To calculate the magnitude of the vector sum of vector A and vector B, we proceed as follows:

Step 1: Resolve each vector into components

• Vector A has a magnitude of $41.3 \, \text{m}$ and a direction of $0^\circ$.

  • $A_x = 41.3 \cdot \cos(0^\circ) = 41.3 \, \text{m}$
  • $A_y = 41.3 \cdot \sin(0^\circ) = 0.0 \, \text{m}$

• Vector B has a magnitude of $66.1 \, \text{m}$ and a direction of $98^\circ$.

  • $B_x = 66.1 \cdot \cos(98^\circ)$
  • $B_y = 66.1 \cdot \sin(98^\circ)$

Step 2: Perform component calculations

Using a calculator:

• $\cos(98^\circ) \approx -0.1392$
• $\sin(98^\circ) \approx 0.9903$

Thus:

• $B_x = 66.1 \cdot (-0.1392) \approx -9.2 \, \text{m}$
• $B_y = 66.1 \cdot (0.9903) \approx 65.4 \, \text{m}$

Step 3: Add components

The resultant vector's components are:

• $R_x = A_x + B_x = 41.3 - 9.2 = 32.1 \, \text{m}$
• $R_y = A_y + B_y = 0.0 + 65.4 = 65.4 \, \text{m}$

Step 4: Compute the magnitude

The magnitude of the resultant vector is given by: $R = \sqrt{R_x^2 + R_y^2}$ Substitute the values: $R = \sqrt{(32.1)^2 + (65.4)^2}$ $R = \sqrt{1030.4 + 4287.2} = \sqrt{5317.6} \approx 72.9 \, \text{m}$

Final Answer:

The magnitude of the vector sum is approximately 72.9 meters.

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Would you like a detailed explanation of each step or further clarifications?
Here are some related questions to consider:

1. What is the angle of the resultant vector relative to the positive x-axis?
2. How do you resolve vectors into components in general?
3. How does the cosine and sine of angles affect vector calculations?
4. Can this method be applied to three-dimensional vectors?
5. What are common applications of vector addition in physics?

Tip: Always double-check your angle measurements (degrees or radians) when using trigonometric functions in calculations!