A body is moving with velocity `30 m//s` towards east. After `10 s` its velocity becomes `40 m//s` towards north. The average acceleration of the body is.

Options

A. `7 ms^(-2)`
B. `sqrt(7)ms^(-2)`
C. `5 ms^(-2)`
D. `1 ms^(-2)`

Explanation

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

### Step 1: Identify Initial and Final Velocities
- The initial velocity \( \vec{u} \) is given as \( 30 \, \text{m/s} \) towards the east. We can represent this in vector form as:
\[ \vec{u} = 30 \hat{i} \, \text{m/s} \]
- The final velocity \( \vec{v} \) is given as \( 40 \, \text{m/s} \) towards the north. We can represent this in vector form as:
\[ \vec{v} = 40 \hat{j} \, \text{m/s} \]

### Step 2: Calculate Change in Velocity
- The change in velocity \( \Delta \vec{v} \) is given by:
\[ \Delta \vec{v} = \vec{v} - \vec{u} \]
- Substituting the values:
\[ \Delta \vec{v} = (40 \hat{j} - 30 \hat{i}) \, \text{m/s} = -30 \hat{i} + 40 \hat{j} \, \text{m/s} \]

### Step 3: Calculate Magnitude of Change in Velocity
- The magnitude of the change in velocity can be calculated using the Pythagorean theorem:
\[ |\Delta \vec{v}| = \sqrt{(-30)^2 + (40)^2} \]
- Calculating:
\[ |\Delta \vec{v}| = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \text{m/s} \]

### Step 4: Calculate Average Acceleration
- The average acceleration \( \vec{a} \) is given by the formula:
\[ \vec{a} = \frac{\Delta \vec{v}}{\Delta t} \]
- Given that the time interval \( \Delta t = 10 \, \text{s} \):
\[ \vec{a} = \frac{50 \, \text{m/s}}{10 \, \text{s}} = 5 \, \text{m/s}^2 \]

### Final Result
- Therefore, the average acceleration of the body is:
\[ \vec{a} = 5 \, \text{m/s}^2 \]