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## Class 9 Physics (India)

### Course: Class 9 Physics (India) > Unit 1

Lesson 4: Acceleration# Airbus A380 take-off time

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)
- The heavier the craft the more force is needed to reach a specific acceleration.(210 votes)

- In which lessons can I learn this way of doing maths, with equations, cancelling out things and so on?(40 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?(22 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

where :

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,

Thank you.

(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)

## Video transcript

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.