Question

Q] A man rows a boat with a speed of 18 km/hr in north-west direction. The shoreline makes an angle of 15° south of west. Obtain the component of the velocity of the boat along the shoreline and perpendicular to the shoreline.​

Answer 1

Answer:

Question :-

• A man rows a boat with a speed of 18 km/hr in north-west direction. The shoreline makes an angle of 15° south of west. Obtain the component of the velocity of the boat along the shoreline and perpendicular to the shoreline.

Given :-

• A man rows a boat with a speed of 18 km/hr in north-west direction.
• The shoreline makes an angle of 15° south of west.

To Find :-

• What is the component of the velocity of the boat along the shoreline and perpendicular to the shoreline.

Solution :-

$\mapsto$ The north-west direction of a boat makes an angle of 60° with the help of shoreline.

Then,

$\leadsto$ Component of the velocity of the boat along the shoreline :

$\implies \sf vcos60^{\circ}$

Given :

• Velocity (v) = 18 km/hr

$\implies \sf 18\: km/hr\: \times cos^{\circ}$

As we know that, [ cos60° = ½ ]

$\implies \sf 18\: km/hr \times \dfrac{1}{2}$

$\implies \sf {\cancel{18}}\: km/hr \times \dfrac{1}{\cancel{2}}$

$\implies \sf 9\: km/hr\: \times \dfrac{1}{1}$

$\implies \sf \dfrac{9}{1}\: km/hr$

$\implies \sf\bold{\red{9\: km/hr}}$

$\therefore$ The component of the velocity of the boat along the shoreline is 9 km/hr .

$\rule{150}{2}$

Again,

$\leadsto$ Component of the velocity of the boat along the perpendicular to the shoreline :

$\implies \sf vsin60^{\circ}$

Given :

• Velocity (v) = 18 km/hr

$\implies \sf 18\: km/hr\: \times sin60^{\circ}$

As we know that, [ sin60° = √3/2 ]

$\implies \sf 18\: km/hr\: \times \dfrac{\sqrt{3}}{2}$

$\implies \sf {\cancel{18}}\: km/hr\: \times \dfrac{\sqrt{3}}{\cancel{2}}$

$\implies \sf 9\: km/hr \times \dfrac{\sqrt{3}}{1}$

$\implies \sf \dfrac{9\sqrt{3}}{1}\: km/hr$

$\implies\sf\bold{\red{9\sqrt{3}\: km/hr}}$

$\therefore$ The component of the velocity of the boat along the perpendicular to the shoreline is 9√3 km/hr .

[Note : Please refer the attachment for the diagram. ]

Answer 2

Question:

A man rows a boat with a speed of 18 km/hr in north-west direction. The shoreline makes an angle of 15° south of west. Obtain the component of the velocity of the boat along the shoreline and perpendicular to the shoreline.

Given:

Speed of the boat= 18km/hr North-west

Angle made by shoreline south of west= 15°

To find:

Component of velocity of the boat along the shoreline and perpendicular to the shoreline.

Solution:

$\tt 9km/hr ~and~ 9 \sqrt{3} ~km/hr$

Explanation:

The boat is along north-west direction i.e it makes 45° with North and the West. (Refer to the attachment)

Now, angle between velocity and shoreline will be (45°+15°)=60°

Break the velocity in two components, one will be v cos 60° and other will be v sin 60°

According to the question, we have to find the velocity along the shoreline, i.e v cos 60°

$\tt v cos 60^\circ = 18 \times \frac{1}{2} \\\\ \tt \cancel{18} ~9\times \frac{1}{\cancel{2}} ~ \\\\ \to\boxed{\bf{9km/hr}}$

Now, similarly, velocity of boat perpendicular to the shore is V sin 60°(see the attachment)

$\tt v sin 60^\circ = 18 \times \frac{\sqrt{3}}{2} \\\\ \tt \cancel{18} ~9\times \frac{\sqrt{3}}{\cancel{2}} ~ \\\\ \to \boxed{\bf{9 \sqrt{3}km/hr}}$