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Calculating Displacement from Velocity-Time Graphs

Patrick Ford
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Hey guys. So sometimes you have to calculate the displacement by using the velocity time graph rather than using your constant velocity equations. So that's we're gonna 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 the Delta X between two points is just going to be the area that is underneath the curve. Now this is an expression that your textbooks and your professors they're going to use. And what under the curve really just means is it's just the area that's between the values of where you are on the velocity time graph and the time access itself. So under the curve just means all of the area inside of this 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 gonna calculate the displacement for the first four seconds. So what you do is you just highlight zero and four seconds and then you calculate, or you just figure out what the area underneath that curve is. So that's all we have to dio. But I might not know what the area formula is for this complicated shape Here. I've got some straight line, some slanted lines. I'm not really sure what that is. So what I can do is I could just break it up into smaller shapes that I do know the area for, so I could break this up. And now I just made a square and a triangle, and guys actually know what the formulas for those you know, for those shapes are I have them down here. Rectangles 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 gonna call that Delta X. So this is just gonna be the area of the square. So I'm gonna call this Delta X. I'm gonna call this Delta X two. I'm gonna call this Delta x three. So this Delta X one is just gonna be Delta X two plus Delta X three. So now I just have to figure out those areas. Will Delta X two is just a rectangle So this is gonna be base times height, the base of this square rectangles to and the height is too. So it's just gonna be two times two and thats four. So I've got 4 m here. Uh, then this triangle here I had I knew the equation for that is one half of base times height. So I'm now I've got one half, and the basis to in the height is to also so I've just got one half of two times to one half of the two will cancel and you'll just get 2 m. So I'm just gonna add those four and two together, and I get a total of 6 m now. One way. You can also think about this 6 m. It's just if you count up all the little boxes that are inside of this curve, I've got 1234 five and then these air, both one half like this is like one half of one half altogether. That makes six. That's kind of another way. You can visualize that. Alright, guys. So let's move on to the part B. Now we're gonna calculate the displacement for the entire motion over here. So the entire motion is actually from zero all the way to six. So you have to figure the area that's under you, concur for all of that stuff, it's gonna include the area that we have for part A. So if I want the total displacement Delta X total, that's gonna be the Delta X one 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. Remember, that's going to be between the values of the graph and the time access. It doesn't matter whether it's above or below, So the area that's underneath the curve here is actually gonna be the area that's in green. So what I'm gonna do is I'm gonna call this. I'm gonna call this Delta X four. Let's say so. I've got Delta X one plus Delta X four. All of that added together is gonna be my total entire displacement So let's just go ahead and do that. So I know Delta X one is six and then I have to figure out this displacement, and that's gonna be my total. So let's just go ahead to the graph. Delta X four is just gonna be one half of base times height because it's a triangle again. The bases to and the height is too. So it's just gonna be one half of two times two. So again, we're gonna cancel and you're just gonna get two out of this. But you have to remember, is that areas above the time axis are going to be positive displacements because you're velocities positive you're moving forward. So these displacement that we gotta here for and to our positive, Whereas the areas below the time axis are going to be negative displacements because you're moving backwards, your velocity is negative there. So that means this is not to This is actually negative too. So this is our Delta X four. So you've got six plus negative to, and you add that together in your total displacement is just going to be positive 4 m and that's really all there is to it. Guys, let's get more practice problems and let me know if you have any