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# Airbus A380 take-off distance

How long of a runway does an A380 need? Learn how to calculate the minimum runway length needed for an Airbus A380 to take off. This tutorial uses principles of constant acceleration and average velocity, demonstrating how these concepts apply to real-world physics. Created by Sal Khan.

## Want to join the conversation?

• 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
• After 1 second it has accellerated to a velocity of 1 m/s. Because it is uniformly accelerating, it thus has moved slower than 1m/s during this first second. On average it has moved (starting velocity + final velocity)/2 which is (0+1)/2=1/2 m/s. Therefore in the first second it has moved 1/2 m.
• why is average velocity the half of the final velocity?
• 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
• 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?
• Let's say that you have something starting at 0 m/s and slowly accelerated so that in 1000 seconds it is going 10 m/s. What seems like a more reasonable average velocity 5 m/s (Vf-Vi/2) or 0.01 m/s (Vf-Vi/1000)?
• If displacement= velocity*time,i.e, 78*78. why is the velocity halved?
• Displacement = AVERAGE velocity * time.
FINAL velocity is halved in order to get average velocity (if initial is 0)
• Where is the first video? where we calculate the 78 seconds answer
• if acceleration is not constant then what should we do?
• This would require calculus to find instantaneous acceleration. Do not worry about that for now.
• I saw the velocity-time slope is constant and it didn't become steeper or lower. Doesn't it mean that the acceleration is 0? I don't know why it is 1m/s^2
• The slope of a velocity vs time graph being constant only means that the acceleration is constant not that it is 0. The slope of a velocity vs time graph is the acceleration so if the slope is 0 then the acceleration is 0 but if the slope is not 0 them neither is the acceleration.
• where is "the last video"