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 particles zero acceleration, then its velocity will be:

We have the displacement of the particle as a function of time, $$x(t)=at+bt^2-ct^3,$$ where the symbols $$a,\;b$$ and $$c$$ are constants.

The velocity is defined as the first derivative of displacement with respect to time, that is $$v(t)=\dfrac{dx}{dt}.$$ Differentiating term-by-term,

$$\begin{aligned} v(t) &= \dfrac{d}{dt}\!\bigl(at\bigr) + \dfrac{d}{dt}\!\bigl(bt^2\bigr) - \dfrac{d}{dt}\!\bigl(ct^3\bigr) \\ &= a + 2bt - 3ct^2. \end{aligned}$$

Acceleration is the derivative of velocity with respect to time, $$a(t)=\dfrac{dv}{dt}.$$ Differentiating the expression for $$v(t)$$ gives

$$\begin{aligned} a(t) &= \dfrac{d}{dt}\!\bigl(a\bigr) + \dfrac{d}{dt}\!\bigl(2bt\bigr) - \dfrac{d}{dt}\!\bigl(3ct^2\bigr) \\ &= 0 + 2b - 6ct \\ &= 2b - 6ct. \end{aligned}$$

The problem asks for the velocity when the acceleration becomes zero. So we set $$a(t)=0$$:

$$2b - 6ct = 0.$$

Solving for $$t$$ yields

$$\begin{aligned} 2b &= 6ct \\ t &= \frac{2b}{6c} \\ t &= \frac{b}{3c}. \end{aligned}$$

Now we substitute this value of $$t$$ back into the velocity expression $$v(t)=a+2bt-3ct^2$$ to find the required velocity:

$$\begin{aligned} v &= a + 2b\left(\frac{b}{3c}\right) - 3c\left(\frac{b}{3c}\right)^2 \\ &= a + \frac{2b^2}{3c} - 3c\left(\frac{b^2}{9c^2}\right) \\ &= a + \frac{2b^2}{3c} - \frac{b^2}{3c} \\ &= a + \frac{2b^2 - b^2}{3c} \\ &= a + \frac{b^2}{3c}. \end{aligned}$$

Hence, the velocity of the particle at the instant when its acceleration is zero is $$a + \dfrac{b^2}{3c}.$$

Hence, the correct answer is Option A.