of. Check out, that's what everybody else got. Okay, that was the easy one because we've just done a problem a lot like that. Then I'll do the acceleration. Because both of these do have an acceleration. What did I say, B relative to A? Why is this different than what we just did with the velocity? Just different numbers, different directions. Why is this one a little bit different? We just handle that with the vectors. We put a minus sign in the middle. Handle that. There's another layer of complexity in this problem. This car is traveling on a circular path. So it's actually going to have a tangential component to the acceleration. In fact, that's what that is. And it's going to have a normal component to the acceleration. Acceleration and B in the normal direction because it's on a circular path. So both of those have to be parts of it. My suggestion, any time you have one or the other or both on a circular path that you actually do this separately. Don't try to do the equation as a whole. Just take them out and do them a piece at a time. It's a lot easier. Then you're working on a couple small problems instead of one big problem. Because as you've seen, we can't lose any minus signs. We can't forget any other types of values we may need. So there's a component of B in the tangential direction. Lucky enough, that's minus J direction so it's pretty easy. But there's also a component of B in the normal direction. It's perpendicular to tangential direction so that will be minus K direction. That's pretty easy to help us make it pretty easy to handle. So you set up those two vectors and then you can just subtract the A acceleration at A right out of it. A is an awful easy one to do since it's the correct way given. It's 2 meters per second squared J direction. So that's the acceleration of A but I want you to find the accelerations of B and then put the two together for the whole piece. We have the tangential acceleration and that, remember, the normal component of a curved path is if it's a circular path so it's a centripetal acceleration. And then put them all together for the relative acceleration. Got it, David? That looks like straight to material. So again, we're looking for the acceleration of B relative to A. I guess if A had a radar gun pointed at B, how the reading on the radar gun would be changing the acceleration effect of this one. Is that? That's just the A vector, the acceleration of A. It's slowing down on the freeway but in the coordinate system I picked that's positive in both directions. The tangential acceleration of B is simply the linear acceleration of the car along its path. Well, not necessarily linear, it's on a circular path but the tangential acceleration. So that's 3 meters per second squared in the minus J direction. Should we assume that's the tangential or would it ever be the magnitude of the tangential plus the voltage? Oh, that you're giving this? Yeah. Well, the way I drew it since its velocity is there that defines the tangential direction. Remember, there is no component of velocity in the normal direction ever. So that immediately defines the tangential direction and I drew them parallel. So you don't have to assume that A and B is both components. It's just the it's got to be the tangential component because it's in that direction. So the normal component we have to add on to it. This is not the full acceleration vector B. This actually is. This one we're coming up with now. How do you find the normal component of acceleration on a circular path? B squared over rho. We use rho in this class for some unknown reason. B squared B of B radius of curvature is 100. And then those are in the minus I direction. We've got to get all the minus signs right. We've got to get the magnitude right. So lots of parentheses. Lots of units. That is meters per second squared in there. And so what's that? Minus 4 4 I minus 3 J meters per second squared. Is that right? Got all the minus signs right there too. Let's add these two together. In fact, it's pretty easy to add them just on the fly. It's minus A. So it's a minus 1 I and a minus 144 I. So that's minus 2.44 I. I've got a minus 1732 and a minus 3 minus 4. 1732 which is 9 there. Agreed? No Chris? I guess I'm just confused because you're right. Given AB is 3, but now we solved AB, B, B, and 3. No. Because this is only the tangential acceleration. We have to know that because it's parallel to the tangential direction, which is the velocity. The velocity can only be there's no normal component for the velocity. So that defines the direction. I guess it just seems like we write AB with other vectors. This is the magnitude. It's the magnitude of the tangential component. You know, this is if we were monitoring out the needle moves on that car, that's what you'd be calculating. Set the taper. Don't read too much into it. Chris, you just have to have to realize that this has got to be the tangential direction since it's on a circular path there has to be a normal component. So you have to find it. So if you're in A, it appears