Một người đứng tại điểm A trên bờ hồ phẳng lặng (hình vẽ

* Hướng dẫn giải

+ Gọi quãng đường DC có độ dài là: x

+ Độ dài quãng đường BD: $\sqrt{d^2+x^2}$

+ Thời gian người này đi từ A đến D rồi đến B là:

                  t = t_AD + t_DB = $\frac{S_{AD}}{{\text{v}}_{\text{2}}}+\frac{S_{DB}}{{\text{v}}_{\text{1}}}=\frac{S-x}{{\text{v}}_{\text{2}}}+\frac{\sqrt{d^2+x^2}}{{\text{v}}_{\text{1}}}$

+ Khi đó:   $t-\frac{S-x}{{\text{v}}_{\text{2}}}=\frac{\sqrt{d^2+x^2}}{{\text{v}}_{\text{1}}}=>t^2-2\frac{S-x}{{\text{v}}_{\text{2}}}t+{\left(\frac{S-x}{{\text{v}}_{\text{2}}}\right)}^2=\frac{d^2+x^2}{{{\text{v}}_{\text{1}}}^2}$

 $$\begin{gathered} {{\text{v}}_{\text{1}}}^2{{\text{v}}_{\text{2}}}^2{t}^2-2{{\text{v}}_{\text{1}}}^2{\text{v}}_{\text{2}}St+2{{\text{v}}_{\text{1}}}^2{\text{v}}_{\text{2}}xt+S^2{{\text{v}}_{\text{1}}}^2-2Sx{{\text{v}}_{\text{1}}}^2+x^2{{\text{v}}_{\text{1}}}^2=d^2{{\text{v}}_{\text{2}}}^2+x^2{{\text{v}}_{\text{2}}}^2 có nghiệm x \\ \left({{\text{v}}_{\text{2}}}^2-{{\text{v}}_{\text{1}}}^2\right)x^2-2\left({{\text{v}}_{\text{1}}}^2{\text{v}}_{\text{2}}t-S{{\text{v}}_{\text{1}}}^2\right)x-\left({{\text{v}}_{\text{1}}}^2{{\text{v}}_{\text{2}}}^2{t}^2+S^2{{\text{v}}_{\text{1}}}^2-2{{\text{v}}_{\text{1}}}^2{\text{v}}_{\text{2}}St-d^2{{\text{v}}_{\text{2}}}^2\right) =0 \mathrm{co} nghiem \end{gathered}$$ 

Khi đó  = $\Delta = {\left({{\text{v}}_{\text{1}}}^2{\text{v}}_{\text{2}}t-S{{\text{v}}_{\text{1}}}^2\right)}^2\left({{\text{v}}_{\text{2}}}^2-{{\text{v}}_{\text{1}}}^2\right)\left({{\text{v}}_{\text{1}}}^2{{\text{v}}_{\text{2}}}^2{t}^2+S^2{{\text{v}}_{\text{1}}}^2-2{{\text{v}}_{\text{1}}}^2{\text{v}}_{\text{2}}St-d^2{{\text{v}}_{\text{2}}}^2\right)\ge 0$

   $tMin =\frac{{\text{Sv}}_{\text{1}}+d\sqrt{{{\text{v}}_{\text{2}}}^2-{{\text{v}}_{\text{1}}}^2}}{{\text{v}}_{\text{1}}{\text{v}}_{\text{2}}}$