 Where did my marker go? Oh, there it is. OK, so back to this example with this little rocket guy. So now let me do a different example with it. And so now I'll just pick up from where I left off. Let's say that we have a velocity of 4 meters per second this way. It's moving already. And now I'm going to fire this rocket right here, so it pushes down. So I have a thrust downward. And let's say for just one second, just to be different. So let's find out, can we find out where it is and how fast it's going? So what if you use the same thing? F net, change in momentum over change in time. So but now, what's my initial velocity? What's my initial momentum? V1 is going to be 4 meters per second, x hat. And F net is just going to be Ft. And it's going to be negative. I said it was 2 newtons, 2 newtons, y hat. So now they're in different directions. You can imagine what's going to happen. It's going to start curving down this way. I'm going to assume it doesn't tilt, so it stays in the same orientation. OK, let's just do the exact same thing. Try it down. Ft is mv2 minus mv1 over delta t. So I want to find, I know v1, I want to find v2. So now v1's not 0, so I get Ft delta t. That's the thrust equals mv2 minus mv1. I can add that to both sides. And I get mv2 equals mv1 plus Ft delta t. OK, and then I can divide both sides by the mass there. So now I have an expression for the final velocity. And it is a vector, OK? OK, now it's not a one-dimensional problem. So we'll do it. I'll do the x, and then I'll do the y. So let's say v2x equals v1x, which is 4 meters per second, plus Ftx delta t over m. Let me just write that Ftx delta t over m. But what's my x component of the thrust? 0. So this whole term is 0. So v2x is 4 meters per second. Let me put that up here, because we'll need it. OK, so here's the important thing. The space robot was moving this way. The thrust, the force was this way. So since the force was only in the y direction, it didn't change the x direction motion at all. The force is only in the y direction, so the motion only changed in the y direction. So at the end over here, even though it's moving down, it still has the same 4 meter per second in the x direction. OK, so now I still have this up here. I still have that. Let's just do the y direction. I'm going to erase that. So now I have the same thing in the y direction, v2y. It's going to be v1y, which is there's an x component with no y components of 0. And then I have f2ft in the y direction is negative 2. So I have negative 2 newtons times, oh, I said one second. One second over the mass of 1 kilogram. And this is going to give me negative 2 meters per second. So putting that together with my x velocity, I have the final velocity v2. It's going to be 4 meters per second x hat minus 2 meters per second y hat. So that's how fast it's going after the one second thrust down. If I want to, I could find the direction for that. I'm not going to. But if I know it's going 4 meters per second that way and 2 meters per second down that way, I can find that angle tangent of theta is going to be opposite over j, so it's going to be vy over vx. And then you could do that. OK, now what about where is it? The last time, which pickup where we left off, and the last time it had a position I'll call this r1, because it's not at the origin anymore, r1, we said it was at 4 meters. So it's going to be 4 meters x hat. That was the vector locating where it is because the origin's over here, so it's over there. It's 4 meters over. So what's r2 after the one second? Well, I can do the same thing I did before. I can use the definition of average velocity. So I can say v average, I'm going to write it as a vector now, is delta r over delta t, where r is just a gen, it's not x, it's not y, it's both x and y. So this is going to be r2 minus r1 over delta t. So multiply both sides by delta t, I get v average delta t equals r2 minus r1. So r2 equals r1 plus v average delta t. And now what's the average velocity? So over here, let me write v average is going to be v1 plus v2 over 2. This is an average because it has a constant changing velocity. You can do that. So this is going to be my initial velocity was 4 x hat. And that was it. And my final is 4 x hat minus 2 y hat. So I'm going to get, let me just write this out as 4. I'm going to leave off the units for now. 4 x hat plus 4 x hat minus 2 y hat over 2. So I can combine these two, right? Because, OK, I'll answer that later. I can combine those two because they have the same direction. Oh, mute. That's what I'll do. OK. And they're in the same direction. So this gives me, and divide by 2, I get 4 x hat plus or net minus. The phone threw me off minus 1 y hat. So that's my average velocity. So now I can write this as x and y directions. x2 equals x1 v average x delta t. So x1 is 4. Average x velocity is 4. And that's 1 second. So this is going to be 4 plus 4 times 1 equals 8 meters. And then I can do the same thing. y2 equals y1, which was 0, minus the average velocity of 1 meter per second times 1 second is going to give me negative 1 meter. So the final position, r2, is going to be 8 meters x hat minus 1 meter y hat. So now I know how fast it's going. And I know where it is after that thrust. A little more complicated situation because it's two-dimensional.