Given the velocity function V = 4t2 − 3t:

The particle stops when its velocity $V(t)$ is equal to zero.
Given the velocity function $V = 4t^2 - 3t$:

$$4t^2 - 3t = 0$$

Factor out $t$:

$$t(4t - 3) = 0$$

This gives two possible times:

$$t = 0 \quad \text{or} \quad 4t - 3 = 0$$

Solving for the second time:

$$4t = 3$$ $$t = \frac{3}{4}$$

The particle starts at $t=0$ and stops at $t=\frac{3}{4}$.

The displacement $s(t)$ is the integral of the velocity function $V(t)$ with respect to time $t$.

$$s(t) = \int V(t)\,dt$$ $$s(t) = \int (4t^2 - 3t)\,dt$$ $$s(t) = \frac{4t^{2+1}}{2+1} - \frac{3t^{1+1}}{1+1} + C$$ $$s(t) = \frac{4t^3}{3} - \frac{3t^2}{2} + C$$

Assuming the particle starts at position $s=0$ at $t=0$, we have $s(0) = 0$, which means $C=0$.
So, the displacement function is:

$$s(t) = \frac{4t^3}{3} - \frac{3t^2}{2}$$

The particle stops at $t = \frac{3}{4}$. Substitute this value into the displacement function:

$$s\left(\frac{3}{4}\right) = \frac{4}{3}\left(\frac{3}{4}\right)^3 - \frac{3}{2}\left(\frac{3}{4}\right)^2$$ $$s\left(\frac{3}{4}\right) = \frac{4}{3}\left(\frac{27}{64}\right) - \frac{3}{2}\left(\frac{9}{16}\right)$$ $$s\left(\frac{1}{4}\right) = \frac{108}{192} - \frac{27}{32}$$

Simplify the first fraction: $\frac{108}{192} = \frac{9 \times 12}{16 \times 12} = \frac{9}{16}$.

$$s\left(\frac{3}{4}\right) = \frac{9}{16} - \frac{27}{32}$$

To subtract these fractions, find a common denominator, which is 32:

$$s\left(\frac{3}{4}\right) = \frac{9 \times 2}{16 \times 2} - \frac{27}{32}$$ $$s\left(\frac{3}{4}\right) = \frac{18}{32} - \frac{27}{32}$$ $$s\left(\frac{3}{4}\right) = -\frac{9}{32}$$

The velocity function $V(t) = t(4t-3)$ is negative for $0 < t < \frac{3}{4}$ (e.g., at $t=0.5$, $V(0.5) = 0.5(2-3) = -0.5$). This means the particle moves in one direction (the negative direction) from $t=0$ until it stops at $t=\frac{3}{4}$.
Therefore, the total distance travelled is the absolute value of the displacement.

$$\text{Distance} = \left|s\left(\frac{3}{4}\right) - s(0)\right|$$ $$\text{Distance} = \left|-\frac{9}{32} - 0\right|$$ $$\text{Distance} = \left|-\frac{9}{32}\right|$$ $$\text{Distance} = \frac{9}{32}$$

The maximum distance travelled by the particle before stopping is $\boxed{\frac{9}{32}}$.