The position of a particle as a function of time t, is given by `x(t)=at+bt^(2)-ct^(3)` where a,b and c are constants. When the particle attains zero acceleration, then its velocity will be :

Options

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

Answer

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

### Step-by-Step Solution:

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

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

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

4. **Find the Velocity at \( t = \frac{b}{3c} \)**:
Now we substitute \( t = \frac{b}{3c} \) back into the velocity function \( v(t) \):
\[ v\left(\frac{b}{3c}\right) = a + 2b\left(\frac{b}{3c}\right) - 3c\left(\frac{b}{3c}\right)^2 \]
Simplifying each term:
- The first term is \( a \).
- The second term is:
\[ 2b\left(\frac{b}{3c}\right) = \frac{2b^2}{3c} \]
- The third term is:
\[ 3c\left(\frac{b^2}{9c^2}\right) = \frac{b^2}{3c} \]
Combining these:
\[ v\left(\frac{b}{3c}\right) = a + \frac{2b^2}{3c} - \frac{b^2}{3c} = a + \frac{b^2}{3c} \]

5. **Final Result**:
Therefore, the velocity of the particle when the acceleration is zero is:
\[ v = a + \frac{b^2}{3c} \]