Ncert Exercises & Solutions

4.14 In a harbour, the wind is blowing at the speed of 72 km/h, and the flag on the mast of a boat anchored in the harbour 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?

NEETprep Answer:

$Velocity of the boat, v_{b} = 51 km / h$

$Velocity of the wind v_{w} = 72 km / h$

$SInce flag is fluttering in north - east direction, so wind is blowing towards north - east direction$

$Now, relative velocity of flag w . r . t . boat is v_{wb}$

$T h e r e f o r e, a n g l e w i t h r e s p e c t t o t h e e a s t d i r e c t i o n = 45 . 11^{0} - 45^{0} = 0 . 11^{0}$

$H e n c e, t h e f l a g w i l l f l u t t e r a l m o s t d u e e a s t .$

The angle between $v_{w}$ and (– $v_{b}$ ) = 90° + 45°

$\begin{array}{l} \tan β = \frac{51 \sin (90^{0} + 45^{0})}{72 + 51 \cos (90^{0} + 45^{0})} \\ = \frac{51 \sin 45^{0}}{72 + 51 (- \cos 45^{0})} = \frac{51 \times \frac{1}{\sqrt{2}}}{72 - 51 \times \frac{1}{\sqrt{2}}} \\ = \frac{51}{72 \sqrt{2} - 51} = \frac{51}{72 \times 1.414 - 51} = \frac{51}{50.800} \\ \Rightarrow β = \tan^{- 1} (1.0038) = {45.11}^{\circ} \end{array}$

$T h e r e f o r e, a n g l e w i t h r e s p e c t t o e a s t d i r e c t i o n 45 . 11^{0} - 45^{0} = 0 . 11^{0}$
$T h e r e f o r e, f l a g w i l l f l u t t e r a l m o s t d u e e a s t .$

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