1. **Problem Statement:** Estimate the bicyclist's average velocity over given intervals, instantaneous velocity at specific times, and maximum velocity from the distance-time graph. 2. **Formulas and Rules:** - Average velocity over interval $[t_1, t_2]$ is given by $$\text{Average velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$$ where $s(t)$ is the distance at time $t$. - Instantaneous velocity at time $t$ is the derivative $s'(t)$, which can be approximated by the slope of the tangent line or by average velocities over very small intervals around $t$. - Maximum velocity is the highest instantaneous velocity value. 3. **Part (a) Average Velocities:** - Over $[0,1]$: Given as approximately 15 mi/hr. - Over $[1,2.5]$: Given as approximately 4 mi/hr. - Over $[2.5,3.5]$: Given as approximately 18 mi/hr. 4. **Part (b) Instantaneous Velocities:** - At $t=0.5$ hr: Given as approximately 15 mi/hr. - At $t=2$ hr: Estimate slope near $t=2$. From graph description, distance changes slowly between 1 and 2.5 hours, so velocity is low. Using average velocity over $[1,2.5]$ as 4 mi/hr, instantaneous velocity at 2 hr is approximately 4 mi/hr. - At $t=3$ hr: Estimate slope near $t=3$. From graph, slope is steep between 2.5 and 3.5 hours, average velocity 18 mi/hr. So instantaneous velocity at 3 hr is approximately 18 mi/hr. 5. **Part (c) Maximum Velocity:** - From the graph, the steepest slope is between 2.5 and 3.5 hours with velocity about 18 mi/hr. - So maximum velocity is approximately 18 mi/hr at $t \approx 3$ hr. **Final answers:** - Instantaneous velocity at $t=2$ hr is approximately 4 mi/hr. - Instantaneous velocity at $t=3$ hr is approximately 18 mi/hr. - Maximum velocity is approximately 18 mi/hr at $t \approx 3$ hr.