Identify the velocity function and check for changes in direction.

The velocity of the particle is given by $v = (-2t + 4) \text{ m/s}$.
To determine if the particle changes direction within the first second ($t=0$ to $t=1$), we find when $v = 0$.

$$-2t + 4 = 0$$ $$2t = 4$$ $$t = 2 \text{ s}$$

Since $t = 2 \text{ s}$ is outside the interval $[0, 1]$ s, the particle does not change direction during the first second. Therefore, the distance moved is simply the absolute displacement.

Position $s$ is the integral of velocity with respect to time $t$.

$$s = \int v \, dt = \int (-2t + 4) \, dt$$ $$s = -t^2 + 4t + C$$

For calculating the distance moved, we can assume the initial position at $t=0$ is $s=0$.

$$0 = -(0)^2 + 4(0) + C \implies C = 0$$

So, the position function is:

$$s(t) = -t^2 + 4t$$

The distance moved is the absolute difference between the position at $t=1$ and the position at $t=0$.

$$s(1) = -(1)^2 + 4(1) = -1 + 4 = 3 \text{ m}$$ $$s(0) = -(0)^2 + 4(0) = 0 \text{ m}$$

Distance moved $= |s(1) - s(0)| = |3 - 0| = 3 \text{ m}$.
The distance moved by the particle during the first second of its motion is $\boxed{\text{3 m}}$.