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Acceleration of aircraft carrier take-off

Using what we know about take-off velocity and runway length to determine acceleration. Created by Sal Khan.

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  • blobby green style avatar for user Cory LaChance
    Wouldn't the initial velocity change with how fast the ship is traveling?
    (108 votes)
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    • blobby green style avatar for user Arkani Vryx
      It depends on your point of reference. In this problem, the velocity is being measured relative to the starting point at the catapult (on the ship's deck). The catapult is point 0 (the initial reference position). During prelaunch, the F-18 is stationary at the catapult... not moving relative to its initial position (i.e. not rolling on the runway), hence its initial velocity is considered 0.

      This example is somewhat simplified, but yes, the aircraft is moving with the ship, so it is moving relative to a fixed point on the globe. There are other factors for an aircraft taking off as well... most importantly, wind speed. Wind speed relative to the aircraft affects lift. Normally (operations allowing) a carrier will turn into the wind for the benefit of lift. For example, the ship could be going 25 knots against a 15 knot wind, effectively generating 40 knots of headwind. Add in the speed of the aircraft from catapult assist (and jet engines) through the wind, and you have some good lift, really fast. Once the aircraft no longer has catapult assist, it relies on its own power from the jet engines pushing against air (like in the prior video talking about rockets). Back to the relative position concept, an aircraft can move at a fixed airspeed (speed relative to air/wind), but relative to the ground, it will travel faster with the wind and slower against a headwind. With the heavier F-14s especially, you'll often notice a bit of a dip in the initial ascent off the deck... then an afterburner kick to get airspeed up.
      (186 votes)
  • leaf green style avatar for user islamic.prince07
    instead of all that can we do 80m/(36m/s) would that be the same thing because i divided the average velocity witch is 36m/s by the distance would that be more common sense?
    (23 votes)
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    • leaf green style avatar for user Kieran
      Yes. I did the same thing. I found the time first by using the fact that the time= (displacement) / (avg velocity)= 2.2s. Then, I used the time to find acceleration from the fact that change in velocity= acceleration x time. Therefore, acceleration= delta v/t or (72m/s)/(2.2s)
      (34 votes)
  • blobby green style avatar for user ebidois
    Im just trying to get a better understanding of acceleration. Sal said that while going 33m/s squared- after 1 second hes going 33m/s, after 2 seconds hes going 66m/s, how fas is he going after the third second? is it 99m/s? or is it 132m/s??
    (14 votes)
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  • blobby green style avatar for user Gio
    I got more or less his answers, but without using squares:

    Displacement = Average Velocity * time; I found time to be 2.22 seconds
    (Displacement is the runway's length, 80 meters and the Average Velocity is about 36 m/s)

    Acceleration = Velocity / Time so I got an acceleration of 32.5 meters/second^2
    (Velocity is 72 m/s and Change in time is 2.22 seconds, as found above)

    I did get the same answers, but I didn't use the same methods... should I be worried? Which method is preferred?
    (14 votes)
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  • leaf green style avatar for user Assemble Academy
    I found a simpler way of finding the acceleration.
    Displacement = Runway length = 80m
    Initial velocity = 0
    Final velocity = Take off speed = 260km/hr = 72m/s
    Time = unknown

