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Anderson Video - Swimming Across River

Professor Anderson
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>> Hello, class. Professor Anderson here. Let's talk about a problem where you are swimming across the river. And we need to figure out something about the motion as you cross the river. And let's say that the following applies. Here's one bank of the river. Here's the other bank of the river. And we know that this distance across the river has some particular value. We'll call it d1. As you swim across the river, you're going to end up downstream and, in fact, we know that distance, d2. And we also know a couple more things. We know the velocity of the river is in that direction. And we know that you are going to point yourself in that direction and swim. So, let's put some numbers in here. Let's say that d1 is 200 meters. D2 is 150 meters. And let's say that the velocity of the river relative to the Earth is 5 meters per second. OK? And now let's see if we can figure out a couple things. We need to figure out how fast you need to swim and at what angle, theta, you should point yourself. All right. This looks like a difficult problem. Has anybody looked at this problem? Yeah. OK. If you have some thoughts on what we should do next, give me some instruction. What should we do next here? Right? We've got a picture. We've got a bunch of givens. What should we do next? >> Doug? >> Find out how long it's going to take to cross the river. >> Find out how long it's going to take to cross the river. I like that. All right. How long to cross the river. So, how do we do that? Doug, what do you think? >> Probably use one of the kinematic equations? >> Excellent. It's almost like it's related to early in the lectures. Maybe. All right. Let's use one of the kinematic equations. So, one of our kinematic equations looks like this. Y-final equals y-initial plus vy-initial times t plus 1/2a-sub-y t-squared. That looks like a good one. Any time we write down a kinematic equation, we should immediately say, "Oh, we need a coordinate system." Let's put our origin right there. We'll say x is to the right and we'll say y is up. And if we do that, then we can simplify this equation quite a bit because what is y-final equal to in this case? y-final is just equal to d1. Y-initial is where we started, which is zero. dy-initial, we're not really sure about that yet. A is zero. OK? That's one of our givens. We're going to say that there's no acceleration in this problem. OK? It's just constant velocity motion. All right. So, it seems like we might have enough. But let's take a look at it and double check. OK. So, let's take a look at this equation right here. We've got d1 equal vy-initial times t. That looks like we might be able to solve that for t. Except, what is vy-initial? Vy-initial is how fast is the swimmer going in this vertical direction, in this y direction? OK? We don't know what that is, but what we do know is if this is the speed of the swimmer in that direction, then vy-initial is right there. OK? And this is a right triangle. And so we can say vy-initial is just equal to the speed of the swimmer relative to the river times the cosine of theta. All right. So, we have one equation now, d1 equals vsr cosine theta. All of that times t. All right. But we're not quite there yet. It seems like we're going to have to worry about the x equations. So, let's try this again, but use the x variables. x-final equals x-initial plus vx-initial times t plus 1/2 a-sub-x t squared. All right. x-initial is zero. X-final is going to be d2. vx-initial, we're not exactly sure yet. There's no acceleration so that is zero. All right. So, we get d2 equals vx-i times t. And now we have to think about this vx-i. OK? We just drew a triangle up here, which had the swimmer. And this was v of the swimmer in the x direction. But that's not the only thing moving in the x direction. The swimmer has some component of that way, but the river has some component that way. And so, in fact, we need to subtract the two. It's going to be v-river relative to the Earth minus this component, v-swimmer relative to the river times sine of theta. And all of that multiplying t. All right. Seems like we are almost there. Right? Now we have an equation for d1. We have an equation for d2. And there's one more thing that we do know is you can calculate this distance, which is just the sum of the squares, square root. All right. So, with this information, hopefully it's enough to figure out the problem. OK? I'm not going to do it since it's one of your homework problems explicitly. But this is sort of how you set up the problem. OK? Any questions about this? Any time you're faced with these relative motion problems, always break it down into x and y and think about which ones you need to add or subtract in terms of the vector. All right. Hopefully that's clear. If not, come see me in office hours.