Rozwiązanie:

$v_w=\sqrt{v_1^2{\sin }^2\alpha +v_1^2{\cos }^2\alpha +2v_1{v}_2\cos \alpha +v_2^2}$

$v_w=\sqrt{{\left(20\frac{\mathrm{km}}{\mathrm{h}}\right)}^2\cdot {\sin }^{2}45^{\circ }+{\left(20\frac{\mathrm{km}}{\mathrm{h}}\right)}^2\cdot {\cos }^{2}45^{\circ }+2\cdot 20\frac{\mathrm{km}}{\mathrm{h}}\cdot 14\frac{\mathrm{km}}{\mathrm{h}}\cdot \cos 45^{\circ }+{\left(14\frac{\mathrm{km}}{\mathrm{h}}\right)}^2}=$

$=\sqrt{400\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}\cdot {\left(\frac{\sqrt{2}}{2}\right)}^2+400\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}\cdot {\left(\frac{\sqrt{2}}{2}\right)}^2+560\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}\cdot \frac{\sqrt{2}}{2}+196\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}}=$

$=\sqrt{400\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}\cdot \frac{1}{2}+400\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}\cdot \frac{1}{2}+\frac{560\sqrt{2}}{2}\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}+196\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}}=$

$=\sqrt{200\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}+200\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}+280\sqrt{2}\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}+196\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}}\approx$

$\approx \sqrt{596\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}+396\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}}=\sqrt{992\frac{{\mathrm{km}}^2}{{\mathrm{h}}^2}}\approx 31,5\frac{\mathrm{km}}{\mathrm{h}}$

Odpowiedź: Statki oddalają się od siebie z szybkością około $31,5\frac{\mathrm{km}}{\mathrm{h}}$.