If velocity of a particle is given by `v=(2t+3)m//s`. Then average velocity in interval `0letle1` s is :

Options

A. `2bhat(j)`
B. `a hat(j)`
C. `-2bhat(j)`
D. None of these

Explanation

To find the average velocity of a particle given its velocity function \( v(t) = 2t + 3 \) m/s over the interval from \( t = 0 \) to \( t = 1 \) seconds, we can follow these steps:

### Step 1: Understand the Formula for Average Velocity
The average velocity \( v_{\text{avg}} \) over a time interval can be calculated using the formula:
\[ v_{\text{avg}} = \frac{\text{Total Displacement}}{\text{Total Time}} \]

### Step 2: Express Displacement in Terms of Velocity
Since velocity is defined as the derivative of displacement with respect to time, we can express displacement \( dx \) as:
\[ dx = v \cdot dt = (2t + 3) \cdot dt \]

### Step 3: Calculate Total Displacement
To find the total displacement over the interval from \( t = 0 \) to \( t = 1 \), we need to integrate the velocity function:
\[ \text{Total Displacement} = \int_{0}^{1} (2t + 3) \, dt \]

### Step 4: Perform the Integration
Now we can compute the integral:
\[ \int (2t + 3) \, dt = \int 2t \, dt + \int 3 \, dt \]
Calculating each part:
- \(\int 2t \, dt = t^2\)
- \(\int 3 \, dt = 3t\)

Thus, we have:
\[ \int (2t + 3) \, dt = t^2 + 3t \]

### Step 5: Evaluate the Integral from 0 to 1
Now we evaluate the integral from 0 to 1:
\[ \left[ t^2 + 3t \right]_{0}^{1} = (1^2 + 3 \cdot 1) - (0^2 + 3 \cdot 0) = (1 + 3) - (0) = 4 \]

### Step 6: Calculate Total Time
The total time for the interval from \( t = 0 \) to \( t = 1 \) is:
\[ \text{Total Time} = 1 - 0 = 1 \text{ second} \]

### Step 7: Calculate Average Velocity
Now we can find the average velocity:
\[ v_{\text{avg}} = \frac{\text{Total Displacement}}{\text{Total Time}} = \frac{4 \text{ m}}{1 \text{ s}} = 4 \text{ m/s} \]

### Final Answer
The average velocity of the particle in the interval from \( t = 0 \) to \( t = 1 \) seconds is:
\[ \boxed{4 \text{ m/s}} \]