A particle is moving eastwards with a velocity `5ms^(-1)`, changes its direction northwards in `10` seconds and moves with same magnitude of velocity. The average acceleration is

Options

A. zero
B. `1/sqrt2ms^(-2)` towards `N-E`
C. `1/sqrt2ms^(-2)` towards `S-E`
D. `1/sqrt2ms^(-2)` towards `N-W`

Explanation

To solve the problem of finding the average acceleration of a particle that changes its direction from east to north while maintaining the same speed, 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.
- The final velocity \( \vec{V_f} \) is \( 5 \, \text{m/s} \) northward.

2. **Represent Velocities as Vectors**:
- In vector form, we can represent the initial velocity as:
\[ \vec{V_i} = 5 \hat{i} \, \text{m/s} \]
- The final velocity can be represented as:
\[ \vec{V_f} = 5 \hat{j} \, \text{m/s} \]

3. **Calculate Change in Velocity**:
- The change in velocity \( \Delta \vec{V} \) is given by:
\[ \Delta \vec{V} = \vec{V_f} - \vec{V_i} = 5 \hat{j} - 5 \hat{i} \]
- This can be expressed as:
\[ \Delta \vec{V} = -5 \hat{i} + 5 \hat{j} \]

4. **Magnitude of Change in Velocity**:
- To find the magnitude of the change in velocity, we use the Pythagorean theorem:
\[ |\Delta \vec{V}| = \sqrt{(-5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \, \text{m/s} \]

5. **Calculate Average Acceleration**:
- The average acceleration \( \vec{a} \) is defined as the change in velocity divided by the time interval:
\[ \vec{a} = \frac{\Delta \vec{V}}{\Delta t} = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2} \, \text{m/s}^2 \]

6. **Direction of Average Acceleration**:
- The direction of the average acceleration is towards the northwest since the change in velocity is from east to north.

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