A particle is moving east wards with a velocity of `15 ms^(-1)`. In a time of 10 s, the velocity changes to `15 ms^(-1)` northwards. Average acceleration during this time is (in `ms^(-2))`

Options

A. `3//sqrt(2)` NE
B. `3sqrt(2)` NE
C. `3//sqrt(2)` NW
D. `3sqrt(2)` NW

Answer

To find the average acceleration of the particle during the given time interval, we can follow these steps:

### Step 1: Identify Initial and Final Velocities
- The initial velocity \( \vec{u} \) is \( 15 \, \text{m/s} \) eastwards.
- The final velocity \( \vec{v} \) is \( 15 \, \text{m/s} \) northwards.

### Step 2: Represent Velocities as Vectors
- We can represent the initial velocity vector \( \vec{u} \) as:
\[ \vec{u} = 15 \hat{i} \, \text{m/s} \quad (\text{eastward direction}) \]
- The final velocity vector \( \vec{v} \) can be represented as:
\[ \vec{v} = 15 \hat{j} \, \text{m/s} \quad (\text{northward direction}) \]

### Step 3: Calculate Change in Velocity
- The change in velocity \( \Delta \vec{v} \) is given by:
\[ \Delta \vec{v} = \vec{v} - \vec{u} = (15 \hat{j} - 15 \hat{i}) \, \text{m/s} \]
- This simplifies to:
\[ \Delta \vec{v} = -15 \hat{i} + 15 \hat{j} \, \text{m/s} \]

### Step 4: Calculate Magnitude of Change in Velocity
- The magnitude of the change in velocity \( |\Delta \vec{v}| \) can be calculated using the Pythagorean theorem:
\[ |\Delta \vec{v}| = \sqrt{(-15)^2 + (15)^2} = \sqrt{225 + 225} = \sqrt{450} = 15\sqrt{2} \, \text{m/s} \]

### Step 5: Calculate Average Acceleration
- Average acceleration \( \vec{a}_{\text{avg}} \) is defined as the change in velocity divided by the time interval \( \Delta t \):
\[ \vec{a}_{\text{avg}} = \frac{\Delta \vec{v}}{\Delta t} = \frac{15\sqrt{2} \, \text{m/s}}{10 \, \text{s}} = \frac{3\sqrt{2}}{2} \, \text{m/s}^2 \]

### Step 6: Determine the Direction of Average Acceleration
- The direction of the average acceleration is in the direction of the resultant vector formed by the change in velocity, which is in the northwest direction.

### Final Answer
The average acceleration during this time is:
\[ \vec{a}_{\text{avg}} = \frac{3\sqrt{2}}{2} \, \text{m/s}^2 \quad \text{(north-west)} \]