Question

In a harbour, the wind is blowing at the speed of 72km/h and flag on the mast of a boat anchors in the harbour flutter along the N-E direction. If the boat starts moving at 51km/h to the north, what is the direction of the flag on the mast of the boat?

Answer 1

Velocity of the boat, $V_b=51km/h$
Velocity of the wind, $V_w=72km/h$
The flag is fluttering in the north – east direction. It shows that the wind is blowing toward the north – east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity ($V_{wb}$) of the wind with respect to the boat.

The angle between $v_{m}and(-v_b)={90}^{\circ }+{45}^{\circ }$
$tan\beta =\frac{51sin(90+45)}{72+51cos(90+45)}$
$=\frac{51sin45}{72+51(-cos45)}=\frac{51\times \frac{1}{\sqrt{2}}}{72-51\times \frac{1}{\sqrt{2}}}$
$=\frac{51}{72\sqrt{2}-51}=\frac{51}{72\times 1.414-51}=\frac{51}{50.800}$
$\therefore \beta =ta{n}^{-1}\left(1.0038\right)=45.11$
Angle with respect to the east direction $={45.11}^{\circ }-{45}^{\circ }={0.11}^{\circ }$
Hence, the flag will flutter almost due east.