Force vs. time graphs | Impacts and linear momentum | Physics | Khan Academy

There's a miniature rocket ship, 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 going to blast off and take their findings home to tell all their friends.

Let's say it's some moment during their ascent. They're moving at 4 m/s, and they're tiny aliens. Their spaceship is only 2.9 kg, but they need to know, are they going to be able to get off this moon or not? So they've got to pay attention to their speed. But instead of using a speedometer, they're clever aliens; they use a force versus time graph.

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, 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 it gives them this force as a function of time, but they want to know what is their velocity going to be after 9 seconds. So they check their force versus time readout, and this is the graph they get. And now they can determine it. Here's how they do it.

They say, "All right, there's a net force of 3 Newtons acting for the first 4 seconds." So during this entire first 4 seconds, there's a constant force of 3 Newtons. And every alien worth his weight knows that 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 3 Newtons that acts, multiply by 4 seconds during which it acts, we get that there's an impulse of 12 Newton seconds. Now you might be like, "Wait, who cares about Newton seconds here? I want the velocity! I don't care about the force and the time; I want to know the velocity at 9 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 want to know something about velocity. So we know momentum is M * V. This net impulse is going to help us get there.

But this 12 Newton seconds was only for the first 4 seconds. How do we figure it out for the next 3 seconds? Look at this: during the next 3 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 going to use a trick. We're going to 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 of 4 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 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 1, 2, 3 seconds, and the height is still 3 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 going to 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 going to be 1/2 the base. This time, it's 2 seconds, and the height is -2, so -2 Newtons, which gives us a net impulse of -2 Newton seconds. And now we can figure out the velocity of this spaceship at 9 seconds.

So assuming that this force readout started at this moment right over here at t = 0 seconds was the moment when it was going 4 m/s, then we can just say the total net impulse should equal the total change in momentum of the spaceship. And we can find the total net impulse by just adding up all the individual impulses.

So during the first 4 seconds, there was 12 Newton seconds of impulse. During the next 3 seconds, there was 4.5 Newton seconds of impulse, and during this last portion, there was -2 Newton seconds of impulse, which if you add all those up, 12 + 4.5 + (-2), you're going to 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 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 * V final minus mass * V initial.

Now, if this were an Earth rocket, this would be hard because Earth rockets, using Earth technology, eject fuel at a huge rate out the back end, and that loses mass. That means this mass isn't going to stay constant. So Earth rockets essentially push fuel down, which causes an 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 going to be the same. Maybe this is where the phrase "It's not rocket science" comes from because 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 got to 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 lets us solve this problem assuming that the mass is constant.

So if we do that—if we assume the mass is constant—we get 14.5 Newton seconds equals; we can pull the mass out. The mass is a constant, so I could just write it as M * V final minus V initial. Since I can pull out a common factor of M, which means I can write this as 2.9 kg multiplied by the final velocity after 9 seconds, and I know it's after 9 seconds because I added up all the impulse during the 9 seconds minus the initial velocity, which was 4 m/s.

So if I divide both sides by 2.9 kg, 14.5 over 2.9 is 5, and that'll be Newton seconds over kilog, which has units of m/s, and that's positive. That's going to equal the leftover over here, which is V final minus 4 m/s.

And now finally, if I add 4 m/s to both sides, I get that V final, somewhere up here, V final of this rocket is going to be 9 m/s. So after 9 seconds, it ended up going 9 m/s. That's just a numerical coincidence in the way you find it.

So recapping the way we did this: we found the area under the curve because the area under our 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 underneath the time axis, the net impulse is going to be negative. We added up all the net impulse, set it equal to the change in momentum, plugged in our values, and solved for our final velocity after 9 seconds.