Answer to Question #269399 in Mechanics | Relativity for pepito

Question

A long distance swimmer is able to swim in calm water at 4.0 km / h. She wants to try swimming from Port Angeles, Washington, directly north to Victoria, British Columbia, a distance of 50 km. An ocean current crosses the Strait of Juan de Fuca from west to east at 3.0 km / h. In what direction (relative to North) must you swim to cross a straight line between the two cities?

Answer 1

Let $\theta$ be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

$\vec{v}_{S/W} = 4.0 (cos \theta \vec{i} + sin \theta \vec{j})$

The current has velocity vector (relative to the Earth)

$\vec{v}_{W/E} = 3.0 \vec{i}$

The swimmer's resultant velocity (her velocity relative to the Earth) is then

$\vec{v}_{S/E} = \vec{v}_{S/W} + \vec{v}_{W/E} \\$

$\vec{v}_{S/E} = 4.0 (cos \theta \vec{i} + sin \theta \vec{j}) + 3.0 \vec{i}$

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

$4.0 \times cos \theta + 3.0 = 0 \\ cos \theta = -\frac{3.0}{4.0} \\ \theta = 138.59°$

which is approximately 41º west of north.