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. Zero
B. `(1)/(sqrt(2)) m//s^2` toward north-west
C. `(1)/(sqrt(2)) m//s^2` toward north-east
D. `(1)/(2) m//s^2` toward north-west

Explanation

To find the average acceleration of the particle, we can follow these steps:

### Step 1: Identify the Initial and Final Velocities
- The particle is initially moving eastward with a velocity of \(5 \, \text{m/s}\). We can represent this as a vector:
\[ \vec{v_1} = 5 \hat{i} \, \text{m/s} \]
- After \(10 \, \text{s}\), the particle is moving northward with a velocity of \(5 \, \text{m/s}\):
\[ \vec{v_2} = 5 \hat{j} \, \text{m/s} \]

### Step 2: Calculate the Change in Velocity
- The change in velocity (\(\Delta \vec{v}\)) can be calculated as:
\[ \Delta \vec{v} = \vec{v_2} - \vec{v_1} = 5 \hat{j} - 5 \hat{i} \]
- This results in:
\[ \Delta \vec{v} = -5 \hat{i} + 5 \hat{j} \]

### Step 3: Calculate the Average Acceleration
- The average acceleration (\(\vec{a}\)) is given by the change in velocity divided by the time interval (\(t\)):
\[ \vec{a} = \frac{\Delta \vec{v}}{t} = \frac{-5 \hat{i} + 5 \hat{j}}{10} \]
- Simplifying this gives:
\[ \vec{a} = -0.5 \hat{i} + 0.5 \hat{j} \, \text{m/s}^2 \]

### Step 4: Calculate the Magnitude of the Average Acceleration
- The magnitude of the average acceleration can be calculated using the Pythagorean theorem:
\[ |\vec{a}| = \sqrt{(-0.5)^2 + (0.5)^2} = \sqrt{0.25 + 0.25} = \sqrt{0.5} = \frac{1}{\sqrt{2}} \, \text{m/s}^2 \]

### Step 5: Determine the Direction of the Average Acceleration
- The average acceleration vector \(-0.5 \hat{i} + 0.5 \hat{j}\) indicates that the acceleration is directed towards the northwest (since it has a negative component in the \(i\) direction and a positive component in the \(j\) direction).

### Final Answer
- The average acceleration is:
\[ \frac{1}{\sqrt{2}} \, \text{m/s}^2 \, \text{towards northwest} \]

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