4. 2D Kinematics

Intro to Motion in 2D: Position & Displacement

# Final Position Vector

Patrick Ford

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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.

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