The image you uploaded depicts a velocity vector $\vec{v} = 100 \, \text{m/s}$, directed due south. However, there is also a coordinate system where the $x$-axis is rotated by $30^\circ$ counterclockwise from the east-west direction.

To find the $x$-component of the velocity vector in this rotated coordinate system, we can use the following steps:

1. The velocity is directed due south, which corresponds to $270^\circ$ from the positive $x$-axis (east) in the standard coordinate system.

2. Since the $x$-axis is rotated $30^\circ$ counterclockwise from the east, we need to adjust the angle accordingly: $\theta' = 270^\circ - 30^\circ = 240^\circ$ This is the angle the velocity vector makes with the rotated $x$-axis.

3. Now, to find the $x$-component of the velocity in this rotated system, we use the cosine function: $v_x = v \cdot \cos(\theta')$ Substituting the known values: $v_x = 100 \, \text{m/s} \cdot \cos(240^\circ)$ $v_x = 100 \, \text{m/s} \cdot (-0.5)$ $v_x = -50 \, \text{m/s}$

Thus, the $x$-component of the velocity vector is $-50 \, \text{m/s}$.

Would you like more details on this calculation or any clarifications?

Here are some follow-up questions for deeper understanding:

1. How would you calculate the $y$-component of the velocity vector in this scenario?
2. What would the angle need to be if the velocity were directed eastward instead of south?
3. How would the calculation change if the vector was not aligned with a cardinal direction (i.e., not due south)?
4. How do you find the magnitude of a vector from its components?
5. How would the scenario change if the coordinate system was rotated by $45^\circ$ instead of $30^\circ$?

Tip: When working with rotated coordinate systems, always adjust the angles carefully relative to the new axes for correct component calculation.