The distance s of a particle in time t is given by `s=t^3-6t^2-4t-8`. Its acceleration vanishes at t=

Options

A. 1
B. 2
C. 3
D. 6

Explanation

To solve the problem, we need to find the time \( t \) at which the acceleration of the particle vanishes. We start with the given distance function:

\[ s(t) = t^3 - 6t^2 - 4t - 8 \]

### Step 1: Find the velocity
The velocity \( v(t) \) is the first derivative of the distance \( s(t) \) with respect to time \( t \):

\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 - 4t - 8) \]

Differentiating term by term:

\[ v(t) = 3t^2 - 12t - 4 \]

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

\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 12t - 4) \]

Differentiating again:

\[ a(t) = 6t - 12 \]

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

\[ 6t - 12 = 0 \]

### Step 4: Solve for \( t \)
Now, solve for \( t \):

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

### Conclusion
The acceleration of the particle vanishes at \( t = 2 \).

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