Hey, guys. So in the last couple of videos, I introduced you to the velocity vector in 2 dimensions. So if the velocity is ever 2D, then just like any other vector, it has x and y components. In this video, I'm going to show you how to calculate those velocity components. There are really 2 sets of equations that we're going to use to jump back and forth between the velocity and its components. It really just comes down to what problems we'll give you. So basically, there are 2 different kinds of scenarios. You're either going to be going back and forth between the components and displacement in time, or you're going to go back and forth between the components and the magnitude and direction of the velocity vector. What we're going to see is that once we fill out this table, once we fill out these equations, these are all just vector and motion equations that we've seen before. Let's go ahead and check it out here. So let's take a look at the first example. We're going to walk 40 meters to the right, 30 meters up in 10 seconds, and we're going to calculate the magnitude and the x and y components of the velocity. So here, we're going to go back and forth between the components and time. So let's go ahead and do that. Right? So we're going to go 40 meters to the right and then 30 meters up, and we're going to calculate the magnitude of the velocity vector. Well, we've seen that equation before. Remember, it's just Δr/Δt (displacement over time). We've seen that equation before. So I know that my time is 10 seconds. And now all I have to do is just calculate my 2-dimensional displacement. Well, I'm going 40 to the right and 30 up, so it means my 2-dimensional displacement, Δr, is going to be the hypotenuse of this triangle. So we're just going to use the Pythagorean theorem. We're just going to use some triangle equations. And really, this is a 3, 4, 5 triangle. So if this is 30 and 40, the legs of the triangle, then that means the hypotenuse is 50. So this is 50 meters. That's our 2-dimensional displacement. And so our velocity is just going to be 5 meters per second. That's the magnitude. Now remember that this velocity vector also points in the same direction as Δr. So that means the velocity vector points in this direction. So we know that v is 5. We're not quite done yet because remember, we have to calculate the velocity's magnitude and its x and y components. So we figured out the first part. Now we're just going to figure out the x and y components. So this velocity vector over here is 2-dimensional. It points at some angle like this. So, we can break it down into a triangle just as we would any other vector. So this is my vx component and this is my vy component over here. So we're going to figure out vx and vy. So what are the equations we're going to use for that? Well, remember that velocity is always displacement over time. In two dimensions, v was just Δr/Δt. But now we're looking for vx, the component of the velocity in the x direction. But it's still going to be displacement over time. It's just going to be the displacement in the x direction over time. So it's Δx/Δt. And in a similar way, vy is going to be the displacement in the y direction over change in time. So it's always displacement over time. That's always going to be what velocity is. Alright. So that means that we're going to use Δx/Δt. So what's our Δx? Well, that's 40, so we're going to use 40 over 10 seconds and this is going to be 4 meters per second. So that's the components in the x direction. You can think of this as how much of this 5 that lies in 2 dimensions lies basically just along the x axis and that's 4. So our vy components are going to be Δy/Δt. So this is going to be 30 over 10 and this is going to be 3 meters per second. So this is basically how much of this vector lies in this direction and sort of like the vertical axis like this. Alright. So notice also how just like we had a 3-4-5 triangle with the displacements, we also ended up with a 3-4-5 triangle with the velocity. And it's because really all of these numbers are getting divided by the time which is 10 seconds. Alright. So let's move on now. Let's take a look at the second example. So in this example, we're going to be walking at 5 meters per second at some angle 37 degrees above the x-axis, and we're going to calculate the x and y components of the velocity. So in this particular problem, we're given a velocity in 2 dimensions. We go in the magnitude and the angle here, and we're going to calculate the x and y components. So, basically, this is just going to break down into a triangle just like any other 2-dimensional vector. So now this is going to be my vx and my vy components here. So basically, the relationship between all these variables, the magnitude, the direction, and your components are all just going to be triangle equations. These are just going to be vector equations. So in general, what happens is the magnitude of this velocity is just going to be the hypotenuse of the triangle. So this is just going to be your Pythagorean Theorem, vx2 + vy2. Your angle θ is going to be the tangent inverse of the absolute value of vy over vx. And if we want to calculate the x and y components, then we're just going to use v cosine θ and v sine θ. So we can see here that all of these equations are either just motion equations that we know, your displacement over time, or they're just vector equations that we already know. Alright. So that means that your velocity component vx is just going to be v times cosine θ, which is just going to be in this example, 5 times the cosine of 37 and this is going to be 4 meters per second. So this is the components of the velocity in the x direction. And then my dy is going to be very similar. It's going to be 5 times the cosine of 37 and that's going to be 3 meters per second. So let me write that a little bit bigger. So this is going to be 3 meters per second. Alright? So this is the components of my x and y velocities. So notice here, the last thing I want to mention, is that we've ended up with the same exact numbers that we did on this right example over here as we did on the left. We ended up with the same exact numbers. So we used 2 different setups, 2 different equations, but these really just describe the same exact problem here. Alright? So let me know if you guys have any questions. That's it for this video.
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4. 2D Kinematics
Velocity in 2D
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