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. `7m//s^(2)`
B. `sqrt(7)m//s^(2)`
C. `5m//s^(2)`
D. `1m//s^(2)`

Answer

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

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

### Step 2: Calculate Change in Velocity
- The change in velocity (\( \Delta \vec{v} \)) can be calculated as:
\[ \Delta \vec{v} = \vec{v_f} - \vec{v_i} = (40 \hat{j} - 30 \hat{i}) \, \text{m/s} \]
This gives us:
\[ \Delta \vec{v} = -30 \hat{i} + 40 \hat{j} \, \text{m/s} \]

### Step 3: Determine Time Interval
- The time interval (\( t \)) during which this change occurs is given as \( 10 \, \text{s} \).

### Step 4: Calculate Average Acceleration
- Average acceleration (\( \vec{a_{avg}} \)) is defined as the change in velocity divided by the time taken:
\[ \vec{a_{avg}} = \frac{\Delta \vec{v}}{t} = \frac{-30 \hat{i} + 40 \hat{j}}{10} \]
Simplifying this gives:
\[ \vec{a_{avg}} = -3 \hat{i} + 4 \hat{j} \, \text{m/s}^2 \]

### Step 5: Calculate Magnitude of Average Acceleration
- The magnitude of the average acceleration can be calculated using the Pythagorean theorem:
\[ |\vec{a_{avg}}| = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{m/s}^2 \]

### Final Answer
- The average acceleration of the body is:
\[ 5 \, \text{m/s}^2 \]