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Sagot :
The acceleration, take off length and possibility of the airplane flying is required.
Acceleration of the plane is [tex]1.76\ \text{m/s}^2[/tex]
Minimum distance required to takeoff is 1277.57 m
It would not be able to fly if the runway was 980 m.
v = Final velocity = [tex]150\times \dfrac{1609.34}{3600}=67.06\ \text{m/s}[/tex]
u = Initial velocity = 0
s = Displacement
t = Time = 38 s
a = Acceleration
From the kinematic equations we get
[tex]v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{67.06-0}{38}\\\Rightarrow a=1.76\ \text{m/s}^2[/tex]
[tex]v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{67.06^2-0}{2\times 1.76}\\\Rightarrow s=1277.57\ \text{m}[/tex]
The minimum distance required to takeoff is 1277.57 m.
So, if the runway was 980 m it would not be able to fly.
Learn more:
https://brainly.com/question/16256332?referrer=searchResults
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