v^2 = 2as, so
s = v^2/2a
now plug in the numbers
s = v^2/2a
now plug in the numbers
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.