Total distance traveled from velocity function

First factor the velocity: $v(t) = t^3 - 6t^2 + 8t = t(t^2 - 6t + 8) = t(t-2)(t-4)$. This shows $v(t) = 0$ at $t = 0, 2, 4$, and testing signs gives $v(t) \geq 0$ on $[0,2]$ and $v(t) \leq 0$ on $[2,4]$. So the particle changes direction at $t=2$, and we must split the distance integral there.

Compute an antiderivative $F(t) = \frac{t^4}{4} - 2t^3 + 4t^2$. On $[0,2]$: $F(2) - F(0) = (4 - 16 + 16) - 0 = 4$. On $[2,4]$: $F(4) - F(2) = (64 - 128 + 64) - 4 = 0 - 4 = -4$, so the displacement there is $-4$ meters.

The total distance is the sum of the absolute values of these two pieces: $|4| + |-4| = 8$ meters. (Note this differs from the net displacement over $[0,4]$, which is $4 + (-4) = 0$.)

The net displacement over the interval is zero, since the particle returns to its starting position, while the total distance traveled is eight meters. Since distance accounts for both legs of the trip without cancellation, it is always at least as large as the magnitude of displacement. Here, net displacement (zero) is smaller than total distance (eight meters).