Considering the initial position of ship A as origin, so the velocity and position of ship will be

$${\overrightarrow v _A} = (30\widehat i + 50\widehat j)$$ and $${\overrightarrow r _A} = (0\widehat i + 0\widehat j)$$

Now, as given in the question, velocity and position of ship B will be, $${\overrightarrow v _B} = - 10\widehat i$$ and $${\overrightarrow r _B} = (80\widehat i + 150\widehat j)$$

Time after which the distance is minimum between A and B can be calculated as

$$t = {{|{{\overrightarrow r }_{BA}}.\,{{\overrightarrow v }_{BA}}|} \over {|{{\overrightarrow v }_{BA}}{|^2}}}$$

where, $${\overrightarrow r _{BA}} = {\overrightarrow r _B} - {\overrightarrow r _A} = 80\widehat i + 150\widehat j$$

and $${\overrightarrow v _{BA}} = - 10\widehat i - (30\widehat i + 50\widehat j)$$

$$ = - 40\widehat i - 50\widehat j$$

$$ \Rightarrow t = {{|(80\widehat i + 150\widehat j)\,.\,( - 40\widehat i - 50\widehat j)|} \over {| - 40\widehat i - 50\widehat j{|^2}}}$$

$$ = {{3200 + 7500} \over {4100}} = {{10700} \over {4100}} = 2.6$$ h