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Force vs. time graphs

David explains how to use a force vs. time graph to find the change in momentum and solves an example problem to find the final velocity of a spaceship. Created by David SantoPietro.

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Video transcript

- [Voiceover] There's a miniature rocketship and it's full of tiny aliens that just got done investigating a new moon with lunar pools and all kinds of organic new life forms. But they're done investigating, so they're gonna blast off and take their findings home to tell all their friends. Let's say at some moment during their ascent, they're moving at four meters per second. And they're tiny aliens, their spaceship is only 2.9 kilograms, but they need to know, are they gonna be able to get off this moon or not? So they gotta pay attention to their speed, but instead of using a speedometer, they clever aliens, they use a force versus time graph. So on their dashboard, they've got a force versus time graph and it tells them what the net force is on them, so let's say this is the net force, not just any force, but this is the total force on them from rocket boosters and the force of gravity and whatever other forces there might be, they've got advanced force sensors. I mean come on, they can determine their net force, let's say, and it gives them this force as a function of time. But they want to know what is their velocity gonna be after nine seconds? So they check their force versus time readout, and this is the graph they get. And now they can determine it. And here's how they do it. They say, "All right there's a force, a net force "of three newtons acting for the first four seconds." So during this entire first four seconds there's a constant force of three newtons. And every alien worth his weight knows that force, the net force multiplied by the time duration during which that force is applied, gives you the net impulse. So this gives us the net impulse. If we take this constant three newtons that acts, multiplied by four seconds during which it acts, we get that there's an impulse of 12 newton seconds. And you might be like, "Wait, who cares "about newton seconds here, I want the velocity. "I don't want the force and the time, "I want to know the velocity at nine seconds." But they teach you at the alien space academy, that the net impulse is not only equal to the net force times the time, it's also equal to the change in momentum of the object that the force was exerted on. And this is good. We know the mass of the object. We wanna know something about velocity. So we know momentum is M times V, this net impulse is gonna help us get there. But his 12 newton seconds was only for the first four seconds. How do we figure it out for the next three seconds? Look at this. During the next three seconds, there's not a constant force, this force is varying, the force is getting smaller. So how do I do this? The force isn't a constant value, so I can't just simply take force times time, because, I mean, what force do I pick? So we're gonna use a trick. We're gonna use a trick, because if you notice for this first section, for the first four seconds, we took the force and multiplied by the time interval, four seconds. So what we did really is we just took the height of this rectangle times the width of this rectangle and that gives us the area of this rectangle. So what we really did is we found the area under the force versus time graph. That gave us the impulse and that's not a coincidence. The impulse equals the area under a force versus time graph, and this is extremely useful to know, because now in this section, where the force was varying, we can still use this, we can just find the impulse by determining the area under that curve. And by area under the curve, we mean from the line curve, in general, to the x-axis, which, in this case, the x-axis is of the time axis. So let's do this. We found the impulse for this first section. That was 12 newton seconds. Now we can find the impulse for this next section by just determining the area. So this is a triangle. We'll do 1/2 base, the base is one, two, three seconds, and the height is still three newtons. So we get a net impulse of 4.5 newton seconds. So we've got one more section to go, but this one's a little weird, this one's located, the area is located below the time axis, so this is still a triangle, but since the forces are negative, this is gonna count as a negative net impulse. So when the area lies above the time axis, it counts as a positive impulse, and when the area lies below the time axis, it counts as a negative net impulse. So how much negative net impulse? We still find the area. So the area of a triangle, again, is gonna be 1/2, the base this time is two seconds, and the height is negative two. So negative two newtons, which gives us a net impulse of negative two newton seconds. And now we can figure out the velocity of the spaceship at nine seconds. So, assuming that this force readout started at this moment right over here at t equals zero seconds was the moment when it was going four meters per second, then we could say the total net impulse should equal the total change in momentum of this spaceship. And we can find the total net impulse by just adding up all the individual impulses. So during the first four seconds, there was 12 newton seconds of impulse. During the next three seconds there was 4.5 newton seconds of impulse. And during this last portion there was negative two newton seconds of impulse, which if you add all of those up, 12 plus 4.5 plus negative two, you're gonna get positive 14.5 newton seconds of impulse. That's good news for our alien buddies over here, they need to get off this moon, which means they need positive impulse, upward impulse. They got some positive impulse. Let's see what their final velocity was. We know that delta P is the change in momentum, so this is final momentum minus initial momentum, which we could write as mass times v final minus mass times v initial. Now, if this were an earth rocket, this would be hard, 'cause earth rockets using earth technology, eject fuel at a huge rate out the backend, and that loses mass. That means this mass isn't gonna stay constant. So earth rockets essentially push fuel down which causes and equal and opposite force back on the rocket upward. But if you're losing mass, this mass doesn't stay constant, and this whole process is a lot harder, because m final and m initial aren't gonna be the same. Maybe this is where the phrase, "It's not rocket science" comes from, 'cause rocket science is a little harder when that mass changes. So let's just say, these clever aliens can eject only a little bit of fuel, they do so. You might say, "How?" Well, there's gotta be a certain amount of momentum, right, that they eject to give themselves momentum up. But let's say they can eject only a small mass at a huge speed, so there's not much fuel that they're losing. The fuel that they eject is ejected at enormous speed, so that they get their momentum upward, but they lose almost no mass. And that let's us solve this problem assuming that the mass is constant. So if we do that, if we assume the mass is constant, we get positive 14.5 newton seconds equals, we can pull the mass out, the mass is a constant, so we could just write it as m times v final minus v initial, since I can pull out a common factor of m, which means I can write this as 2.9 kilograms multiplied by the final velocity after nine seconds and I know it's after nine seconds, 'cause I added up all the impulse during the nine seconds, minus the initial velocity was four meters per second. So if I divide both sides by 2.9 kilograms, 14.5 over 2.9 is 5 and that will be newton seconds over kilograms which has units of meters per second, and it's positive. And that's gonna equal the leftover over here is gonna be v final minus four meters per second. And now finally if I add four meters per second to both sides, I get the v final somewhere up here, v final of this rocket is gonna be nine meters per second. So after nine seconds it ended up going nine meters per second. That's just a numerical coincidence. The way you find it, so recapping the way we did this, we found the area under the curve, 'cause the area under a curve under our force versus time graph represents the impulse on the object. We found it for the entire trip, noting that underneath the time axis, when this curve goes under the time axis, the net impulse is gonna be negative. We added up all the net impulse, we set it equal to the change in momentum, we plugged in our values and we solved for our velocity after nine seconds.