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In physics, vector decomposition is a technique used to break a vector into its component parts. Think of a vector as an arrow that has both a direction and a magnitude. When you want to understand how this vector behaves in a certain scenario, it helps to break it down into its vertical and horizontal components. This makes it easier to analyze and solve problems.

For instance, in the example provided, Ship B's velocity is a vector because it has both a speed and a direction. To analyze this, we decompose it into two separate components:

• The northward component, which shows how much of Ship B's velocity is directed towards the north.
• The eastward component, which indicates how much of Ship B's velocity is directed toward the east.

By expressing these components mathematically, you can use trigonometry to solve problems regarding directions, such as keeping Ship B always due north of Ship A.

Trigonometry is a branch of mathematics that studies the relationships between the angles and sides of triangles. In physics, it is especially useful to solve problems involving vectors and directional components.

In the context of the provided exercise, we applied trigonometry to determine the components of Ship B's velocity. Ship B is moving 30° east of north, which forms a right triangle when observed as vector decomposition.

The northward component of the velocity can be calculated using the cosine function: \[ v_n = v \cdot \cos(30^{\circ}) \]

And, the eastward component using the sine function:\[ v_e = v \cdot \sin(30^{\circ}) \]

Here:

• \(\cos(30^{\circ})\) gives us the portion of Ship B's velocity directed northward.
• \(\sin(30^{\circ})\) provides the part of Ship B's velocity directed eastward.

Use these calculations to set conditions for relative motion, keeping one object in a particular position relative to another.

To maintain the relative position of objects moving in space, understanding their directional components of velocity is crucial. Each component of velocity represents how fast an object is moving in a particular direction.

In the exercise, Ship A moves strictly east at 10 km/h. Meanwhile, Ship B, moving at an unspecified speed initially, needs its components adjusted to maintain it due north of Ship A. Specifically, the condition is:

• The eastward component of Ship B's velocity should equal Ship A's velocity.

Given the calculation process reveals:\[ v \cdot \sin(30^{\circ}) = 10 \]

Where:

• \(\sin(30^{\circ})\) is known to be 0.5, allowing us to conclude that Ship B must travel at 20 km/h to satisfy this condition.

This ensures that the eastward motion of Ship B equals that of Ship A, thus keeping Ship B directly north of Ship A, with its northward component allowing it to maintain this position dynamics.