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**Hint:**In order to find the direction in which the flag flutters we need to know the direction of the resultant vector of the two velocities given. That is the velocity of wind with respect to man will give the direction in which the flag will flutter. Write the given velocities in vector form and subtract the velocity of man from the velocity of wind to find the velocity of wind with respect to man.

**Complete step by step solution:**

It is given that a man holding a flag is running in the north east direction. The speed with which he is running is $10\,m/s$ .

We need to find the direction in which the flag flutters when the wind is blowing in the east direction and when the wind is blowing in the north direction.

Let us first analyze the case where the wind is towards the east.

The magnitude of wind is given as ${v_w} = 5\sqrt 2 \,m/s$ .

Since it is blowing along the x direction, we can write it in vector form as

$\overrightarrow {{v_w}} = 5\sqrt 2 \,\,\widehat i$

Now let us write the vector form of the velocity of the man holding the flag ${v_m}$ .

It is given that he is moving in the north east direction. Which means this vector is making an angle ${45^ \circ }$ with the east direction.

The component of this vector along x-axis will be ${v_m}\cos \theta $

That is ${v_m}\cos \,\,{45^ \circ } = 10 \times \dfrac{1}{{\sqrt 2 }} = 5\sqrt 2 $

And the component along y axis will be ${v_m}\sin \theta $ .

That is,

${v_m}\sin \,{45^ \circ } = 10 \times \dfrac{1}{{\sqrt 2 }} = 5\sqrt 2 $

Therefore, the velocity vector of the man can be written as

$\overrightarrow {{v_m}} = 5\sqrt 2 \,\widehat i + 5\sqrt 2 \,\widehat j$

Now we need to find the direction in which the flag will flutter. For that let us find the resultant velocity and its direction. If we find the velocity of wind with respect to man it will be the direction in which the flag flutters.

The resultant velocity of wind with respect to man is given as,

$\overrightarrow {{v_{wm}}} = \overrightarrow {{v_w}} - \overrightarrow {{v_m}} $

Let us substitute the value of these vectors. Then we get ,

$\overrightarrow {{v_{wm}}} = 5\sqrt 2 \,\widehat i - \left( {5\sqrt 2 \,\widehat i + 5\sqrt 2 \,\widehat j} \right)$

$ \Rightarrow \overrightarrow {{v_{wm}}} = - 5\sqrt 2 \,\widehat j$

Since we got negative $ - \widehat j$ we can say that this resultant is along the south direction.

**Therefore, option A is correct.**

Now let us consider the second case in which the wind is blowing towards the north direction.

Since it is blowing along the positive y direction, we can write it in vector form as

$\overrightarrow {{v_w}} = 5\sqrt 2 \,\,\widehat j$ .

The velocity of the man is the same as in the first case.

$\overrightarrow {{v_m}} = 5\sqrt 2 \,\widehat i + 5\sqrt 2 \,\widehat j$

The velocity of wind with respect to man in this case is,

$\overrightarrow {{v_{wm}}} = \overrightarrow {{v_w}} - \overrightarrow {{v_m}} $

On substituting the values, we get,

$\overrightarrow {{v_{wm}}} = 5\sqrt 2 \,\widehat j - \left( {5\sqrt 2 \,\widehat i + 5\sqrt 2 \,\widehat j} \right)$

$\overrightarrow {{v_{wm}}} = - 5\sqrt 2 \,\widehat i$

Since we got negative $ - \widehat i$ we can say that this resultant is along the west direction.

**Therefore, option D is correct.**

**Note:**While doing the problems with vectors involved always take care of the directions. In order to write a vector which is along the x-axis we need to use the unit vector $\widehat i$ . If the vector is along the negative x direction it should be denoted using $ - \widehat i$ . Similarly, the direction of positive y-axis will be denoted using unit vector $\widehat j$ and the direction of the negative y axis will be denoted using $ - \widehat j$ .

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