To solve for the magnitude and direction of a vector given its components $V_x$ and $V_y$, we can follow these steps:

1. Break down the components:

• $V_x = -4.5 \cos 60^\circ + 3.0$
• $V_y = 4.5 \sin 60^\circ$

Now, let's calculate these values.

Step 1: Calculate $V_x$

$V_x = -4.5 \cos 60^\circ + 3.0$

Since $\cos 60^\circ = 0.5$: $V_x = -4.5 \times 0.5 + 3.0 = -2.25 + 3.0 = 0.75$

Step 2: Calculate $V_y$

$V_y = 4.5 \sin 60^\circ$

Since $\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866$: $V_y = 4.5 \times 0.866 = 3.897$

2. Find the magnitude of the vector:

The magnitude of the vector $V$ is given by: $|V| = \sqrt{V_x^2 + V_y^2}$ Substituting the values of $V_x$ and $V_y$: $|V| = \sqrt{(0.75)^2 + (3.897)^2} = \sqrt{0.5625 + 15.18} = \sqrt{15.7425} \approx 3.969$

3. Find the direction of the vector:

The direction $\theta$ of the vector is given by: $\theta = \tan^{-1}\left(\frac{V_y}{V_x}\right)$ Substituting the values of $V_x$ and $V_y$: $\theta = \tan^{-1}\left(\frac{3.897}{0.75}\right) = \tan^{-1}(5.196) \approx 79.04^\circ$

Final Answer:

• Magnitude: $|V| \approx 3.97$
• Direction: $\theta \approx 79.04^\circ$ (above the positive x-axis)

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Would you like more details or have any questions?

Here are 5 related questions:

1. How does vector addition affect magnitude and direction?
2. Can you find the resultant of two vectors with different angles?
3. How would the vector change if $V_x$ were negative?
4. What happens to the magnitude if the vector's components are scaled by a factor?
5. How is the direction of a vector determined in different quadrants?

Tip: The magnitude of a vector always remains positive, while its direction depends on the signs of $V_x$ and $V_y$.