The distances moved by a particle in time t seconds is given by `s=t^(3)-6t^(2)-15t+12`. The velocity of the particle when acceleration becomes zero, is

Options

A. 15
B. `-27`
C. `6//5`
D. none of these

Answer

To find the velocity of the particle when the acceleration becomes zero, we can follow these steps:

### Step 1: Write down the displacement function
The displacement \( s \) of the particle is given by:
\[ s = t^3 - 6t^2 - 15t + 12 \]

### Step 2: Differentiate the displacement function to find velocity
The velocity \( v \) is the first derivative of displacement with respect to time \( t \):
\[ v = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 - 15t + 12) \]
Differentiating term by term:
\[ v = 3t^2 - 12t - 15 \]

### Step 3: Differentiate the velocity function to find acceleration
The acceleration \( a \) is the derivative of velocity with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t - 15) \]
Differentiating term by term:
\[ a = 6t - 12 \]

### Step 4: Set the acceleration to zero and solve for \( t \)
To find when the acceleration is zero:
\[ 6t - 12 = 0 \]
Solving for \( t \):
\[ 6t = 12 \implies t = 2 \]

### Step 5: Substitute \( t = 2 \) into the velocity function
Now we need to find the velocity at \( t = 2 \):
\[ v = 3(2^2) - 12(2) - 15 \]
Calculating:
\[ v = 3(4) - 24 - 15 \]
\[ v = 12 - 24 - 15 \]
\[ v = -27 \]

### Final Answer
The velocity of the particle when the acceleration becomes zero is:
\[ \boxed{-27} \]

---