Question

The velocity of water in a river is 2kmph, while width is 400m. A boat is rowed from a point rowing always aiming opposite point at 8kmph of still water velocity. On reaching the opposite bank the drift obtained is

Answer 1

$$\begin{gathered} \mathrm{Speed}\mathrm{of}\mathrm{river}={\mathrm{v}}_{\mathrm{r}}=2\mathrm{Km}/\mathrm{h}=2\times 5/18=5/9\mathrm{m}/\mathrm{s}. \\ \mathrm{and}\mathrm{speed}\mathrm{of}8\mathrm{Km}/\mathrm{h}=8\times 5/18=\frac{20}{9}\mathrm{m}/\mathrm{s}. \\ \mathrm{as}\mathrm{these}\mathrm{two}\mathrm{velocities}\mathrm{are}\mathrm{perpendicular}\mathrm{to}\mathrm{each}\mathrm{other}\mathrm{so} \\ \mathrm{their}\mathrm{relative}\mathrm{velocity}\mathrm{v}=\sqrt{\frac{25}{81}+\frac{400}{81}}=\frac{\sqrt{425}}{9}=2.29\mathrm{m}/\mathrm{s}. \\ \mathrm{let}\mathrm{time}\mathrm{taken}\mathrm{to}\mathrm{cross}\mathrm{the}\mathrm{river}\mathrm{be}\mathrm{t}\mathrm{then}, \\ \frac{400}{\mathrm{t}}=\frac{\sqrt{425}}{9}. \\ \mathrm{or}\mathrm{t}=\frac{400\times 9}{\sqrt{425}}=\frac{3600}{\sqrt{425}}\mathrm{second}. \\ \mathrm{or}\mathrm{the}\mathrm{horizontal}\mathrm{drift}\mathrm{during}\mathrm{this}\mathrm{time}=\mathrm{x}. \\ \mathrm{therefore}\mathrm{x}={\mathrm{v}}_{\mathrm{r}}\times \mathrm{t}=\frac{5}{9}\times \frac{3600}{\sqrt{425}}=2000/20.61=97.04\mathrm{m}. \end{gathered}$$