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Airbus A380 take-off distance

How long of a runway does an A380 need? Created by Sal Khan.

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  • male robot donald style avatar for user Dok Schwammkopf
    why do we have to get the average velocity?
    (32 votes)
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  • blobby green style avatar for user Daniel
    I don't understand when you say the plane doesn't move a full meter in the first second, rather a half meter. Could you please explain for me more in details. Thanks
    (13 votes)
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  • orange juice squid orange style avatar for user Sebastian Ls
    why is average velocity the half of the final velocity?
    (9 votes)
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    • piceratops ultimate style avatar for user Roman Rubtsov
      Becouse initial velocity (when plane starts his movement) is 0, and final velocity (when plane take off) is 78 m/s. So to find average velosity we should sum initial and final velocity and divided it by 2 (becouse we have only 2 magnitude of velocity). If it was more magnitudes, for example 3, we sum all 3 magnitudes of velocity and divide it by 3. Sorry English not native language =) Hope it helps
      (21 votes)
  • leafers tree style avatar for user Palaash Tarapore
    Can you please explain why we divide Vf-Vi by 2 to find the average velocity?
    Don't we have to divide Vf-Vi by the total time it took to reach to Vf to find the average velocity?
    (6 votes)
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  • leaf orange style avatar for user Aman Parekh
    If displacement= velocity*time,i.e, 78*78. why is the velocity halved?
    (4 votes)
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  • leafers tree style avatar for user mamillagunasekhar
    if acceleration is not constant then what should we do?
    (3 votes)
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  • leaf blue style avatar for user tombo5462
    Where did you get the two
    (3 votes)
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  • aqualine ultimate style avatar for user Jeff Thawn
    Couldn't you just times ( d= t x v ) and divide by 2 to get the average?
    For example: (78 seconds * 78 m/s) = (6084 / 2) = 3042
    (2 votes)
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  • aqualine ultimate style avatar for user Martin
    Does the average velocity always have to be in between the initial and final velocities on a graph? Also, doesn't Sal mean that displacement is the area under the velocity timeline?
    (2 votes)
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  • mr pink red style avatar for user Alexei
    I'm finding a slightly different result. Does the following add up?

    Instead of assuming that vf (takeoff velocity) is 78 m/s, and that ∆t (takeoff time) is 78s, I went back to the original parameters (technically, that 78 was calculated from 700/9, which comes to 77.77... which Sal rounded to 78). So vf = 280 km/h = 700/9 m/s
    - Starting with the premise that Vavg = ∆s (displacement) / ∆t (time) we are looking for ∆s as the total displacement, i.e. the minimum runway length.
    - We also know that a (acceleration) = ∆v (change in velocity) / ∆t ; therefore: ∆t = (vf-v0)/a
    - Based on the above formulas:
    - ∆s = Vavg*∆t
    - ∆s = (vf - v0)/2 * (vf-v0)/a
    - ∆s = (vf)^2 / 2a -- since v0 = 0 m/s
    - ∆s = (700/9)^2 / 2
    - ∆s = 3,024 meters

    Thoughts? Does this make sense, with Sal using a rounded up quantity as the explanation for the difference?
    (3 votes)
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Video transcript

In the last video, we figured out that given a takeoff velocity of 280 kilometers per hour-- and if we have a positive value for any of these vectors, we assume it's in the forward direction for the runway-- given this takeoff velocity, and a constant acceleration of 1 meter per second per second, or 1 meter per second squared, we figured out that it would take an Airbus A380 about 78 seconds to take off. What I want to figure out in this video is, given all of these numbers, how long of a runaway does it need, which is a very important question if you want to build a runway that can at least allow Airbus A380s to take off. And you probably want it to be a little bit longer than that just in case it takes a little bit longer than expected to take off. But what is the minimum length of the runway given these numbers? So we want to figure out the displacement, or how far does this plane travel as it is accelerating at 1 meter per second squared to 280 kilometers per hour, or to 78-- or where did I write it over here-- to 78. I converted it right over here. As it accelerates to 78 meters per second, how much land does this thing cover? So let's call this, the displacement is going to be equal to-- So displacement is equal to-- You could view it as velocity times time. But the velocity here is changing. If we just had a constant velocity for this entire time, we could just multiply that times however long it's traveling, and it would give us the displacement. But here our velocity is changing. But lucky for us, we learned-- and I encourage you to watch the video on why distance, or actually the video on average velocity for constant acceleration-- but if you have constant acceleration, and that is what we are assuming in this example-- so if you assume that your acceleration is constant, then you can come up with something called an average velocity. And the average velocity, if your acceleration is constant, if and only if your acceleration is constant, then your average velocity will be the average of your final velocity and your initial velocity. And so in this situation, what is our average velocity? Well, our average velocity-- let's do it in meters per second-- is going to be our final velocity, which is-- let me calculate it down here. So our average velocity in this example is going to be our final velocity, which is 78 meters per second, plus our initial velocity. Well, what's our initial velocity? We're assuming we're starting at a standstill. Plus 0, all of that over 2. So our average velocity in this situation, 78 divided by 2, is 39 meters per second. And the value of an average velocity in this situation-- actually, average velocity in any situation-- but in this situation, we can calculate it this way. But the value of an average velocity is we can figure out our displacement by multiplying our average velocity times the time that goes by, times the change in time. So we know the change in time is 78 seconds. We know our average velocity here is 39 meters per second, just the average of 0 and 78, 39 meters per second. Another way to think about it, if you want think about the distance traveled, this plane is constantly accelerating. So let me draw a little graph here. This plane's velocity time graph would look something like this. So if this is time and this is velocity right over here, this plane has a constant acceleration starting with 0 velocity. It has a constant acceleration. This slope right here is constant acceleration. It should actually be a slope of 1, given the numbers in this example. And the distance traveled is the distance that is the area under this curve up to 78 seconds, because that's how long it takes for it to take off. So the distance traveled is this area right over here, which we cover in another video, or we give you the intuition of why that works and why distance is area under a velocity timeline. But what an average velocity is, is some velocity, and in this case, it's exactly right in between our final and our initial velocities, that if you take that average velocity for the same amount of time, you would get the exact same area under the curve, or you would get the exact same distance. So our average velocity is 39 meters per second times 78 seconds. And let's just get our calculator out for this. We have 39 times 78 gives us 3,042. So this gives us 3,042. And then meters per second times second just leaves us with meters. So you need a runway of over 3,000 meters for one of these suckers to take off, or over 3 kilometers, which is like about 1.8 or 1.9 miles, just for this guy to take off, which I think is pretty fascinating.