An aircraft has a lift-off speed of 120 km/hr. What minimum constant accelerationdoes this require if the aircraft is to be airborn after a take-off run of 240 m?
change 120km/hr to m/s
V^2=2a*240 find a
120 km/h = 33.3 m/s
V = sqrt(2aX) = 33.3 m/s
Solve for the acceleration, a.
X = 240 m.
To find the minimum constant acceleration required for the aircraft to take off, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (lift-off speed)
u = initial velocity (0, as the aircraft starts from rest)
a = acceleration
s = displacement (take-off run)
In this case, the lift-off speed (v) is given as 120 km/hr. We need to convert it to m/s:
120 km/hr = 120 * 1000 m / (60 * 60) s ≈ 33.33 m/s
The initial velocity (u) is 0 m/s since the aircraft starts from rest.
The take-off run (s) is given as 240 m.
Plugging these values into the kinematic equation, we have:
(33.33 m/s)^2 = (0 m/s)^2 + 2a * 240 m
1111.1 m^2/s^2 = 480a
Now, isolate the acceleration (a) by dividing both sides of the equation by 480:
a = 1111.1 m^2/s^2 / 480
a ≈ 2.31 m/s^2
Therefore, the minimum constant acceleration required for the aircraft to be airborne after a take-off run of 240 m is approximately 2.31 m/s^2.