A swimmer can swim in still water with speed v and the river is flowing with speed `v//2`. What is the ratio of the time taken to swimming across the river in shortest time to that of swimming across the river over the shortest distance ?

Options

A. `sqrt(3)/2`
B. 1
C. 2
D. `sqrt(3)`v

Explanation

To solve the problem, we need to analyze the swimmer's motion in two scenarios: swimming across the river in the shortest time and swimming across the river over the shortest distance.

### Step 1: Define the Variables
- Let the speed of the swimmer in still water be \( v \).
- Let the speed of the river be \( \frac{v}{2} \).
- Let the width of the river be \( d \).

### Step 2: Swimming Across the River in the Shortest Time
To swim across the river in the shortest time, the swimmer must swim directly perpendicular to the flow of the river. In this case, the effective velocity of the swimmer across the river is \( v \) (the swimmer's speed), and the time taken \( T_1 \) to cross the river is given by:

\[ T_1 = \frac{d}{v} \]

### Step 3: Swimming Across the River Over the Shortest Distance
When swimming across the river over the shortest distance, the swimmer must angle his swimming direction upstream to counteract the current of the river. The effective velocity of the swimmer perpendicular to the river flow can be found using the cosine of the angle \( \theta \) formed between the swimmer's direction and the river flow.

Using the relationship of the velocities:
- The swimmer's effective speed across the river is \( v \cos \theta \).
- The speed of the river is \( \frac{v}{2} \).

From the triangle formed by the velocities, we can use the sine function:
\[ \sin \theta = \frac{\text{speed of river}}{\text{speed of swimmer}} = \frac{\frac{v}{2}}{v} = \frac{1}{2} \]

This implies that \( \theta = 30^\circ \).

Using the cosine of the angle:
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \]

Now, the time taken \( T_2 \) to swim across the river over the shortest distance is given by:

\[ T_2 = \frac{d}{v \cos \theta} = \frac{d}{v \cdot \frac{\sqrt{3}}{2}} = \frac{2d}{v \sqrt{3}} \]

### Step 4: Calculate the Ratio of Times
Now, we can find the ratio of the time taken to swim across the river in the shortest time \( T_1 \) to the time taken to swim across the river over the shortest distance \( T_2 \):

\[ \frac{T_2}{T_1} = \frac{\frac{2d}{v \sqrt{3}}}{\frac{d}{v}} = \frac{2}{\sqrt{3}} \]

### Conclusion
The ratio of the time taken to swim across the river in the shortest time to that of swimming across the river over the shortest distance is:

\[ \frac{T_1}{T_2} = \frac{\sqrt{3}}{2} \]