A jogger travels a route that has two parts. The first is a displacement `vecA` of `2.50 km` due south, and the second involves a displacement `vecB` that points due east, Suppose that `vecA-vecB` had a magnitude of `3.75km.` What then would be the magnitude of `vecB,` and what is the direction of `vecA-vecB` relative to due south ?

Options

A. `4.5 km,63^(@)` east of south
B. `4.5 km, 56 ^(@)` west of south
C. `3.7 km , 56 ^(@)` west of south
D. `2.8 km, 48^(@)` west of south

Answer

To solve the problem, we need to find the magnitude of vector \( \vec{B} \) and the direction of \( \vec{A} - \vec{B} \) relative to due south. We will use the Pythagorean theorem and some trigonometric functions to find the required values.

### Step-by-Step Solution:

1. **Identify the vectors and their magnitudes**:
- The displacement \( \vec{A} \) is \( 2.50 \, \text{km} \) due south.
- The displacement \( \vec{B} \) is due east, and we need to find its magnitude.

2. **Set up the equation for \( \vec{A} - \vec{B} \)**:
- We know that the magnitude of \( \vec{A} - \vec{B} \) is given as \( 3.75 \, \text{km} \).
- We can express this mathematically as:
\[ |\vec{A} - \vec{B}| = 3.75 \, \text{km} \]

3. **Use the Pythagorean theorem**:
- Since \( \vec{A} \) is directed south and \( \vec{B} \) is directed east, we can treat this as a right triangle where:
- \( |\vec{A}| = 2.50 \, \text{km} \) (south)
- \( |\vec{B}| = b \, \text{km} \) (east)
- The resultant vector \( |\vec{A} - \vec{B}| = 3.75 \, \text{km} \)

- According to the Pythagorean theorem:
\[ |\vec{A} - \vec{B}|^2 = |\vec{A}|^2 + |\vec{B}|^2 \]
- Substituting the known values:
\[ (3.75)^2 = (2.50)^2 + b^2 \]

4. **Calculate the squares**:
- Calculate \( (3.75)^2 \):
\[ 3.75^2 = 14.0625 \]
- Calculate \( (2.50)^2 \):
\[ 2.50^2 = 6.25 \]

5. **Set up the equation**:
- Now, substituting these values into the equation:
\[ 14.0625 = 6.25 + b^2 \]

6. **Solve for \( b^2 \)**:
- Rearranging gives:
\[ b^2 = 14.0625 - 6.25 = 7.8125 \]

7. **Take the square root to find \( b \)**:
- Thus, the magnitude of \( \vec{B} \) is:
\[ b = \sqrt{7.8125} \approx 2.79 \, \text{km} \]

8. **Determine the direction of \( \vec{A} - \vec{B} \)**:
- We can find the angle \( \theta \) relative to due south using the tangent function:
\[ \tan(\theta) = \frac{|\vec{B}|}{|\vec{A}|} = \frac{b}{|\vec{A}|} = \frac{2.79}{2.50} \]

9. **Calculate \( \theta \)**:
- Using the inverse tangent function:
\[ \theta = \tan^{-1}\left(\frac{2.79}{2.50}\right) \approx 48.37^\circ \]

### Final Answers:
- The magnitude of \( \vec{B} \) is approximately \( 2.79 \, \text{km} \).
- The direction of \( \vec{A} - \vec{B} \) is \( 48.37^\circ \) east of due south.