Related Rates: One airplane is approaching an airport from the north at 136km/hr. A second airplane approaches

Question

from the east at 285km/hr. Find the rate at which the distance between the planes changes when the southbound plane is 26km away from the airport and the westbound plane is 15 km from the airport.
*please explain step; any help is greatly appreciated*
possible answers:
A. 2142
B. 1982
C. 396
D. 82
E. none of the above

Answer 1

The answer, strictly speaking, is "E", none of the above, as A through D are simply pure numbers. (The answer has to be a SPEED!)

I dare you to tell THAT to whomever set you this question.

(But you should really try to work it out for yourself. After all, you are actually given the numerical values, one of which, despite not having "km/sec" attached, must surely be what you were intended to pick. Of course, given the initial information you could always use the time-honoured method of U.S students, namely striking out the ludicrous answers. Just look at the two initial speeds given, 285 and 136 km/hr. Do you REALLY think that any possible combination of those could give you as much as 2142 km/hr. ["Of course not," I hope to hear you say!] Ditto 1982 km/hr. Also, 82 km/hr just looks too small.

That only leaves the reasonable looking 396 km/hr to be checked.)

P.S. Sidharth's method is correct but he/she seems to have made a mistake in working out the square root. Clearly sqrt (36^2 + 15^2) can NOT be about 30! In fact, (5, 12, 13) is a famous Pythagorean triple, so the sqrt is 3 x 13 = 39.

So, correcting the above result one gets ~200.3 km/hr. So it would appear that "E" is truly the answer, even without my picky point about the missing units.

Answer 2

let the distance o soth.plane be x
let the distance o west.plane be y
distance btw two any time is=s=(x^2+y^2)^.5
dx/dt=136
dy/dt=285
we have to find
ds/dt=(xdx/dt +ydy/dt)/(x^2+y^2)^.5
x=36,y=15
=(136*26+285*15)/(30.01)=260.22(decreasing)
therefore answer is none