Hey, guys. Let's work out this example together. We're shown the velocity-time diagram. So that's \( v \) versus \( t \) or the velocity-time graph of a moving car, and we're told that the initial position is negative 21, we're supposed to find out what's the car's final position at \( t = 5 \) seconds, so final position. So the variable that we are looking for is going to be \( x \) or \( x_{\text{final}} \). So that's going to be what we're looking for here. So how do we find the final position? Well, we find position by finding the displacements. Remember \( \Delta x \). \( \Delta x \) can be written as \( x_{\text{final}} - x_{\text{initial}} \). But when we have a velocity-time graph, \( \Delta x \), the displacement, is also the area that is under the curve. And so, what's the variable I'm actually looking for here? I'm looking for the final position. So I can rearrange this formula really quickly and say that the final position is going to be the initial position plus \( \Delta x \). In other words, plus the area that's under the curve. So if this is what I'm looking for, I already have the initial position, which is -21 meters. So now all I have to do is I just have to figure out the area that's under the curve. I have to figure out what displacement is. So let's go ahead and do that over here.

The displacement, \( \Delta x \), is just going to be the area that's under the curve between the two times I'm looking for. It says what's the car's final position at \( t = 5 \) seconds, and we're also told that the initial position is at \( t = 0 \). So what this really means is we're looking for the areas under the curve of this entire graph over here. So the area that's under the curve is just going to be all of this stuff that's highlighted in yellow. So let me see if I could do that real quick. Yeah. It's pretty good. Alright. So how do we find now the area of this really complicated shape? Remember, anytime we do this, we just have to break it up into a bunch of little smaller shapes that are easy, that are a little bit more manageable. So, I always like to break it up into the smallest number of shapes. I see a big triangle over here, and if I keep this line going, I actually have a smaller triangle and then a big rectangle. So it's only 3 shapes I need to worry about and that's good. So if I want to figure out the total displacement over here, now I've got these 3 smaller shapes, so I'm just gonna label them. \( \Delta x_1 \). Let's call this guy \( \Delta x_2 \), and then we'll call the smaller one \( \Delta x_3 \). Okay? So let's just get to it. And the total displacement is just gonna be by adding up all of those areas or all of those smaller displacements up. And so \( \Delta x_1 \) is going to be a triangle, so we're going to use \(\frac{1}{2}\) base times height. Now we just have to figure out what the base and height is. And so we're going from 0 all the way to 3, so the base here is 3, and the height is going to be going from 2 all the way up to 10. So that means that this height of this triangle is going to be 8. So it means we have \(\frac{1}{2} \times 3 \times 8\) and that's 12. Let's move on to part, the second one. The second one is going to be from 0 all the way to 5. It's a rectangle, which means the equation we're going to use is base times height. And we have that the base is 5 and the height is going to be 2, right here. So we have just displacement is 10. So we've already got these first two. Now, we just have to look at the last one, \( \Delta x_3 \). It's a small triangle, so we're still going to use the same equation, \(\frac{1}{2}\), \(\frac{1}{2}\) wow. Okay. \(\frac{1}{2}\) base times height. So we've got \(\frac{1}{2}\), the base is going to be 2 because going from 3 to 5, and the height is also going to be from 2 because we're going from 2 up to 4. So we have \(\frac{1}{2} \times 2 \times 2\). \( \frac{1}{2} \) and 2 will cancel, just leaving one factor of 2 there. And so now we just put everything together, all of these areas, and then add everything up. The \( \Delta x \) is going to be \( 12 + 10 + 2 \), and so that's just going to be 24. So now the question is, are we done? Is this our answer? No, it's not. It's not our answer. Remember, this just represents the displacement. We actually have to take this number and we have to plug it back into this formula over here to figure out the final position. So you can't forget that last step there. So we're going to take this number over here, and we're going to pop it into this equation. And so, for the final answer, the final position is going to be the initial position of negative 21 plus 24, which is the displacement. And so what we get is positive 3 meters. That is our final position. Alright, guys. Let me know if you have any questions, and I'll see you in the next session.