The motion of a particle is described as `s = 2-3t + 4t^(3)`. What is the acceleration of the particle at the point where its velocity is zero?

Options

A. A)0
B. B)4
C. C)8 unit
D. D)12 unit

Answer

To find the acceleration of the particle at the point where its velocity is zero, we will follow these steps:

### Step 1: Write down the position function
The motion of the particle is described by the equation:
\[ s(t) = 2 - 3t + 4t^3 \]

### Step 2: Find the velocity function
Velocity is the first derivative of the position function with respect to time \( t \). Therefore, we differentiate \( s(t) \):
\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(2 - 3t + 4t^3) \]
Differentiating term by term:
- The derivative of \( 2 \) is \( 0 \).
- The derivative of \( -3t \) is \( -3 \).
- The derivative of \( 4t^3 \) is \( 12t^2 \).

So, the velocity function is:
\[ v(t) = -3 + 12t^2 \]

### Step 3: Set the velocity function to zero and solve for \( t \)
To find when the velocity is zero, we set \( v(t) = 0 \):
\[ -3 + 12t^2 = 0 \]
Rearranging gives:
\[ 12t^2 = 3 \]
Dividing both sides by \( 12 \):
\[ t^2 = \frac{3}{12} = \frac{1}{4} \]
Taking the square root:
\[ t = \frac{1}{2} \quad \text{(since time cannot be negative)} \]

### Step 4: Find the acceleration function
Acceleration is the derivative of the velocity function with respect to time \( t \):
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(-3 + 12t^2) \]
Differentiating term by term:
- The derivative of \( -3 \) is \( 0 \).
- The derivative of \( 12t^2 \) is \( 24t \).

Thus, the acceleration function is:
\[ a(t) = 24t \]

### Step 5: Evaluate the acceleration at \( t = \frac{1}{2} \)
Now we substitute \( t = \frac{1}{2} \) into the acceleration function:
\[ a\left(\frac{1}{2}\right) = 24 \times \frac{1}{2} = 12 \]

### Final Answer
The acceleration of the particle at the point where its velocity is zero is:
\[ \boxed{12} \text{ units} \]