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Review 1: A plane is located x = 40 miles (horizontally) away from an airport at an altitude of h miles. Radar at the airport detects that the distance s(t) between the plane and airport is changing at the rate of s(t) = −240 mph. If the plane flies toward the airport at the constant altitude h = 4, what is the speed |x(t)| of the airplane?

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.

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