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8 Jesctions IeisuoJ 3 # motion thtouen Indicated:18 Yost Yoen Dack 1 6 J E 1 motlon Molionconstant You 1...

Question

8 Jesctions IeisuoJ 3 # motion thtouen Indicated:18 Yost Yoen Dack 1 6 J E 1 motlon Molionconstant You 1

8 Jesctions IeisuoJ 3 # motion thtouen Indicated: 18 Yost Yoen Dack 1 6 J E 1 motlon Molion constant You 1



Answers

A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time $t$ is proportional to (i) $t^{1 / 2}$ (ii) $t$ (iii) $t^{3 / 2}$ (iv) $t^{2}$

So we want to find the length and direction when defined of you cross me And if you cross you so in the top, right, we're given that we can find across part by building everything into that matrix there. And also we have that once we find you cross V, be cross you is just going to be the negative of you Cross V. And we know that the magnitudes of each of these alliance will be safe now. We could go ahead and use that matrix if we wanted, but that's kind of overkill for what we have here. Because when we do, you cross V while this is going to be too. I cross minus three J and one of the properties is we can pull the scale er's out. So too times negative three and they'll just be I cross J. And in the chapter, it tells us that I cross J is just k So you cross b is going to be negative six. Okay. And so Kay is a unit vector. So we already know the direction of this. So let's go ahead and see. Profess what we get. So first, the magnitude of you cross a V? Well, this is going to be the magnitude of negative six K. But we can pull that six out by making it positive. And then the magnitude of K is just one since the unit. So the magnitude of that will be six. And we also know that this will be the same magnitude of be cross you. So we have that for our magnitudes and the direction of you cross B. So the direction of this well, that's just going to be okay, since we already know Kay is a unit vector. But the direction of are other when we do it in the opposite direction is going to be the negative off the direction of you cross be. So that's going to be negative. Negative. K. And I almost forgot the negative right here for this direction. Ah, so then we can go ahead and just get that the direction of be cross you is going to be okay

We want to fine dealing and direction wants to find of you cross bmp cross you So in Destructor I have some of the equations that were given. So in Green's we have the definition of cross product using the determinant matrix. This second equation shows that if we find the cross part of you and B, if we just take a negative of that, we will get the cross you And we also know that the magnitudes of you cross the end, be cross you will peopie say so this time, using that matrix will be a little bit overkill. Something else they say in this chapter is if we write. So you cross v So we were to do too. I cross minus three j. Well, one of the properties of this chapter is weekend factor. The Constance helped. So we can write this Abbs six. Her negative six. I cross j and I cross J is going to be kicked a negative six. Okay. And now we can break this up because we know that kay is a unit. Ah, Victor. So we can really rewrite this if we wanted as negative six 001 and then distribute that negative and keep the six on the outside. And now this here is still our unit says our unit. So this is really our direction. So this is the direction of you across the and on the outside. That will be the magnitude of you, Crosby. You cross B magnitude. And that top right equation says that will also be be cross you not be crusty of the cross. You all right? So we have we have this or the magnitudes of each and at least the direction of you. Cross B will be 00 negative one and now defined the direction, the direction of the cross. You will use that middle equation there and just take the negative of the direction of you. Cross be so 00 minus one, which is 00 What? So this would be our direction of the cross blocked in the other direction

So for this problem were asked to find the cross product of these two vectors and find the magnitude and the direction. So there are a couple of ways we can do this one way. Uh, just part of the most convenient way in general is just to use the following formula which can be written out in terms of a three by three determinant. So you do is you take you write your unit vectors I, j and K in the first row. Then your components of your first vector. You go in the second room, that would be 23 and zero in the case component. And then your second victor goes in the third row. So here we have minus 11 and zero. Okay. And so now what is the, uh, cross product formula? Say, Well, if you know anything about determinants, we're just going to expand this determinant. Ah, by co factors along the first row. So if you don't know what that means, that's fine. But basically, here's the idea. So the ice component of our cross product is gonna be the two by two determinant you get when you erase the first column So we want to evaluate this two by two determinant. Okay? And then what about the second component while you delete the second column and compute that two by two? Except if you know about determinants, there will be a factor of minus one here from the expansion. Okay. And in the last column, we delete and get the last component. Okay, so this is going to be our determinant. And now So now we just have to evaluate this. So maybe down here in the corner, I'll remind everybody what a two by two determinant is. It's just a D minus BC. So we're doing that in each component here, So I get three times zero minus zero times one is just zero. Also, get zero for the second component. And then here we have two minus and negative threes. That gives me a five case. Another way we could write. This is just five times K. So what does that tell us? While the direction of this factor is in the K direction, So it's like pointing up the way we only draw the standard coordinate axes que J ai. Okay. And then what's the magnitude or here? we can just read it off. It's five. Okay. And then what about were also asked about V cross you. But in General V cross, you is always gonna be same. Uh, vector is You cross be just reversed. So it's in the same line of direction. It's just magnitude is in the magnitude of the same. It's just pointing in the opposite direction.

So we want to calculate you cross be in this instance it's probably easiest just to use the determinant formula something you write that out. Put you in the second row The in the third row. Okay. And so now here and my first component, I'm goingto have minus 1/2 times two minus one. So let's see. Minus 1/2 times too. It's minus one. I have minus one minus one gives me a negative to in the first component. In the second component, I delete the second calm. So I have three minus one gives me too. And then for the third component, I have three halves minus and negative half. So I have three house plus 1/2 which is to okay. And so now we want to find the magnitude of this guy. Well, that's just factor out to maybe so I can write. This is two times negative. 111 And then the magnitude of you cross ity is going to be two times the magnitude of this vector. So it's one squared plus one. Sorry. This should be negative One. Technically, of course it it's the same as one squared. So the magnitude of a vector given in components is just the square root of the sum of the squares of the components. So this is two squares of three is the magnitude.


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