A river is flowing from west to east at a speed of `5m//s`. A man on the south bank of the river capable of swimming at `10m//s` in a still water wants to swim, across the river in a shortest time. He should swim in a direction

Options

A. due North
B. `30^(@)` East of North
C. `30^(@)` West of North
D. `60^(@)` East of North

Explanation

To solve the problem of the man swimming across the river in the shortest time, we can break down the solution into clear steps:

### Step 1: Understand the Problem
The river flows from west to east at a speed of 5 m/s. The man can swim at a speed of 10 m/s in still water. We need to determine the angle at which he should swim to cross the river in the shortest time.

### Step 2: Set Up the Components of Motion
- Let \( V_r = 5 \, \text{m/s} \) (velocity of the river).
- Let \( V_m = 10 \, \text{m/s} \) (velocity of the man in still water).
- The man swims at an angle \( \theta \) with respect to the direction directly across the river (north).

### Step 3: Resolve the Man's Velocity
The man's velocity can be resolved into two components:
1. **Across the river (northward)**: \( V_{mr} \cos(\theta) \)
2. **Downstream (eastward)**: \( V_{mr} \sin(\theta) \)

### Step 4: Determine the Time to Cross the River
The time taken to cross the river (width \( w \)) can be expressed as:
\[ t = \frac{w}{V_{mr} \cos(\theta)} \]
To minimize the time \( t \), we need to maximize \( V_{mr} \cos(\theta) \).

### Step 5: Maximize the Cosine Component
To maximize \( V_{mr} \cos(\theta) \):
- The maximum value of \( \cos(\theta) \) is 1, which occurs when \( \theta = 0^\circ \).
- This means the man should swim directly north (perpendicular to the riverbank).

### Step 6: Conclusion
Thus, the man should swim in the direction due north (0 degrees) to cross the river in the shortest time.

### Final Answer
The man should swim at an angle of \( 0^\circ \) (due north).

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