If a vector has an x-component of -250units and a y-component of 40.0 units, then the magnitude and direction of this vector is

Options

A. `5sqrt(89)` units, `sin^(-1)((5)/sqrt(89))` with +ve-x-axis
B. `5sqrt(89)` units, `cos^(-1)((5)/sqrt(89))` with -ve-x-axis
C. 45 units, `cos^(-1)((-5)/(9))` with x-axis
D. 45 units, `sin^(-1)((-5)/(9))` with x-axis

Explanation

To find the magnitude and direction of a vector given its x and y components, we can follow these steps:

### Step 1: Identify the components
The x-component of the vector is given as:
- \( p_x = -250 \) units (which is in the negative x-direction)
The y-component of the vector is given as:
- \( p_y = 40.0 \) units (which is in the positive y-direction)

### Step 2: Calculate the magnitude of the vector
The magnitude \( |p| \) of the vector can be calculated using the Pythagorean theorem:
\[ |p| = \sqrt{p_x^2 + p_y^2} \]
Substituting the values:
\[ |p| = \sqrt{(-250)^2 + (40)^2} \]
Calculating the squares:
\[ |p| = \sqrt{62500 + 1600} \]
\[ |p| = \sqrt{64100} \]
Now, simplifying:
\[ |p| \approx 253.0 \text{ units} \]

### Step 3: Calculate the direction of the vector
To find the direction (angle \( \theta \)) of the vector relative to the negative x-axis, we can use the tangent function:
\[ \tan(\theta) = \frac{p_y}{|p_x|} \]
Substituting the values:
\[ \tan(\theta) = \frac{40}{250} \]
Calculating:
\[ \tan(\theta) = \frac{8}{50} = \frac{4}{25} \]
Now, we find the angle \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{4}{25}\right) \]

### Step 4: Determine the angle
Using a calculator:
\[ \theta \approx 9.0^\circ \]
Since the vector is in the second quadrant (negative x and positive y), the angle with respect to the negative x-axis is:
\[ \theta \approx 180^\circ - 9.0^\circ = 171.0^\circ \]

### Final Result
- **Magnitude of the vector**: \( \approx 253.0 \) units
- **Direction of the vector**: \( \approx 171.0^\circ \) from the positive x-axis or \( 9.0^\circ \) from the negative x-axis in the clockwise direction.

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