- What is acceleration?
- Airbus A380 take-off time
- Airbus A380 take-off distance
- Why distance is area under velocity-time line
- What are velocity vs. time graphs?
- Acceleration vs. time graphs
- What are acceleration vs. time graphs?
- Acceleration and velocity
Figuring how long it takes an A380 to take off given a constant acceleration. Explore the physics of an Airbus A380 take-off with Khan Academy. Learn how to calculate the time it takes for the aircraft to reach takeoff velocity using principles of acceleration and velocity. This practical example brings the concepts of speed, time, and acceleration to life. Created by Sal Khan.
Want to join the conversation?
- Quick question, does the weight of the Airbus A380 play a part in how fast the aircraft accelerates?(145 votes)
- In which lessons can I learn this way of doing maths, with equations, cancelling out things and so on?(39 votes)
- Type in 'dimensional analysis' in the search bar near the top of your page and it will bring you to the section in algebra 1 that deals with these types of conversions. Developing a solid understanding of this concept while in algebra will really help you in many subjects.(23 votes)
- This may be covered later, or just sloppy thinking on my part, but is there some sort of something beyond acceleration? like, meters/second/second/SECOND? that would be an increase in the rate of acceleration? or is that just change in acceleration and I'm being confusing and dramatic? I mean, mathematically obviously we can do this, but does it make sense in the real world?(21 votes)
- That is exponential acceleration. Not only is velocity increasing, but the acceleration that increases the velocity is increasing!(10 votes)
- If an aircraft is smaller will the speed be more faster? For ex. a Boeing 737/Airbus 320.(5 votes)
- The aerodynamic shape of the plane and its engines are more important to its speed than the size. For example, a Concorde can travel faster than a F-104 Nighthawk eve though it is nearly 10 times larger because it is sleek and light and has 4 Rolls-Royce engines powering it to supersonic speed whereas the Nighthawk is built for spying so it has heavier "cloaking" material and crooked sides that disperse RADAR waves so it cannot be detected thus increasing its drag. Size doesn't necessarily matter!(24 votes)
- Gee, still can't understand why there is square in the m/s^2 formula. if every second smth accelerates x m/s, why then there is not 2s or m/2s?(5 votes)
- velocity tells you how many meters you move per second: hence m/s
acceleration tells you how much the velocity (m/s) changes per second: hence (m/s)/s which is m/s^2!(20 votes)
- I don't understand why would you multiply 280 by 1/3600 and 1000/1(5 votes)
- For example, suppose we wish to convert 15.0 in. to centimetrse.
Because 1 in. is defined as exactly 2.54 cm, we find that
15.0 in. =15.0 in. *2.54 cm/1 in.
= 38.1 cm
where the ratio in parentheses is equal to 1. We express 1 as 2.54 cm/1 in. (rather than 1 in./2.54 cm) so that the unit “inch” in the denominator cancels with the unit
in the original quantity. The remaining unit is the centimeter, our desired result.(6 votes)
- Why velocity over acceleration gives us the time of take off? How do we know that?(5 votes)
- Because acceleration is how much velocity you gain per second. So, assuming the initial velocity is 0 and a constant acceleration, dividing the final velocity by the acceleration give you the time it took to reach this final velocity :
Vf = Vi + a*t
and because the initial velocity is 0 :
Vf = a*t
and because the acceleration is constant :
t = Vf/a
Vi = initial velocity
Vf = final velocity
a = acceleration(4 votes)
- I was wondering, the 78 seconds Mr. Khan got is just an estimate of the total answer. So if you would want to know the exact answer, would you need more information or would you use a different equation?(3 votes)
- I believe you would need to use a different equation because the actual acceleration of the airplane is not constant, but you need a constant acceleration to use the equations shown in this video.(5 votes)
- Can we find the distance traveled by and object if the acceleration of the object is given.
I know that Acceleration = change in velocity / total time taken.
So, If a car accelerated from 5 m/s to 25 m/s in 10 seconds, how far will it travel ?
Acceleration = 20/10 = 2m/s^2
But from the acceleration how can we find the Distance ?(3 votes)
- basically its simple! I am in class seven still though but i can i can help you out in this!
