A body moves with a velocity of `3 m//s` due east and then turns due north to travel with the same velocity. If the total time of travel is 6s. The acceleration of the body is

Options

A. `sqrt(3) m//s^(2)` towards north west
B. `(1)/(sqrt(2)) m//s^(2)` towards north west
C. `sqrt(2)m//s^(2)` towards north east
D. all the above

Explanation

To find the acceleration of the body that changes direction while maintaining the same speed, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial and Final Velocities**:
- The body moves with an initial velocity \( \vec{V_i} = 3 \, \text{m/s} \) due east.
- After turning, the body moves with a final velocity \( \vec{V_f} = 3 \, \text{m/s} \) due north.

2. **Determine Change in Velocity**:
- The change in velocity \( \Delta \vec{V} \) is given by:
\[ \Delta \vec{V} = \vec{V_f} - \vec{V_i} \]
- Since the velocities are in perpendicular directions (east and north), we can represent them as vectors:
\[ \vec{V_i} = 3 \hat{i} \quad (\text{east direction}) \]
\[ \vec{V_f} = 3 \hat{j} \quad (\text{north direction}) \]
- Therefore, the change in velocity becomes:
\[ \Delta \vec{V} = 3 \hat{j} - 3 \hat{i} = -3 \hat{i} + 3 \hat{j} \]

3. **Calculate the Magnitude of Change in Velocity**:
- The magnitude of \( \Delta \vec{V} \) can be calculated using the Pythagorean theorem:
\[ |\Delta \vec{V}| = \sqrt{(-3)^2 + (3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \, \text{m/s} \]

4. **Determine the Time Interval**:
- The total time of travel is given as \( \Delta t = 6 \, \text{s} \).

5. **Calculate Acceleration**:
- Acceleration \( \vec{a} \) is defined as the change in velocity over time:
\[ \vec{a} = \frac{\Delta \vec{V}}{\Delta t} \]
- Substituting the values we have:
\[ \vec{a} = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2} \, \text{m/s}^2 \]

6. **Direction of Acceleration**:
- Since the change in velocity is from east to north, the direction of acceleration is towards the northwest.

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