1. **State the problem:** We need to find the time interval during which the teenager's velocity changes at the greatest rate, i.e., the greatest acceleration. 2. **Recall the formula:** Acceleration is the rate of change of velocity with respect to time, mathematically given by $$a = \frac{\Delta v}{\Delta t}$$ where $\Delta v$ is the change in velocity and $\Delta t$ is the change in time. 3. **Analyze the graph:** The graph shows velocity increasing steeply from 0 ft/s at 0 minutes to about 6 ft/s at 4 minutes, then decreasing gradually until 9 minutes. 4. **Calculate acceleration for each interval:** - Interval A (0 to 2 minutes): Velocity increases from 0 to approximately 3 ft/s. $$a = \frac{3 - 0}{2 - 0} = \frac{3}{2} = 1.5\, \text{ft/s}^2$$ - Interval B (3 to 4.5 minutes): Velocity increases from about 4.5 ft/s to 6 ft/s. $$a = \frac{6 - 4.5}{4.5 - 3} = \frac{1.5}{1.5} = 1.0\, \text{ft/s}^2$$ - Interval C (6.7 to 9 minutes): Velocity decreases from about 3 ft/s to 1 ft/s. $$a = \frac{1 - 3}{9 - 6.7} = \frac{-2}{2.3} \approx -0.87\, \text{ft/s}^2$$ 5. **Interpretation:** The greatest acceleration in magnitude is during interval A with $1.5\, \text{ft/s}^2$. **Final answer:** The teenager was moving with the greatest acceleration during the interval from 0 minutes to 2 minutes (Option A).