I. E. Irodov Solution 1.6

Solution to I. E. Irodov in General Physics and H. C. Verma in Concept of Physics is like a bible for student who are appearing for IIT-JEE, JEE/Main and JEE/Advance, UG NEET, AIIMS or any other Engineering and Medical entrance examination. All the questions in these books are of high level, which requires all basics to applied concept of physics. We here makes your task very easy, we have presented complete solution with detailed explanation step by step. The solution of these books teaches students also teachers in a suitable manner and then tests you with some tricky questions. To answer these questions you need to have thorough understanding of the concepts and this is where most students falter.

Problem: 1.6
A ship moves along the equator to the east with velocity ${v_0} = 30km/h$. The southeastern wind blows at an angle $\phi = 60^\circ$ to the equator with velocity $v = 15km/h$. Find the wind velocity $v'$ relative to the ship and the angle $\phi '$ between the equator and the wind direction in the reference frame fixed to the ship.

Solution: 1.6

If ${v_0}$ is the velocity vector of the ship and ${v_{wind}}$ is the velocity vector of the wind, then the velocity of the wind relative to the ship is simply $\left( {{v_{wind}} - {v_0}} \right)$.
This is indicated in the above figure. The magnitude $\left| {{v_{wind}} - {v_0}} \right|$ is then given by:
$= \sqrt {v_0^2 + v_{wind}^2 + 2{v_0}{v_{wind}}\cos \left( {180^\circ - \phi } \right)}$
$= \sqrt {v_0^2 + v_{wind}^2 + 2{v_0}{v_{wind}}\cos \phi }$
$= \sqrt {{{15}^2} + {{30}^2} + 2.15.30\cos 60^\circ }$
$= \sqrt {225 + 900 + 450}$
$= \sqrt {1575} \approx 40km/h$

The angle of the direction of the wind will be
$\tan \phi ' = \left( {\frac{{{v_{wind}}\sin \phi }}{{{v_0} + {v_{wind}}\cos \phi }}} \right)$
$\tan \phi ' = \left( {\frac{{15\sin 60^\circ }}{{30 + 15\cos 60^\circ }}} \right)$
$\phi ' \approx 19^\circ$