The position of a body with respect to time is given by `x = 4t^(3) – 6t^(2) + 20 t + 12` . Acceleration at t = 0 will be-

Options

A. `- 12` units
B. 12 units
C. 24 units
D. `-24` units

Explanation

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:
\[ x(t) = 4t^3 - 6t^2 + 20t + 12 \]

### Step 2: Differentiate the position function to find velocity
The velocity \( v(t) \) is the first derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(4t^3 - 6t^2 + 20t + 12) \]
Using the power rule for differentiation:
\[ v(t) = 12t^2 - 12t + 20 \]

### Step 3: Differentiate the velocity function to find acceleration
The acceleration \( a(t) \) is the derivative of the velocity function:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(12t^2 - 12t + 20) \]
Again, applying the power rule:
\[ a(t) = 24t - 12 \]

### Step 4: Evaluate acceleration at \( t = 0 \)
Now, we substitute \( t = 0 \) into the acceleration function:
\[ a(0) = 24(0) - 12 = -12 \]

### Conclusion
The acceleration of the body at \( t = 0 \) is:
\[ \boxed{-12 \text{ units}} \]

---