Physics

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The airplane can take off when its airspeed (speed of the air flowing over the wing) is equal to 65 knots. What is the length of runway required for the plane to take off if there is a 22 knots head wind? The runway at the Tallahassee Regional Airport has a length of 8000 ft.

What equation do I need to use for this? The head wind part throws me off...

  • Physics -

    With a 22kt headwind, you have a windspeed over the wings of 22kt even before you start moving. How many more kts of wind speed over the wing are needed for takeoff?

  • Physics -

    43 knots, right?

  • Physics -

    There is missing data. You will need to know the acceleration of the plane.

  • Physics -

    I'm sorry. There was a problem before it where I had to figure out the average acceleration, and that was 1.35 m/s^2. Which fomula can I use to get the length of the runway?

  • Physics -

    The plane needs to accelerate to a ground speed of 43kts (to obtain an airspeed of 65kts).

    How many seconds are needed with an acceleration of 1.5m/s^2 to get to a ground speed of 43kts?

    Now, when you have found the seconds it takes to reach the required speed, you can put that into the familiar formulat:
    distance = (1/2) a * t^2

    There is a direct forumla, but it derives from this.

  • Physics -

    To get how many seconds, would I have to take the 43 knots and divide it by 1.5?

  • Physics -

    Be careful, you have the correct idea, but 43kts is not the same distance units as 1.35m/s^2. You need to get the speed and the acceleration into the same units.

  • Physics -

    43 knots is 22.097 m/s. So could I now do 22.097 divided by 1.35?

  • Physics -

    Yes, that will work.

  • Physics -

    Ok, so I came up with 16.37 s as my answer for the time.

    So I do I simply plug that number into the equation 'distance=1/2 x a x t^2' and that is my runway length?

  • Physics -

    Yes, again remember that you will be getting the answer in meters. Don't forget to convert to feet if the problems requires the solution in feet.

  • Physics -

    THANKS! :)

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