4. 2D Kinematics

Intro to Motion in 2D: Position & Displacement

# Position and Displacement in 2D

Patrick Ford

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Hey, guys, We've already seen how position and displacement work in one dimension along a flat line, like in the X or y axis. But what are you talking about two dimensional motion? Now I want to show you this video how position and displacement work in two dimensions so along an angle like this. And here's the whole idea. If we ever have position and displacement that air two dimensional vectors and they just turn into a bunch of triangles. So that means we can use all of our triangle or vector equations to jump back and forth between two dimensional vectors. And they're one dimensional X and Y components. That's the whole thing. It really is just Mawr triangle math. So let's get started. So let's talk about the position first. Now the position when we talked about in one dimension was basically just an X or Y value. So it was either a point along the X or the Y axis. But now we're talking about two dimensional motions now, so it's not X or y. It's actually X and y. So it's a coordinate X coming. Why that court that that basically described where you are now, another way you can think about the position is that the position is a vector in two D. So it gets a letter r with an arrow on top of it. So it's a vector or an arrow that points from the origin to the point where you are. Let's take a look at an example. So a position is, uh, in this diagram here, 3.6 at 33.7 degrees and then later on our position is something else. We're gonna calculate the X and Y components of our positions at A and B, so we want the components of these vectors. But first I have to draw out the position vectors in the first place. So remember, it's the origin. It's arrow from origin, two points. So that means that by vector at a it's just a narrow. This is our A and then it be this is just an arrow pointing from the origin to be so this is our be so notice how these vectors here don't just point in the X or the y axis. They actually point at an angle, which means that we could break them down into triangles. So this turns into a triangle like this. And this is my X and y components X a y A This is an angle fate A same thing happens for be I'm gonna break it down into a triangle And these my components X b y b and the angle theta be So in this case, in this particular problem here, I want to figure out the components of these two dimensional vectors. So I'm just gonna use all of my vector equations. I'm just gonna use my decomposition and composition equations to figure out my components. And so those equations are my X and y equals R CoSine. Data are signed data. That's really all there is to it. So, for example, my position at A is just gonna be our eight times the cosine of data A. So this is just gonna be 3.6 times the cosine of 33.7 and you'll get three. And if you do the same thing for why you're just gonna get 3.6 times the sign of 33.7 and you'll get to so basically these are my ex and why positions or my X and y coordinates you can think about. This is the legs of the triangle three and two. Let's do the same thing for be so X B is just gonna be RB times cosine thing to be. So this is just gonna be 8.49. That's what I'm told in the problem and times the cosine of 45 degrees. So these air magnitudes and directions, you'll get six. And you do the same thing for why you're gonna do 8.49 times the sign of 45 and you'll also get six. So my why is six and my ex is six. So notice how we can take any two dimensional vector. And then we could just use vector equations to break them down into their X and Y components. So that's it. Just vector equations. Let's move on now. So another thing when you talk about is the displacement. So the displacement, um, the difference between the displacement and position is that whereas the position is an arrow from the origin to a point, the displacement is the shortest path from one point to another point. And so basically, you can think about this as a change in your position. So instead of the symbol are like we use for position, It's Delta are remember Delta always means change. So let's take a look at this example. So we're gonna use the same, you know, dots A and B s before, and we calculate the magnitude and direction of the displacement from A to B. So now the displacement is the shortest path between this point Point A to point B. So this is my delta arts, the change in position. So notice how this vector also points at an angle, not just in the X or y. So it also gets broken down into a triangle like this. Except now the symbols are gonna be Delta X and Delta y. And so we're just gonna use again vectors equations to figure out the magnitude of the direction. The magnitude is just the Pythagorean theorem that's just your square root in your X squared plus y squared, and then your angle here, your direction is just the tangent inverse. So if I want to figure out the magnitude of my displacement, then I'm just gonna have to use the Pythagorean theorem. So this is Delta X squared plus Delta y squared. And then my angle theta is gonna be the tangent inverse of the absolute value of Delta Y over Delta X. Alright, So notice how both of these equations involve me having Delta X and Delta y, but I actually don't know what those are. I don't know what the legs of this triangle are, so I'm gonna have to go figure them out to figure out the magnitude and direction. So how do we do that? Well, one way you can think about the displacement in the X direction is that remember? It's just a change in your position. So the Delta X here is gonna be remember that my exposition at a was three and then my wife my exposition at B when I calculated this was six. And in a similar way, we had white position over here, which was two. And then our Y position it be was equal to six. So what happens is my Delta X is really just the change in the exposition. So it's XB minus X a. So that's six minus three and that's three. And then similarly for the Y direction is just gonna be the difference. So six minus two is four. So now I actually have the legs of the triangle. I know this is just three and four, so I can plug in into my Pythagorean theorem. So it means the magnitude of the displacement is just gonna be Pythagorean theorem. Three squared plus four squared. And that's 5 m. So this delta Rs five and the direction Fada is gonna be the tangent inverse of my absolute value of 4/3. So you plug this into your calculator, you are going to get 53 degrees. All right, so that is the magnitude and the direction. Alright, guys, that's all there is to it again. It's a bunch of just vectors, things we've already seen before. So let's move on. Thanks for watching.

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