that B is has two negative components it appears to be accelerating away from you. Mostly due to the fact that you're slowing down but also because its vector is coming around on some object. Okay. Any questions on that? Before I clear it up Joe. What do you have to your velocity vector with respect to A? Oh, I already erased it. Velocity of B with respect to A was BB is that in the negative direction. And it all reduces to that was 9 and 3.6. Is that right? Yeah. Both are positive. So if you're in car A it appears as if B is coming towards you. Okay. Another problem. One more. And then we can take another look at relative motion. Similar problem for you. Imagine a path like that. Is an airplane moving at 30 kilometers per hour. 50 kilometers per hour squared. So that's got to be then the acceleration. Not in the usual units we're used to but fair enough. And then another plane passes it following a circular path. So right when they're right beside each other it has a velocity of 600 kilometers per hour. A tangential acceleration anticipating 100 kilometers. No reason you can't leave this in kilometers and hours. And the radius 400 kilometers. One other piece. The 4 kilometers of heart. At the instant they pass. Well it's very common for fighter jet pilots to fly a passenger plane just to scare the heck out of the passenger because they're still awake. So it's a very standard maneuver when a hot dog in the government pay for a jet. So fine. Find the velocity and acceleration of B relative to A. Chris when you take this class again next year I won't actually leave all that acceleration vector A, B, C. I avoided that here. So a blue is acceleration. A blue is force. That's coming up. What's red is, red is force. Blue is divergence. And green is auxiliary information. Nobody's taking notes. Colored pencils. When I first had a professor I started using colored chalk. It took me about two weeks to take these colored pencils. I guess if you got a little bit of a monochromatic world. So fairly similar to the last problem just some slightly different type numbers and jets instead of cars that are taking what they were given finding the components that make it up. The relative motion will become very, very useful to us in a couple weeks when we get into the complex problems. Is there going to throw an orange to each other or something? They don't actually like people throwing oranges between planes which is why the windows don't roll down. Like open up the window on a plane and get some pressure. Just leave them in kilometers and hours no reason to convert to meters and seconds here. It's real useful to give it to kilometers per hour squared but it looks like lots of people on the velocity one. So while you're working on the acceleration one I'll put up the velocity one 600 and if we pick a coordinate direction maybe you did already maybe it's that one it's not. It just means a minus sign usually. So B is 600 I minus A 700 J per hour. The velocity one was easy. If you're in plane A it appears as if B is going backwards in 100 kilometers per hour. Look out the window and it'll be slipping back. The acceleration of A is pretty easy just the 50 kilometers per hour squared J. The acceleration of B very similar to the last problem. Isn't it? Not a tie fighter faster than that don't they? Well that's embarrassing. Looks like birds. Pretty amazing birds. Superbirds. Kind of a minimalist airplane. That's a fun part. John's flying in a delta plane fighter. X wing. Actually the Millennium Falcon. Alright. Again this is very similar to the last problem we did. There's a normal antitangential component to the acceleration of B. And you have to come up with both of them in vector form. So what's the tangential acceleration of B? That's what? Minus 100 J. Add to that the normal component of the acceleration of B. Because it's on a circular path E squared over r that's kilometers per hour squared squared over kilometers 600 Oh yeah B. If you go on A and here on B humans didn't work out. And that's in the plus I direction per hour squared. And so just picking out the numbers what is that? 900 Oh yeah this is 900 that's 100 Yeah. Yeah that's what I meant. So 900 J minus 100 I Yeah. I switched the number. I didn't switch the number. And then you can finish because we already have the acceleration of A. And so that's in the J direction so it doesn't affect the I component as the 50 minus 100 minus look about right? If you're in A looking out at B it appears as if it's accelerating you're in A looking at B it's accelerating away and back a little bit. Okay any questions? Don't forget this business went on a circular path. It actually holds for any path it's just a straight path the normal component to zero. We only have tangential acceleration on a linear path but don't forget that especially on these curved paths. Alright we're going to start then Monday this was unconstrained motion we're going to look at constrained motion on Monday and it's not worth starting it especially since I can't do a meeting. So that's it.