An object moves from position (6,8) to (12,10) in the x-y plane. Magnitude and direction of displacement is

Options

A. `sqrt(40)` and `18.43^(@)`
B. `sqrt(40)` and `61.56^(@)`
C. 10 and `53^(@)`
D. `sqrt(244)` and `53^(@)`

Explanation

To find the magnitude and direction of the displacement of an object moving from position (6,8) to (12,10) in the x-y plane, we can follow these steps:

### Step 1: Define the Initial and Final Position Vectors
- The initial position vector \( \mathbf{r_1} \) is given by the coordinates (6, 8).
\[ \mathbf{r_1} = 6 \hat{i} + 8 \hat{j} \]
- The final position vector \( \mathbf{r_2} \) is given by the coordinates (12, 10).
\[ \mathbf{r_2} = 12 \hat{i} + 10 \hat{j} \]

### Step 2: Calculate the Displacement Vector
- The displacement vector \( \mathbf{r} \) is calculated by subtracting the initial position vector from the final position vector.
\[ \mathbf{r} = \mathbf{r_2} - \mathbf{r_1} = (12 \hat{i} + 10 \hat{j}) - (6 \hat{i} + 8 \hat{j}) \]
\[ \mathbf{r} = (12 - 6) \hat{i} + (10 - 8) \hat{j} = 6 \hat{i} + 2 \hat{j} \]

### Step 3: Calculate the Magnitude of the Displacement
- The magnitude of the displacement vector \( |\mathbf{r}| \) can be calculated using the Pythagorean theorem:
\[ |\mathbf{r}| = \sqrt{(6^2 + 2^2)} = \sqrt{36 + 4} = \sqrt{40} \]

### Step 4: Calculate the Direction of the Displacement
- To find the direction (angle \( \theta \)) of the displacement vector with respect to the x-axis, we can use the tangent function:
\[ \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{6} = \frac{1}{3} \]
- Therefore, we find \( \theta \) by taking the arctangent:
\[ \theta = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.43^\circ \]

### Conclusion
- The magnitude of the displacement is \( \sqrt{40} \) and the direction is approximately \( 18.43^\circ \) from the x-axis.

### Final Answer
- Magnitude: \( \sqrt{40} \)
- Direction: \( 18.43^\circ \)

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