    The solution-
    1st part: Finding the time.
    Vavg = s/t
    t = s / Vavg
    t = s / ((Vf+Vi)/2)
    t = s / ((72m/s+0m/s)/2)
    t = s / (72/2)
    t = 80/36
    t= 2.2 sec
    2nd part: Finding the acceleration
    a = Δv/t
    a = ((72-0)m/s)/2.2sec
    a = 72m/s / 2.2sec
    a = 32.7m/(s*s)
    So, isn't this method right?
    (12 votes)
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  • orange juice squid orange style avatar for user Michael Highers
    Thinking of aerofoils, would a tailwind during flight increase the minimum speed needed to maintain lift? Let me think out loud: the wind would also reduce the needed energy to maintain the momentum vs air resistance and if the energy exerted is constant, it would accelerate the plane. I guess the answer depends on if the tailwind and force exerted cancel out?
    (3 votes)
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    • leaf green style avatar for user Mark Zwald
      The lift from a wing is created by the relative velocity of the wing to the air. If the air is moving at the same velocity as the wing, the relative velocity will be zero and there will be no lift. So a tail wind would increase the wing velocity required for take-off and a head wind would reduce it.
      (6 votes)
  • piceratops tree style avatar for user alex smith
    would the movement of the ship on the sea make the plane have a different take off each time
    (3 votes)
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  • hopper cool style avatar for user olipet
    Does Vavg = x/t, equal to Vavg = (vf+vi)/2?
    (3 votes)
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    • stelly green style avatar for user The #1 Pokemon Proponent
      Only under constant acceleration conditions. This is because v_avg = total disp/total time which can be written as (if initial velocity is u and accn a, time interval t) [ut+(1/2)at^2] / [t] which is equal to u+1/2at and if final velocity is v, then v = u+at. Therefore, at = v-u, Substituting, v_avg=u+(v-u)/2 = (v+u)/2.

      Hence, both the average velocities are one and the same.
      (3 votes)
  • boggle yellow style avatar for user Anuja
    In general, how do you know which formula to use to derive another one by looking at a problem? (Like how would you know whether you should try to use the formula of displacement to figure out the acceleration or whether to use the formula of acceleration itself. Trying it with all of them takes a lot of time.)
    Sorry if this question is stupid.
    (2 votes)
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    • male robot hal style avatar for user Charles LaCour
      I wouldn't say that it is a stupid question. The best way I have found to figure out what equations to use is to look at the types of information you are given and what answer you are supposed to give. Make a list of things like you are given starting velocity, distance and time and are asked and asked for acceleration. Then look at the equations and find the one (one ones) that have the types of values you are given are being asked for.
      (4 votes)
  • blobby green style avatar for user Lina Tasama
    At what is the difference between initial velocity minus final velocity over two and initial velocity minus final velocity over acceleration
    (3 votes)
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Video transcript

