Question

Two ships are $60\text{km}$ apart on North-South vertical at an instant as shown in the figure. The one farther North is streaming South at $30\sqrt{3}\text{km/hr}$ and the other is streaming East at $30\text{km/hr}$. What is the time taken by the ships to reach the closest approach?

• A. $60\text{min}$
• B. $52\text{min}$
• C. $30\text{min}$
• D. $45\text{min}$

Answer 1

The correct option is B $52\text{min}$

As per the question taking everything in $B$ frame we have
$\overrightarrow{v_{AB}}=\overrightarrow{v_A}-\overrightarrow{v_B}$
So, we have from above equation
$|\overrightarrow{v_{AB}}|=\sqrt{(30\sqrt{3}{)}^2+(-30{)}^2}=60\text{km/hr}$
and, angle $\theta$ is given by
$\theta ={\tan }^{-1}\left(\frac{30}{30\sqrt{3}}\right)={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)={30}^{\circ }$
Thus, the relative motion can be shown as
So, from the above figure we can say that the relative distance moved by ship $A$ to raech the closest approach is $AC$, which is given by
$AC=60\cos \theta =60\cos {30}^{\circ }\approx 52\text{km}$
Hence, time taken by the ships is given by
$t=\frac{52}{|\overrightarrow{v_{AB}}|}=\frac{52}{60}=52\text{min}$