1. **State the problem.** We have ship $A$ moving due west at $19$ knots, and ship $B$ starting $5$ nautical miles southwest of $A$ and moving $N30^\circ E$ at $17$ knots. We want the distance between the ships when they are nearest together, and the time it takes to get there. 2. **Set up coordinates and the relative-position idea.** Let ship $A$ start at the origin, so its position is $A(t)=(-19t,0)$. Since $B$ is $5$ nautical miles southwest of $A$, its initial position is $\left(-\frac{5}{\sqrt{2}},-\frac{5}{\sqrt{2}}\right)$. A heading of $N30^\circ E$ means the velocity of $B$ has east component $17\sin 30^\circ$ and north component $17\cos 30^\circ$. So $$v_B=\left(17\sin 30^\circ,17\cos 30^\circ\right)=\left(\frac{17}{2},\frac{17\sqrt{3}}{2}\right).$$ 3. **Write the relative position vector.** The position of $B$ after $t$ hours is $$B(t)=\left(-\frac{5}{\sqrt{2}}+\frac{17}{2}t,-\frac{5}{\sqrt{2}}+\frac{17\sqrt{3}}{2}t\right).$$ So the vector from $A$ to $B$ is $$R(t)=B(t)-A(t)=\left(-\frac{5}{\sqrt{2}}+\frac{55}{2}t,-\frac{5}{\sqrt{2}}+\frac{17\sqrt{3}}{2}t\right).$$ 4. **Use the distance-squared formula.** The distance between the ships is minimized when $|R(t)|^2$ is minimized. So define $$D^2(t)=\left(-\frac{5}{\sqrt{2}}+\frac{55}{2}t\right)^2+\left(-\frac{5}{\sqrt{2}}+\frac{17\sqrt{3}}{2}t\right)^2.$$ Expand it: $$D^2(t)=250.5t^2-\left(\frac{350}{\sqrt{2}}+\frac{85\sqrt{3}}{\sqrt{2}}\right)t+25.$$ 5. **Find the time of closest approach.** For a quadratic $at^2+bt+c$, the minimum occurs at $t=-\frac{b}{2a}$. Here, $$a=250.5,$$ $$b=-\left(\frac{350}{\sqrt{2}}+\frac{85\sqrt{3}}{\sqrt{2}}\right).$$ So $$t=\frac{\frac{350}{\sqrt{2}}+\frac{85\sqrt{3}}{\sqrt{2}}}{2(250.5)}=\frac{350+85\sqrt{3}}{\sqrt{2}\cdot 501}.$$ Numerically, $$t\approx 0.602\text{ hours}.$$ This is about $36.1$ minutes. 6. **Find the minimum distance.** The minimum value of a quadratic is $$D^2_{\min}=c-\frac{b^2}{4a}.$$ Substituting gives $$D_{\min}\approx 1.30\text{ nautical miles}.$$ 7. **Final answer.** The ships are nearest together after about $0.602$ hours, or about $36.1$ minutes, and the minimum distance between them is about $1.30$ nautical miles.