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Airbus A380 take-off distance

How long of a runway does an A380 need? Created by Sal Khan.
Video transcript
In the last video we figured out that given a takeoff velocity of 280 km/h and if we have a positive value for any of these vectors we assume that it's in the forward direction for the runway Given this takeoff velocity, and a constant acceleration of 1 m/s^2 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 that given all these numbers how long of a runway does it need? Which is a very important question if you want to build a runway that can at least allow Airbus A380 to take off you probably want it to be a little 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's accelerating at 1 m/s^2 to 280 km/h...or to 78 m/s...I converted it right over here as it accelerates to 78 m/s how much land does this thing cover? So we can say, so let's call this the displacement is going to be equal to... So displacement is equal to...you can 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... (I would encourage you to watch the video on average velocity for constant acceleration) but if you have constant acceleration and that is what we are assuming in this example If you assume that your acceleration is constant then you could 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 m/s is going to be our final velocity, which is (let me calculate it down here) so our average velocity in this example Velocity_average in this example... is going to be our final example...78 m/s plus our initial velocity,. But 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 m/s and the value of an average velocity in this situation, in any situation the value of an average velocity is that we could figure out our displacement by multiplying our *average velocity* times the <i>time</i> 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 m/s... (just the average of 0 and 78..39 m/s) Another way to think about it if you want to think about the distance traveled this plane is constantly accelerating so let me draw a little graph here this plane's velocity-time graph will look something like this so 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 its constant acceleration it should actually be a slope of 1, given the numbers in this example And the distance traveled 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 will cover in another video) Or let me give you the intuition of why that works... why distance is area under a velocity-time line What an average velocity is, is some velocity and in this case it's exactly right between our two... our final and initial velocities But 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 m/s times 78 seconds and let's just get our calculator out for this we have 39*78...gives us 3042 so this gives us 3042... and then m/s times second just leaves us with meters so you need a runway of over 3000 meters for one of these suckers to take off or over 3 kilometers, which is about 1.8 or 1.9 miles just for this guy to take off which I think is pretty fascinating