- Average velocity for constant acceleration
- Acceleration of aircraft carrier take-off
- Airbus A380 take-off distance
- Deriving displacement as a function of time, acceleration, and initial velocity
- Plotting projectile displacement, acceleration, and velocity
- Projectile height given time
- Deriving max projectile displacement given time
- Impact velocity from given height
- Viewing g as the value of Earth's gravitational field near the surface
- What are the kinematic formulas?
- Choosing kinematic equations
- Setting up problems with constant acceleration
- Kinematic formulas in one-dimension
Airbus A380 take-off distance
How long of a runway does an A380 need? Created by Sal Khan.
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- 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)
- When I do the practice problems I continually get them wrong. I'm not sure what terms (m, s, m/s, m per s squared) to keep in the equation when solving and how to solve when the value of something is zero. Any help?(2 votes)
- Dimensional Analysis is the act of adding & removing the units of your equation (m, s, m/s, etc).
It is important to take your time so that you understands the dimensional analysis properly. It also helps you visualize what exactly is going on.
Distance is measured in lots of measurements in everyday life but in physics meters (m) is the most common one you'll see (in my experience).
Time of course is measured in seconds (s).
Velocity is a vector quantity measuring how fast we are moving in a direction and has units of (m/s). We read this as meters per second. So for every second how many meters do we travel?
Acceleration is how fast we are speed up in a direction, aka how our velocity is changing. Acceleration is measured in units of m/s^2, which is meters per second squared. Acceleration isn't quite as intuitive as velocity, but here is the way to think of it:
Velocity = meters per second = (m/s)
Acceleration = velocity per second = vel / s = (m/s) / s = m/s^2.
Hope this helps,
- Convenient Colleague(3 votes)
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.