A particle is moving eastwards with a velocity of ` 5 ms_(-1)`. In `10 seconds` the velocity changes to `5 ms^(-1)` northwards. The average acceleration in this time is

Options

A. Zero
B. `1//sqrt(2) ms^(-2)` towards north-west
C. `1//2 ms^(-2)` towards north-west
D. `1//2 ms^(-2)` towards north.

Explanation

To solve the problem, we need to find the average acceleration of a particle that changes its velocity from 5 m/s east to 5 m/s north over a time interval of 10 seconds.

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities:**
- Initial velocity (\( \vec{v_i} \)) = 5 m/s east
- Final velocity (\( \vec{v_f} \)) = 5 m/s north

2. **Express Velocities in Vector Form:**
- We can represent the initial velocity as \( \vec{v_i} = 5 \hat{i} \) m/s (where \( \hat{i} \) is the unit vector in the east direction).
- The final velocity can be represented as \( \vec{v_f} = 5 \hat{j} \) m/s (where \( \hat{j} \) is the unit vector in the north direction).

3. **Calculate Change in Velocity (\( \Delta \vec{v} \)):**
- The change in velocity 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:**
- The magnitude of the change in velocity can be calculated using 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 (\( \vec{a_{avg}} \)):**
- The average acceleration is given by:
\[ \vec{a_{avg}} = \frac{\Delta \vec{v}}{\Delta t} \]
- Here, \( \Delta t = 10 \) seconds.
- Thus,
\[ \vec{a_{avg}} = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2} \text{ m/s}^2 \]

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

### Final Answer:
The average acceleration is \( \frac{\sqrt{2}}{2} \) m/s² in the northwest direction.

---