Calculating Displacement from Velocity-Time Graphs

2. 1D Motion / Kinematics

Calculating Displacement from Velocity-Time Graphs - Video Tutorials & Practice Problems

On a tight schedule?

Get a 10 bullets summary of the topic

1

concept

Calculating Displacement from Velocity-Time Graphs

Video duration:

4m

Play a video:

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

2

Problem

Problem

A moving box's motion is described by the velocity graph below. The box's initial position is x = 0. (a) How far has the box moved at t = 4? (b) What is the displacement of the box from t=4 to t=10?

A

(a) 16 m; (b) 0 m

B

(a) 16 m; (b) 24 m

C

(a) 24 m; (b) 0 m

D

(a) 24 m; (b) 24 m

E

(a) 32 m; (b) 0 m

F

(a) 32 m; (b) 24 m

3

example

Displacement of a Car

Video duration:

4m

Play a video:

Hey, guys, let's work out this example. Together we're showing the velocity time diagram. So that's V versus Tear the velocity time graph of a moving car. And we're told that the initial position is negative. 21 were supposed to find out what's the cars? Final position AT T equals five seconds of final position. So the variable that we're looking for is going to be X or X final. So that's gonna be what we're looking for here. So how do we find the final position? Well, we find position by finding the displacement. Remember, Delta X. Delta X can be written as X final minus X 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 amongst 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, already have the initial position, which is 21 or negative 21 m. So now all I have to dio is I just have to figure out what the area that's under the curve, I have to figure out. The 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. Four between the two times I'm looking for it says, What's the cars? Final position AT T equals five seconds. And we're also told that the initial positions is Teoh t equals zero. So what this really means is we're looking for the areas are the curve off this entire graph over here? So the area that's under under the curve is just gonna 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. All right. So how do we find now the area of this really complicated shape? Remember, any time 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 was 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 three 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 three smaller shapes, so I'm just gonna label them Delta X one. Let's call this guy Delta X two and then we'll call the smaller one Delta X three. Okay, so let's just get to it. Uh, in the total displacement is just gonna be by adding up all of those all of those areas, they're all of those smaller displacements up. And so Delta X one is gonna be a triangles. We're gonna use one half base times height. Now we just have to figure out the base and height is, and so we're going from zero all the way to three. So the base here is three, and the height is gonna be going from to all the way up to 10. So that means that this height of this triangle is gonna be eight. So it means we have one half of three times eight and that's 12. Let's move on to part the second one. The second one is gonna be from zero all the way to five. It's a rectangle, which means the equation we're gonna use is base times, height, and we're gonna use s so we have that. The base is five and the height is gonna be too 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 three. It's a small triangle. So we're still going to use the same equation. One half, one half. Wow. Okay, one half base times height. So we've got one half the base is going to be, too, because going from 3 to 5 and the height is also going to be from two, because we're going from two up to four. So we have one half of two times to one half and two will cancel Lee, just leaving one factor of two there. And so now we just put everything together, all of these areas, and then add everything up. The Delta X is gonna be 12 plus 10 plus two. And so that's just gonna 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 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 remember that last. You can't forget that last step there. So we're gonna take this number over here and we'll pop it into this equation. And so, for the final answer, the final position is gonna be the initial position of negative 21 plus 24 which is the displacement. And so what we get is positive. 3 m. That is our final position. Alright, guys, let me know if you have any questions and I'll see in the next one.

Do you want more practice?

We have more practice problems on Calculating Displacement from Velocity-Time Graphs

Additional resources for Calculating Displacement from Velocity-Time Graphs