To solve for the resultant magnitude and angle of the vectors $V_1$ and $V_2$, we first need to determine their components in Cartesian form and then combine them.

Step 1: Determine the Cartesian components of the vectors.

For $V_1$:

• $V_1$ x-component = 2.46 (directed West, hence negative in Cartesian form: -2.46)
• $V_1$ y-component = 0.43 (directed North, positive in Cartesian form: +0.43)

So the Cartesian form of $V_1$ is: $V_1 = (-2.46, +0.43)$

For $V_2$:

• $V_2$ x-component = 1.93 (directed East, positive in Cartesian form: +1.93)
• $V_2$ y-component = 2.30 (directed North, positive in Cartesian form: +2.30)

So the Cartesian form of $V_2$ is: $V_2 = (+1.93, +2.30)$

Step 2: Add the components of $V_1$ and $V_2$.

Resultant x-component:

$R_x = (-2.46) + (+1.93) = -0.53$

Resultant y-component:

$R_y = (+0.43) + (+2.30) = +2.73$

Step 3: Calculate the magnitude of the resultant vector.

The magnitude $R$ of the resultant vector is given by the Pythagorean theorem: $R = \sqrt{R_x^2 + R_y^2}$

Substitute the values: $R = \sqrt{(-0.53)^2 + (2.73)^2} = \sqrt{0.2809 + 7.4529} = \sqrt{7.7338}$

$R \approx 2.78$

Step 4: Calculate the direction (angle) of the resultant vector.

The angle $\theta$ with respect to the positive x-axis is given by: $\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

Substitute the values: $\theta = \tan^{-1}\left(\frac{2.73}{-0.53}\right)$

$\theta \approx \tan^{-1}(-5.15) \approx -79^\circ$

Since the resultant vector lies in the second quadrant (negative x, positive y), the actual angle is: $\theta = 180^\circ - 79^\circ = 101^\circ$

Final Answer:

• Magnitude of the resultant vector $R \approx 2.78$
• Direction of the resultant vector $\theta \approx 101^\circ$ (measured counterclockwise from the positive x-axis)

Would you like more details or clarifications on any step?

5 Related Questions:

1. How would the result change if the second vector had a negative y-component?
2. Can you calculate the unit vector of the resultant vector?
3. How can we decompose the resultant vector back into its x- and y-components geometrically?
4. How would the vector calculation change in 3D space?
5. Can you explain how vectors are represented in polar coordinates?

Tip:

Always be mindful of the quadrant when determining angles, as the arctangent function might return a negative angle that needs adjustment based on the vector's components.