question_answer

Two ships A and B are sailing straight away from a fixed point O along routes such that \[\angle AOB\] is always \[120{}^\circ \]. At a certain instance, OA = 8 km, OB = 6 km and the ship A is sailing at the rate of 20 km/hr while the ship B sailing at the rate of 30 km/hr. Then the distance between A and B is changing at the rate (in km/ hr):   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

A) \[\frac{260}{\sqrt{37}}\]              

B) \[\frac{260}{37}\]

C) \[\frac{80}{\sqrt{37}}\]                 

D) \[\frac{80}{37}\]

Correct Answer: A

Solution :

Let \[OA=xkm,OB=y\,km,AB=R\] \[{{(AB)}^{2}}={{(OA)}^{2}}+{{(OB)}^{2}}-2(OA)(OB)cos{{120}^{o}}\] \[{{R}^{2}}={{x}^{2}}+{{y}^{2}}-2xy\left( -\frac{1}{2} \right)={{x}^{2}}+{{y}^{2}}+xy\]?.(1) R at x = 6 km, and y = 8 km \[R=\sqrt{{{6}^{2}}+{{8}^{2}}+6\times 8}=2\sqrt{37}\] Differentiating equation (1) with respect to t \[2R\frac{dR}{dt}=2\frac{dx}{dt}+2y\frac{dy}{dx}+\left( x\frac{dy}{dt}+y\frac{dx}{dt} \right)\] \[=\frac{1}{2R}[2\times 8\times 20+2\times 6\times 30+(8\times 30+6\times 20)]\] \[\frac{dR}{dt}=\frac{1}{2\times 2\sqrt{37}}[1040]=\frac{260}{\sqrt{37}}\]