Answer :
Explanation:
Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,
[tex]s(t)^2={h^2+x^2}[/tex]...........(1)
Given, h = 4, x = 40 and s(t) = -20 mph
Differentiate equation (1) wrt t
[tex]2s(t)s'(t)=2x(t)x'(t)[/tex]
[tex]x'(t)=\dfrac{s(t)s'(t)}{x(t)}[/tex]
When x = 40, [tex]s(t)=\sqrt{40^2+4^2}=40.19\ m[/tex]
[tex]x'(t)=\dfrac{-240s(t)}{x(t)}[/tex]
[tex]x'(t)=\dfrac{-240\times 40.19}{40}[/tex]
[tex]x'(t)=-241.14\ m/s[/tex]
So, the speed of the airplane is 241.14 m/s. Hence, this is the required solution.