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

Options

A. zero
B. `1//sqrt (2) m//s^(2)` towards no rth - west
C. `1//sqrt (2) m//s^(2)` towards no rth - east
D. `1//2 m//s^(2)` towards no rth - west

Explanation

To solve the problem of finding the average acceleration of a particle that changes its velocity from 5 m/s eastwards to 5 m/s northwards in 10 seconds, we can follow these steps:

### Step 1: Identify Initial and Final Velocities
- The initial velocity \( \vec{V_i} \) is \( 5 \, \text{m/s} \) east, which can be represented as \( \vec{V_i} = 5 \hat{i} \, \text{m/s} \).
- The final velocity \( \vec{V_f} \) is \( 5 \, \text{m/s} \) north, represented as \( \vec{V_f} = 5 \hat{j} \, \text{m/s} \).

### Step 2: Calculate the Change in Velocity
- The change in velocity \( \Delta \vec{V} \) is given by:
\[ \Delta \vec{V} = \vec{V_f} - \vec{V_i} = (5 \hat{j} - 5 \hat{i}) \, \text{m/s} \]
- This results in:
\[ \Delta \vec{V} = -5 \hat{i} + 5 \hat{j} \, \text{m/s} \]

### Step 3: Calculate the Magnitude of the Change in Velocity
- The magnitude of \( \Delta \vec{V} \) can be calculated using the Pythagorean theorem:
\[ |\Delta \vec{V}| = \sqrt{(-5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \, \text{m/s} \]

### Step 4: Calculate the Average Acceleration
- The average acceleration \( \vec{a} \) is defined as the change in velocity divided by the time interval \( T \):
\[ \vec{a} = \frac{\Delta \vec{V}}{T} = \frac{5\sqrt{2} \, \text{m/s}}{10 \, \text{s}} = \frac{\sqrt{2}}{2} \, \text{m/s}^2 \]

### Step 5: Determine the Direction of the Average Acceleration
- The direction of the change in velocity \( \Delta \vec{V} \) is in the northwest direction since it combines both negative east and positive north components.

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

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