A particle is moving eastwards with velocity of `5 m//s`. In `10 sec` the velocity changes to `5 m//s` northwards. The average acceleration in this time is.

Options

A. `(1)/(sqrt(2))m//s^(2)` (North-West)
B. `(1)/(sqrt(2))m//s^(2)` (North-East)
C. `sqrt(2)m//s^(2)` (North-West)
D. `sqrt(2)m//s^(2)` (North-East)

Explanation

To solve the problem of finding the average acceleration of a particle moving from an eastward velocity to a northward velocity, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities**:
- The initial velocity \( \vec{V_i} \) is \( 5 \, \text{m/s} \) eastward. In vector form, this can be represented as:
\[ \vec{V_i} = 5 \hat{i} \, \text{m/s} \]
- The final velocity \( \vec{V_f} \) is \( 5 \, \text{m/s} \) northward. In vector form, this can be represented as:
\[ \vec{V_f} = 5 \hat{j} \, \text{m/s} \]

2. **Calculate Change in Velocity**:
- The change in velocity \( \Delta \vec{V} \) is given by:
\[ \Delta \vec{V} = \vec{V_f} - \vec{V_i} \]
- Substituting the values:
\[ \Delta \vec{V} = 5 \hat{j} - 5 \hat{i} = -5 \hat{i} + 5 \hat{j} \]

3. **Determine the Time Interval**:
- The time interval \( t \) is given as \( 10 \, \text{s} \).

4. **Calculate Average Acceleration**:
- The average acceleration \( \vec{a}_{avg} \) is defined as:
\[ \vec{a}_{avg} = \frac{\Delta \vec{V}}{t} \]
- Substituting the change in velocity and time:
\[ \vec{a}_{avg} = \frac{-5 \hat{i} + 5 \hat{j}}{10} \]
- This simplifies to:
\[ \vec{a}_{avg} = -0.5 \hat{i} + 0.5 \hat{j} \, \text{m/s}^2 \]

5. **Calculate the Magnitude of Average Acceleration**:
- The magnitude of the average acceleration can be calculated using the Pythagorean theorem:
\[ |\vec{a}_{avg}| = \sqrt{(-0.5)^2 + (0.5)^2} = \sqrt{0.25 + 0.25} = \sqrt{0.5} = \frac{1}{\sqrt{2}} \, \text{m/s}^2 \]

6. **Determine the Direction**:
- The average acceleration vector \( \vec{a}_{avg} = -0.5 \hat{i} + 0.5 \hat{j} \) indicates that the acceleration is directed towards the northwest.

### Final Answer:
The average acceleration of the particle is:
\[ \vec{a}_{avg} = \frac{1}{\sqrt{2}} \, \text{m/s}^2 \text{ in the northwest direction.} \]