For a particle moving in a straight line, the displacement of the particle at time `t` is given by
`S=t^(3)-6t^(2) +3t+7`
What is the velocity of the particle when its acceleration is zero?

Options

A. `- 9 ms^(-1)`
B. `-12 ms^(-1)`
C. `3 ms^(-1)`
D. `42 ms^(-1)`

Explanation

To solve the problem, we need to find the velocity of a particle when its acceleration is zero. The displacement of the particle is given by the equation:

\[ S = t^3 - 6t^2 + 3t + 7 \]

### Step 1: Find the velocity of the particle.

The velocity \( V \) is the first derivative of the displacement \( S \) with respect to time \( t \):

\[ V = \frac{dS}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 3t + 7) \]

Calculating the derivative:

\[ V = 3t^2 - 12t + 3 \]

### Step 2: Find the acceleration of the particle.

The acceleration \( A \) is the derivative of the velocity \( V \) with respect to time \( t \):

\[ A = \frac{dV}{dt} = \frac{d}{dt}(3t^2 - 12t + 3) \]

Calculating the derivative:

\[ A = 6t - 12 \]

### Step 3: Set the acceleration to zero and solve for \( t \).

To find the time when the acceleration is zero, we set \( A = 0 \):

\[ 6t - 12 = 0 \]

Solving for \( t \):

\[ 6t = 12 \\ t = 2 \]

### Step 4: Find the velocity at \( t = 2 \).

Now that we have the time when the acceleration is zero, we will substitute \( t = 2 \) into the velocity equation:

\[ V = 3(2^2) - 12(2) + 3 \]

Calculating this:

\[ V = 3(4) - 24 + 3 \\ V = 12 - 24 + 3 \\ V = -9 \, \text{m/s} \]

### Final Answer:

The velocity of the particle when its acceleration is zero is:

\[ \boxed{-9 \, \text{m/s}} \]

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