A man walks for some time 't' with velocity(v) due east. Then he walks for same time 't' with velocity (v) due north. The average velocity of the man is:-

Options

A. 2v `cos 40^(@)`
B. `sqrt(2)v`
C. v
D. `(v)/sqrt(2)`

Explanation

To solve the problem of finding the average velocity of a man who walks due east and then due north, we can follow these steps:

### Step 1: Understand the Motion
The man walks for time 't' with velocity 'v' due east and then for the same time 't' with velocity 'v' due north.

### Step 2: Calculate the Displacement
1. **Displacement in East Direction**:
- Distance covered in the east direction = velocity × time = \( v \times t \).

2. **Displacement in North Direction**:
- Distance covered in the north direction = velocity × time = \( v \times t \).

3. **Resultant Displacement**:
- Since the man moves in two perpendicular directions (east and north), we can use the Pythagorean theorem to find the resultant displacement \( S \):
\[ S = \sqrt{(vt)^2 + (vt)^2} = \sqrt{2(vt)^2} = vt\sqrt{2} \]

### Step 3: Calculate the Total Time
- The total time taken for the journey is the sum of the time spent walking in both directions:
\[ \text{Total Time} = t + t = 2t \]

### Step 4: Calculate the Average Velocity
- The average velocity \( V_{avg} \) is defined as the total displacement divided by the total time:
\[ V_{avg} = \frac{\text{Displacement}}{\text{Total Time}} = \frac{S}{2t} = \frac{vt\sqrt{2}}{2t} \]
- Simplifying this gives:
\[ V_{avg} = \frac{v\sqrt{2}}{2} \]

### Final Answer
Thus, the average velocity of the man is:
\[ \frac{v\sqrt{2}}{2} \]