The distance moved by the particle in time t is given by `x=t^(3)-12t^(2)+6t+8`. At the instant when its acceleration is zero, the velocity, is

Options

A. 42
B. `- 42`
C. 48
D. `- 48`

Explanation

To solve the problem, we need to find the velocity of the particle at the instant when its acceleration is zero. The distance moved by the particle is given by the equation:

\[ x(t) = t^3 - 12t^2 + 6t + 8 \]

### Step 1: Find the velocity function
The velocity \( v(t) \) is the first derivative of the position function \( x(t) \) with respect to time \( t \).

\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 12t^2 + 6t + 8) \]

Using the power rule of differentiation:

\[ v(t) = 3t^2 - 24t + 6 \]

### Step 2: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity function \( v(t) \).

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 24t + 6) \]

Again, using the power rule:

\[ a(t) = 6t - 24 \]

### Step 3: Set the acceleration to zero
To find the time when the acceleration is zero, set \( a(t) \) equal to zero:

\[ 6t - 24 = 0 \]

Solving for \( t \):

\[ 6t = 24 \\ t = 4 \]

### Step 4: Find the velocity at \( t = 4 \)
Now that we have the time when the acceleration is zero, we can find the velocity at that time by substituting \( t = 4 \) into the velocity function \( v(t) \):

\[ v(4) = 3(4^2) - 24(4) + 6 \]

Calculating each term:

\[ v(4) = 3(16) - 96 + 6 \\ v(4) = 48 - 96 + 6 \\ v(4) = -42 \]

### Final Answer
The velocity of the particle at the instant when its acceleration is zero is:

\[ \boxed{-42} \]