A particle is moing with a velocity of `10m//s` towards east. After 10s its velocity changes to `10m//s` towards north. Its average acceleration is:-

Options

A. zero
B. `sqrt(2) m//s^(2)` towards N-W
C. `(2)/sqrt(2) m//s^(2)` towards N-W
D. `(1)/sqrt(2)m//s^(2)` towards N-W

Explanation

To find the average acceleration of the particle, we need to determine the change in velocity and then divide it by the time interval. Let's break down the problem step by step.

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities:**
- Initial velocity (\(\vec{v_i}\)): \(10 \, \text{m/s}\) towards east. In vector form, this is \(10 \, \hat{i}\).
- Final velocity (\(\vec{v_f}\)): \(10 \, \text{m/s}\) towards north. In vector form, this is \(10 \, \hat{j}\).

2. **Calculate the Change in Velocity (\(\Delta \vec{v}\)):**
\[ \Delta \vec{v} = \vec{v_f} - \vec{v_i} \]
Substituting the given values:
\[ \Delta \vec{v} = 10 \, \hat{j} - 10 \, \hat{i} \]
So,
\[ \Delta \vec{v} = -10 \, \hat{i} + 10 \, \hat{j} \]

3. **Determine the Time Interval (\(\Delta t\)):**
\[ \Delta t = 10 \, \text{s} \]

4. **Calculate the Average Acceleration (\(\vec{a_{avg}}\)):**
\[ \vec{a_{avg}} = \frac{\Delta \vec{v}}{\Delta t} \]
Substituting the values:
\[ \vec{a_{avg}} = \frac{-10 \, \hat{i} + 10 \, \hat{j}}{10} \]
Simplifying:
\[ \vec{a_{avg}} = -1 \, \hat{i} + 1 \, \hat{j} \]

5. **Find the Magnitude of the Average Acceleration:**
The magnitude of \(\vec{a_{avg}}\) is given by:
\[ |\vec{a_{avg}}| = \sqrt{(-1)^2 + (1)^2} \]
\[ |\vec{a_{avg}}| = \sqrt{1 + 1} \]
\[ |\vec{a_{avg}}| = \sqrt{2} \]
\[ |\vec{a_{avg}}| = \sqrt{2} \, \text{m/s}^2 \]

6. **Determine the Direction of the Average Acceleration:**
The direction is given by the vector components \(-1 \, \hat{i} + 1 \, \hat{j}\), which points towards the northwest.

### Final Answer:
The average acceleration of the particle is \(\sqrt{2} \, \text{m/s}^2\) towards the northwest.