let th initial position of the ships are denoted by $\ (x_{1},y_{1}),(x_{2},y_{2})$ and velocity vector $\ (u_{1},v_{1}),(u_{2},v_{2})$ here the unit is miles and for velocity miles/minute now the squared distance at time t is $ \ f(t)=((x_{1}+u_{1}t)-(x_{2}+u_{2}t))^{2}+((y_{1}+v_{1}t)-(y_{2}+v_{2}t))^{2} $ take derivatiove and set =0 to get $ \ t_{min}=\frac{-[(x_{1}-x_{2})(u_{1}-u_{2})+(y_{1}-y_{2})(v_{1}-v_{2})]}{(u_{1}-u_{2})^{2}+(v_{1}-v_{2})^{2}}$ =$ \frac{-a}{b} $ then we have to get a and b. now using the f(0),f(35) and f(55) we get the following $ \ 14a+245b=-35 $ and $ \ 110a+3025b=-231 $ solving them we get$ \ b=0.04,a=-3.2 $ so $ \ t_{min}=80 \ minute$ hence $ \ Time_{min}=10:20 $ and $ \ d_{min}=12 \ mile$