River crossing

The velocities $\vec v_i$ and $\vec u$ of the boat and the flow add up to the total velocity $\vec w_i$ , that is aligned parallel to the line $\overline{AB}$ (with index $i = 1,2$ ). This can be graphically represented by the correspondings vector parallelograms:

With the help of angles $\alpha_1$ and $\alpha_2$ we can represent the vector addition by the follows sets of equations: $\begin{aligned} \vec w_1 = \left( \begin{array}{c} w_1/\sqrt{2} \\ w_1/\sqrt{2} \end{array}\right) &= \left( \begin{array}{c} v \cos \alpha_1 \\ v \sin \alpha_1 \end{array}\right) + \left( \begin{array}{c} 0 \\ u \end{array}\right) = \vec v_1 + \vec u \qquad (1) \\ \vec w_2 = \left( \begin{array}{c} - w_2/\sqrt{2} \\ -w_2/\sqrt{2} \end{array}\right) &= \left( \begin{array}{c} -v \cos \alpha_2 \\ -v \sin \alpha_2 \end{array}\right) + \left( \begin{array}{c} 0 \\ u \end{array}\right) = \vec v_2 + \vec u \qquad (2) \end{aligned}$ By elimination of the variable $w_i$ , we obtain the following equations: $\begin{aligned} v \cos \alpha_1 &= v \sin \alpha_1 + u & \Rightarrow \qquad \cos \alpha_1 - \sin \alpha_1 &= \frac{u}{v} \\ v \cos \alpha_2 &= v \sin \alpha_2 - u & \Rightarrow \qquad \cos \alpha_2 - \sin \alpha_2 &= -\frac{u}{v} \end{aligned}$ With the help of the Pythagorean theorem $\sin^2 x + \cos^2 y = 1$ and the addition theorem $\sin(x + y) = \sin x \cos y + \cos x \sin y$ we can simplify the left hand side of both formulas: $\begin{aligned} (\cos \alpha - \sin \alpha)^2 &= \cos^2 \alpha - 2 \cos \alpha \sin \alpha + \sin^2 \alpha = 1 - \sin 2\alpha = \left(\frac{u}{v}\right)^2 \\ \Rightarrow \qquad \alpha &= \frac{1}{2} \text{arcsin}\left( 1 - \left(\frac{u}{v}\right)^2 \right) = \frac{1}{2} \text{arcsin} \frac{16}{25} = 19.9^\circ \text{ or } 70.1^\circ \end{aligned}$ The different solutions corresponds to the different angles $\alpha_1$ and $\alpha_2$ . Substituting these numbers into the equations (1) and (2) yields the total velocities. $\begin{aligned} w_1 &= \sqrt{2} v \cos \alpha_1 = 6.65 \frac{\text{m}}{\text{s}} \\ w_2 &= \sqrt{2} v \cos \alpha_2 = 2.41\frac{\text{m}}{\text{s}} \\ \end{aligned}$ Therefore, the total traveling time $T$ results $T = t_1 + t_2 = \frac{s}{w_1} + \frac{s}{w_2} = \frac{\sqrt{2} \cdot 200\,\text{m}}{6.65\,\text{m}/\text{s}} + \frac{\sqrt{2} \cdot 200\,\text{m}}{2.41 \,\text{m}/\text{s}} \approx 160 \,\text{s}$