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 21 knots head wind? The runway at the Tallahassee Regional Airport has a length of 8000 ft.
change (65-21)knots to m/s. You can do that by putting this in the google search window:
(65-21)knots in m/s
Then
speedabove^2=2ad where d is the length of the runway (in meters), and I assume you messed up the problem, you are looking for acceleration a.
and, once you can find a, you can figure the thrust of the engines:
Thrust= mass*a
so how would i go about finding acceleration?
a= .5*speed^2/distance
make certain your units are in meters, and seconds.
thank u so much!! ^w^
To determine the length of runway required for the plane to take off, we need to consider the effect of the headwind on the airplane's groundspeed (actual speed with respect to the ground).
The groundspeed is calculated by subtracting the headwind (21 knots) from the airspeed (65 knots).
So, the groundspeed would be 65 - 21 = 44 knots.
To convert the groundspeed from knots to feet per minute (ft/min), we can use the conversion factor of 1 knot = 101.3 ft/min.
Therefore, the groundspeed in ft/min would be 44 knots * 101.3 ft/min = 4,465.2 ft/min.
Next, we need to calculate the time it takes for the airplane to travel the length of the runway, which is 8,000 ft.
The time (in minutes) can be obtained by dividing the length of the runway by the groundspeed: 8,000 ft / 4,465.2 ft/min = 1.79 minutes.
Finally, there are 60 seconds in a minute, so to convert 1.79 minutes to seconds, we multiply by 60: 1.79 minutes * 60 seconds/minute = 107.4 seconds.
Therefore, the length of runway required for the plane to take off is approximately 107.4 seconds.