So I'm curious about how much acceleration does a pilot, or the pilot and the plane, experience when they need to take off from an aircraft carrier? So I looked up a few statistics on the Internet, this right here is a picture of an F/A-18 Hornet right over here. It has a take-off speed of 260 kilometers per hour. If we want that to be a velocity, 260 km/hour in this direction, if it's taking off from this Nimitz class carrier right over here. And I also looked it up, and I found the runway length, or I should say the catapult length, because these planes don't take off just with their own power. They have their own thrusters going, but they also are catapulted off, so they can be really accelerated quickly off of the flight deck of this carrier. And the runway length of a Nimitz class carrier is about 80 meters. So this is where they take off from. This right over here is where they take off from. And then they come in and they land over here. But I'm curious about the take-off. So to do this, let's figure out, well let's just figure out the acceleration, and from that we can also figure out how long it takes them to be catapulted off the flight deck. So, let me get the numbers in one place, so the take-off velocity, I could say, is 260 km/hour, so let me write this down. So that has to be your final velocity when you're getting off, of the plane, if you want to be flying. So your initial velocity is going to be 0, and once again I'm going to use the convention that the direction of the vector is implicit. Positive means going in the direction of take-off, negative would mean going the other way. My initial velocity is 0, I'll denote it as a vector right here. My final velocity over here has to be 260 km/hour. And I want to convert everything to meters and seconds, just so that I can get my, at least for meters, so that I can use my runway length in meters. So let's just do it in meters per second, I have a feeling it'll be a little bit easier to understand when we talk about acceleration in those units as well. So if we want to convert this into seconds, we have, we'll put hours in the numerator, 1 hour, so it cancels out with this hour, is equal to 3600 seconds. I'll just write 3600 s. And then if we want to convert it to meters, we have 1000 meters is equal to 1 km, and this 1 km will cancel out with those kms right over there. And whenever you're doing any type of this dimensional analysis, you really should see whether it makes sense. If I'm going 260 km in an hour, I should go much fewer km in a second because a second is so much shorter amount of time, and that's why we're dividing by 3600. If I can go a certain number of km in an hour a second, I should be able to go a lot, many many more meters in that same amount of time, and that's why we're multiplying by 1000. When you multiply these out, the hours cancel out, you have km canceling out, and you have 260 times 1000 divided by 3600 meters per second. So let me get my trusty TI-85 out, and actually calculate that. So I have 260 times 1000 divided by 3600 gets me, I'll just round it to 72, because that's about how many significant digits I can assume here. 72 meters per second. So all I did here is I converted the take-off velocity, so this is 72 m/s, this has to be the final velocity after accelerating. So let's think about what that acceleration could be, given that we know the length of the runway, and we're going to assume constant acceleration here, just to simplify things a little bit. But what does that constant acceleration have to be? So let's think a little bit about it. The total displacement, I'll do that in purple, the total displacement is going to be equal to our average velocity while we're accelerating, times the difference in time, or the amount of time it takes us to accelerate. Now, what is the average velocity here? It's going to be our final velocity, plus our initial velocity, over 2. It's just the average of the initial and final. And we can only do that because we are dealing with a constant acceleration. And what is our change in time over here? What is our change in time? Well our change in time is how long does it take us to get to that velocity? Or another way to think about it is: it is our change in velocity divided by our acceleration. If we're trying to get to 10 m/s, or we're trying to get 10 m/s faster, and we're accelerating at 2 m/s squared, it'll take us 5 seconds. Or if you want to see that explicitly written in a formula, we know that acceleration is equal to change in velocity over change in time. You multiply both sides by change in time, and you divide both sides by acceleration, so let's do that, multiply both sides by change in time and divide by acceleration. Multiply by change in time and divide by acceleration. And you get, that cancels out, and then you have that cancels out, and you have change in time is equal to change in velocity divided by acceleration. Change in velocity divided by acceleration. So what's the change in velocity? Change in velocity, so this is going to be change in velocity divided by acceleration. Change in velocity is the same thing as your final velocity minus your initial velocity, all of that divided by acceleration. So this delta t part we can re-write as our final velocity minus our initial velocity, over acceleration. And just doing this simple little derivation here actually gives us a pretty cool result! If we just work through this math, and I'll try to write a little bigger, I see my writing is getting smaller, our displacement can be expressed as the product of these two things. And what's cool about this, well let me just write it this way: so this is our final velocity plus our initial velocity, times our final velocity minus our initial velocity, all of that over 2 times our acceleration. Our assumed constant acceleration. And you probably remember from algebra class this takes the form: a plus b times a minus b. And so this equal to -- and you can multiply it out and you can review in our algebra playlist how to multiply out two binomials like this, but this numerator right over here, I'll write it in blue, is going to be equal to our final velocity squared minus our initial velocity squared. This is a difference