Consider a ship traveling due east along the equator with velocity v0. If southeastern wind blows at an angle of ' φ ' to the equator with velocity 'v', the wind velocity relative to the ship v' and the angle between the equator and the wind direction in the reference frame fixed to the ship are

Question

Consider a ship traveling due east along the equator with velocity v0. If southeastern wind blows at an angle of ' φ ' to the equator with velocity 'v', the wind velocity relative to the ship v' and the angle between the equator and the wind direction in the reference frame fixed to the ship are (A) v' = √v02 + v2 + v0 v cos φ , sin-1 ((v sin φ/v') ) (B) v' = √v02 + v2 + v0 v sin φ , cos-1 ((v cos φ/v') ) (C) v' = √v02 + v2 + v0 v cos φ , sin-1 ((v/v') ) (D) v' = √v02 + v2 + v0 v cos φ , cos-1 ((v /v') )

Tardigrade Admin · Accepted Answer

Correct answer is (a) v' = √v02 + v2 + v0 v cos φ , sin-1 ((v sin φ/v') )

$v^{'} = v_{0}^{2} + v^{2} + v_{0} v cos ϕ, sin^{- 1} (\frac{v s i n ϕ}{v ^{'}})$

$v^{'} = v_{0}^{2} + v^{2} + v_{0} v sin ϕ, cos^{- 1} (\frac{v c o s ϕ}{v ^{'}})$

$v^{'} = v_{0}^{2} + v^{2} + v_{0} v cos ϕ, sin^{- 1} (\frac{v}{v ^{'}})$

$v^{'} = v_{0}^{2} + v^{2} + v_{0} v cos ϕ, cos^{- 1} (\frac{v}{v ^{'}})$

Correct answer is (a) $v^{'} = v_{0}^{2} + v^{2} + v_{0} v cos ϕ, sin^{- 1} (\frac{v s i n ϕ}{v ^{'}})$

v′=v02​+v2+v0​vcosϕ​,sin−1(v′vsinϕ​)

v′=v02​+v2+v0​vsinϕ​,cos−1(v′vcosϕ​)

v′=v02​+v2+v0​vcosϕ​,sin−1(v′v​)

v′=v02​+v2+v0​vcosϕ​,cos−1(v′v​)

Correct answer is (a) v′=v02​+v2+v0​vcosϕ​,sin−1(v′vsinϕ​)

$v^{'} = v_{0}^{2} + v^{2} + v_{0} v cos ϕ, sin^{- 1} (\frac{v s i n ϕ}{v ^{'}})$

$v^{'} = v_{0}^{2} + v^{2} + v_{0} v sin ϕ, cos^{- 1} (\frac{v c o s ϕ}{v ^{'}})$

$v^{'} = v_{0}^{2} + v^{2} + v_{0} v cos ϕ, sin^{- 1} (\frac{v}{v ^{'}})$

$v^{'} = v_{0}^{2} + v^{2} + v_{0} v cos ϕ, cos^{- 1} (\frac{v}{v ^{'}})$

Correct answer is (a) $v^{'} = v_{0}^{2} + v^{2} + v_{0} v cos ϕ, sin^{- 1} (\frac{v s i n ϕ}{v ^{'}})$