To find the x-component $V_x$ of the vector $\vec{V}$, we use the following formula:

$V_x = |\vec{V}| \cdot \cos(\theta)$

Given:

• $|\vec{V}| = 9.77 \, \text{m}$
• $\theta = 154^\circ$

Now, calculate:

$V_x = 9.77 \cdot \cos(154^\circ)$

First, evaluate $\cos(154^\circ)$:

$\cos(154^\circ) \approx -0.8988$

Thus:

$V_x = 9.77 \cdot (-0.8988) \approx -8.78 \, \text{m}$

Therefore, the x-component of the vector is approximately:

$V_x \approx -8.78 \, \text{m}$

Would you like more details or further clarification on this?

Here are 5 related questions to extend your understanding:

1. How is the x-component affected by the angle's reference quadrant?
2. What is the y-component of the same vector?
3. How would the x-component change if the angle were in the first quadrant?
4. Can the magnitude of a vector ever be negative?
5. What is the relationship between the x-component and the overall direction of the vector?

Tip: The cosine of angles greater than 90 degrees is always negative because they lie in the second or third quadrants.