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 (a) `42`      (b) `-42`          (c) `48`           (d) `-48`

Options

A. (a) `42`
B. (b) `-42`
C. (c) `48`
D. (d) `-48`

Explanation

To solve the problem step by step, we will first find the velocity and acceleration of the particle based on the given distance function.

### Step 1: Write down the distance function
The distance moved by the particle in time `t` is given by:
\[ x(t) = t^3 - 12t^2 + 6t + 8 \]

### Step 2: Find the velocity function
The velocity \( v(t) \) is the first derivative of the distance function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 12t^2 + 6t + 8) \]
Using the power rule for differentiation:
\[ v(t) = 3t^2 - 24t + 6 \]

### Step 3: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity function:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 24t + 6) \]
Again applying the power rule:
\[ a(t) = 6t - 24 \]

### Step 4: Set the acceleration to zero
To find the time when the acceleration is zero, we set the acceleration function equal to zero:
\[ 6t - 24 = 0 \]
Solving for \( t \):
\[ 6t = 24 \implies t = 4 \]

### Step 5: Find the velocity at \( t = 4 \)
Now we will substitute \( t = 4 \) into the velocity function to find the velocity at that instant:
\[ v(4) = 3(4^2) - 24(4) + 6 \]
Calculating each term:
\[ v(4) = 3(16) - 96 + 6 = 48 - 96 + 6 = -42 \]

### Conclusion
Thus, the velocity of the particle at the instant when its acceleration is zero is:
\[ \text{Velocity} = -42 \]

The correct answer is (b) -42.