Velocity of ship relative to the water is given as
[tex]v_{r} = 7m/s[/tex] North
velocity of water is given as
[tex]v_w = 1.50 m/s[/tex]
in direction 40 degree North of East
now the resultant speed with respect to earth is given as
[tex]v = \vec v_r + \vec v_w[/tex]
now the resultant magnitude is given as
[tex]v = \sqrt{v_r^2 + v_w^2 + 2v_r v_w cos\theta}[/tex]
here angle between relative speed and water speed is 50 degree
[tex]v = \sqrt{7^2 + 1.5^2 + 2*1.5*7* cos50}[/tex]
[tex]v = 8.05 m/s[/tex]
now in order to find its direction we have
[tex]\theta = tan^{-1}\frac{v_wsin50}{v_wcos50 + v_r}[/tex]
[tex]\theta = tan^{-1}\frac{1.5sin50}{1.5cos50 + 7}[/tex]
[tex]\theta = 8.2 degree[/tex]
so it will move with speed 8.05 m/s in the direction 8.2 degree East of North
