 Okay, so let me go ahead and do an example with relative velocity. This is a pretty classic problem. It's all over the place, so I'm just going to make up my own small variation of it. We're doing real physics. I've got to take off my jacket. It's not like Mr. Rodgers, you know, he always takes off his jacket where he puts on a sweater. I don't remember. Okay, so here is situation. Let's say I have a river, and let's say it's ten meters wide, I'm completely just making up stuff here, and the water's flowing. So let's say the magnitude of the velocity of water with respect to the ground is one meter per second, and so here I'm using the notation of the water with respect to the ground. Okay? And right here when you look at the water, it's going one meter per second that way. Now you have a boat, that's my boat, and the velocity of the boat with respect to the water has a maximum velocity of, let's say, three meters per second, okay? And so the question is, how would I have to aim at what angle so that I would travel straight across the river? That's the question. Okay, so I have my magnitude of the velocity of the boat with respect to the water and the magnitude of the water with respect to the ground. Now what do I want? Let's say if I look at this boat from the ground, I want to see it going straight across. You know, do that, it'll be like this, aimed this way and going straight across. Okay? That's what I want. So if I want to see what's the velocity of the boat with respect to the ground, okay? And so I know that's going to be the velocity of the boat with respect to the water as a vector, plus the velocity of the, okay, so if I add those two things, the trick to remember is if these subscripts are in the middle, then I get the ones on the outside. And I show you how to do that in the book. Okay, so what do I know about the velocity of the boat with respect to the ground? Let me call this xy axis. Then I know that, let me say, the velocity of the boat with respect to the ground in the x direction is going to be zero meters per second. I want that to be true. So that's going to be the velocity of the boat with respect to the water in the x direction plus the velocity of the water with respect to the ground in the x direction, which is zero, right? Because it's, oh no, I'm sorry, not zero. Okay, so that's my x direction. Then I have zero equals velocity of the boat with respect to the water in the x direction plus one meter per second. And if I, let me redraw my boat here, here's my boat vector. This is the velocity of the boat with respect to the water, theta. So this is going to be the velocity of the boat with respect to the water in the x direction right there, that component right there. So I can write this as zero equals VB water, the magnitude, the whole thing, negative, because it's in the negative direction. Now I have the opposite side of that triangle, so I need to have, say the sign of theta is opposite over j, so that means, I mean opposite over hypotenuse. So the hypotenuse times the sign of theta would be BW, the velocity of the boat with respect to the water in the x direction, so that's it right there. And then I have this plus one. So right there, now I can go ahead and solve for theta. So I can say VBW, sine theta equals one meter per second, sine theta equals one meter per second, the magnitude of the velocity of the boat with respect to the water is three meters per second, and the units cancel. You can't have the sign of something, have something with units. So theta is going to be the inverse sine of one third. And let me see, so 19.5 degrees, so that's that angle right there. So you have to aim 19.5 degrees to the left in order to go straight across. Now another common extended part to this is, well how long does it take you to get across? Okay, so now if I just, if I look at the velocity of the boat with respect to the ground in the y direction, then I can, I know the distance in the y direction, I can find it. I can do this. The velocity of the boat with respect to the ground in the y direction equals delta y ground over delta t. So I can, I know delta y ground is ten meters, so I can find delta t. What's the velocity of the boat with respect to the ground in the y direction? This is the same part of this, v b g y equals v b w y plus v water ground y, that's zero, so these are the same thing. So if I look back over here at the y direction, v b y, v b ground y, it's going to be this, the cosine of that angle times the hypotenuse. So it's going to be three meters per second times cosine of nineteen point five degrees. Okay, let me erase this. So now I can solve for delta t. It's going to be ten meters, my change in y with respect to the ground divided by the y component three meters per second cosine nineteen point five degrees. And if I put that in, I get ten divided by, I get three point five three seconds. Does that make sense? We can, well the units work, right, meters cancel and one divided by one over seconds does give me seconds. And it does make sense because what if there was no, what if there's a pond? If there was a pond, you'd just aim straight across because the velocity of the boat with respect to the, the velocity of the water with respect to the ground would be zero, so these two would be the same. So now if I'm just going three meters per second across a ten meter pond, it would take me three seconds, ten, no wait, ah, if that's three and that's ten, then it'd be ten divided by three, so three point three three. Now I'm aiming a little to the left, so I'm, I'm not going straight across, it's going to take me longer. Okay, it's going to take me longer so that is longer than going straight across in still water, so that's good. Another question that you could ask that I'm not going to ask, but you could do this. How far is this path like that and, and how long, and how far downstream does the person travel? If they aimed, oh actually, no that's a different question. The question is what if they aimed straight across? Would they get across faster or not? And how far downstream this way would they go? Yeah, so I'll leave that one for you. Okay.