Question

a helicopter flies parallel to the ground at an altitude of 1/2 kilometer and at a speed of 2 kilometers per minute. if the helicopter passes directly over the white house, at what rate is the distance between the helicopter and the white house changing 1 minute after the helicopter flies over the white house?

Answer 1

The Pythagoras' theorem and the uniform motion allows to find the answer for the rate of change of distance is

               v = 32.3 m / s

The distance is the length of the segment that a white house wing with the helicopter, this can be found using the Pythagoras' theorem

               R = $\sqrt{x^2 + y^2}$

Where R is the distance, x is the horizontal distance where the helicopter flies and y is the vertical distance that corresponds to the flight height

y = 500 m

Since the helicopter flies at constant speed, we can use the uniform motion relation

              $v_h$ = $\frac{x}{t}$

              x = $v_h$t

Where $v_h$ is the speed of the helicopter and t is the time

We substitute  

                r = $\sqrt {(V_h \ t)^2 + y^2}$

Kinematics defines velocity as the change of position in the unit of time

            $v = \frac{dR}{dt}$

           $v = \frac{1}{2} \ \frac{2 ( v_h t) v_h}{ \sqrt{(v_h t)^2 + y^2} }$

           $v = \frac{v_h^2 \ t}{ \sqrt{(v_h t)^2 + y^2} }$

This expression is the change in distance as a function of time for a given speed. In the exercise, this speed is requested for a time of one minute, for the helicopter speed of

        v = 2km/min ($\frac{1000m}{1km}$ ) ( $\frac{1 min}{60s}$) = 33.33 m / s

Let's calculate

            v = $\frac{33.33^2 \ 60 }{ \sqrt{ )33.3 \ 60)^2 +500^2} }$

         

            v = 32.3 m / s

In conclusion using the Pythagoras' theorem and the uniform motion we can find the answer for the rate of change in distance is

               v = 32.3 m / s

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