Question

A man holding a flag is running in the north-east direction with speed \[10\,{\text{m/s}}\]. Wind is blowing in the east direction with speed \[5\sqrt 2 \,{\text{m/s}}\]. Find the direction in which the flag will flutter.
A. South
B. North
C. East
D. West

Answer 1

Hint: First write the velocity vector of the wind. Then determine the velocity vector of the man considering that the man is running exactly between the north and east direction making an angle of \[45^\circ \] with the X and Y axes. Then determine the velocity vector for the flag by calculating the relative velocity of the wind with respect to the man.

Complete step by step answer:
We have given that a man is holding a flag in his hands and running in a north-east direction with a speed \[10\,{\text{m/s}}\].
\[{v_m} = 10\,{\text{m/s}}\]
The wind is blowing in the east direction with a speed \[5\sqrt 2 \,{\text{m/s}}\].The velocity vector of the wind can be written as
\[{\vec v_w} = 5\sqrt 2 \hat i\,{\text{m/s}}\]

Let us now determine the velocity vector of the man in the north-east direction.Since the man is running in the north-east direction, the angle made by the velocity vector of the man makes an angle of \[45^\circ \] with the x and Y axes.Hence, the velocity vector of the man can be written as
\[{\vec v_m} = {v_m}\cos 45^\circ \hat i + {v_m}\sin 45^\circ \hat j\]
\[ \Rightarrow {\vec v_m} = \left( {10\,{\text{m/s}}} \right)\dfrac{1}{{\sqrt 2 }}\hat i + \left( {10\,{\text{m/s}}} \right)\dfrac{1}{{\sqrt 2 }}\hat j\]
\[ \Rightarrow {\vec v_m} = \left( {5\sqrt 2 \hat i + 5\sqrt 2 \hat j} \right)\,{\text{m/s}}\]

The direction in which the flag will flutter is the direction of velocity of the wind with respect to the wind.
\[{\vec v_{wm}} = {\vec v_w} - {\vec v_m}\]
Substitute \[5\sqrt 2 \hat i\,{\text{m/s}}\] for \[{\vec v_w}\] and \[\left( {5\sqrt 2 \hat i + 5\sqrt 2 \hat j} \right)\,{\text{m/s}}\] for \[{\vec v_m}\] in the above equation.
\[{\vec v_{wm}} = \left( {5\sqrt 2 \hat i\,{\text{m/s}}} \right) - \left( {5\sqrt 2 \hat i + 5\sqrt 2 \hat j} \right)\,{\text{m/s}}\]
\[ \therefore {\vec v_{wm}} = - 5\sqrt 2 \hat j\,{\text{m/s}}\]
The velocity of the flag is \[5\sqrt 2 \hat j\,{\text{m/s}}\] in the vertical direction and the negative sign indicates that the direction of the flag is in the south direction.Therefore, the flag will flutter in south direction.

Hence, the correct option is A.

Note: The students may get confused about how we have determined the velocity vector of the man directly. But we know that the horizontal component of the velocity is velocity into cosine of angle between the velocity and X axis and vertical component of velocity is velocity into sine of angle between the velocity and X axis. The students should not get confused between these components.