Professor Anderson

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>> Hello class Professor Anderson here. Let's take a look at an example problem of a boat crossing a river. This is one of the relative motion problems and let's say that we have the following scenario. We have a river that is flowing in this direction, here's our river and we'll call that speed of the river VRE speed of the river relative to the earth and now we got a boat here and we want our boat to go straight across the river. If our boat's going to go straight across the river should I point my boat directly across the river? What do you guys think? Probably not right everybody knows that if you point it straight across the river the river is going to carry it down stream. So we can't point our boat in that direction, we in fact have to point our boat up stream in this direction. Ok and we need to figure out what direction I need to point the boat and how fast do I have to move the boat in order to get straight across the river. Ok so here's our two relative velocities. We have the velocity of the river relative to the earth and the velocity of the boat relative to the river and let's give you some numbers. Let's say that the speed of the river relative to the earth is 5 kilometers per hour and it is heading east and the velocity of the boat relative to the river the maximum speed that it can do is 10 kilometers per hour but we don't know what the angle is yet. Ok so RV means river relative to earth, VR means boat relative to river and we want to find what is the speed of the boat relative to the earth and at what angle do we need to orientate it, namely this angle right here. Ok and the goal is to go straight across the river, that's the caveat here. Ok there's a couple of different ways to think about this problem but one way to think about it is the following: We can just draw some vectors, this is the vector of the boat relative to the river, this is the river relative to the earth, and if I add those two up it looks like I will get the velocity of the boat relative to the earth. This is a right angle, this is our angle theta and I think that's just about all of the information we need. And now look we've drawn a triangle we know that Pythagoras holds for this right triangle, we've got a 5, we've got a 10, we can solve this for VBE. VBR squared equals VRE squared plus VBE squared and now we can solve this for VBE. VBE equals square root of VBR squared minus VRE squared and we know all of those numbers right? We've got 10 squared, we're going to subtract 5 squared, that's the square root of 75, and we did this earlier, square root of 75 was 8.66 kilometers per hour. Alright, so that's the first part. Before we think about the second part theta, is there some other way we could do this problem? Anybody have another thought on how to do this problem? [ pause in speaking ] Anybody? Ok Ian yeah what do you think how else could we do this problem? >> I would use trigonometry. >> Ok so what would you do with trigonometry? >> I would take the velocity of the boat relative to the river and either use sine or cosine. >> Ok, excellent idea, let's try that. >> Ok so Ian's suggesting that we go back to the original picture and we do a little trig using sines or cosines. Let's try that. Right. VBR is pointing up like that, which means there's two components to it right? There's some component in that direction-VBR we'll call it Y and there is some component in that direction-VBRX. This one is of course the sine component if this is theta, this one is the cosine component so this is VBR sine of theta, this is going to be VBR cosine of theta and if we're just thinking about speeds there you can go like that. Alright, if I'm pointing this thing slightly up stream then my component here up stream has to be what if I'm going to end up moving straight across the river? It has to be exactly the speed of the river right? And whatever the component of the velocity of the boat going up stream that has to exactly match a component of the river going down stream and so our condition now is the following: VBRX has to equal the speed of the river going down stream. That makes sense right? If I turn my boat all the way up stream such that I was going up stream at 5 kilometers per hour and the river is going down stream at 5 kilometers per hour, I wouldn't go anywhere right? We would be exactly still and so whatever component I have going up stream as long as it equals the velocity of the river, the speed of the river, then any remaining vertical component is just going to carry me straight across the river. Alright, so look we've got VBR times the sine of theta equals VRE and now you can calculate theta. Theta is the arc sine of VRE over VBR and so using trig it allowed us to in fact first get theta and once you get theta then you can go back and calculate this quantity VBE the velocity of the boat going across the river. Ok hopefully that's clear and if you want to plug in the numbers you can double check my answer. We've got a theta of I think 30 degrees. Ok any questions about that one? Alright, if that's not clear come see me in office hours. Cheers.

