Hey, guys. So now that we've covered all the motion variables in 2 dimensions, you're going to run into problems in which you have to solve kinematics problems in 2 dimensions. So what I want to show you in this video is that solving problems with constant acceleration in 2 dimensions is done the exact same way that we did this for 1-dimensional motion. We're going to use the same list of steps with just some minor differences and the same list of equations here to solve these kinds of problems. Because remember that the whole point of 2-dimensional motion is that we can always break it up into two one-dimensional motions and then tackle them separately. So we have everything we need. Let's just jump in and solve this problem.

So we've got a hockey puck that is sliding along the lake with 8 meters per second east. There's a strong wind that accelerates at 3 meters per second in some direction. We're going to calculate the magnitude and direction of the displacement after 5 seconds. So the first step is always we're going to draw the diagram. Now, the only new thing is we're going to have to decompose any 2-dimensional vectors into x and y. So let's check it out.

So we've got a hockey puck that is sliding along the east. So this is my v_{0}, and we know this is 8 meters per second. Now, there's an acceleration that is at some angle 37 degrees northeast. So we've got an acceleration vector, a=3, and we know that this angle here is 37 degrees. So, now, we're going to figure out what the displacement after 5 seconds is. Well, what happens is the acceleration in that northeast direction is going to start curving the hockey puck like this. So eventually, at some point after 5 seconds, we're going to try to figure out the magnitude and the direction of that displacement. So if we're starting from here, then the magnitude is going to be the length of this hypotenuse. This is the magnitude Δr and the direction is going to be given by the angle relative to the axis. Right? So now we have a bunch of 2-dimensional vectors. We're going to decompose them all into x and y.

So what I'm going to do is this is my displacement vector like this. So, I'm going to call this Δy. I've got an acceleration, and then, the x direction here is going to be Δx, and then I do the same thing for my a vector, right? So I've got a component in the y direction like this, a_{y}, and a component in the x direction like this, a_{x}. And then my velocity vector is already in the x direction, so I don't have to do anything to that.

So that's the first step. Now I'm going to need to list the 5 variables for Δx for x and y. I'm going to identify known and target variables. But first, let's take a look at what I'm actually asked for in the problem. I'm asked for the magnitude and the direction, which is the Δr and the θΔr. Right? So let's take a look at my equations. I know the magnitude for Δr is just going to be the Pythagorean theorem, Δx^{2} + Δy^{2}. And the direction θΔr is going to be the tangent inverse of my Δy over Δx with the absolute values. So notice how both of these equations actually involve Δy and Δx. So I'm actually going to have to solve for those first before I can solve for the displacement, for the magnitude and the direction of the displacement.

So let's move on to the second step now. We're going to have to list the 5 variables for x and y. So I'm going to have my x direction here, my y direction here. Remember, there are 5 variables. I need Δx, v_{0} in the x direction, v_{final} in the x direction, a_{x}, and then t. And then the same thing for y. Δy. Then I've got v_{initial} in the y then v_{final}, and then a_{y} and a t and t. So we're going to have to identify the known variables and our target variables. Well, we know that in order to solve for the magnitude and the direction, I'm actually going to need Δx and Δy to basically put them into the 2-dimensional displacement vectors or equations. So these are my target variables.

So what about V_{0}s? What can I figure out the initial velocity? Well, I'm told that the initial velocity of the puck is 8 meters per second and it lies purely in the east direction. So what that means is that this vector over here, all of it lies along the x direction. So v_{0} in the x component is 8. So I know what this is. This is 8. And because all that lies purely in the east direction, then that means the initial velocity in the y direction is 0. There is no component that points north if you will. So this is going to be 0.

What about the final velocity? Don't know anything about the final velocity. What about the acceleration? I'm told the magnitude and direction, but I don't have the components. So let's move on to time. Well, I know that the time is equal to 5 seconds and it's the same for both. You don't have to distinguish x and y. So basically, what happens is I have 2 out of my 5 variables in the x and y, which is not enough. I'm going to need one more of these equations in either x and y to pick an equation. So, between my final velocity and my acceleration vector, let's take a look at which one I can solve. I'm not told anything about the final velocity, but I do have the magnitude and the direction of the acceleration vector. So I have a, and I have θ, which means that I can use vector equations to solve for a_{x} and a_{y}.

So my a_{x} component is just going to be a times the cosine of θ. So it's going to be magnitude which is 3 times the cosine of 37 degrees. And so what I get is I get 2.4. And if I do the same thing for the y, I'm going to get 3 times the sine of 37 and that's 1.8. So basically, I know this a_{x} is 2.4 and this a_{y} is 1.8. So now I actually have a_{x} and a_{y}.

Alright. So now I have 3 out of my 5 variables, which means I can move on to the next step. So, I can just pick equations without the ignored variable. So, in this case, my ignored variable is going to be my final velocity. I don't know anything about it, and I don't care about it. So this is my ignored variable, and I'm trying to solve for Δx and Δy. So for the x-axis, what I'm going to do over here is I'm going to use equation number 3. Notice how all of the others have final velocity, and we can't use that one. So we're going to use equation 3, which is that Δx, which just tells us directly what we're looking for. We're looking for Δx = b_{0} x t + 1/2 a in the x direction times t squared.

Alright? So this is just going to be, 8 times 5, so 8 times 5 + 1/2 a_{x} is 2.4, as we calculated, times 5 squared. When you plug all of this in, you're going to get 70 meters. Alright? Remember, this isn't our final answer. This is only the displacement in the x direction, but this is one of the numbers that we need to plug into the magnitude formula.

Let's do the exact same thing over here for the y-axis. So the y axis, I'm going to use the same exact equation, right, because I have the same set of variables. I'm going to use equation 3, which is that Δy = b_{0}y t + 1/2 a_{y}t squared. Alright? And the only difference here is that there actually is no initial velocity in the y direction, so this whole thing actually just ends up being 0. So, really, all that happens is we only just have this term over here, which is just going to be 1/2, 1.8, that's the acceleration in the y axis we calculated, times again that 5 squared. When you work this out, you're going to get 22.5 meters.

So that's the displacement in the y axis. Alright? So now let's bring all this back to what we originally set out to do, which is calculate our magnitude and direction in 2 dimensions. Alright? So, I'm going to bring this down over here, and the magnitude of my Δr is going to be the Pythagorean theorem. So, this is going to be a Pythagorean theorem of 70 squared + 22.5 squared. Alright? So just make sure you square root these things and have all your parentheses, all that stuff. What you should get is a magnitude of 73.5 meters, and this is actually one of your final answers.

Alright? So this is basically how far the hockey puck lands or how far it is away at 5 seconds from where it started from in 2 dimensions. Alright? So, notice that it's also bigger than either of these numbers, so that's like the hypotenuse of the triangle. Now, let's go ahead and figure out the direction of this, this displacement over here, which is just going to be the tangent inverse of the absolute value of Δy over Δx. Both of these are positive numbers, so we could just drop the absolute value. This is just going to be 22.5, that's the Δy, divided by 70. And when you take the tangent inverse, what you should get is an angle of 17.8 degrees, and that is basically, presumably north of east. Right? So that's the direction it was traveling. Alright?

So those are your 2 final answers. Hopefully, this made sense. Thanks for watching, and I'll see you in the next one.