Displacement (x) of a particle is related to time (t) as
`x = at + b t^(2) - c t^(3)`
where a,b and c are constant of the motion. The velocity of the particle when its acceleration is zero is given by:

Options

A. `a + (b^(2))/(c)`
B. `a + (b^(2))/(2c)`
C. `a + (b^(2))/(3c)`
D. `a + (b^(2))/(4c)`

Explanation

To find the velocity of the particle when its acceleration is zero, we can follow these steps:

### Step 1: Write down the displacement equation
The displacement \( x \) of the particle is given by:
\[ x = at + bt^2 - ct^3 \]

### Step 2: Find the velocity
The velocity \( v \) is the first derivative of displacement with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(at + bt^2 - ct^3) \]
Calculating the derivative:
\[ v = a + 2bt - 3ct^2 \]

### Step 3: Find the acceleration
The acceleration \( a \) is the derivative of velocity with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(a + 2bt - 3ct^2) \]
Calculating the derivative:
\[ a = 2b - 6ct \]

### Step 4: Set acceleration to zero
To find the time when acceleration is zero, set the acceleration equation to zero:
\[ 2b - 6ct = 0 \]
Solving for \( t \):
\[ 6ct = 2b \quad \Rightarrow \quad t = \frac{b}{3c} \]

### Step 5: Substitute \( t \) back into the velocity equation
Now we substitute \( t = \frac{b}{3c} \) back into the velocity equation:
\[ v = a + 2b\left(\frac{b}{3c}\right) - 3c\left(\frac{b}{3c}\right)^2 \]

### Step 6: Simplify the velocity expression
Calculating each term:
1. The first term is \( a \).
2. The second term is:
\[ 2b\left(\frac{b}{3c}\right) = \frac{2b^2}{3c} \]
3. The third term is:
\[ -3c\left(\frac{b^2}{9c^2}\right) = -\frac{b^2}{3c} \]

Combining these terms:
\[ v = a + \frac{2b^2}{3c} - \frac{b^2}{3c} = a + \frac{b^2}{3c} \]

### Final Answer
Thus, the velocity of the particle when its acceleration is zero is:
\[ v = a + \frac{b^2}{3c} \]

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