To find the magnitude and direction of vector A with components $A_x = -5.50 \, \text{m}$ and $A_y = 8.50 \, \text{m}$, we can use the following steps:

Magnitude of the vector:

The magnitude $A$ of vector A is calculated using the Pythagorean theorem: $A = \sqrt{A_x^2 + A_y^2}$ Substituting the given values: $A = \sqrt{(-5.50)^2 + (8.50)^2} = \sqrt{30.25 + 72.25} = \sqrt{102.5}$ $A \approx 10.12 \, \text{m}$

Direction of the vector:

The direction $\theta$ of vector A (counterclockwise from the positive $x$-axis) is found using the tangent function: $\theta = \tan^{-1}\left(\frac{A_y}{A_x}\right)$ Substituting the values: $\theta = \tan^{-1}\left(\frac{8.50}{-5.50}\right)$ $\theta \approx \tan^{-1}(-1.545) \approx -57.5^\circ$

Since the vector lies in the second quadrant (because $A_x$ is negative and $A_y$ is positive), we add 180° to the angle: $\theta_{\text{final}} = 180^\circ - 57.5^\circ = 122.5^\circ$

Final Results:

• Magnitude: $10.12 \, \text{m}$
• Direction: $122.5^\circ$ counterclockwise from the positive $x$-axis.

Would you like further details or explanations?

Here are some related questions:

1. How do we calculate the components of a vector given its magnitude and direction?
2. What are the quadrant rules for angles and their adjustments?
3. What is the geometric interpretation of the tangent inverse function?
4. Can you explain the significance of vector direction in physical problems?
5. How do we add vectors in component form?

Tip: When using inverse trigonometric functions, always check which quadrant the vector is in to adjust the angle properly.