Any aircraft will takeoff at different speeds, consider the a380 with different amounts of passengers, cargo etc. unless you are trying to get an average takeoff time this is too simplistic. Usually an aircraft's speed is measured in knots, roughly between 140 knots and 160 knots for an a380 depending on weight. There are so many parameters to this that it is unreasonable to try to calculate the time, wind speed and direction play a part (aircrafts tend to takeoff into the wind), the effect temperature might have on the engine would also need to be considered. Pilots may also derate the engines on takeoff to prevent noise, engine damage and maintenance costs. I see what you're getting at here, but takeoff time is not a given, and with the everything that can contribute to takeoff time this is too simple. The truth is nobody knows, the computers on board can't even give you a time, the only thing a pilot knows is the speed at which they can takeoff.
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 take-off velocity,
and the specs I got were 280 km/h.
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.
So when we're talking about acceleration,
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
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.
Acceleration of 1.0 m/s per second
So after every second,
it can go one m/s faster
than it was going
at the beginning of that second.
Or, another way to write this is
1.0 m/s per second,
which can also be written as:
I find this a little bit more intuitive,
a little bit neater to write.
So let's figure this out.
So the first thing
that 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 take-off velocity
in terms of kilometers and hours.
So let's just convert
this take-off velocity into m/s,
and then it might simplify
answering this question.
So if we have 280 km/h,
how do we convert that to m/s?
So let's convert it to km/s first.
So we want to get rid of this 'hour'.
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 in 1 hour there are 3600 seconds.
60 seconds in a minute,
60 minutes in an hour
And so 1 of the larger unit
is equal to 3600 of the smaller unit.
And so we can multiply by that,
And if we do that,
The 'hours' will cancel out.
And we'll get 280 divided by 3600
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 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 when you multiply this out,
the kilometers are going to cancel out,
and you're going to be left with
280 times 1,000 all over 3600,
And the units we have left are:
meters per second.
So let's get my trusty TI-85 out
and actually calculate this.
So we have 280 * 1,000,
which is obviously 280,000,
but let me just divide that by 3600.
And it gives me 77.7 repeating.
And it looks like I had 2 significant digits
in each of these original things,
I had 1.0 over here,
not 100% clear how many
significant digits I have over here.
Was the spec rounded
to the nearest 10 kilometers,
or was it exactly 280 km/h?
Just to be safe,
I'll assume that it's rounded
to the nearest 10 kilometers,
so we only have 2 significant digits here.
So we should only have 2 significant digits
in our answer,
so we're gonna round this to 78 m/s.
So this is going to be 78 m/s,
which is pretty fast!
For this thing to take off,
every second that goes by,
it has to travel 78 meters,
roughly 3/4 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 m/s per second,
which tells us:
after every second,
it's going 1 m/s faster.
So, if you start at a velocity of 0,
and then after 1 second,
it will be going 1 m/s.
After 2 seconds,
it will be going 2 m/s.
After 3 seconds,
it will be going 3 m/s.
So how long will it get to 78 m/s?
Well, it will take 78 seconds.
It'll take 78 seconds, or roughly
a minute and 18 seconds.
And just to verify this
with our definition of 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 Δt * acceleration
is equal to
change in velocity.
And to solve for change in time,
divide both sides by the acceleration,
you get 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 the velocity of 0 m/s,
and we're getting up to 78 m/s,
so our change in velocity is
the 78 m/s.
So this is equal,
in our situation.
78 m/s is our change in velocity.
I'm taking the final velocity,
and subtracting from that
the initial velocity,
which is 0 m/s,
and you just get this
divided by the acceleration,
divided by 1 m/s per second,
or 1 m/s^2.
So the numbers part is 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 m/s^2,
that's the same thing as multiplying by
seconds squared per meter.
Dividing by something is 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 sec^2 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.
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