The position is described in the sketch given below:
At noon they are #130# miles apart. After #t# hours, Ship A would have moved #25t# miles east and Ship B would have moved #10t# miles north.
Hence distance #D(t)# between them after #t# hours would be given by
#D(t)=sqrt((130-25t)^2+(10t)^2)=sqrt(16900+625t^2-6500t+100t^2)#
= #sqrt(725t^2-6500t+16900)#
and #(dD(t))/(dt)=(1450t-6500)/(2sqrt(725t^2-6500t+16900))#
= #(290t-1300)/(2sqrt(29t^2-260t+676))#
and at #4:00pm# i.e. after #4# hours distance is changing at a speed of
#(dD(4))/(dt)=(290xx4-1300)/(2sqrt(29xx16-260xx4+676))#
= #(1160-1300)/(2sqrt(464-1040+676))=-140/(2sqrt100)=-7#
Hence, after #4# hours distance is reducing at a speed of #7# miles per hour
