Answer :
Answer:
[tex]\vec{v} = 1.15 \hat{i} + 7.964 \hat{j}[/tex]
[tex]|v| = 8.05 m/s[/tex]
[tex]\alpha = 81.78^o[/tex] north of east
Explanation:
Let i and j be the unit components of east and north directions, respectively. We can calculate the velocity components of ocean current:
[tex]\vec{v_o} = 1.5cos40^0 \hat{i} + 1.5sin40^0\hat{j}[/tex]
[tex]\vec{v_o} = 1.15 \hat{i} + 0.964 \hat{j}[/tex]
The velocity vector for ship with respect to water is
[tex]\vec{v_s} = 7\hat{j}[/tex]
So the velocity vector of the ship with respect to Earth is the sum vector of velocity of ship with respect to water and velocity of water with respect to Earth
[tex]\vec{v} = \vec{v_s} + \vec{v_o} = 7\hat{j} + 1.15 \hat{i} + 0.964 \hat{j} = 1.15 \hat{i} + 7.964 \hat{j}[/tex]
This vector would have a magnitude and direction of:
[tex]|v| = \sqrt{v_j^2 + v_i^2} = \sqrt{7.964^2 + 1.15^2} = \sqrt{63.425296 + 1.3225} = \sqrt{64.747796} = 8.05[/tex]
[tex]tan\theta = \frac{v_j}{v_i} = \frac{7.964}{1.15} = 6.93[/tex]
[tex]\alpha = tan^{-1}6.93 = 1.43 rad \approx 81.78^o[/tex] north of east