Acceleration
Airbus A380 Take-off Distance How long of a runway does as a380 need
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- In the last video we figured out that
- given a takeoff velocity of 280 km/h
- and if we have a positive value for any of these vectors
- we assume that it's in the forward direction for the runway
- Given this takeoff velocity, and a constant acceleration of 1 m/s^2
- 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 that given all these numbers
- how long of a runway does it need?
- Which is a very important question
- if you want to build a runway that can at least allow Airbus A380 to take off
- you probably want it to be a little 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's accelerating at 1 m/s^2
- to 280 km/h...or to 78 m/s...I converted it right over here
- as it accelerates to 78 m/s
- how much land does this thing cover?
- So we can say, so let's call this the displacement
- is going to be equal to...
- So displacement is equal to...you can 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...
- (I would encourage you to watch the video on average velocity for constant acceleration)
- but if you have constant acceleration
- and that is what we are assuming in this example
- If you assume that your acceleration is constant
- then you could 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 m/s
- is going to be our final velocity, which is
- (let me calculate it down here)
- so our average velocity in this example
- Velocity_average in this example...
- is going to be our final example...78 m/s
- plus our initial velocity,. But 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 m/s
- and the value of an average velocity
- in this situation, in any situation
- the value of an average velocity
- is that we could 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 m/s...
- (just the average of 0 and 78..39 m/s)
- Another way to think about it
- if you want to think about the distance traveled
- this plane is constantly accelerating
- so let me draw a little graph here
- this plane's velocity-time graph will look something like this
- so 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 its constant acceleration
- it should actually be a slope of 1, given the numbers in this example
- And the distance traveled
- 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 will cover in another video)
- Or let me give you the intuition of why that works...
- why distance is area under a velocity-time line
- What an average velocity is, is some velocity
- and in this case it's exactly right between our two...
- our final and initial velocities
- But 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 m/s times 78 seconds
- and let's just get our calculator out for this
- we have 39*78...gives us 3042
- so this gives us 3042...
- and then m/s times second just leaves us with meters
- so you need a runway of over 3000 meters
- for one of these suckers to take off
- or over 3 kilometers, which is about 1.8 or 1.9 miles
- just for this guy to take off
Be specific, and indicate a time in the video:
At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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