Calculating Velocity Components

by Patrick Ford
239 views
Was this helpful ?
0
Hey, guys. So in the last couple videos, I introduce you the velocity vector into dimensions. So if the velocity is ever to d then just like any other vector, it has X and y components. So in this video, I'm gonna show you how to calculate those velocity components. And there's really two sets of equations that we're gonna use to jump back and forth between the velocity and its components and really just comes down to what problems will give you. So basically, there's two different kinds of scenarios. You're either gonna be going back and forth between the components and displacement and time, or you're gonna go back and forth between the components and the magnitude and direction of the velocity vector. And really, what we're gonna see is that once we fill out this table, what's thes? 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 gonna walk 40 m to the right, 30 m up in 10 seconds, and we're gonna calculate the magnitude and the X and Y components of the velocity. So here we're gonna go back and forth between the components and displacement and time. So let's go ahead and do that right. So we're gonna go 40 m to the right and then 40 and then 30 m up and we're gonna calculate the magnitude of the velocity vector. What? We've seen that equation before, remember, That's just Delta are over Delta T 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 two dimensional displacement. Well, im going 40 to the right and 30 up. So that means my two dimensional displacement Delta are is gonna be the high pot news of this triangle. So we're just gonna use Pythagorean theorem. We're just gonna use, um, triangle equations. And really, this is a 345 triangle. So if this is 30 and 40 the legs of the triangle, then that means the high pot news is 50. So this is 50 m. That's our two dimensional displacement. And so our velocity is just gonna be 5 m per second. That's the magnitude. Now, remember that this velocity vector also points in the same direction as Delta are. So that means the velocity vector points in this direction. So we know that V is five. But we're not quite done yet because remember, we have to calculate the velocities magnitude, and it's X and Y components. So we figure out the first part now we're just gonna figure out the X and Y components, So this velocity vector over here is two dimensional. It pointed some angle like this so we can break it down into a triangle just like we would any other vector. So this is my V X component, and this is my V y component over here. So we're gonna figure out VX and VY y. So what do the equations we're gonna use for that? Well, remember that velocity is always displacement Overtime in two dimensions. V was just Delta are over Delta T. But now we're looking for Vieques, the component of the velocity in the X direction, But it's still gonna be displacement over time. It's just gonna be the displacement in the X direction over time. So it's Delta X over Delta T and in a similar way V Y is gonna be the displacement in the Y direction over change in time. So it's always displacement over time. That's always gonna be what velocity is. All right, So that means we're gonna use Delta X over Delta T. So what's our Delta X? Well, that's 40. So we're gonna use 40/10 seconds and this is gonna be 4 m per second. So that's the component in the X direction. You could think of this as how much of this five that lies in two dimensions lies basically just along the X axis. And it's four. So our B Y components is gonna be Delta y over Delta T. So there's gonna be 30/10 and this is gonna be 3 m 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 345 triangle with the displacements. We also ended up with a 345 triangle with the velocity. And it's because, really, all 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 gonna be walking at 5 m per second at some angle 37 degrees above the X axis on, we're gonna calculate the X and Y components of the velocity. So in this particular problem, were given a velocity in two dimensions were given the magnitude and the angle here, and we're gonna calculate the X and Y components. So basically, this is just gonna break down into a triangle just like any other two dimensional vector. So now this is gonna be my V x and my V Y components here. So basically, the relationship between all these variables the magnitude, the direction and your components are all just gonna be triangle equations. These they're just gonna be vector equations. So in general, what happens is the magnitude of this velocity is just gonna be the hypothesis of the triangle. So this is just gonna be your Pythagorean theorem. V X squared plus v y squared your angle theta is gonna be the tangent inverse of the absolute value of ey over Vieques. And if we want to calculate the X and y components, then we're just gonna use V cosign data and V sign data 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. All right, So that means that your velocity component V X is just gonna be v times the cosine data, which is just gonna be in this example five times the coastline of 37. And this is gonna be 4 m per second. So this is the components of the velocity in the X direction, and then my d y is gonna be very similar. It's gonna be five times the sign of 37 and that's gonna be 3 m per second. So let me write that a little bit bigger, so there's gonna be three meters per second. All right, so this is the components of my X and Y velocities. So notice here last week I mentioned is that we've ended up with 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 use two different setups, two different equations. But he's really just described the same exact problem here. All right, so let me know if you guys have any questions, that's it for this one.