Question

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, $OA=8km,OB=6km$ and the ship $A$ is sailing at the rate of $20km/hr$ while the ship $B$ sailing at the rate of $30km/hr$. Then the distance between $A$ and $B$ is changing at the rate (in km/hr) :

• A. $\frac{260}{\sqrt{37}}$
• B. $\frac{280}{\sqrt{37}}$
• C. $\frac{80}{\sqrt{37}}$
• D. $\frac{180}{\sqrt{37}}$

Answer 1

Answer: (A)

Let at any time t
$OA=xOB=y$
$\frac{dx}{dt}=20\frac{dx}{dt}=30$
$cos\left(120^{\circ }\right)=\frac{x^2+y^2-A{B}^2}{2xy}$
$A{B}^2=x^2+y^2+xy\dots \left(1\right)$
d. w. r. to . t
$2\left(AB\right)\frac{d}{dt}\left(AB\right)=2x\frac{dx}{dt}+2y\frac{dy}{dt}+2\frac{dy}{dt}+y\frac{dx}{dt}\dots \left(1\right)$
when $x=8y=6$ then $AB=\sqrt{148}$ from $\left(1\right)$
So $\frac{d}{dt}\left(AB\right)=\frac{\left(2x\frac{dx}{dt}+2y\frac{dy}{dt}+\frac{xdy}{dt}+y\frac{dx}{dt}\right)}{2AB}$
use $x=8y=6AB=\sqrt{148}$
$\frac{d}{dt}\left(AB\right)=260/\sqrt{37}$