Topic 8.2: Connecting Position, Velocity, and Acceleration of Functions Using Integrals

One try

It’s the middle of the night, hundreds of kilometers out in the Atlantic, and this ship, with nobody on board, is waiting for something to fall out of the sky. That’s the first stage of a Falcon 9 rocket. Less than ten minutes ago it lifted a satellite off a pad in Florida and climbed past the edge of space, and now it’s falling back, tail first, at more than a thousand kilometers an hour.

It gets exactly one try at this, so keep your eyes on the deck. And it’s down, standing on a ship about the size of a football field, in the dark, in the middle of the ocean.

What makes that so hard is that this booster can’t hover, because even one engine at its lowest setting out-pushes the nearly empty booster. So its speed has to reach zero at the exact moment it reaches the deck.

Let’s fly it again with its velocity drawn beside it, and watch this glowing region fill in while the height drains away. They run out together, because the area of that region is exactly the height where the booster lit its engines, and that one fact is the whole landing.

By the end of this lesson, you’ll be able to read a velocity graph like this one and work out exactly how high a booster has to light its engines. Then you’ll get to light one yourself and land it. Suppose all we could read was the booster’s velocity. Could we still work out how much sky it has left?

The area is the distance

Let’s go back up to six kilometers, where the booster is falling at a steady three hundred sixty meters per second and its engines haven’t lit yet. We’ll call up the positive direction, which means anything heading down has a negative velocity, so the booster’s velocity right now is negative three hundred sixty meters per second.

If it keeps falling at that same steady speed for ten more seconds, how high above the deck will it be? Without noticing, you just did an integral, because that rectangle’s area, ten seconds times three hundred sixty meters per second, is exactly how far the booster fell. Seconds times meters per second leaves you with meters, so an area on a velocity graph isn’t square anything, it’s a distance.

Once the engines light, the speed doesn’t stay steady anymore, so we do what you already know how to do. We slice the time into thin strips, and inside each strip the velocity barely changes, so each one is almost a rectangle. Add up all those thin rectangles, let them get thinner and thinner, and you know exactly what you get.

It’s the definite integral. The definite integral of velocity from a to b is the displacement, how much the position changed over that stretch of time. The reason it works is the Fundamental Theorem of Calculus. Velocity is the derivative of height, so integrating the velocity from a to b gives you the height at b minus the height at a, the net change. When the graph sits below the axis like ours does, the integral comes out negative, which is exactly right, because the booster’s height is going down.

The integral only tells you how much the height changed, and never where the booster actually is. For that you need a starting point, so its height at any time t is its starting height plus the integral of the velocity from zero to t.

Go back to our steady fall from six thousand meters and forget the starting height, and you’d get a booster at negative thirty-six hundred meters, three point six kilometers under the ocean, which is a pretty good sign you’ve left something out. With the start put back, it’s six thousand plus negative thirty-six hundred, which is twenty-four hundred meters, so whenever a question asks where something is, think starting value plus accumulated change.

Light it high enough

Now we can answer the only question the booster really cares about, which is how high it has to light its engines. Let’s start with the simplest plan, where the engines brake at a constant thirty meters per second every second. Then the velocity climbs from negative three hundred sixty up to zero in exactly twelve seconds.

The change in height is the integral from zero to twelve. The antiderivative is negative three hundred sixty t plus fifteen t squared, and plugging in twelve gives negative forty-three twenty plus twenty-one sixty, which is negative twenty-one sixty meters. We can check that with geometry, because the region is a triangle, twelve seconds wide and three hundred sixty deep. Half of twelve times three sixty is twenty-one sixty, the same answer.

The booster has to finish at zero, so zero equals the ignition height plus negative twenty-one sixty, and it has to light at twenty-one hundred and sixty meters. If you took precalc with us, you might recognize that number, because it’s exactly where you lit this same booster last time. Real braking isn’t constant, though. The booster gets lighter as it burns fuel, so the same engines slow it down harder and harder, and in our model of the burn the braking climbs steadily from twenty-four meters per second every second up to thirty-six.

That’s an acceleration of twenty-four plus t. It’s positive even though the booster is falling, because it’s pushing the velocity up toward zero, and that’s what slowing down looks like when your velocity is negative. First we need the velocity, and it works exactly the way the height did. The velocity at time t is the starting velocity plus the integral of the acceleration.

The integral of twenty-four plus x from zero to t is twenty-four t plus one half t squared, so the velocity is negative three sixty, plus twenty-four t, plus one half t squared. Let’s make sure it really stops at twelve seconds. Negative three sixty, plus two eighty-eight, plus seventy-two, is zero, so it does.

Then comes the landing itself. The height has to come out to zero at twelve seconds, when the burn ends. So zero equals the ignition height plus the integral of this velocity from zero to twelve, and now it’s your turn. How high does the booster need to be when it lights its engines?

That makes sense when you put the two graphs together. The ramped braking starts out gentler, so its curve stays deeper for longer, and that extra sliver of area is a hundred forty-four extra meters of falling.

What happens if you light this booster at the old height? It runs out of sky about nine seconds into the burn and hits the deck at more than three hundred fifty kilometers an hour. Light it at twenty-three oh four instead, and it touches down with nothing left over. A hundred forty-four meters of sky was the whole difference, and you found it with an integral.

Now suppose a booster is already committed to lighting at twenty-four hundred meters, a little higher than ours. Its computer can’t change where it lights, only how the braking ramps up during the twelve seconds. Bend the braking ramp until you think this booster lands, and then press light to test it.

Forward, then back

A falling booster only ever goes one way, but a car can drive forward and then back up, and so can a particle on the AP exam. This car rolls past the line at eight meters per second and brakes, and the graph shows its velocity for the next six seconds. It hits zero at four seconds, and after that it’s negative, which means it’s backing up.

Drag the cone to the spot where you think the car will be at six seconds, and then press go to see if you’re right. Which one is the answer? It depends on the question, because those two numbers answer two different questions. Displacement asks how far the car ended up from where it started, which is just the integral of the velocity, with the backing-up part counting as negative. Total distance asks how far it actually drove, so we flip the negative part up and integrate the speed instead.

On paper, the displacement is the integral of v from zero to six, sixteen minus four, which is twelve meters. The total distance is the integral of the absolute value of v, sixteen plus four, which is twenty.

Divide each one by the six seconds and you get two different averages, an average velocity of two meters per second and an average speed of about three point three. Without a calculator, you find where the velocity changes sign, which is exactly where the car turns around, then split the integral there and add up the sizes of the pieces. Taking the absolute value of the whole integral at the end doesn’t work, because by then the backing up has already canceled out part of the trip.

Let’s try one where a car backs out of a parking space and then pulls away. Its velocity is t squared minus four t for six seconds, so how far does it travel in all, to the nearest tenth of a meter?

Now you drive. Get the car back to the bay line with at least twenty meters on the odometer, and watch your own velocity graph draw itself as you go.