Question

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 $v=15km/h$. The angle ${\phi }^{\prime }$ between the equator and the wind direction in the reference frame fixed to the ship is (in degrees)

Answer 1

We have
${\overrightarrow{v}}^{\prime }=\overrightarrow{v}-\overrightarrow{v_0}$ (1)
From the vector diagram [of equation (1)] and using properties of triangle
$v^{\prime }=\sqrt{v_0^2+v^2+2v_{0}v\cos \phi }=39.7km/h$ (2)
and $\frac{v^{\prime }}{\sin (\pi -\phi )}=\frac{v}{\sin \theta }$ or, $\sin \theta =\frac{v\sin \phi }{v^{\prime }}$
or $\theta ={\sin }^{-1}\left(\frac{v\sin \phi }{v^{\prime }}\right)$
Using (2) and putting the values of $v$ and $d$
$\theta ={19.1}^{\circ }$