(of course, Silverpig's method is correct)
Here's the explanation... and link to picture to follow if I AT will let me post it before JRunning me outta town..
solution.jpg
magnitude of horizontal velocity = wind velocity. 150 km/hr is constant. Speed toward airport is, of course, decreasing, but doesn't matter for the problem.
Here's the equations:
tan(theta) = horiz/150
take the derivative of both sides with respect to t...
sec^2(theta)*(d_theta/dt) = 1/150 (d_horiz/dt)
we know that d_horiz/dt = 20 km/h/h
d_theta/dt (what you're looking for) = 20/150 cos^2(theta)
just need a substitution for cos^2(theta)
cos^2(theta) = (speed toward airport)^2 / 150^2
speed toward airport (pythag thm) = sqrt(150^2 - horiz^2)
so, cos^2(theta) = [150^2 - horiz^2] / 150^2
t=0 is the starting time,
horiz(t) = 20t (t in hours, I'd prefer minutes, but don't want 1/3 or .333 in here)
thus, cos^2(theta) = [150^2 - 400t^2] / 150^2
And, finally, d_theta/dt = 20/150 * ^^^^
which simplified =
20[150^2 - 400 t^2] / 150^3
I didn't write this on paper, thus I'm quite prone to errors while typing (especially since I can't get a reply box to open today and have to do all this in a tiny little quick reply box)
Since, without paper and pen, I've lost the "flow" that I'm more used to, I make no guarantees of this solution. But, I will guarantee if it was on paper, I wouldn't have made any mistakes that I may have made above.
Nonetheless, I propose this as a somewhat more challenging follow-up question:
What is the maximum distance from the airport that the plane could travel in this manner and still reach the airport (excluding landing, of course)
