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

Options

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

Explanation

To find the change in velocity after 3 seconds given the acceleration function \( a(t) = 3t^2 + 2t + 1 \, \text{m/s}^2 \), we will follow these steps:

### Step 1: Understand the relationship between acceleration and velocity
Acceleration is defined as the rate of change of velocity with respect to time. Mathematically, this can be expressed as:
\[ a(t) = \frac{dv}{dt} \]
where \( a(t) \) is the acceleration, \( v \) is the velocity, and \( t \) is time.

### Step 2: Set up the integral to find the change in velocity
To find the change in velocity, we can rearrange the equation:
\[ dv = a(t) \, dt \]
Substituting the expression for acceleration, we have:
\[ dv = (3t^2 + 2t + 1) \, dt \]

### Step 3: Integrate both sides
To find the total change in velocity from \( t = 0 \) to \( t = 3 \), we integrate:
\[ \int_{0}^{v} dv = \int_{0}^{3} (3t^2 + 2t + 1) \, dt \]
The left side simplifies to \( v - 0 = v \).

### Step 4: Calculate the integral on the right side
Now we need to compute the integral:
\[ \int (3t^2 + 2t + 1) \, dt = t^3 + t^2 + t + C \]
We will evaluate this from \( t = 0 \) to \( t = 3 \):
\[ \left[ t^3 + t^2 + t \right]_{0}^{3} = \left( 3^3 + 3^2 + 3 \right) - \left( 0^3 + 0^2 + 0 \right) \]
Calculating the values:
\[ = (27 + 9 + 3) - 0 = 39 \]

### Step 5: Conclusion
Thus, the change in velocity after 3 seconds is:
\[ \Delta v = 39 \, \text{m/s} \]