Intro to Motion in 2D: Position & Displacement - Online Tutor, Practice Problems & Exam Prep

Intro to Motion in 2D

Hey, guys. So for the next couple of videos, we're going to see how motion in a two-dimensional plane works, or this is sometimes called motion at an angle because you're going to need to know how to solve these kinds of problems. So what I want to do in this first video is just give you a brief overview of what motion in 2D is all about. Let's check it out.

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 to either the x or the y axis. So we're kind of locked. I can only move on the x-axis or the horizontal axis or on the y-axis or the vertical axis. Now, both of these are examples of one-dimensional motion because, again, you're 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 θ. And we saw how this worked when we studied vectors. So this is just 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 two one-dimensional motions. It's as if you went from A to B and then B to C at the same time. So, really, this just turns into a bunch of triangles, so we're just going to combine motion with vector's equations.

So here's the deal. Whenever we have motion in 2 dimensions, we're going to break it down into x and y, and then we're going to use combinations of all of the motion equations, basically, all of our average velocities and our Uniformly Accelerated Motion (UAM) equations with our vectors equations. Basically, everything 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.

Position and Displacement in 2D

Hey, guys. We've already seen how position and displacement work in one dimension along a flat line like on the x or y axis. But we're already talking about 2-dimensional motion now. So I want to show you in this video how position and displacement work in 2 dimensions, so along an angle like this. And here's the whole idea. If we ever have position and displacement that are 2-dimensional vectors, then 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 2-dimensional vectors and their one-dimensional x and y components. That's the whole thing. It really is just more 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 2-dimensional motions now. So it's not x or y. It's actually x and y. So it's a coordinate, x, y that cord, that basically describes where you are. Now, another way you can think about the position is that the position is a vector in 2-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, in this diagram here, 3.6 at 33.7 degrees and then later on, our position is something else. We're going to 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 the arrow from origin to point, so that means that my vector at a is just an arrow. This is r_a. And then at b, this is just an arrow pointing from the origin to b, so this is r_b. 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 can break them down into triangles. So this turns into a triangle like this and this is my x and y components x_ay_a, this is an angle theta_a. The same thing happens for b. I'm going to break it down into a triangle and these are my components, x_b, y_b and the angle theta_b. So in this case, in this particular problem here, I want to figure out the components of these 2-dimensional vectors. So I'm just going to use all of my vector equations. I'm just going to use my decomposition and composition equations to figure out my components. And so those equations are my x and y equals r cosine theta r sine theta. That's really all there is to it. So for example, my position at a is just going to be r_a times the cosine of theta_a. So this is just going to be 3.6 times the cosine of 33.7, and you'll get 3. And if you do the same thing for y, you're just going to get 3.6 times the sine of 33.7 and you'll get 2. So basically, these are my x and y positions or my x and y coordinates. You can think about this as the legs of the triangle 32. Let's do the same thing for b. So x_b is just going to be r_b times the cosine of theta_b. So, this is just going to be 8.49. That's what I'm told in the problem and times the cosine of 45 degrees. These are magnitudes and directions. You'll get 6 and you do the same thing for y, you're going to do 8.49 times the sine of 45 and you'll also get 6. So my y is 6 and my x is 6. So notice how we can take any 2-dimensional vector and then we can 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 we need to talk about is the displacement. So the displacement, 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 r, like we use for position, it's delta r. Remember, delta always means change. So let's take a look at this example. So we're going to use the same dots a and b as before, and we're going to 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 r. It's 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 going to be delta x and delta y. And so we're just going to use again vectors equations to figure out the magnitude and the direction. The magnitude is just the Pythagorean theorem. That's just your square root and your x squared plus y squared. And then your angle here, your direction is just the_TAG_will_inverse. So if I want to figure out the magnitude of my displacement, then I'm just going to have to use the Pythagorean theorem. So this is delta x squared plus delta y squared, and then my angle theta is going to 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 going to have to go figure them out to figure out the magnitude and the direction. So how do we do that? Well, one way you can think about the displacement in the x direction is that and remember, it's just a change in your position. So the delta x here is going to be remember that my x position at a was 3. And then my x position at b, when I calculated this, was 6. And in a similar way, we had our y position over here, which was 2, and then our y position at b was equal to 6. So what happens is my delta x is really just the change in the x position. So it's x_b minus x_a. So that's 6 minus 3 and that's 3. And then similarly for the y direction, it's just going to be the difference. So 6 minus 2 is 4. So now I actually have the legs of the triangle. I know this is just 34, so I can plug in into my Pythagorean theorem. So it means the magnitude of the displacement is just going to be Pythagorean theorem, 3 squared plus 4 squared, and that's 5 meters. So this delta r is 5. And the direction theta is going to be the tangent inverse of my absolute value of 4 over 3. So if you plug this into your calculator, you are going to get, 53 degrees. Alright? 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.

At point A, a hiker is 10m east from the origin. After 35s, the hiker arrives at point B 40m at 60° north of east from the origin. Calculate the magnitude and direction of the hiker's displacement.

• A. 50m; 60° north of east
• B. 36m; 73.9° north of east
• C. 36m; 60° north of east
• D. 36m; 16.1°north of east

Final Position Vector

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 r_a, we have 6.2. We know this angle here, theta_a, 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 r_c. And really, what we're interested in is the magnitude and the direction of r_c, 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, x_c, y_c. 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 2

For magnitude,

θ C = tan - 1 y x

The direction is given by the inverse tangent of y over x. And so notice how both of these equations involve you calculating x_c and y_c, 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 x_c and y_c, 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 r_a 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 r_B to _c is 2, and I know this delta r_AB 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 r_c. Right? So how do we get this, r_a? 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 r_a times the cosine of theta_a. 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 R_AB, 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 x_c and y_c 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.