The acceleration of paticle varies with time as :
`a(t) = 3t^(2) +4`
If the initial velocity of particle is `2 m//s`, find the velocity of particle at `t = 3sec`.

Options

A. `41 m//s`
B. `4 m//s`
C. `39 m//s`
D. `27 m//s`

Explanation

To find the velocity of the particle at \( t = 3 \) seconds given the acceleration function \( a(t) = 3t^2 + 4 \) and the initial velocity \( v(0) = 2 \, \text{m/s} \), we can follow these steps:

### Step 1: Write down the relationship between acceleration and velocity
Acceleration \( a(t) \) is defined as the derivative of velocity \( v(t) \) with respect to time \( t \):
\[ a(t) = \frac{dv}{dt} \]

### Step 2: Substitute the given acceleration function
We substitute the given acceleration function into the equation:
\[ \frac{dv}{dt} = 3t^2 + 4 \]

### Step 3: Rearrange the equation for integration
We can rearrange the equation to separate variables:
\[ dv = (3t^2 + 4) dt \]

### Step 4: Integrate both sides
Now we integrate both sides. The left side integrates to \( v \) and the right side integrates to:
\[ \int dv = \int (3t^2 + 4) dt \]
This gives us:
\[ v = \int (3t^2 + 4) dt = t^3 + 4t + C \]
where \( C \) is the constant of integration.

### Step 5: Determine the constant of integration
To find the constant \( C \), we use the initial condition \( v(0) = 2 \):
\[ v(0) = 0^3 + 4(0) + C = 2 \implies C = 2 \]
Thus, the velocity function becomes:
\[ v(t) = t^3 + 4t + 2 \]

### Step 6: Calculate the velocity at \( t = 3 \) seconds
Now we substitute \( t = 3 \) into the velocity function:
\[ v(3) = (3)^3 + 4(3) + 2 \]
Calculating this gives:
\[ v(3) = 27 + 12 + 2 = 41 \, \text{m/s} \]

### Final Answer
The velocity of the particle at \( t = 3 \) seconds is \( \boxed{41 \, \text{m/s}} \).