Hey guys. So maybe we're asked to calculate the acceleration of an object that's moving in 2 dimensions. So not in the x or y, but at an angle like this. What we're gonna see is that it works very similar to how velocity in 2 dimensions works. We're gonna end up with a very similar set of 2 equations to jump back and forth between the acceleration and its components. These equations are gonna look almost identical. It's really only just a letter that's different. So let's just talk about the differences and then do a quick example. Alright, guys. Remember that acceleration always causes a change in an object's velocity. It changes either the magnitude or the direction of the velocity or sometimes even both. And this actually works for 1 dimensional motion or 2 dimensional motion at an angle. So the equation that we use was a=ΔvΔt and it's gonna work the exact same way in 2 dimensions. It's just that now we have things at angles.

For the velocity vector in 2 dimensions, we had 2 different equations to calculate the magnitude. We either were given the displacement in time, we can calculate v, or we could use the components of the vector. So we would take this 2 dimensional vector and break it up into its components vx and vy. Now up until now, we've always visualized these sort of magnitudes, directions, and components as triangles because they help us visualize all the triangle equations that we're gonna use. Well, from now on, we're actually gonna start drawing all the components starting from the same point like this. So this is the vy component. The math is the same. It's really just only the notation that's different. And so the hypotenuse of the magnitude is just the Pythagorean theorem, vx2 + vy2. The same exact thing works for acceleration. So you have 2 different equations. You could either always relate this to the change in velocity of a change in time. So if you're given Δv and Δt, you can calculate the magnitude of a. Or if you're given the components, ax and ay, then you could calculate the hypotenuse of the triangle by using the Pythagorean theorem. ax2 + ay2. Alright. Let's move on. So the angle of this vector here is just related to the components by the tangent inverse equation. Right? The angle is always tangent inverse of vy over vx. It's the same thing for the angle for the acceleration. It's just the tangent inverse of ay over ax. So there's nothing really new there.

Okay. So let's look at the components now of the acceleration vector. Now for the velocity components, we also had 2 equations. Remember velocity is always displacement over time. So velocity in the x direction was either calculated if you have Δx and Δt or if you had the magnitude and the direction of that two dimensional vectors. Those 2 different equations. Well, it's the same kind of idea for the acceleration vector. There's 2 different equations. Remember that if the two dimensional acceleration is change in velocity over change in time, then the acceleration in the x direction is just the change in velocity in the x direction over time. And, the change in velocity and the ay is gonna be the change in the y direction over change in time. So it's the same idea there. Now we could also calculate these ax and ay components by using the magnitude and the direction of the acceleration vector. So if we have a and θ, then ax and ay are just a cosine θ and a sine θ. Alright, guys. So the equations are very similar against really just the letters that are different. They work the same exact way. So let's just go ahead and jump right into an example.

Alright. So we have a toy car that's moving initially at 20 meters per second and then later on it's moving at 67 at some angle. So in this first part of the problem we're gonna get, we're gonna calculate the x and y components of the car's acceleration. So for part a, we're asked for ax and ay. But before we get to that, I just wanna draw a quick sketch of what's going on here. So we have this toy car that is initially moving, so this is my initial velocity, at 20 meters per second. It's purely in the x axis like this. So basically, this is my initial. This is my initial. And then my final is where the velocity is not moving just along the x axis. Now it's actually moving at some angle above. So it's basically moving at some angle like this. We know the final velocity here is 67, and we we know this angle here is 26.5 degrees. So there's a change in velocities. The magnitude and the direction both changed, so there were some acceleration. So how do we calculate the components of this acceleration vector over here? Well, let's just go through our equations. Remember, we have 2 equations. We could either use the change in the velocity in time, or we can use the magnitude and the direction of the of the acceleration vector. But if we take a look here, we actually don't have the, the acceleration vector, the magnitude or the direction. The only thing we have is the magnitude and direction of the final velocity vector. Final velocity vector. Those are not the same thing, so don't get confused there. So let's see. I also am told some information about the time. I know that's 10 seconds. And I have the initial and final velocities here. So I don't have any of these, you know, I don't have the magnitude or direction, but I can I might be able to figure out what the the change in velocity in the x and y directions are by using v initial and v final? So I'm gonna use these equations over here. So my ax component is gonna be the change in velocity in the x direction over change in time. So it's basically just v final in the x minus v initial in the x divided by Δt, and then ay is gonna be very similar. So it's gonna be change in velocity in the y direction over change in time or velocity final minus velocity initial in the y divided by Δt. If I can figure out basically what all these components are of these initial and final velocities, then I can figure out the acceleration. So how do I do that? Well, let's take a look at my initial velocity vector. It's purely in the x direction, which means that technically it's a one-dimensional vector, but that also means I know the components. So I know that the x direction component is just 20 meters per second. It's all in basically in the x axis. And there's no components that lie in the y axis. So it means that my v initial in the y is 0 meters per second. It's not moving in the y axis. But the final velocity actually does point in both directions, so I'm gonna have to break it down into its x and y components. So this is gonna be my vy and this is gonna be my vx. And how do I get the vy and vx? Well, I actually have the magnitude and the direction of that two-dimensional velocity vector. So I'm just gonna use these equations over here, v cosine θ and v sine θ. So my v final in the x direction is gonna be my v final times the cosine of θ. So it's basically just gonna be 67 times the cosine of 26.5, and I get 60. So I know v final in the x is 60. My vy is gonna be, in a very similar way, 67 times the sine of 26.5, and I get 30. So I know my v final in the y direction is 30. So now I actually have all the components that I need to solve the acceleration equation. I know that this is gonna be my v final in the x is 60. My v initial in the x is 20 over 10. So my acceleration component in the x direction is 4 meters per second squared. Now in a similar way, the v final on the y is 30. My v initial on the y is 0. Remember, it's not moving along the x or y, or sorry, along the y direction. So 30 minus 0 over 10 is gonna be 3 meters per second squared. So these are my acceleration components, 4 and 3. So we're done with part a. So now move on to part b. We're gonna calculate the magnitude and the direction of the car's acceleration. So this is basically gonna be what's the magnitude of a and what is the direction of a. So we know that the, that the, let's take a look at our equations. Right? So we have 2 different ways of calculating the magnitude of a. We can use the change in velocity over change in time. Remember, this is the change in velocity in 2 dimensions, or we can just use the Pythagorean theorem if we know the components. So if we take a look here, we actually just calculated the components ax and ay, so we're just gonna use the Pythagorean theorem. So our a is just gonna be the Pythagorean theorem of ax2 + ay2. So this is just, this is gonna be the square root of 4 squared plus 3 squared and we know that's equal to 5 meters per second squared. So that is the magnitude of the acceleration. Now, for the direction, we're just gonna use the tangent inverse of and then we're going to use 3 over 4 and we have to put the absolute value signs over there. And this is gonna be 37 degrees. So 37 degrees. And that's the direction. Alright, guys. That's it for this one. Let me know if you guys have any questions.