To find the x and y components of the velocity vector, we can use the following equations:

• $V_x = V \cdot \cos(\theta)$
• $V_y = V \cdot \sin(\theta)$

where:

• $V = 23.7 \, \text{m/s}$ (the magnitude of the velocity vector),
• $\theta = 18^\circ$ (the angle south of west),
• The x-component (West-East) will be negative since it's directed to the west,
• The y-component (North-South) will be negative since it's directed to the south.

Now, let's calculate:

1. $V_x = 23.7 \cdot \cos(18^\circ)$
2. $V_y = 23.7 \cdot \sin(18^\circ)$

I'll compute these values for you.The x and y components of the velocity vector are:

• $x$-component: $-22.5 \, \text{m/s}$
• $y$-component: $-7.3 \, \text{m/s}$

Would you like further details or have any questions?

Here are some related questions:

1. How do you interpret a vector's direction in relation to the x and y components?
2. How would the components change if the angle were 18° North of West?
3. What is the magnitude of the vector if the components were given instead?
4. How do the signs of the components reflect the direction of the velocity?
5. How can the Pythagorean theorem verify the magnitude of a vector from its components?

Tip: Always remember that the trigonometric functions $\cos$ and $\sin$ help resolve vectors along horizontal and vertical axes.