Intro to Motion in 2D: Position & Displacement

Hey, guys. So for the next couple videos, we're gonna see how motion in a two dimensional plane works or sometimes called motion at an angle, because you're gonna need to know how to solve these kinds of problems. So I want to do in this first video is just give you a brief overview of what motion in two D is really all about. Let's check it out. So let's say I had a problem in which I had to move from Point A to Point C. Now, when we studied one dimensional motion, we were restricted into either the X or the Y axis. So we're kind of locked. I could only move in the X axis or the horizontal axis or in the y axis or the vertical axis. Now, both of these examples off one dimensional motion because you're again locked into either the X or the Y. Now, if you weren't restricted to the X or Y, if you could actually move freely, then you could just move straight from a to C at some angle feta. And we saw how this worked when we studied vectors. So this is just a two dimensional motion and notice how we just come up with another triangle. So really motion at an angle, like going from A to C, is really just combining to one dimensional motions. It's as if you went from A to B and then B two c at the same time. So really, this just turns into a bunch of triangles, so we're just gonna combine motion with vectors equations. So here's the deal. Whenever we have motion in two dimensions, we're gonna break it down into X and Y, and then we're gonna use combinations off all of the motion equations, basically all of our average velocities and our, um, equations with our vectors equations, basically everything that that tells us how to deal with triangles. That's really all there is to it. So let's go ahead and check it out.

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.

Hey, guys, let's check out this problem here. We're getting an initial position. 6.2 At 25 we're gonna travel some distance, 9.9 78 degrees and then another 2 m in the negative X direction, and we're gonna figure out the magnitude in the direction of our final position Vector. So let's go ahead and draw out a diagram to see, you know, just to see what was going on over here. So let's just make a big X and Y coordinate system over here. So where this is gonna be the plus y direction and to the plus X. So our initial position is 6.2 25 degrees below the X axis. That's important. Make sure you read. You know, those Let those words carefully. So this this means that our initial position is gonna be somewhere in this direction. So let's label this are we have 6.2. We know this angle here. Theta A is 25 degrees. So now we're gonna travel 9.9 at an angle of 78 degrees above the positive X axis. So that's gonna look like this over here. So what? I'm gonna do is I'm gonna label this point over here. I'm gonna label this point a and this is gonna be point B. And then what? We get to point B, we're gonna have another 2 m in the negative X direction. That's the next displacement. So from point B, we're gonna travel in this direction to point C. And then what happens here is we're gonna have a final position. We're gonna figure out the magnitude the direction of that final position Vector. So remember the vector of the position Vector points from the origin to the point. So that means that your final position is gonna be looking like this. So we're gonna call this R C. And we're really what we're interested in is we're interested in the magnitude and the direction of our C. Right? So we want the high pot news of the triangle and the angle that it makes. Remember, with these two dimensional vectors, you're gonna have to break them up into their components. This is gonna be the X coordinate and y coordinates X c Y c. Let's get to the equations. So you got magnitude direction, remember, that's just a vectors. equation, right? You're just gonna use your Pythagorean theorem and tension in verse so you're r C is just gonna be the Pythagorean theorem of X squared X plus y c both squared, square rooted, right? And then the direction fantasy is going to be the inverse tangents of your why See, over XY and so notice how both of these equations involve you calculating x e n y c So we're gonna need to know what those components are. We need to figure out the legs of the triangle before we can figure out the magnitude in the direction. So that's really what the the whole problem is about. How do we actually figure out that those coordinates or those, um, those components there, Right. So if you think about what's going on here, we've actually got these three vectors, and we're kind of adding them tip to tail. So if you think about this, your initial position gets added onto the displacement vector from eBay to be and then from B to C, so you really just adding all of these vectors tip to tail. So the way we solve this problem to get the final coordinates is gonna be exactly the same way that we solve vector addition problems to get X e and Y c We're just gonna build a table of X and Y coordinates. And they were just gonna calculate all the components and then add them all up together. Same thing what we did with that table. So when we have our X and y table like this, all I've got to do is just make this little table nice and big like that. And so this is gonna be my r a. And then what happens is I'm gonna label these vectors over here. So what happens is because this is the displacement from A to B. I'm gonna call this Delta are from A to B and this is gonna be from B to C. So I'm gonna call this Delta are from B to C. Now, I know this Delta R B to C is too, and I know this Delta are Abe is equal to 9.9. This is just purely along the negative X direction. So what happens if this gets a negative too? And this goes at some angle over here, right? This this is gonna be 78 degrees. That's, uh, you know, 78 degrees above the horizontal or the positive. Like that. So, really, again, all these things just turn into a bunch of triangles. We're just gonna break them all down to their components, and we're just gonna add them in this table over here. So we've got Delta. Are, uh, this is gonna be dealt are from a to B. Delta are from B to C. And once we add all those three things up together, all their components, then we're gonna get our c. All right, so how do we get this? Are a We're basically just have the magnitude and the directions. We're just gonna use all of our component equations. So this is just gonna be ara times the cosine of data A. That's how we get those components. So this is gonna be 6.2 times the co sign of 25. And what this is gonna equal is gonna equal 5.2 62 and we do the same thing over here 6.2 times the sign of 25 we get 2.62. The one thing we have to be very careful about when we do these kinds of problems is keeping track of your positives and negatives. So, for example, we know this vector here points in this direction. So if you break it up into its components, that what happens is one component points in the right direction and we know that's positive. And the other component points in the negative direction in the downward direction. And that's negative. So what happens is this picks up a negative sign this to 62. This one's positive. Let's move on. So now for Delta R A B, we do the same exact thing. This component's gonna be positive and positive, so we don't have to worry about that. And this is basically just gonna be 9.9 times the cosine of 78. If you were one and work this out, you're gonna get, um, you're gonna get 2.6 and then you 9.9 times the sign of 78 you're gonna get 9.68 and then Delta are from B to C, so delta for me to see again, remember, just lies in the negative X directions purely along the X axis. So what happens here is whenever we have a one dimensional vector like this, then all of the all of the vector is gonna lie in one direction either the X or the Y. So this Delta X from A to B from B C is gonna be negative two. And there is no component Delta y in. There's no component in the Y direction. So basically what happens is all of the vector lies purely in the X direction. So what happens is we just put this is negative two, and this is gonna be zero. And now we feel that the table now it's just gonna we're just gonna add straight down. So if you add the 5 62 in the 2.6, you're gonna get 7.68. Oh, I'm sorry. I'm sorry. 5.62 2.6, and then subtract. You're gonna get for your final value here 5.68 do the same thing over here. This negative with the 9.68 becomes 7. Yep. So now these are the components these air basically my x and Y C components. So these are the numbers. I'm just gonna plug into my magnitude and direction equation, and that's it. That's really all there is to it. So we've got the magnitude, which is the Pythagorean theorem, and we've got 5.68 squared plus 7. squared. And if you plug this into your calculators, you're gonna get 9.6 So what that does is it eliminates answer choices A and B. And now, to figure out the direction we just have to use the tangent inverse. So that the tangent inverse of 7.6 divided by 5.68 you know, authority about the square roots because they're both positive, and you're gonna get 51.2 degrees. So these are the magnitude and direction. Which leaves us with answer choice. See, that's all there is to guys. Let me know if you have any questions.