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 solve the problem, we need to find the average acceleration of a particle that changes its velocity from 10 m/s towards the east to 10 m/s towards the north over a time interval of 10 seconds.

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities:**
- The initial velocity \( \vec{u} \) is \( 10 \, \text{m/s} \) towards the east. We can represent this in vector form as:
\[ \vec{u} = 10 \hat{i} \, \text{m/s} \]
- The final velocity \( \vec{v} \) is \( 10 \, \text{m/s} \) towards the north. In vector form, this is:
\[ \vec{v} = 10 \hat{j} \, \text{m/s} \]

2. **Calculate Change in Velocity:**
- The change in velocity \( \Delta \vec{v} \) is given by:
\[ \Delta \vec{v} = \vec{v} - \vec{u} = (10 \hat{j} - 10 \hat{i}) \, \text{m/s} = -10 \hat{i} + 10 \hat{j} \, \text{m/s} \]

3. **Calculate Average Acceleration:**
- Average acceleration \( \vec{a}_{\text{avg}} \) is defined as the change in velocity divided by the time interval \( \Delta t \):
\[ \vec{a}_{\text{avg}} = \frac{\Delta \vec{v}}{\Delta t} = \frac{-10 \hat{i} + 10 \hat{j}}{10 \, \text{s}} = -1 \hat{i} + 1 \hat{j} \, \text{m/s}^2 \]

4. **Magnitude of Average Acceleration:**
- To find the magnitude of the average acceleration, we use the Pythagorean theorem:
\[ |\vec{a}_{\text{avg}}| = \sqrt{(-1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} \, \text{m/s}^2 \]

5. **Direction of Average Acceleration:**
- The direction of the average acceleration is towards the northwest since it has components in the negative x-direction (west) and positive y-direction (north).

6. **Final Answer:**
- Therefore, the average acceleration is:
\[ \vec{a}_{\text{avg}} = \sqrt{2} \, \text{m/s}^2 \text{ towards northwest} \]