SPEAKER 1: Now, we're going to look at a trajectory, where you have acceleration in one dimension. And, of course, we're going to use gravity. Usually, you do this just by throwing chalk, but here, we're in Texas. So, we're going to do it with a gun. We're going to have this Nerf gun shoot its bullet and see if we can watch the trajectory. All right, we're loaded and mounted. Let's see if we can see a nice trajectory here. Yeah, that looked OK. It kind of came down like I wanted, but I don't know. The curvature-- it's so fast this way, and it falls so slow that way. I don't feel like you could see it. I guess I could increase gravity or compress space. Which should we do? Let's go with the latter. Here we go. That's a lot of work, hold on. OK, that's going to hold. Give me a minute. Oh, I look good, real young. All right. So, let's come over here and try this again. Let me cock the gun. And now, you might be able to see the curvature we're looking for a little bit better. Let's see. Yeah. Now, I think it looked better that time. I feel like maybe you could really see how it's supposed to go down. Oh, I forgot. Let me-- (grunts) OK. Let me get this out of the way, and now, let's analyze this kind of motion. Here we go. So, what we had was an object here, x y. We're plotting the trajectory, not the kinematics plot. And it started out flat, and it just kind of went down like that. To analyze this, let's draw delta r vectors, some displacements. We know we started like this, and then there, and then there, and then there. We already talked about how you look at the difference between two proper r vectors, from the origin, and they give you delta r. So, there's one delta r, and here is the next delta r, if these are uniform in time. And then, there is the next delta r. So, what you see here-- we could have labeled them one, two, three, for three, uniform changes in time. And what you see is that, delta r 1 does not equal delta r 2. That's because the path is curving. If you think about, what does that mean in terms of physics, well that means that delta v 1-- better call it an average. The average velocity on this one does not equal the average velocity on this one, for 2. I'm sorry, you don't need the deltas in there. Just the average velocities in those regions are different, v average 1 and v average 2. If the velocity is changing in time, that means the a does not equal 0. It has acceleration. And, of course it does. Gravity. It was accelerating down. Makes perfect sense. So, it was moving, gravitationally decelerate-- it was falling this way, moving uniformly that way. Let's see if we can do the constant acceleration kinetics, but in terms of the position vector r. OK. Let's write r. We could start out-- do you want to be real general? Let's be real general. Let's start out and just say, it's always x initial plus vx initial t plus one half axt squared i hat plus the y initial plus vy initial t plus one half ayt squared j hat. Because this is constant acceleration kinetics for each case. Now, we just have to go to our problem and plug in the right values. Some of these are zero. So, in this case, an x-- let's see, we don't have an initial x position. We started out at x equals zero, and we have no acceleration in the x, because gravity is just in the y. So, x, really, is just that initial vxi velocity. We shot the bullet with an initial velocity vxi. That's the only thing that goes on the x. I hat plus y. Direction, we did have an initial yi, so yi would go here. We had no initial velocity in the y. We shot it straight. All the initial velocity was in the x, so nothing in the y. That term is zero. The acceleration in the y was negative. It was negative 9.8 meters per second squared. This time, I'm not defining down as positive. I'm defining up as positive, since I wrote that up with a y on it, right? We have to actually plug in the negative value here. We're going to put minus one half 9.8 We'll assume we're in MKS and won't write the unit. And t squared j hat. That tells you everything you need to know about the position. You know the position at all time, if you know the initial velocity in the x, and you know the initial position in the y. And, you can further use this. So, you want to know the velocity at all time, the velocity vector? That would be vxi hat. I took that derivative. All that's left is the initial velocity. Plus, derivative of that is zero, and then, this is minus one half gt squared. So, let's see, take the derivative, 2, cancels that 2, minus 9.8 tj hat. So, the velocity is just a constant velocity, and the y velocity going down. Let's do the acceleration. Why not? The acceleration is just another derivative. Derivative in the x is zero, no acceleration in the x. That makes sense. Derivative of that is minus 9.8, minus 9.8j hat. Sure enough, it recovers what you would expect. We have acceleration down. How do you use this? What is the point of this? One way you could use it is, say you were asked what is the angle of the trajectory at a certain time? The angle of the trajectory has to do with the velocity vector, so you need the angle of the velocity vector at a certain time. Well, you plug in that time, and then you have velocity as a vector, x component y component. If you want the angle, inverse tangent of y over x. So, we'll show you some problems where you can use these things. But, the key is, if you can just write that position vector, everything usually comes out of that.