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, $OA=8$ 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. $\frac{260}{37}$
• C. $\frac{80}{37}$
• D. $\frac{80}{\sqrt{37}}$

Answer 1

The correct option is A $\frac{260}{\sqrt{37}}$

If we take $X$-axis in the direction of $OA$, we get at the given instance
$\overrightarrow{V_{OA}}=20\hat{i}$
$\overrightarrow{V_{OB}}=30(-\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{i})$
${\overrightarrow{V}}_{AB}={\overrightarrow{V}}_{OB}-{\overrightarrow{V}}_{OA}=-35\hat{i}+15\sqrt{3}\hat{j}$

Let at any instant $\overrightarrow{OA}=a\hat{i}$ and $\overrightarrow{OB}=b_x\hat{i}+b_y\hat{j}$
So, $\overrightarrow{AB}=(b_x-a)\hat{i}+b_y\hat{j},$
$|\overrightarrow{AB}|=\sqrt{(b_x-a{)}^2+b_y^2}$
$\frac{d|\overrightarrow{AB}|}{dt}=\frac{1}{2|\overrightarrow{AB}|}\left(2\right(b_x-a\left)|\right(V_{b_x}-V_a)|+2b_y\cdot V_{b_y})$
$=\frac{1}{2\sqrt{148}}\times \left[2\right(11)\cdot (35)+2\cdot 15\sqrt{3}\cdot 3\sqrt{3}]$
$=\frac{1}{2.2\cdot \sqrt{37}}[22\cdot (35)+270]=\frac{1040}{4.\sqrt{37}}=\frac{260}{\sqrt{37}}$