Question

Two ships are $80\text{km}$ apart on North-South vertical at an instant as shown in the figure. The one farther North is streaming South at $40\text{km/hr}$ and the other is streaming East at $30\text{km/hr}$. What is their distance of closest approach?

• A. $48\text{km}$
• B. $32\text{km}$
• C. $60\text{km}$
• D. $52\text{km}$

Answer 1

The correct option is A $48\text{km}$

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
$|\overrightarrow{v_{AB}}|=\sqrt{(40{)}^2+(-30{)}^2}=50\text{km/hr}$
and, angle $\theta$ is given by
$\theta ={\tan }^{-1}\left(\frac{30}{40}\right)={\tan }^{-1}\left(\frac{3}{4}\right)={37}^{\circ }$
Thus, the relative motion can be shown as

So, from the above figure we can say that the distance of closest approach is
$d_{short}=80\sin \theta =80\sin {37}^{\circ }=48\text{km}$