The position `x` of a particle varies with time `t` as `x=at^(2)-bt^(3)`. The acceleration at time `t` of the particle will be equal to zero, where (t) is equal to .`

Options

A. `(2a)/(3b)`
B. `(a)/(b)`
C. `(a)/(3b)`
D. zero

Explanation

To find the time \( t \) when the acceleration of the particle is zero, we start with the position function given by:

\[ x = at^2 - bt^3 \]

### Step 1: Find the velocity \( V \)
The velocity \( V \) is the first derivative of the position \( x \) with respect to time \( t \):

\[ V = \frac{dx}{dt} = \frac{d}{dt}(at^2 - bt^3) \]

Using the power rule for differentiation, we get:

\[ V = 2at - 3bt^2 \]

### Step 2: Find the acceleration \( a \)
The acceleration \( a \) is the derivative of the velocity \( V \) with respect to time \( t \):

\[ a = \frac{dV}{dt} = \frac{d}{dt}(2at - 3bt^2) \]

Again, using the power rule, we find:

\[ a = 2a - 6bt \]

### Step 3: Set the acceleration to zero
To find the time when the acceleration is zero, we set the equation for acceleration equal to zero:

\[ 0 = 2a - 6bt \]

### Step 4: Solve for \( t \)
Rearranging the equation gives us:

\[ 6bt = 2a \]

Dividing both sides by \( 6b \) yields:

\[ t = \frac{2a}{6b} = \frac{a}{3b} \]

### Final Answer
Thus, the time \( t \) when the acceleration of the particle is zero is:

\[ t = \frac{a}{3b} \]

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