The formula is "s=ut+1/2at^2"
I hope you can easily understand the cause,
(Forgive me for any mistakes)(3 votes)
- why not just say 1m/s squared (acceleration) whats the point of the ".0"(2 votes)
- 1.0 is not the same as 1, it is more precise. When you say 1.0, we know you don't mean 1.1 and you don't mean 0.9. When you say 1, you are indicating that the value is between 0.6 and 1.5.(4 votes)
This right here is a picture of an Airbus A380 aircraft. And I was curious how long would it take this aircraft to take off? And I looked up its takeoff velocity. And the specs I got were 280 kilometers per hour. And to make this a velocity we have to specify a direction as well, not just a magnitude. So the direction is in the direction of the runway. So that would be the positive direction right over there. So when we're talking about acceleration or velocity in this, we're going to assume it's in this direction, the direction of going down the runway. And I also looked up its specs, and this, I'm simplifying a little bit, because it's not going to have a purely constant acceleration. But let's just say from the moment that the pilot says we're taking off to when it actually takes off it has a constant acceleration. Its engines are able to provide a constant acceleration of 1.0 meters per second per second. So after every second it can go one meter per second faster than it was going at the beginning of that second. Or another way to write this is 1.0-- let me write it this way-- meters per second per second can also be written as meters per second squared. I find this a little bit more intuitive. This is a little bit neater to write. So let's figure this out. So the first thing we're trying to answer is, how long does take off last? That is the question we will try to answer. And to answer this, at least my brain wants to at least get the units right. So over here we have our acceleration in terms of meters and seconds, or seconds squared. And over here we have our takeoff velocity in terms of kilometers and hours. So let's just convert this takeoff velocity into meters per second. And then it might simplify answering this question. So if we have 280 kilometers per hour, how do we convert that to meters per second? So let's convert it to kilometers per second first. So we want to get rid of this hours. And the best way to do that, if we have an hour in the denominator, we want an hour in the numerator, and we want a second in the denominator. And so what do we multiply this by? Or what do we put in front of the hours and seconds? So one hour, in one hour there are 3,600 seconds, 60 seconds in a minute, 60 minutes in an hour. And so you have one of the larger unit is equal to 3,600 of the smaller unit. And that we can multiply by that. And if we do that, the hours will cancel out. And we'll get 280 divided by 3,600 kilometers per second. But I want to do all my math at once. So let's also do the conversion from kilometers to meters. So once again, we have kilometers in the numerator. So we want the kilometers in the denominator now. So it cancels out. And we want meters in the numerator. And what's the smaller unit? It's meters. And we have 1,000 meters for every 1 kilometer. And so when you multiply this out the kilometers are going to cancel out. And you are going to be left with 280 times 1, so we don't have to write it down, times 1,000, all of that over 3,600, and the units we have left are meters per-- and the only unit we have left here is second-- meters per second. So let's get my trusty TI-85 out and actually calculate this. So we have 280 times 1000, which is obviously 280,000, but let me just divide that by 3,600. And it gives me 77.7 repeating indefinitely. And it looks like I had two significant digits in each of these original things. I had 1.0 over here, not 100% clear how many significant digits over here. Was the spec rounded to the nearest 10 kilometers? Or is it exactly 280 kilometers per hour? Just to be safe I'll assume that it's rounded to the nearest 10 kilometers. So we only have two significant digits here. So we should only have two significant digits in our answer. So we're going to round this to 78 meters per second. So this is going to be 78 meters per second, which is pretty fast. For this thing to take off every second that goes by it has to travel 78 meters, roughly 3/4 of the length of a football field in every second. But that's not what we're trying to answer. We're trying to say how long will take off last? Well we could just do this in our head if you think about it. The acceleration is 1 meter per second, per second. Which tells us after every second it's going 1 meter per second faster. So if you start at a velocity of 0 and then after 1 second it'll be going 1 meter per second. After 2 seconds it will be going 2 meters per second. After 3 seconds it'll be going 3 meters per second. So how long will it take to get to 78 meters per second? Well, it will take 78 seconds, or roughly a minute and 18 seconds. And just to verify this with our definition of our acceleration, so to speak, just remember acceleration, which is a vector quantity, and all the directions we're talking about now are in the direction of this direction of the runway. The acceleration is equal to change in velocity over change in time. And we're trying to solve for how much time does it take, or the change in time. So let's do that. So let's multiply both sides by change in time. You get change in time times acceleration is equal to change in velocity. And to solve for change in time, divide both sides by the acceleration. So divide both sides by the acceleration you get a change in time. I could go down here, but I just want to use all this real estate I have over here. I have change in time is equal to change in velocity divided by acceleration. And in this situation, what is our change in velocity? Well, we're starting off with the velocity, or we're assuming we're starting off with a velocity of 0 meters per second. And we're getting up to 78 meters per second. So our change in velocity is the 78 meters per second. So this is equal, in our situation, 78 meters per second is our change in velocity. I'm taking the final velocity, 78 meters per second, and subtract from that the initial velocity, which is 0 meters per second. And you just get this. Divided by the acceleration, divided by 1 meter per second per second, or 1 meter per second squared. So the numbers part are pretty easy. You have 78 divided by 1, which is just 78. And then the units you have meters per second. And then if you divide by meters per second squared, that's the same thing as multiplying by seconds squared per meter. Right? Dividing by something the same thing as multiplying by its reciprocal. And you can do the same thing with units. And then we see the meters cancel out. And then seconds squared divided by seconds, you're just left with seconds. So once again, we get 78 seconds, a little over a minute for this thing to take off.