A particle moves along the x-axis with position given by s(t) = t^3 - 6t^2 + 9t, where t is measured in seconds and s is measured in meters. What is the total distance traveled by the particle from t = 0 to t = 4?

Core Concept

First, find the velocity function: v(t) = s'(t) = 3t^2 - 12t + 9. Set v(t) = 0 to find when the particle changes direction: 3t^2 - 12t + 9 = 0, which simplifies to t^2 - 4t + 3 = 0. Factoring gives (t - 1)(t - 3) = 0, so critical points are at t = 1 and t = 3. Calculate the position at these critical points and endpoints: s(0) = 0, s(1) = 4, s(3) = 0, s(4) = 4. The particle moves from 0 to 4 (distance 4), then back to 0 (distance 4), then to 4 again (distance 4). Total distance = 4 + 4 + 4 = 16 meters.