Figuring how long it takes an a380 to take off given a constant acceleration. Created by Sal Khan.

Watch the next lesson:

Missed the previous lesson?

Physics on Khan Academy: Physics is the study of the basic principles that govern the physical world around us. We’ll start by looking at motion itself. Then, we’ll learn about forces, momentum, energy, and other concepts in lots of different physical situations. To get the most out of physics, you’ll need a solid understanding of algebra and a basic understanding of trigonometry.

About Khan Academy: Khan Academy offers practice exercises, instructional videos, and a personalized learning dashboard that empower learners to study at their own pace in and outside of the classroom. We tackle math, science, computer programming, history, art history, economics, and more. Our math missions guide learners from kindergarten to calculus using state-of-the-art, adaptive technology that identifies strengths and learning gaps. We’ve also partnered with institutions like NASA, The Museum of Modern Art, The California Academy of Sciences, and MIT to offer specialized content.

For free. For everyone. Forever. #YouCanLearnAnything

Subscribe to Khan Academyâ€™s Physics channel:

Subscribe to Khan Academy:

is it really necessary to convert the units first before the actual computation?

where did you get that value for acceleration? It feels too low to me

It's fantastic!

km/h to m/s, just divide by 3.6.

ØªÙ…Ø§Ø§Ø§Ø§Ø§Ø§Ù…

0=3

Did you calculate the acceleration based on the mass of the aircraft and the thrust provided by the engines? I'm curious to see that calculation.

Unfortunately, there is no way this plane maintains a constant acceleration! Due to kinetic energy being a function of the square of velocity, it turns out it takes MUCH longer to go from, say, 275 to 280 kmh, than it does from 0 to 5 kmh. And that is without the additional air resistance at the higher velocities.

Fun problem to think about, but bad example for constant a.

The mathematics is correct and your information of your 'take of velocity' 280km/151nm per hour is correct but this is the Air Speed not ground or actual speed.

So if the aircraft has a head wind of 20nm the A380 would only have to accelerate to 131nm

Or

Head wind of 30nm then Acceleration would only be 121nm ground/actual speed

So there are lots of variations, your theory is correct for an A380 in what they call ISA CONDITIONS, international Standard atmosphere with pressure being 1013mb/29.92hmg and +15 degree c….. If the temperature is say +5 then the air will be more dense, so more air particals around the wing, there for its True Airspeed will be less again.

So there are many factors to be accounted for when answering this question.

Why not just use the formula x=1/2at^2? You square 78 (time)= 6084 and then multiply it by 1/2 making it = 3042. Same answer much faster, it seems

Completely false in reality, but the calculation is right.

When a plane takes off the pilot doesn't simply slam the throttles open – he needs to stabilise the engines first at only 40% thrust. The acceleration is far from constant. Also, take-off speed varies depending on weight. Many A380 flights are overweight for take-off, meaning that it takes far longer to actually lift off the ground.

I'm getting 5,248-ft (Vr) @ 170-mph (150-kts) which comes to 31-seconds from take-off roll to rotate. That's assuming it's loaded normal. This an assumption since I've not read the POH for this a/c. Watching this video is a bit confusing since you are demonstrating in British math instead of American. I only know 1-meter = 3.28-ft. (distance), and 1-kilogram = 2.2-pounds (weight). Why can't there be only one system?

How can he write that fast? ^^

No its not. You can clearly see that the flaps aren't fully extended as they would be for landing. Also, the pitch up angle is far too great for a landing as the tail would strike the ground at the angle shown in the picture. This is definitely a picture of an A380 taking-off…

I Lol'd when he took out the ti-85

you must clarify and be focused on what you know

an airplane take off time is not calculated in this way you show here how long takes to reach that speed,if you know the take off speed a X weight.

Singaporean =3

Thats cuz its very light. A fully loaded airplane can take up to over 50s to takeoof while A light plane (low fuel and passenger) will take much less time.

I've seen a a380 take off in just over 30 seconds

in higher physics, you don't even need to know the mass or force to figure out the acceleration. just need 3 different locations of the plane at different times.

@userwI2850

but the weight is not needed simply because the weight doesn't slow down the speed of the aircraft. If anything, that would be more or less an energy or force problem.

yes, but surely if F=ma then a = f/m?

And through my whole explanation, I meant velocity as the the true air speed, which is the speed of the aircraft through the air, which is different from the conventional view of speed in relation to the ground. And for your subsequent video on the distance, the distance must be factored by the wind direction (component of head or tail wind). Because speed through ground = Air speed +/- wind component.

Again I apologize if all this is irrelevant to your video.

Lift is a product of 1/2.dynamic pressure.area of aero foil.coefficient of lift. The varying runway conditions (slope, wetness etc.) will also need to be factored, and atmospheric conditions will govern the air density, which will change the rotation velocity of the aircraft. So the rotation velocity differs.

I don't mean to be a smart alec. I still think your video is intuitive in a mathematical point of view. However I thought this might interest you and the others who are watching this.

Hello Salman Khan. I am a student pilot and I'm half way though my course so I half quite an idea on how take-off time and distance are calculated.

There are different types of speeds, relevant to different situations and calculations. In this case, 'indicated airspeed' is usually used. Indicated airspeed actually dynamic pressure, which is = 1/2.density.velocity(squared). This is used rather than regular speed because it is the pressure that creates lift to govern the aircraft's flight.

I think there is a problem. Acceleration is the rate of change of velocity divided by time and not the "rate of change" in time? Other than that! Awesome work!

Yeah, but also considering many factors including take-off weight, wind speed, direction, temperature, pressure & humidity. And for constants for this particular airliner, like total lift area, engine thrust & output graph, and loss of weight, and many things just how much you like to complicate it.

But believe me, isn't it so sweet to close your eyes at the passenger seat, and feel the power of taking-off?!

Why is everybody taking this so literally? It's a physics problem intended to teach..

Well firstly weather is pre calculated into the any pilots Take Off profile. Next, the safety margin is what Vr represents it a pre calculated speed that is computed that's why we have a V1 V2 Vr to work out. Last 50 sec or that extra 9 sec you have added will have the A380 GS too fast and run distance off the chart… oh well.

In real life it takes about 50 seconds with a full passenger plane from 0 km/h to lift off.

Remember that the lift off speed is not exactly 278 km/h, since it depends on many factors like the weather for instance. And some pilots to have some safety margin after "rotate" til they actually rotate so they have more margin speed during lift off in case any thing would happen.

i don't get how T=V/A is the seconds it takes for it to take off? I get 78m/s with a at 1m/s^2…how do we know its take off at that point and not some other random time. confused here

What software do you. Use?

@jnrolf thank you

Khan i love these new HD Physics, however, i miss the "intuition" part from your old videos. I found the idea of NEVER memorizing Physics formulas fascinating and how all these new projectile motion formuals are derived from the basic formula D=va*t. Anyways i hope you incorporate this idea(Polite)

However, in real life the A380 takes;

41sec to go from 0 to 150 Kts (278km/h) over 1600m for Vr (Velocity of rotation) = Takeoff. At 595 961 kg (1313869 lb) gross weight, Flaps 1

can you do a video on a hypothetical star cannon and how tall it would have to be (given a certain constant acceleration) to reach escape velocity? I've always wondered…

However, in real life the A380 takes;

41sec to go from 0 to 150 Kts (278km/h) over 1600m for Vr (Velocity of rotation) = Takeoff.