Question

In a harbor, wind is blowing at the speed of $72{{km/h}}$ and the flag on the mast of a boat anchored in the harbor flutters along the N-E direction. If the boat starts moving at a speed of $51{{km/h}}$ to the north, what is the direction of the flag on the mast of the boat?

Answer 1

Hint: When the boat moves in the north direction, the flag will flutter in the direction of the relative velocity of wind with respect to the boat. Consider the angle between the velocity vector of wind and the velocity vector of wind with respect to the boat. From that angle the direction of fluttering of the flag can be found.

Complete step by step answer:
Given the wind is blowing at the speed of $72{{km/h}}$ and the flag flutters along the N-E direction. This means the wind is blowing in N-E direction. Also given the boat starts moving at a speed of $51{{km/h}}$ to the north. When the boat starts moving the flag will flutter in the direction of the relative velocity of wind with respect to the boat.
The following figure can illustrate the direction.

Here ${\vec v_b}$ is the velocity of boat, ${\vec v_w}$ is the velocity of wind and ${\vec v_{wb}}$ is the velocity of wind with respect to boat.
The angle between ${\vec v_w}$ and ${\vec v_{wb}}$ is given as $\beta $ .
The angle between ${\vec v_w}$and $ - {\vec v_b}$is ${45^\circ } + {90^\circ } = {135^\circ }$
Velocity of wind with respect to boat can be defined as,
 $
  {{\vec v}_{wb}} = {{\vec v}_w} - {{\vec v}_b} \\
\Rightarrow{{\vec v}_w} + \left( { - {{\vec v}_b}} \right) \\
$
Therefore the angle between ${\vec v_w}$ and ${\vec v_{wb}}$can be called as angle between ${\vec v_w}$ and ${\vec v_w} + \left( { - {{\vec v}_b}} \right)$.
$\tan \beta = \dfrac{{{v_b}\sin {{135}^\circ }}}{{{v_w} + {v_b}\cos {{135}^\circ }}}$
Substituting the values in the above expression,
$
  \tan \beta = \dfrac{{51{{km/h}} \times \sin {{135}^\circ }}}{{72{{km/h}} + 51{{km/h}} \times \cos {{135}^\circ }}} \\
\Rightarrow\dfrac{{36.0624}}{{72 - 36.0624}} \\
\Rightarrow1.00347 \\
$
$
  \beta = {\tan ^{ - 1}}\left( {1.00347} \right) \\
 \beta = {45.0992^\circ } \\
$
Thus the angle between ${\vec v_w}$ and ${\vec v_{wb}}$is ${45.0992^\circ }$.
Subtracting ${45^\circ }$ from $\beta $ , we get
${45.0992^\circ } - {45^\circ } = {0.0992^\circ }$
Hence, the flag flutters almost in the direction of the East.

Note: We had taken the angle between the velocity and perpendicular of east as ${45^\circ }$. Therefore the angle between the velocity vector of wind and the velocity vector of wind respect to the boat would be compared with ${45^\circ }$. If the difference is small then it is almost close to perpendicular to the east direction.