A ship moves along the equator to the east with velocity `v_0=30km//hour`. The southeastern wind blows at an angle `varphi=60^@` to the equator with velocity `v=15km//hour`. Find the wind velocity `v^'` relative to the ship and the angle of `varphi^'` between the equator and the wind direction in the reference frame fixed to the ship.

Answer

`v^'=sqrt(v_0^2+v^2+2v_0vcosvarphi)~~40km` per hour, `varphi^'=19^@`

Explanation

We have
`vecv=vecv-vecv_0` (1)
From the vector diagram [of Eq. (1)] and using properties of triangle
`v^'=sqrt(v_0^2+v^2+2v_0vcosvarphi)=39.7km//hr` (2)
and `(v^')/(sin(pi-varphi))=(v)/(sintheta)` or, `sin theta=(vsin varphi)/(v^')`
or `theta=sin^-1((vsintheta)/(v^'))`
Using (2) and putting the values of v and d
`theta=19.1^@`