1. **State the problem:** Calculate the total distance travelled given the position function values at times 0, 4, 10, and 15. 2. **Formula:** Total distance $d$ is the sum of the absolute changes in position between these times: $$d=|x(4)-x(0)| + |x(10)-x(4)| + |x(15)-x(10)|$$ 3. **Substitute the given values:** $$d=\left|\frac{208}{3} - 0\right| + \left|\frac{100}{3} - \frac{208}{3}\right| + \left|150 - \frac{100}{3}\right|$$ 4. **Calculate each term:** $$= \frac{208}{3} + \left|\frac{100 - 208}{3}\right| + \left|150 - \frac{100}{3}\right| = \frac{208}{3} + \frac{108}{3} + \left|150 - 33\frac{1}{3}\right|$$ 5. **Simplify the last absolute value:** $$150 - 33\frac{1}{3} = 116\frac{2}{3} = \frac{350}{3}$$ 6. **Sum all terms:** $$d = \frac{208}{3} + \frac{108}{3} + \frac{350}{3} = \frac{666}{3} = 222$$ **Final answer:** The total distance travelled is $222$ units.