Hey, guys. So sometimes you'll have to calculate the displacement by using the velocity-time graph rather than using your constant velocity equations. So that's what we're going to cover in this video, but we're going to see that all of these problems will just break down to just using really simple equations for really simple shapes like rectangles and triangles. So let's get to it.

Guys, in a velocity-time graph, the displacement, the change in the position, that's Δx between two points, is just going to be the area that is underneath the curve. This is an expression that your textbooks and your professors are going to use. And what under the curve really just means, is it's the area that's between the values of where you are on the velocity-time graph and the time axis itself. So under the curve just means all of the area inside of the shape that I've made between the red lines. Guys, that's all there is to it. We just have to figure out the area of this shape.

So let's just go ahead and take a look at this example. We've got a velocity-time graph for this moving object. We're going to calculate the displacement for the first 4 seconds. So what you do is you just highlight 0 and 4 seconds. And then you calculate or you just figure out what the area underneath that curve is. So that's all we have to do. But I might not know what the area formula is for this complicated shape here. I've got some straight lines, some slanted lines. I'm not really sure what that is. So what I can do is I can just break it up into smaller shapes that I do know the area for. So I can break this up and now I've just made a square and a triangle. And guys, I actually know what the formulas for those shapes are. A rectangle is just base times height, and a triangle is just one-half base times height.

So for this first part here, if I want to calculate the displacement, I'm going to call that Δx1. This is just going to be the area of the square. So I'm going to call this Δx2, and I'm going to call this Δx3. So this Δx1 is just going to be Δx2 plus Δx3. So now I just have to figure out those areas. Well, Δx2 is just a rectangle, so this is going to be base times height. The base of this rectangle is 2 and the height is 2, so it's just going to be 2 times 2 and that's 4. So I've got 4 meters here. And then this triangle here, I knew the equation for that is 12 of base times height. So now I've got 12, and now the base is 2 and the height is 2 also. So I've just got 12×2×2. The 12 and the 2 will cancel, and you'll just get 2 meters. So I'm just going to add those 4 and 2 together, and I get a total of 6 meters. Now, another way you can think about this 6 meters is just if you count up all the little boxes that are inside of this curve. I've got 1, 2, 3, 4, 5, and then these are both one-half, like, this is like one-half and one-half. Altogether, that makes 6. That's another way you can visualize that.

Alright, guys. So let's move on to part B. Now we're going to calculate the displacement for the entire motion. So the entire motion is actually from 0 all the way to 6. So we have to figure out the area that's underneath the curve for all of that. It's going to include the area that we have for part A. So if I want the total displacement,Δxtotal, that's going to be Δx1 that I just found out over here, which I already know the answer for, plus now the area that's under the curve from 4 to 6. So I'm going to call thisΔx4. I've got Δx1 plus Δx4. All of that added together is going to be my total entire displacement. So let's just go ahead and do that.

So I know Δx1 is 6, and then I have to figure out this displacement, and that's going to be my total. So let's just go ahead to the graph, Δx4 is just going to be one-half of base times height because it's a triangle again, the base is 2 and the height is 2. So this is just going to be 12×2×2. So again, we're going to cancel and you're just going to get 2 out of this. But what you have to remember is that areas above the time axis are going to be positive displacements because your velocity's positive. You're moving forward. So these displacements that we got over here, 4 and 2, are positive, whereas the areas below the time axis are going to be negative displacements because you're moving backward. Your velocity is negative there. So that means this is not +2, this is actually -2. So this is our Δx4. So we've got 6 +(-2), and you add that together, and your total displacement is just going to be positive 4 meters.

And that's really all there is to it, guys. Let's get some more practice problems, and let me know if you have any other questions!