A man can swim in still water at a speed of 6 kmph and he has to cross the river and reach just opposite point on the other bank. If the river is flowing at a speed of 3 kmph, and the width of the river is 2 km, the time taken to cross the river is (in hours)

Options

A. `2/27`
B. `2/sqrt(27)`
C. `2/3`
D. `2/sqrt(45)`

Answer

To solve the problem of a man swimming across a river with a current, we can break it down into a few steps:

### Step 1: Identify the given values
- Speed of the man in still water (Vm) = 6 km/h
- Speed of the river (Vr) = 3 km/h
- Width of the river (D) = 2 km

### Step 2: Understand the scenario
The man wants to swim directly across the river to reach a point directly opposite on the other bank. To do this, he must swim at an angle upstream to counteract the downstream current of the river.

### Step 3: Calculate the effective speed across the river
To find the effective speed at which the man can swim directly across the river, we can use the Pythagorean theorem. The man's swimming speed and the river's speed form a right triangle where:
- The man's speed (Vm) is the hypotenuse.
- The river's speed (Vr) is one leg of the triangle.
- The effective speed across the river (Vd) is the other leg.

Using the Pythagorean theorem:
\[ Vm^2 = Vr^2 + Vd^2 \]
Substituting the known values:
\[ 6^2 = 3^2 + Vd^2 \]
\[ 36 = 9 + Vd^2 \]
\[ Vd^2 = 36 - 9 \]
\[ Vd^2 = 27 \]
\[ Vd = \sqrt{27} = 3\sqrt{3} \text{ km/h} \]

### Step 4: Calculate the time taken to cross the river
The time taken (t) to cross the river can be calculated using the formula:
\[ t = \frac{D}{Vd} \]
Substituting the values:
\[ t = \frac{2 \text{ km}}{3\sqrt{3} \text{ km/h}} \]
To simplify:
\[ t = \frac{2}{3\sqrt{3}} \text{ hours} \]

### Step 5: Rationalize the denominator (optional)
To express the time in a more standard form, we can rationalize the denominator:
\[ t = \frac{2}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{9} \text{ hours} \]

### Final Answer
The time taken to cross the river is:
\[ t = \frac{2}{3\sqrt{3}} \text{ hours} \text{ or } \frac{2\sqrt{3}}{9} \text{ hours} \]

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