A Cessna aircraft has a lift-off speed of
120 km/h.
What minimum constant acceleration does
this require if the aircraft is to be airborne
after a take-off run of 265 m?
Answer in units of m/s2
b) How long does it take the aircraft to become
airborne?
Answer in units of s
To find the minimum constant acceleration required for the Cessna aircraft to become airborne, we can use the following kinematic equation:
v^2 = u^2 + 2as
Where:
v is the final velocity (lift-off speed) = 120 km/h = (120 * 1000) / 3600 m/s ≈ 33.33 m/s
u is the initial velocity (0 m/s at the start of take-off run) = 0 m/s
a is the constant acceleration we need to find
s is the distance traveled during the take-off run = 265 m
Plugging in the given values, the equation becomes:
(33.33)^2 = 0 + 2a * 265
1111.09 = 530a
To find the value of a, we need to divide both sides of the equation by 530:
a = 1111.09 / 530 ≈ 2.1 m/s^2
Therefore, the minimum constant acceleration required for the aircraft to become airborne is approximately 2.1 m/s^2.
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To calculate the time it takes for the aircraft to become airborne, we can use another kinematic equation:
v = u + at
Where:
v is the final velocity (lift-off speed) = 33.33 m/s
u is the initial velocity (0 m/s at the start of take-off run) = 0 m/s
a is the acceleration we found earlier = 2.1 m/s^2
t is the time we need to find
Plugging in the given values, the equation becomes:
33.33 = 0 + 2.1 * t
Simplifying the equation:
33.33 = 2.1t
To find t, we can divide both sides of the equation by 2.1:
t = 33.33 / 2.1 ≈ 15.9 s
Therefore, it takes approximately 15.9 seconds for the aircraft to become airborne.