A man walks for some time 't' with velocity (V) due east. Then he walks for the same time 't' with velocity (V) due north. The average velocity of the man is:
(A) 2V
(B) √2 V
(C) V
(D) V / √2

To find the average velocity of the man, we can use vector addition because the man is initially moving east and then north. The average velocity is the total displacement divided by the total time.

Let's consider the eastward motion first. The displacement in the east direction can be represented as a vector:

Δx = V * t (displacement in the east direction)

Now, let's consider the northward motion. The displacement in the north direction can be represented as a vector:

Δy = V * t (displacement in the north direction)

To find the total displacement (Δr), we can use the Pythagorean theorem because the east and north displacements form a right triangle:

Δr² = Δx² + Δy²
Δr² = (V * t)² + (V * t)²
Δr² = 2 * (V * t)²

Δr = √(2 * (V * t)²)
Δr = √2 * V * t

Now, the total time (T) is 2t since he walks for 't' time in each direction.

Average velocity (A) is given by:

A = Δr / T
A = (√2 * V * t) / (2t)
A = (V * √2) / (2)

Now, simplify the expression:

A = (V * √2) / 2

So, the average velocity of the man is option D: (V * √2) / 2.