Hey, guys. Let's check out this problem here. We're getting an initial position, 6.2 at 25 degrees. We're going to travel some distance 9.9 at 78 degrees and then another 2 meters in the negative x direction. And we're going to figure out the magnitude and the direction of our final position vector. So let's go ahead and draw out a diagram just to see what's going on over here. So let's just make a big x and y coordinate system over here. So where this is going to be the plus y direction and to the plus x. So our initial position is 6.2 at 25 degrees below the x axis. That's important. Make sure you read those words carefully. So this means that our initial position is going to be somewhere in this direction. So let's label this ra, we have 6.2. We know this angle here, thetaa, is 25 degrees. So now we're going to travel 9.9 at an angle of 78 degrees above the positive x axis. So that's going to look like this over here. So what I'm going to do is I'm going to label this point over here. I'm going to label this point A and this is going to be point B. And then when we get to point B, we're going to have another 2 meters in the negative x direction. That's the next displacement. So from point B, we're going to travel in this direction to point C. And then what happens here is we're going to have a final position, and we're going to figure out the magnitude and the direction of that final position vector. So remember, the position vector points from the origin to the point, so that means that your final position is going to be looking like this. So we're going to call this rc. And really, what we're interested in is the magnitude and the direction of rc, right? So we want the hypotenuse of the triangle and the angle that it makes. So remember, with these two-dimensional vectors, you're going to have to break them up into their components. This is going to be the x coordinate and y coordinate, xc, yc. Now let's get to the equations. So if you got magnitude and direction, remember, that's just a vector's equation. Right? You're just going to use your Pythagorean theorem and tangent inverse.
x 2 + y 2For magnitude,
θ C = tan - 1 y xThe direction is given by the inverse tangent of y over x. And so notice how both of these equations involve you calculating xc and yc, so we're going to need to know what those components are. We need to figure out the legs of the triangle before we can figure out the magnitude and the direction. So that's really what the whole problem is about. How do we actually figure out those coordinates or those components there? Right? So if we think about what's going on here, we've actually got these 3 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 A to B and then from B to C. So you're really just adding all of these vectors tip to tail. The way that we solve this problem to get the final coordinates is going to be exactly the same way that we solve vector addition problems. To get xc and yc, we're just going to build a table of x and y coordinates, and then we're just going to calculate all the components and then add them all up together. Same thing 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 going to be my ra and then what happens is I'm going to label these vectors over here. So what happens is because this is the displacement from A to B, I'm going to call this delta r from A to B. And this is going to be from B to C, so I'm going to call this delta r from B to C. Now, I know this delta rB to c is 2, and I know this delta rAB is equal to 9.9. This is just purely along the negative x direction. So what happens if this gets a negative 2? And this goes at some angle over here. Right? So this is going to be 78 degrees. That's, you know, 78 degrees above the horizontal or the positive direction like that. So really, again, all these things just turn into a bunch of triangles. We're just going to break them all down to their components and we're just going to add them in this table over here. So we've got delta r, this is going to be delta r from A to B, delta r from B to C. And once we add all those three things up together, all their components, then we're going to get rc. Right? So how do we get this, ra? We're basically just have the magnitude and the directions. We're just going to use all of our component equations. So this is just going to be ra times the cosine of thetaa. That's how we get those components. So this is going to be 6.2 times the cosine of 25. And what this is going to equal is it's going to equal 5.62. And we do the same thing over here, 6.2 times the sine of 25, and 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, then what happens is one component points in the right direction and we know that's positive. And then the other component points in the downward direction, and that's negative. So what happens is this picks up a negative sign, this 2.62. This one's positive. Let's move on. So now for delta RAB, we do the same exact thing. This component is going to be positive and positive, so we won't have to worry about that. And this is basically just going to be 9.9 times the cosine of 78. If you work this out, you're going to get 2.06. And then you could do 9.9 times the sine of 78, you're going to get 9.68. And then delta r from B to C. So delta r from B to C again, remember, just lies in the negative x directions purely along the x axis. So what happens here is wherever we have a one-dimensional vector like this, then all of the vector is going to lie in one direction either the x or the y. So this delta x from A to B or from B to C is going to be negative 2 and there is 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 as negative 2 and this is going to be 0. And now we filled up the table. Now it's just going to we're just going to add straight down. So if you add the 5.62 and the 2.06, you're going to get 7.68. Oh, I'm sorry. Yeah. Sorry. 5.62, 2.06 and then you subtract, you're going to get for your final value here, 5.68. And if you do the same thing over here, this negative with the 9.68 becomes 7.06. Yep. So now these are the components. These are basically my xc and yc components. So these are the numbers I'm just going to plug into my magnitude and the 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 7.06 squared. And if you plug this into your calculators, you're going to get 9.06. 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 the tangent inverse of 7.06 divided by 5.68. You don't have to worry about the square roots because they're both positive, and you're going to get 51.2 degrees. So these are the magnitude and direction which leaves us with answer choice C. That's all there is to it, guys. Let me know if you have any questions.