A position of particle is `x=`at`+bt^(2)-ct^(3)` find out velocity when acceleration is zero (1) `v=a+b^2/(3c)` (2) `v=a-b^2/(3c)` (3) `v=2a-b^2/(3c)` (4) None of these

Options

A. `v-a+(b^(2))/(3c)`
B. `v-a-(b^(2))/(3c)`
C. None of these
D. 1

Answer

To solve the problem, we need to find the velocity of a particle when its acceleration is zero, given the position function \( x = at + bt^2 - ct^3 \).

### Step-by-step Solution:

1. **Find the velocity function**:
The velocity \( v \) is the first derivative of the position \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = \frac{d}{dt}(at + bt^2 - ct^3) \]
Differentiating term by term:
\[ v = a + 2bt - 3ct^2 \]

2. **Find the acceleration function**:
The acceleration \( a \) is the derivative of the velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}(a + 2bt - 3ct^2) \]
Differentiating term by term:
\[ a = 0 + 2b - 6ct = 2b - 6ct \]

3. **Set the acceleration to zero**:
To find the time when the acceleration is zero, set the acceleration equation to zero:
\[ 2b - 6ct = 0 \]
Rearranging gives:
\[ 6ct = 2b \quad \Rightarrow \quad t = \frac{b}{3c} \]

4. **Substitute \( t \) back into the velocity equation**:
Now substitute \( t = \frac{b}{3c} \) into the velocity equation:
\[ v = a + 2b\left(\frac{b}{3c}\right) - 3c\left(\frac{b}{3c}\right)^2 \]
Simplifying each term:
\[ v = a + \frac{2b^2}{3c} - 3c\left(\frac{b^2}{9c^2}\right) \]
\[ v = a + \frac{2b^2}{3c} - \frac{b^2}{3c} \]
Combining the terms:
\[ v = a + \frac{2b^2}{3c} - \frac{b^2}{3c} = a + \frac{b^2}{3c} \]

5. **Final result**:
The velocity when the acceleration is zero is:
\[ v = a + \frac{b^2}{3c} \]

### Conclusion:
The correct answer is:
\[ \text{(1) } v = a + \frac{b^2}{3c} \]