Two forces `vec(F)_(1)=500N` due east and `vec(F)_(2)=250N` due north have their common initial point. `vec(F)_(2)-vec(F)_(1)` is

Options

A. `250sqrt(5)N,tan^(-1)(2) W of N`
B. `250 N, tan^(-1) (2) W` of N
C. Zero
D. `750N,tan^(-1)(3//4) N of W`

Explanation

To solve the problem of finding \(\vec{F}_2 - \vec{F}_1\), we will follow these steps:

### Step 1: Identify the Forces
We have two forces:
- \(\vec{F}_1 = 500 \, \text{N}\) due east
- \(\vec{F}_2 = 250 \, \text{N}\) due north

### Step 2: Represent the Forces as Vectors
In a Cartesian coordinate system:
- \(\vec{F}_1\) can be represented as \( (500, 0) \) (since it acts along the x-axis).
- \(\vec{F}_2\) can be represented as \( (0, 250) \) (since it acts along the y-axis).

### Step 3: Calculate \(\vec{F}_2 - \vec{F}_1\)
Now we will calculate \(\vec{F}_2 - \vec{F}_1\):
\[ \vec{F}_2 - \vec{F}_1 = (0, 250) - (500, 0) = (0 - 500, 250 - 0) = (-500, 250) \]

### Step 4: Calculate the Magnitude of the Resultant Vector
To find the magnitude of the resultant vector \(\vec{F}_2 - \vec{F}_1\):
\[ |\vec{F}_2 - \vec{F}_1| = \sqrt{(-500)^2 + (250)^2} \]
Calculating this:
\[ |\vec{F}_2 - \vec{F}_1| = \sqrt{250000 + 62500} = \sqrt{312500} = 250\sqrt{5} \, \text{N} \]

### Step 5: Calculate the Direction of the Resultant Vector
To find the direction (angle \(\theta\)) of the resultant vector, we can use the tangent function:
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{250}{500} = \frac{1}{2} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \]
This angle is measured from the negative x-axis (due west) towards the positive y-axis (due north).

### Final Answer
The resultant vector \(\vec{F}_2 - \vec{F}_1\) has a magnitude of \(250\sqrt{5} \, \text{N}\) and an angle of \(\tan^{-1}(0.5)\) north of west.

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