1. **State the problem:** We need to calculate the total distance travelled by the graphic from time $t=0$ to $t=15$ seconds. 2. **Understand the velocity function:** The velocity is given by $v(t) = t^2 - 14t + 40$. The graphic is at rest when $v(t) = 0$. 3. **Find when the graphic is at rest:** Solve $v(t) = 0$: $$t^2 - 14t + 40 = 0$$ Factorizing: $$(t - 4)(t - 10) = 0$$ So, $t = 4$ or $t = 10$ seconds. 4. **Calculate position values:** Given positions at these times: - $x(0) = 0$ - $x(4) = 69 \frac{1}{3} = \frac{208}{3}$ - $x(10) = 33 \frac{1}{3} = \frac{100}{3}$ - $x(15) = 150$ 5. **Calculate total distance travelled:** Total distance is the sum of absolute changes in position between these times: $$d = |x(4) - x(0)| + |x(10) - x(4)| + |x(15) - x(10)|$$ Substitute values: $$= \left|\frac{208}{3} - 0\right| + \left|\frac{100}{3} - \frac{208}{3}\right| + |150 - \frac{100}{3}|$$ Calculate each term: $$= \frac{208}{3} + \frac{108}{3} + \left|150 - 33 \frac{1}{3}\right|$$ $$= 69 \frac{1}{3} + 36 + 116 \frac{2}{3}$$ 6. **Sum the distances:** $$d = 69 \frac{1}{3} + 36 + 116 \frac{2}{3} = 222$$ **Final answer:** The total distance travelled by the graphic is $222$ cm. This method correctly finds the times when the graphic stops (velocity zero), then sums the absolute position changes between these times to get total distance, not just displacement.