Answer:
The acceleration is [tex]a =51945 \ km/h^2[/tex]
Explanation:
From the question we are told that
The lift up speed is [tex]v = 147 \ km/h[/tex]
The distance covered for the take off run is [tex]s = 208 m = 0.208 \ km[/tex]
Generally from kinematic equation we have that
[tex]v^2 = u^2 + 2as[/tex]
Here u is the initial speed of the aircraft with value 0 m/ s give that the aircraft started from rest
So
[tex]147^2 = 0^2 + 2* a* 0.208[/tex]
=> [tex]a =51945 \ km/h^2[/tex]
