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Question

Physics Question on Relative Velocity

Consider a ship traveling due east along the equator with velocity v0v_0. If southeastern wind blows at an angle of ϕ' \phi ' to the equator with velocity v'v', the wind velocity relative to the ship vv' and the angle between the equator and the wind direction in the reference frame fixed to the ship are

A

v=v02+v2+v0vcosϕ,sin1(vsinϕv)v' = \sqrt{v_0^2 + v^2 + v_0 v \cos \phi} , \sin^{-1} \left(\frac{v \sin \phi}{v'} \right)

B

v=v02+v2+v0vsinϕ,cos1(vcosϕv)v' = \sqrt{v_0^2 + v^2 + v_0 v \sin \phi} , \cos^{-1} \left(\frac{v \cos \phi}{v'} \right)

C

v=v02+v2+v0vcosϕ,sin1(vv)v' = \sqrt{v_0^2 + v^2 + v_0 v \cos \phi} , \sin^{-1} \left(\frac{v}{v'} \right)

D

v=v02+v2+v0vcosϕ,cos1(vv)v' = \sqrt{v_0^2 + v^2 + v_0 v \cos \phi} , \cos^{-1} \left(\frac{v }{v'} \right)

Answer

v=v02+v2+v0vcosϕ,sin1(vsinϕv)v' = \sqrt{v_0^2 + v^2 + v_0 v \cos \phi} , \sin^{-1} \left(\frac{v \sin \phi}{v'} \right)

Explanation

Solution

Answer (a) v=v02+v2+v0vcosϕ,sin1(vsinϕv)v' = \sqrt{v_0^2 + v^2 + v_0 v \cos \phi} , \sin^{-1} \left(\frac{v \sin \phi}{v'} \right)