The position of a body wrt time is given by `x = 3t^(3) -6t^(2) + 12t + 6`.
If t = 0, the acceleration is___________

Answer

To find the acceleration of the body at \( t = 0 \), we will follow these steps:

### Step 1: Write down the position function
The position of the body with respect to time is given by:
\[ x(t) = 3t^3 - 6t^2 + 12t + 6 \]

### Step 2: Find the velocity function
The velocity \( v(t) \) is the first derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(3t^3 - 6t^2 + 12t + 6) \]
Using the power rule of differentiation:
- The derivative of \( 3t^3 \) is \( 9t^2 \)
- The derivative of \( -6t^2 \) is \( -12t \)
- The derivative of \( 12t \) is \( 12 \)
- The derivative of a constant \( 6 \) is \( 0 \)

Thus, the velocity function is:
\[ v(t) = 9t^2 - 12t + 12 \]

### Step 3: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity function:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(9t^2 - 12t + 12) \]
Again, applying the power rule:
- The derivative of \( 9t^2 \) is \( 18t \)
- The derivative of \( -12t \) is \( -12 \)
- The derivative of a constant \( 12 \) is \( 0 \)

Thus, the acceleration function is:
\[ a(t) = 18t - 12 \]

### Step 4: Calculate acceleration at \( t = 0 \)
Now, we substitute \( t = 0 \) into the acceleration function:
\[ a(0) = 18(0) - 12 = -12 \]

### Final Answer
The acceleration at \( t = 0 \) is:
\[ \boxed{-12 \, \text{m/s}^2} \]

---