Question

Two ships $A$ and $B$ are sailing straight away from a fixed point $O$ along routes such that $?\,AOB$ 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) :

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

Answer 1

The Correct Option is A

Let at any time t $OA = x \quad OB = y$ $\frac{dx}{dt} = 20\quad\frac{dx}{dt} = 30$ $cos\left(120^{\circ}\right) = \frac{x^{2}+y^{2}-AB^{2}}{2xy}$ $AB^{2} = x^{2} + y^{2} + xy\quad\quad\ldots\left(1\right)$ D.w.R. To . t $2\left(AB\right) \frac{d}{dt} \left(AB\right) = 2x \frac{dx}{dt} +2y \frac{dy}{dt}+2 \frac{dy}{dt}+y \frac{dx}{dt}\quad \quad \ldots \left(1\right)$ when $x = 8 \,y = 6$ then $AB = \sqrt{148}$ from $\left(1\right)$ So $\frac{d}{dt} \left(AB\right) = \frac{\left(2x \frac{dx}{dt} +2y \frac{dy}{dt}+ \frac{xdy}{dt}+y \frac{dx}{dt}\right)}{2\,AB}$ use $x = 8 \quad y = 6 \,AB = \sqrt{148}$ $\frac{d}{dt} \left(AB\right) =260 / \sqrt{37}$