The problem describes a particle with an initial velocity of 12.0 m/s, directed 60.0 degrees above the negative x-axis. You're asked to determine the x and y components of this velocity.

We can decompose the velocity vector into its components using trigonometric functions:

1. For Part A (the x-component, $v_x$):
   • Since the angle is measured above the negative x-axis, the x-component will be negative.
   • $v_x = - v \cdot \cos(\theta)$
   • Given $v = 12.0 \, \text{m/s}$ and $\theta = 60.0^\circ$, we can calculate:

$v_x = - 12.0 \cdot \cos(60.0^\circ)$

$v_x = - 12.0 \cdot 0.5 = -6.0 \, \text{m/s}$

2. For Part B (the y-component, $v_y$):
   • The y-component is positive because the velocity is above the x-axis.
   • $v_y = v \cdot \sin(\theta)$
   • Using $v = 12.0 \, \text{m/s}$ and $\theta = 60.0^\circ$:

$v_y = 12.0 \cdot \sin(60.0^\circ)$

$v_y = 12.0 \cdot \frac{\sqrt{3}}{2} \approx 12.0 \cdot 0.866 = 10.4 \, \text{m/s}$

Summary:

• Part A (x-component): $v_x = -6.0 \, \text{m/s}$
• Part B (y-component): $v_y = 10.4 \, \text{m/s}$

Would you like more details on this process or further clarifications?

Here are 5 follow-up questions:

1. What would change if the angle were 30 degrees instead of 60 degrees?
2. How can the components be calculated if the particle were moving in 3D space?
3. What happens to the velocity components if the velocity magnitude increases?
4. What is the general approach to solving vector decomposition problems in physics?
5. Can you express the velocity as a vector using its components?

Tip: Always remember to check the sign of each component based on the direction of the vector.