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

Answer

To find the time \( t \) at which the acceleration of the particle is equal to zero, we start with the given position function:

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

### Step 1: Find the velocity function
The velocity \( v \) of the particle 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:

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

### Step 2: Find the acceleration function
The acceleration \( a \) of the particle 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:

\[ a = 2a - 6bt \]

### Step 3: Set the acceleration to zero
According to the problem, we need to find the time \( t \) when the acceleration is zero:

\[ 0 = 2a - 6bt \]

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

\[ 6bt = 2a \]

Dividing both sides by \( 6b \) (assuming \( b \neq 0 \)):

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

Thus, the time \( t \) at which the acceleration of the particle is zero is:

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

### Final Answer
The time \( t \) at which the acceleration is zero is \( \frac{a}{3b} \).

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