Asked by Lindsay
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...
What equation do I need to use for this? The head wind part throws me off...
Answers
Answered by
Anonymous
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?
Answered by
Lindsay
43 knots, right?
Answered by
Anonymous
There is missing data. You will need to know the acceleration of the plane.
Answered by
Lindsay
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?
Answered by
Anonymous
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.
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.
Answered by
Lindsay
To get how many seconds, would I have to take the 43 knots and divide it by 1.5?
Answered by
Anonymous
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.
Answered by
Lindsay
43 knots is 22.097 m/s. So could I now do 22.097 divided by 1.35?
Answered by
Anonymous
Yes, that will work.
Answered by
Lindsay
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?
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?
Answered by
Anonymous
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.
Answered by
Lindsay
THANKS! :)
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