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)`

Answer

To solve the problem, we need to find the velocity of a particle when its acceleration is zero, given the displacement function \( S(t) = t^3 - 6t^2 + 3t + 7 \).

### Step-by-Step Solution:

1. **Find the Velocity Function**:
The velocity \( V(t) \) is the first derivative of the displacement \( S(t) \) with respect to time \( t \).
\[ V(t) = \frac{dS}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 3t + 7) \]
Differentiating term by term:
\[ V(t) = 3t^2 - 12t + 3 \]

2. **Find the Acceleration Function**:
The acceleration \( A(t) \) is the derivative of the velocity \( V(t) \) with respect to time \( t \).
\[ A(t) = \frac{dV}{dt} = \frac{d}{dt}(3t^2 - 12t + 3) \]
Differentiating term by term:
\[ A(t) = 6t - 12 \]

3. **Set the Acceleration to Zero**:
To find when the acceleration is zero, set \( A(t) \) equal to zero:
\[ 6t - 12 = 0 \]
Solving for \( t \):
\[ 6t = 12 \implies t = 2 \text{ seconds} \]

4. **Find the Velocity at \( t = 2 \)**:
Now, substitute \( t = 2 \) into the velocity function \( V(t) \):
\[ V(2) = 3(2^2) - 12(2) + 3 \]
Calculating each term:
\[ V(2) = 3(4) - 24 + 3 = 12 - 24 + 3 = -9 \text{ m/s} \]

### Final Answer:
The velocity of the particle when its acceleration is zero is \( -9 \text{ m/s} \).