Given that,
Initial velocity of the airplane = 66 m/s
Distance = 560 m
Final velocity of the airplane = 0 (stops)
To find,
Acceleration of the airplane.
Solution,
Let us consider that a is the acceleration of the airplane. We can find it using third equation of motion as follows :
[tex]v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{0^2-66^2}{2(560)}\\\\a=-3.9\ m/s^2[/tex]
Therefore, the acceleration of the airplane is [tex]-3.9\ m/s^2[/tex]
