Total distance traveled via velocity sign analysis

Differentiate to get $v(t) = s'(t) = 4 - 2t$. Setting $v(t) = 0$ gives $t = 2$, so the particle changes direction at $t=2$. Since $v(t) > 0$ on $(0,2)$ and $v(t) < 0$ on $(2,6)$, the particle moves in the positive direction first, then reverses.

Compute positions: $s(0) = 0$, $s(2) = 4(2) - 2^2 = 4$, and $s(6) = 4(6) - 6^2 = -12$. The displacement on $[0,2]$ is $s(2)-s(0) = 4$, and on $[2,6]$ it is $s(6)-s(2) = -16$.

Total distance is the sum of the absolute values of these displacements: $|4| + |-16| = 4 + 16 = 20$. So the particle travels a total distance of $20$ meters over $[0,6]$.

The particle changes direction at t equals two, where velocity equals zero. For t greater than two, v of t equals four minus two t is negative. A negative velocity means the particle is moving backward, that is, in the negative direction, for the rest of the interval.