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>> Hello class Professor Anderson here. Let's take a look at an example problem of a boat crossing a river. This is one of the relative motion problems and let's say that we have the following scenario. We have a river that is flowing in this direction, here's our river and we'll call that speed of the river VRE speed of the river relative to the earth and now we got a boat here and we want our boat to go straight across the river. If our boat's going to go straight across the river should I point my boat directly across the river? What do you guys think? Probably not right everybody knows that if you point it straight across the river the river is going to carry it down stream. So we can't point our boat in that direction, we in fact have to point our boat up stream in this direction. Ok and we need to figure out what direction I need to point the boat and how fast do I have to move the boat in order to get straight across the river. Ok so here's our two relative velocities. We have the velocity of the river relative to the earth and the velocity of the boat relative to the river and let's give you some numbers. Let's say that the speed of the river relative to the earth is 5 kilometers per hour and it is heading east and the velocity of the boat relative to the river the maximum speed that it can do is 10 kilometers per hour but we don't know what the angle is yet. Ok so RV means river relative to earth, VR means boat relative to river and we want to find what is the speed of the boat relative to the earth and at what angle do we need to orientate it, namely this angle right here. Ok and the goal is to go straight across the river, that's the caveat here. Ok there's a couple of different ways to think about this problem but one way to think about it is the following: We can just draw some vectors, this is the vector of the boat relative to the river, this is the river relative to the earth, and if I add those two up it looks like I will get the velocity of the boat relative to the earth. This is a right angle, this is our angle theta and I think that's just about all of the information we need. And now look we've drawn a triangle we know that Pythagoras holds for this right triangle, we've got a 5, we've got a 10, we can solve this for VBE. VBR squared equals VRE squared plus VBE squared and now we can solve this for VBE. VBE equals square root of VBR squared minus VRE squared and we know all of those numbers right? We've got 10 squared, we're going to subtract 5 squared, that's the square root of 75, and we did this earlier, square root of 75 was 8.66 kilometers per hour. Alright, so that's the first part. Before we think about the second part theta, is there some other way we could do this problem? Anybody have another thought on how to do this problem? [ pause in speaking ] Anybody? Ok Ian yeah what do you think how else could we do this problem? >> I would use trigonometry. >> Ok so what would you do with trigonometry? >> I would take the velocity of the boat relative to the river and either use sine or cosine. >> Ok, excellent idea, let's try that. >> Ok so Ian's suggesting that we go back to the original picture and we do a little trig using sines or cosines. Let's try that. Right. VBR is pointing up like that, which means there's two components to it right? There's some component in that direction-VBR we'll call it Y and there is some component in that direction-VBRX. This one is of course the sine component if this is theta, this one is the cosine component so this is VBR sine of theta, this is going to be VBR cosine of theta and if we're just thinking about speeds there you can go like that. Alright, if I'm pointing this thing slightly up stream then my component here up stream has to be what if I'm going to end up moving straight across the river? It has to be exactly the speed of the river right? And whatever the component of the velocity of the boat going up stream that has to exactly match a component of the river going down stream and so our condition now is the following: VBRX has to equal the speed of the river going down stream. That makes sense right? If I turn my boat all the way up stream such that I was going up stream at 5 kilometers per hour and the river is going down stream at 5 kilometers per hour, I wouldn't go anywhere right? We would be exactly still and so whatever component I have going up stream as long as it equals the velocity of the river, the speed of the river, then any remaining vertical component is just going to carry me straight across the river. Alright, so look we've got VBR times the sine of theta equals VRE and now you can calculate theta. Theta is the arc sine of VRE over VBR and so using trig it allowed us to in fact first get theta and once you get theta then you can go back and calculate this quantity VBE the velocity of the boat going across the river. Ok hopefully that's clear and if you want to plug in the numbers you can double check my answer. We've got a theta of I think 30 degrees. Ok any questions about that one? Alright, if that's not clear come see me in office hours. Cheers.