The speed of a swimmer in still water is $20 ext{ m/s}$. The speed of river water is $10 ext{ m/s}$ and is

Question

The speed of a swimmer in still water is $20 	ext{ m/s}$. The speed of river water is $10 	ext{ m/s}$ and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes with respect to the north is given by:

Accepted Answer

1. **Identify the Goal:** To cross the river along the **shortest path**, the swimmer must land at the point directly opposite to the starting point. This means the resultant velocity vector (velocity with respect to the ground, $\vec{v}_g$) must be directed due North (perpendicular to the river flow).
2. **Vector Analysis:** 
   - Velocity of swimmer w.r.t. water: $\vec{v}_s = 20 	ext{ m/s}$.
   - Velocity of river: $\vec{v}_r = 10 	ext{ m/s}$ (due East).
   - Let $	heta$ be the angle the swimmer makes with the North direction towards the West.
3. **Condition for Zero Drift:** The horizontal component of the swimmer's velocity must cancel out the river's velocity to prevent any drift downstream .
   $$ v_s \sin 	heta = v_r $$
4. **Calculation:**
   $$ \sin 	heta = \frac{v_r}{v_s} = \frac{10}{20} = \frac{1}{2} $$
   $$ 	heta = \sin^{-1}(0.5) = 30^{\circ} $$
5. **Direction:** Since the river flows East, the swimmer must direct his strokes towards the West to counteract the flow. Thus, the angle is $30^{\circ}$ west of north.

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