If `a=(3t^2+2t+1)` `m/s^2` is the expression according to which the acceleration of a particle varies. Then-
The change in velocity after 3 seconds of its start is:

Options

A. 30 m/s
B. 39 m/s
C. 3 m/s
D. 20 m/s

Explanation

To solve the problem, we need to find the change in velocity of a particle after 3 seconds, given the acceleration function \( a(t) = 3t^2 + 2t + 1 \) m/s².

### Step-by-step Solution:

1. **Understand the relationship between acceleration and velocity**:
The acceleration \( a \) is defined as the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
This implies:
\[ dv = a \cdot dt \]

2. **Set up the integral to find velocity**:
We can find the change in velocity by integrating the acceleration function over the time interval from \( t = 0 \) to \( t = 3 \) seconds:
\[ \Delta v = \int_{0}^{3} a(t) \, dt = \int_{0}^{3} (3t^2 + 2t + 1) \, dt \]

3. **Calculate the integral**:
We will compute the integral term by term:
- The integral of \( 3t^2 \) is \( t^3 \).
- The integral of \( 2t \) is \( t^2 \).
- The integral of \( 1 \) is \( t \).

Therefore:
\[ \int (3t^2 + 2t + 1) \, dt = t^3 + t^2 + t \]

4. **Evaluate the definite integral from 0 to 3**:
Now we will evaluate the integral from 0 to 3:
\[ \Delta v = \left[ t^3 + t^2 + t \right]_{0}^{3} = \left(3^3 + 3^2 + 3\right) - \left(0^3 + 0^2 + 0\right) \]
Simplifying this gives:
\[ \Delta v = (27 + 9 + 3) - 0 = 39 \, \text{m/s} \]

5. **Final result**:
The change in velocity after 3 seconds is:
\[ \Delta v = 39 \, \text{m/s} \]