If a particle moves according to the law `s=6t^(2)-(t^(3))/(2)`, then the time at which it is momentarily at rest is

Options

A. t=0 only
B. t=8 only
C. t=0,8
D. none of these

Answer

To find the time at which the particle is momentarily at rest, we need to determine when the velocity of the particle is zero. The position of the particle is given by the equation:

\[ s = 6t^2 - \frac{t^3}{2} \]

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

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

### Step 2: Differentiate the position function
Now, we differentiate each term:

\[ v = \frac{d}{dt}(6t^2) - \frac{d}{dt}\left(\frac{t^3}{2}\right) \]

Using the power rule, we get:

\[ v = 12t - \frac{3t^2}{2} \]

### Step 3: Set the velocity to zero
To find when the particle is momentarily at rest, we set the velocity \( v \) to zero:

\[ 12t - \frac{3t^2}{2} = 0 \]

### Step 4: Factor the equation
We can factor out \( t \):

\[ t\left(12 - \frac{3t}{2}\right) = 0 \]

This gives us two cases to solve:

1. \( t = 0 \)
2. \( 12 - \frac{3t}{2} = 0 \)

### Step 5: Solve for \( t \)
For the second case, we solve for \( t \):

\[ 12 = \frac{3t}{2} \]

Multiplying both sides by 2:

\[ 24 = 3t \]

Now, divide by 3:

\[ t = 8 \]

### Conclusion
The times at which the particle is momentarily at rest are:

\[ t = 0 \quad \text{and} \quad t = 8 \]

### Final Answer
The particle is momentarily at rest at \( t = 0 \) seconds and \( t = 8 \) seconds.

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