Catch the Thief!

Sorry I'am replying late , Since I was little Busy in my JEE prep. , and Gautam , thanks for adding Diagram ,It is easy, It is actually Similar Problem Like as dog-cat chase , but in that problem finally dog will catch the cat means x=0 , but here It is not possible Since Velocities are same , But okay For Solution to this problem , Let at time t=t : distance b/w ship-A and Ship-B as ' $r$ ' . Now Since their separation is changing
continuously . So we use relative velocity concept , Also Let Line Joining their separation and the Line of motion of ship -A (Vertical Line) has angle $\theta$ b/w them.Also Let final time is t=T.

$\displaystyle{\cfrac { -dr }{ dt } =v\cos { \theta } -v\\ \int _{ d }^{ x }{ -dr } =\int _{ 0 }^{ T }{ (v\cos { \theta } -v)dt } .\quad .\quad .\quad .\quad (1)}$

Now , Also difference b/w horizontal distance travelled by the two ships is equal to 'x' , So :

$\displaystyle{x\equiv \int _{ 0 }^{ T }{ vdt } -\int _{ 0 }^{ T }{ v\cos { \theta dt } } \quad .\quad \quad .\quad .\quad .(2)}$

using 1 and 2 we get ,

$\boxed { x=\cfrac { d }{ 2 } }$

Hopes this helps you!

This question can be solved using polar coordinates.

$\frac{dr}{dt}=v\sin\theta-v$

and $v\cos\theta=r\frac{d\theta}{dt}$

Now dividing both the equations we get:-

$\frac{dr}{rd\theta}=\frac{\sin\theta-1}{\cos\theta}$

Integrating this expression: $\int_d^r{\frac{dr}{r}}=\int_0^{\pi/2}{\frac{\sin\theta-1}{cos\theta}}d\theta$

We get $\ln\frac{r}{d}=-\ln 2$

hence answer is d/2