A particle moves in a straight line. Its velocity, v m/s, at time t seconds is given by v = 3t² − 10t + 8. Find the distance traveled by the particle between t = 1 second and t = 3 seconds.

We had:

$$s(1) = 1^3 - 5(1)^2 + 8(1) = 1 - 5 + 8 = 4 \text{ m}$$

Continuing with $s(4/3)$:

$$s(4/3) = \left(\frac{4}{3}\right)^3 - 5\left(\frac{4}{3}\right)^2 + 8\left(\frac{4}{3}\right)$$ $$s(4/3) = \frac{64}{27} - 5\left(\frac{16}{9}\right) + \frac{32}{3}$$ $$s(4/3) = \frac{64}{27} - \frac{80}{9} + \frac{32}{3}$$

To combine these, find a common denominator, which is 27:

$$s(4/3) = \frac{64}{27} - \frac{80 \times 3}{9 \times 3} + \frac{32 \times 9}{3 \times 9}$$ $$s(4/3) = \frac{64}{27} - \frac{240}{27} + \frac{288}{27}$$ $$s(4/3) = \frac{64 - 240 + 288}{27} = \frac{112}{27} \text{ m}$$

Now, calculate $s(2)$ and $s(3)$:

$$s(2) = 2^3 - 5(2)^2 + 8(2) = 8 - 5(4) + 16 = 8 - 20 + 16 = 4 \text{ m}$$ $$s(3) = 3^3 - 5(3)^2 + 8(3) = 27 - 5(9) + 24 = 27 - 45 + 24 = 6 \text{ m}$$

The critical points are $t = 4/3$ and $t = 2$. The intervals are $[1, 4/3]$, $[4/3, 2]$, and $[2, 3]$.
Displacement for $[1, 4/3]$:

$$|s(4/3) - s(1)| = \left|\frac{112}{27} - 4\right| = \left|\frac{112}{27} - \frac{108}{27}\right| = \left|\frac{4}{27}\right| = \frac{4}{27} \text{ m}$$

Displacement for $[4/3, 2]$:

$$|s(2) - s(4/3)| = \left|4 - \frac{112}{27}\right| = \left|\frac{108}{27} - \frac{112}{27}\right| = \left|-\frac{4}{27}\right| = \frac{4}{27} \text{ m}$$

Displacement for $[2, 3]$:

$$|s(3) - s(2)| = |6 - 4| = |2| = 2 \text{ m}$$

Total distance traveled:

$$\text{Distance} = \frac{4}{27} + \frac{4}{27} + 2$$ $$\text{Distance} = \frac{8}{27} + 2$$ $$\text{Distance} = \frac{8}{27} + \frac{54}{27}$$ $$\text{Distance} = \frac{62}{27} \text{ m}$$

The total distance traveled by the particle between $t = 1$ second and $t = 3$ seconds is $\boxed{\frac{62}{27} \text{ m}}$.