Aship sets sail from rotterdam, the netherlands, heading due north at 6.55 m/s relative to the water. the local ocean current is 1.46 m/s in a direction 36.0° north of east.

(a) in what direction (in degrees west of north) would the ship have to travel in order to have a velocity straight north relative to the earth, assuming its speed relative to the water remains 6.55 m/s? ° west of north?

(b)
what would its speed (in m/s) be relative to the earth? m/s?

A) Consider a 2D Cartesian coordinate system representing the ocean's surface (just within the ship's local area) with the ship located at the origin. The x-axis runs west to east and the y-axis runs north to south. It helps to imagine a bird's-eye view looking directly down upon the ship and the ocean's surface.

The ship's resultant velocity can be thought of as the vector sum of the vectors representing the ocean current and the ship's velocity relative to the water. Imagine drawing the ocean current vector with the tail starting at the origin, 1.46 units long, and oriented such that it forms a 36° angle with respect to the +x-axis (36° north of east). The tip of this vector will wind up somewhere within the first quadrant of the coordinate system. The ocean current vector has a positive x-component and positive y-component. We want the ship's resultant velocity vector to have no x-component and a positive y-component (traveling straight north). Therefore we want the vector representing the ship's velocity relative to the water to have a negative x-component and a positive y-component such that the two vectors' x-components will cancel out and the y-components will add upon each other. Visually, the vector representing the ship's velocity relative to the water should start at the origin and its tip should wind up somewhere within the second quadrant.

Let θ be the angle that the vector representing the ship's velocity relative to the water forms with respect to the +y-axis (north).

Since the angle giving the direction of the ocean current vector is measured off the +x-axis, the x-component of the ocean current vector is given by:

1.46cos(36.0°)

Since the angle giving the direction of the vector representing the ship's velocity relative to the water is measured off the +y-axis, the x-component of this vector is given by:

-6.55sin(θ)

The sum of the x-components of the vectors must sum to 0:

1.46cos(36.0°) - 6.55sin(θ) = 0

Solve for θ

6.55sin(θ) = 1.46cos(36.0°)

6.55sin(θ) = 1.18

sin(θ) = 0.18

θ = 10.4°

Therefore the ship must travel at 6.55m/s at a direction 10.4° west of north.

B) Remember that the ship's resultant velocity is the sum of the vectors representing the ocean current and the ship's velocity relative to the water, so the resultant velocity's y-component is the sum of the y-components of the two vectors' y-components.

The y-component of the ocean current vector is given by:

1.46sin(36.0°)

The y-component of the vector representing the ship's velocity relative to the water is given by:

6.55cos(θ)

Sum up these y-components:

1.46sin(36.0°) + 6.55cos(θ)

We already solved θ = 10.4, so substitute this value into the expression

1.46sin(36.0°) + 6.55cos(10.4°)

The result is 7.30

Therefore the ship's resultant velocity will be 7.30m/s north relative to the earth.