The position of a particle moving along x-axis given by `x=(-2t^(3)-3t^(2)+5)m`. The acceleration of particle at the instant its velocity becomes zero is

Options

A. `12 m//s^(2)`
B. `-12 m//s^(2)`
C. `-6 m//s^(2)`
D. Zero

Answer

To find the acceleration of the particle at the instant its velocity becomes zero, we will follow these steps:

### Step 1: Write down the position function
The position of the particle is given by:
\[ x(t) = -2t^3 - 3t^2 + 5 \, \text{m} \]

### Step 2: Find the velocity function
The velocity \( v(t) \) is the first derivative of the position function with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = \frac{d}{dt}(-2t^3 - 3t^2 + 5) \]
Calculating the derivative:
\[ v(t) = -6t^2 - 6t \]

### Step 3: Set the velocity to zero and solve for \( t \)
To find the time when the velocity is zero:
\[ -6t^2 - 6t = 0 \]
Factoring out \( -6t \):
\[ -6t(t + 1) = 0 \]
This gives us two solutions:
1. \( t = 0 \)
2. \( t + 1 = 0 \) → \( t = -1 \) (not physically meaningful in this context)

Thus, the relevant solution is:
\[ t = 0 \]

### Step 4: Find the acceleration function
The acceleration \( a(t) \) is the derivative of the velocity function:
\[ a(t) = \frac{dv}{dt} = \frac{d}{dt}(-6t^2 - 6t) \]
Calculating the derivative:
\[ a(t) = -12t - 6 \]

### Step 5: Calculate the acceleration at \( t = 0 \)
Now we substitute \( t = 0 \) into the acceleration function:
\[ a(0) = -12(0) - 6 = -6 \, \text{m/s}^2 \]

### Conclusion
The acceleration of the particle at the instant its velocity becomes zero is:
\[ \boxed{-6 \, \text{m/s}^2} \]

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