of squares, you can factor it out into the sum of the two terms times the difference of the two terms, so that when you multiply these two out you just get that over there, over 2 times the acceleration. Now what's really cool here is we were able to derive a formula that just deals with the displacement, our final velocity, our initial velocity, and the acceleration. And we know all of those things except for the acceleration. We know that our displacement is 80 meters. We know that this is 80 meters. We know that our final velocity, just before we square it, we know that our final velocity is 72 meters per second. And we know that our initial velocity is 0 meters per second. And so we can use all of this information to solve for our acceleration. And you might see this formula, displacement, sometimes called distance, if you're just using the scalar version, and really we are thinking only in the scalar, we're thinking about the magnitudes of all of these things for the sake of this video. We're only dealing in one dimension. But sometimes you'll see it written like this, sometimes you'll multiply both sides times the 2 a, and you'll get something like this, where you have 2 times, really the magnitude of the acceleration, times the magnitude of the displacement, which is the same thing as the distance, is equal to the final velocity, the magnitude of the final velocity, squared, minus the initial velocity squared. Or sometimes, in some books, it'll be written as 2 a d is equal to v f squared minus v i squared. And it seems like a super mysterious thing, but it's not that mysterious. We just very simply derived it from displacement, or if you want to say distance, if you're just thinking about the scalar quantity, is equal to average velocity times the change in time. So, so far we've just derived ourselves a kind of a neat formula that is often not derived in physics class, but let's use it to actually figure out the acceleration that a pilot experiences when they're taking off of a Nimitz class carrier. So we have 2 times the acceleration times the distance, that's 80 meters, times 80 meters, is going to be equal to our final velocity squared. What's our final velocity? 72 meters per second. So 72 meters per second, squared, minus our initial velocity. So our initial velocity in this situation is just 0. So it's just going to be minus 0 squared, which is just going to be 0, so we don't even have to write it down. And so to solve for acceleration, to solve for acceleration, you just divide, so this is the same thing as 160 meters, well, let's just divide both sides by 2 times 80, so we get acceleration is equal to 72 m/s squared over 2 times 80 meters. And what we're gonna get is, I'll just write this in one color, it's going to be 72 divided by 160, times, we have in the numerator, meters squared over seconds squared, we're squaring the units, and then we're going to be dividing by meters. So times, I'll do this in blue, times one over meters. Right? Because we have a meters in the denominator. And so what we're going to get is this meters squared divided by meters, that's going to cancel out, we're going to get meters per second squared. Which is cool because that's what acceleration should be in. And so let's just get the calculator out, to calculate this exact acceleration. So we have to take, oh sorry, this is 72 squared, let me write that down. So this is, this is going to be 72 squared, don't want to forget about this part right over here. 72 squared divided by 160. So we have, and we can just use the original number right over here that we calculated, so let's just square that, and then divide that by 160, divided by 160. And if we go to 2 significant digits, we get 33, we get our acceleration is, our acceleration is equal to 33 meters per second squared. And just to give you an idea of how much acceleration that is, is if you are in free fall over Earth, the force of gravity will be accelerating you, so g is going to be equal to 9.8 meters per second squared. So this is accelerating you 3 times more than what Earth is making you accelerate if you were to jump off of a cliff or something. So another way to think about this is that the force, and we haven't done a lot on force yet, we'll talk about this in more depth, is that this pilot would be experiencing more than 3 times the force of gravity, more than 3 g's. 3 g's would be about 30 meters per second squared, this is more than that. So an analogy for how the pilot would feel is when he's, you know, if this is the chair right here, his pilot's chair, that he's in, so this is the chair, and he's sitting on the chair, let me do my best to draw him sitting on the chair, so this is him sitting on the chair, flying the plane, and this is the pilot, the force he would feel, or while this thing is accelerating him forward at 33 meters per second squared, it would feel very much to him like if he was lying down on the surface of the planet, but he was 3 times heavier, or more than 3 times heavier. Or if he was lying down, or if you were lying down, like this, let's say this is you, this is your feet, and this is your face, this is your hands, let me draw your hands right here, and if you had essentially two more people stacked above you, roughly, I'm just giving you the general sense of it, that's how it would feel, a little bit more than two people, that squeezing sensation. So his entire body is going to feel 3 times heavier than it would if he was just laying down on the beach or something like that. So it's very very very interesting, I guess, idea, at least to me. Now the other question that we can ask ourselves is how long will it take to get catapulted off of this carrier? And if he's accelerating at 33 meters per second squared, how long would it take him to get from 0 to 72 meters per second? So after 1 second, he'll be going 33 meters per second, after 2 seconds, he'll be going 66 meters per second, so it's going to take, and so it's a little bit more than 2 seconds. So it's going to take him a little bit more than 2 seconds. And we can calculate it exactly if you take 72 meters per second, and you divide it by 33, it'll take him 2.18 seconds, roughly, to be catapulted off of that carrier.