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## Physics library

### Unit 1: Lesson 3

Acceleration- Acceleration
- 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

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# Airbus A380 take-off distance

How long of a runway does an A380 need? Created by Sal Khan.

## Want to join the conversation?

- why do we have to get the average velocity?(32 votes)
- it seems like the easiest way to calculate final position. You could also use X = Xi + Vi*t + 1/2*a*t^2 but thats alot longer :](39 votes)

- 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(13 votes)
- 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.(31 votes)

- why is average velocity the half of the final velocity?(9 votes)
- 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(21 votes)

- 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?(6 votes)- 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)?(4 votes)

- If displacement= velocity*time,i.e, 78*78. why is the velocity halved?(4 votes)
- Displacement = AVERAGE velocity * time.

FINAL velocity is halved in order to get average velocity (if initial is 0)(6 votes)

- if acceleration is not constant then what should we do?(3 votes)
- This would require calculus to find instantaneous acceleration. Do not worry about that for now.(5 votes)

- Where did you get the two(3 votes)
- When is the initial is not 0? And how is Initial velocity not 0?(2 votes)

- Couldn't you just times ( d= t x v ) and divide by 2 to get the average?

For example: (78 seconds * 78 m/s) = (6084 / 2) = 3042(2 votes) - Does the average velocity always have to be in between the initial and final velocities on a graph? Also, doesn't Sal mean that displacement is the area under the velocity timeline?(2 votes)
- No, imagine a car that speeds up and then slows down back to the original speed.(3 votes)

- I'm finding a slightly different result. Does the following add up?

Instead of assuming that vf (takeoff velocity) is 78 m/s, and that ∆t (takeoff time) is 78s, I went back to the original parameters (technically, that 78 was calculated from 700/9, which comes to 77.77... which Sal rounded to 78). So vf = 280 km/h = 700/9 m/s

- Starting with the premise that Vavg = ∆s (displacement) / ∆t (time) we are looking for ∆s as the total displacement, i.e. the minimum runway length.

- We also know that a (acceleration) = ∆v (change in velocity) / ∆t ; therefore: ∆t = (vf-v0)/a

- Based on the above formulas:

- ∆s = Vavg*∆t

- ∆s = (vf - v0)/2 * (vf-v0)/a

- ∆s = (vf)^2 / 2a -- since v0 = 0 m/s

- ∆s = (700/9)^2 / 2

- ∆s = 3,024 meters

Thoughts? Does this make sense, with Sal using a rounded up quantity as the explanation for the difference?(3 votes)

## Video transcript

In the last video,
we figured out that given a takeoff velocity
of 280 kilometers per hour-- and if we have a positive
value for any of these vectors, we assume it's in the forward
direction for the runway-- given this takeoff velocity,
and a constant acceleration of 1 meter per second per second,
or 1 meter per second squared, we figured out that it
would take an Airbus A380 about 78
seconds to take off. What I want to figure
out in this video is, given all of these
numbers, how long of a runaway does it need, which is a very
important question if you want to build a runway that
can at least allow Airbus A380s to take off. And you probably want it to
be a little bit longer than that just in case it takes a
little bit longer than expected to take off. But what is the minimum
length of the runway given these numbers? So we want to figure
out the displacement, or how far does
this plane travel as it is accelerating at
1 meter per second squared to 280 kilometers per
hour, or to 78-- or where did I write it
over here-- to 78. I converted it right over here. As it accelerates to
78 meters per second, how much land does
this thing cover? So let's call this,
the displacement is going to be equal
to-- So displacement is equal to-- You could view
it as velocity times time. But the velocity
here is changing. If we just had a constant
velocity for this entire time, we could just multiply
that times however long it's traveling, and it
would give us the displacement. But here our
velocity is changing. But lucky for us,
we learned-- and I encourage you to watch the video
on why distance, or actually the video on average velocity
for constant acceleration-- but if you have
constant acceleration, and that is what we are
assuming in this example-- so if you assume that your
acceleration is constant, then you can come
up with something called an average velocity. And the average velocity, if
your acceleration is constant, if and only if your
acceleration is constant, then your average velocity
will be the average of your final velocity
and your initial velocity. And so in this situation,
what is our average velocity? Well, our average
velocity-- let's do it in meters per
second-- is going to be our final velocity,
which is-- let me calculate it down here. So our average velocity
in this example is going to be our
final velocity, which is 78 meters per second,
plus our initial velocity. Well, what's our
initial velocity? We're assuming we're
starting at a standstill. Plus 0, all of that over 2. So our average velocity in this
situation, 78 divided by 2, is 39 meters per second. And the value of an average
velocity in this situation-- actually, average velocity
in any situation-- but in this situation, we
can calculate it this way. But the value of
an average velocity is we can figure
out our displacement by multiplying our average
velocity times the time that goes by, times the
change in time. So we know the change
in time is 78 seconds. We know our average
velocity here is 39 meters per second,
just the average of 0 and 78, 39 meters per second. Another way to think
about it, if you want think about the
distance traveled, this plane is
constantly accelerating. So let me draw a
little graph here. This plane's velocity time graph
would look something like this. So if this is time and this
is velocity right over here, this plane has a
constant acceleration starting with 0 velocity. It has a constant acceleration. This slope right here is
constant acceleration. It should actually
be a slope of 1, given the numbers
in this example. And the distance traveled
is the distance that is the area under this
curve up to 78 seconds, because that's how long it
takes for it to take off. So the distance traveled is
this area right over here, which we cover in another video, or
we give you the intuition of why that works and why distance is
area under a velocity timeline. But what an average velocity
is, is some velocity, and in this case, it's exactly
right in between our final and our initial
velocities, that if you take that average velocity
for the same amount of time, you would get the exact
same area under the curve, or you would get the
exact same distance. So our average
velocity is 39 meters per second times 78 seconds. And let's just get our
calculator out for this. We have 39 times
78 gives us 3,042. So this gives us 3,042. And then meters per second
times second just leaves us with meters. So you need a runway
of over 3,000 meters for one of these
suckers to take off, or over 3 kilometers, which is
like about 1.8 or 1.9 miles, just for this guy
to take off, which I think is pretty fascinating.