Hey, guys. So sometimes, you'll have to calculate changes in velocity, from the acceleration time graphs instead of using your equations. So that's what we're gonna check out in this video, but we're gonna see that it is exactly like how we calculated displacements from velocity time graphs. Let's check it out.

So guys, when we had velocity time graphs, the area under the curve, remember, was just the area that's enclosed between the values of the graph and the time axis. So you just make this little shape here, and then your area under the curve represents your change in the position or the displacement, Δx. Well, for an acceleration-time graph, it's the same exact principle. The area under the curve is just the area between the acceleration values and the time graph, except now the area that's enclosed within the shape here doesn't represent the displacement. It represents your change in the velocity. So it's the same idea, it's just the variable that's different here. So that's really all there is to it, guys. It's the same exact procedures. We're just gonna be calculating a bunch of areas. So let's go ahead and check out this example.

We've got an acceleration-time graph for a moving box, and we're told it's initially at rest. And so part a says we're gonna calculate the box's velocity at t=3. So we know how to do this. We just figure out t=0, t=3, and then we just have to split this up, and we have to calculate the area that's inside of this shape over here. This is not a very simple shape, and I don't know what the formula off the top of my head, but I can break it up into a rectangle or square and a triangle. And so I can calculate these areas, which should be ΔV1, I'm gonna call that, and ΔV2. And now I just have to figure out the total velocity by calculating and adding up these two things.

Now I'm also told in part a that it's not the change in velocity that I'm looking for, it's the actual velocity. So what does that mean? Well, what it's asking for is v, but I know how to calculate the velocity by first calculating the change in the velocity by looking at the area of the graph. Now remember that the change in the velocity, Δv, is just v_{final}−v_{initial}. And so, and I'm told that initially the box is at rest. So I know that this v0 is just equal to 0. So the box's velocity and the change in the velocity are the exact same thing.

So that's really all it's asking me for. So I'm gonna look for the velocity and I'm gonna call this v3, and all I have to do is just add up these 2, shapes together, ΔV1 and ΔV2. So ΔV1 is just a rectangle. It's just base times height, that's the area. Remember the base is 2 and the height is 2. So that means that's just gonna be 2 times 2, which is 4. Now the units for this are gonna be in meters per second. Remember, because it is a change in the velocity. So you have to be careful that it's meters per second.

Now ΔV2 is just triangle. So it's gonna be 12base×height. That's the formula. The base of this is 1, the height of this is 2. So it's gonna be 12×1×2. The one-half and the 2 will cancel, and we'll just get 1 meter per second. So now we just have to add these 2 sort of like smaller values and that's gonna be the total velocity at 3 seconds. So it's just gonna be 4 plus 1, and that is 5 meters per second.

That's all there is to it. So now let's figure out what the velocity is at t=5. So all we have to do is now figure out the area under the curve from 0 to 5 seconds. And so that's gonna include everything from 0 to 3, plus now from 3 to 5. So that's all we have to do is we just have to tack on this extra area that we have between this little triangle here and the graph. So I'm gonna highlight this in green. So this is gonna be the area that I have to include now. It's not a very pretty shape, so what I can do is just break it up into 2 small right triangles here. I'll call this 1, ΔV3, or sorry, ΔV4. But what you'll see is that they're symmetrical. They're the exact same pieces. So whatever I find for one value, it's gonna be the same for the other one.

Okay. So the value or the velocity at 5, I'm gonna call that v5, is just gonna be delta v. It's actually going to be, whoops. It's gonna be the velocity that I found out in part a. It's gonna be the 5 meters per second plus, now it's gonna be ΔV3 plus ΔV4. And so, I already have what this is. I know this is 5 meters per second. And now plus, I can just add up together these other two velocities here, and this is gonna be my total velocity in meters per second. So that's the plan.

So, I've got ΔV3. I know this is triangle, so I'm gonna use 12base×height. The base is 1, and the height is 2. So I'm gonna have 12×1×2, and that's again just going to give me 1. But again, remember that when you have areas that are above the time axis, those are gonna be positive changes in your velocity because your acceleration is positive. Whereas if you have negative, or sorry below the time axis, then those are gonna be negative changes in ΔV because your acceleration is negative. So your velocity is becoming more negative. So that's really all there is to it. So that means this is a negative one, and ΔV4 is the same exact thing. It's also negative one. So this is gonna be negative one and negative one, and so your total velocity at 5 seconds is 5 minus 2, which is going to be 3 meters per second, and it's gonna be positive.

So those are our two answers. Let me know if you guys have any questions. Let's get some more practice.