A river is 120 m wide and flows at $3 \mathrm{~m} / \mathrm{s}$. If the person swims with a velocity

Question

A river is 120 m wide and flows at $3 \mathrm{~m} / \mathrm{s}$. If the person swims with a velocity $5 \mathrm{~m} / \mathrm{s}$ in such a way that he reaches the opposite bank by the shortest path. The direction in which he is to swim makes an angle $	heta$ with the direction of the river where $	heta$ is\

Professor PC Thomas Classes · Answer

River flowing with a velocity $u$ and the person wanting to cross the river in the shortest path\ He must swim along AD making an angle $	heta$ with AB such that $\sin 	heta=\frac{\mathrm{u}}{\mathrm{v}}$. His resultant velocity $=\sqrt{v^{2}-u^{2}}$ along $A B$. So he will reach $B$ along $A B$, the shortest path. Time taken $t_{3}=\frac{d}{\sqrt{v^{2}-u^{2}}}$. For a round trip he must swim back along BD. Time for a round $\operatorname{trip} \frac{2 \mathrm{~d}}{\sqrt{\mathrm{v}^{2}-\mathrm{u}^{2}}}$ $u=3 . v=5 \sin 	heta=\frac{3}{5}=0.6 	heta=37^{\circ}$\ Required angle is $90+37$\