Question

A ship sets sail Cape Ann, Massachusetts, heading due north at 7.00 m/s relative to the water. The local ocean current is 1.50 m/s in a direction 40.0o north of east. What is the velocity of the ship relative to the Earth

Answer 1

Answer:

$\vec{v} = 1.15 \hat{i} + 7.964 \hat{j}$

$|v| = 8.05 m/s$

$\alpha = 81.78^o$ 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:

$\vec{v_o} = 1.5cos40^0 \hat{i} + 1.5sin40^0\hat{j}$

$\vec{v_o} = 1.15 \hat{i} + 0.964 \hat{j}$

The velocity vector for ship with respect to water is

$\vec{v_s} = 7\hat{j}$

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

$\vec{v} = \vec{v_s} + \vec{v_o} = 7\hat{j} + 1.15 \hat{i} + 0.964 \hat{j} = 1.15 \hat{i} + 7.964 \hat{j}$

This vector would have a magnitude and direction of:

$|v| = \sqrt{v_j^2 + v_i^2} = \sqrt{7.964^2 + 1.15^2} = \sqrt{63.425296 + 1.3225} = \sqrt{64.747796} = 8.05$

$tan\theta = \frac{v_j}{v_i} = \frac{7.964}{1.15} = 6.93$

$\alpha = tan^{-1}6.93 = 1.43 rad \approx 81.78^o$ north of east