the required takeoff speed for a large airplane is 180mi/hr(80m/s)the plane is capable of accelerating at a rate of 4m/s^2 assume the plane starts at rest, determine the minimum runway length required for this plane to safely take off, a smaller airplane requires a takeoff speed that is half as much as the large airplane determine the minimum runway length required for this smaller plane to take off
v^2 = 2as, so
s = v^2/2a
now plug in the numbers
To determine the minimum runway length required for a large airplane to safely take off, we need to use the equations of motion. We will assume that the acceleration remains constant during the takeoff.
We're given:
Takeoff speed (v) = 80 m/s
Acceleration (a) = 4 m/s^2
To find the minimum runway length (d), we can use the following equation of motion:
v^2 = u^2 + 2ad
Where:
v = final velocity (takeoff speed)
u = initial velocity (0 m/s since the plane starts at rest)
a = acceleration
d = distance (runway length)
Rearranging the equation, we get:
d = (v^2 - u^2) / (2a)
Substituting the values we have:
d = (80^2 - 0^2) / (2 * 4)
d = 6400 / 8
d = 800 meters
Therefore, the minimum runway length required for the large airplane to safely take off is 800 meters.
To determine the minimum runway length required for the smaller plane, which requires half the takeoff speed of the large airplane, we can follow a similar procedure.
Takeoff speed for the smaller plane = 80 m/s / 2 = 40 m/s
Using the same equation:
d = (40^2 - 0^2) / (2 * 4)
d = 1600 / 8
d = 200 meters
Therefore, the minimum runway length required for the smaller plane to safely take off is 200 meters.