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Anderson Video - 2D Motion and Derivatives

Professor Anderson
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 >> Hello, class, Professor Anderson here. Let's talk a little bit about calculating velocity in two dimensions, and let's say we have the following. Let's say we are given the position vector, okay, and the position vector r is going to look like something. Let's just make something up. Let's say that it is 5t i hat minus 4.9 t squared j hat. Okay, this is a position vector as a function of time, and we could, of course, map this out, if we like. Here's an x,y coordinate space. Let's see if we can map out what r looks like as a function of time. If t equals 0, where are we? Yeah? >> We're at 0. >> We're at 0, right? We start at the origin. t equals 0. We are at 0, and now, as time marches on, we have two competing properties here. We have 5t, and we also have minus 4.9t squared. So how do we figure out what's going to happen? Well, we can just plug in some numbers. Let's say that t is going to equal 1. Okay, so at t equals 1, let's see what we have. We have 5i hat, and then we're going to have minus 4.9 j hat. And later on, at t equals 2, what will we have? We will have 10i hat minus 4.9 times 2 squared plus 4. So 4.9 times 4 is what? Well, that's pretty close to 5 times 4 is 20, and then we need to fix up a little bit, take off 0.4, and so we should get 19.6j hat. Okay, and then we can keep thinking about more points along the way. All right, so right off the bat, it looks like I need to change my coordinate system. So let's move this x axis up to here. All right, t equals 0. This guy is right there. At t equals 1, it is at 5i hat minus 4.9j hat. So let's mark this off. One, two, three, four, five, six, seven, eight, nine, 10. One, two, three, four, five, six, seven, eight, nine, 10, and at 5i hat, we move over five, and we're going to go down 4.9j hat. So down to there. The next point, we are at 10i hat minus 19.6j hat, which is way down here off the scale. So it's going to be down there somewhere, and so look what happens. This thing is going to go like this. Okay? That's the trajectory, r, as a function of time. Let's see if we can figure out what the velocity is doing. Velocity is, of course, the derivative, dr/dt, and now we can do that, right? That's just a derivative of this thing. What do we get? Derivative of 5t is just 5. Derivative of 4.9t squared is minus 9.8t j hat. Okay, and so that's the velocity as a function of time, and you can plot it out similarly to how we did right here. Let's say we take it one more step, and we want to calculate the acceleration. What we know is that acceleration is the derivative of the velocity. If I take a derivative of 5i hat, what do I get? Yeah? >> Zero. >> Zero. It's a constant. There's no time in there at all. So the derivative of that is 0. If I take a derivative of 9.8t, I, of course, just get 9.8, and then there is still a j hat hanging on there. Okay, so the acceleration is negative 9.8j hat. So this trajectory looks a little familiar, doesn't it? What sort of trajectory are we actually drawing here? If the acceleration is negative 9.8, that sounds like a familiar number, right? What is negative 9.8? Yeah, on the right. >> It's acceleration. So it's gravity. >> It's gravity, right? This example that we drew here works out to be gravity. So this is, in fact, just free fall, which is projectile motion. Okay, acceleration is always negative 9.8 meters per second square. There's no acceleration in the x direction. This projectile that we launched, of course, had some initial x velocity. It had a value of 5. If we're in SI units, then this would be 5 meters per second, and so that would be constant velocity in the x direction, and it would be increasing its speed in the y direction as it falls, and this would be the position. So our picture should sort of level out a little bit more up here, but you get the idea. Okay, questions about that? Okay.