the takeoff speed for a large airplane is 180 mi/hr(roughly 80m/s0 the plane is capable of accelerating at a rate of 4m/s^2 assume that 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
vf^2=2ad
80^2=2*4*d solve for d.
if takeoff speed is 1/2, then the distance is (1/2)^2=1/4 as far
To determine the minimum runway length required for both the larger and smaller airplanes to take off, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
- v is the final velocity (takeoff speed)
- u is the initial velocity (0 for both planes since they start at rest)
- a is the acceleration
- s is the distance (runway length)
1. For the larger airplane:
Given:
- Takeoff speed, v = 80 m/s
- Acceleration, a = 4 m/s^2
Using the kinematic equation, we can solve for the runway length, s:
v^2 = u^2 + 2as
(80 m/s)^2 = (0 m/s)^2 + 2 * 4 m/s^2 * s
6400 m^2/s^2 = 8 m/s^2 * s
s = 6400 m^2/s^2 / 8 m/s^2
s = 800 m
Therefore, the minimum runway length required for the larger airplane to safely take off is 800 meters.
2. For the smaller airplane:
Given:
- Takeoff speed (half of the larger airplane) = 40 m/s
- Acceleration, a = 4 m/s^2 (same as the larger airplane)
Again, using the same kinematic equation, we can solve for the runway length, s:
v^2 = u^2 + 2as
(40 m/s)^2 = (0 m/s)^2 + 2 * 4 m/s^2 * s
1600 m^2/s^2 = 8 m/s^2 * s
s = 1600 m^2/s^2 / 8 m/s^2
s = 200 m
Therefore, the minimum runway length required for the smaller airplane to safely take off is 200 meters.