The two aircraft form a right triangle with the right ange at the airport.

Suppose the westbound aircraft distance from the airport is x and the southbound aircraft distance is y. Then we can find the distance between the aircraft using the Pythagorean Theorem.

`s^2 = x^2 + y^2`

Now we need to implicitly differentiate with respect to time to get

` 2s(ds)/(dt) = 2x(dx)/(dt) + 2y(dy)/(dt)`

.

Now `(dx)/(dt) = -600 "km/hr"` and `(dy)/(dt) = -250 "km/hr"` if we consider this on a coordinate plane.

So `(ds)/(dt) = (x(dx)/(dt) + y(dy)/(dt))/s`

The first aircraft (y) is 60km and (x) is 25km.

`(ds)/(dt) = (25(-600)+(60)(-250))/(sqrt(60^2+25^2)) ~~ -462 "km/hr"`

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