Question #269399
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?





1
Expert's answer
2021-11-21T17:31:00-0500

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

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

The current has velocity vector (relative to the Earth)

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

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

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

vS/E=4.0(cosθi+sinθj)+3.0i\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×cosθ+3.0=0cosθ=3.04.0θ=138.59°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.


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