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 solve the problem, we need to find the time `t` at which the acceleration of the particle is equal to zero. The position of the particle is given by the equation:

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

### Step 1: Find the velocity
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 of differentiation:

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

### Step 2: Find the acceleration
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, applying the power rule:

\[ a = 2a - 6bt \]

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

\[ 2a - 6bt = 0 \]

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

\[ 6bt = 2a \]

Now, we can solve for \( t \):

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

### Conclusion
The time \( t \) at which the acceleration of the particle is equal to zero is:

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