Let's break down the problem and solve it step-by-step. ### **Step 1: Understand the Problem** We are given the velocities (in mph) of Chris (\(v_C\)) and Kelly (\(v_K\)) at different seconds during the first 10 seconds of a race. We are to use the **midpoint rule** to estimate how much farther Kelly travels than Chris during these 10 seconds. ### **Step 2: Organize the Data** | t (s) | v_C | v_K | |-------|-----|-----| | 0 | 0 | 0 | | 1 | 22 | 24 | | 2 | 33 | 37 | | 3 | 47 | 53 | | 4 | 55 | 62 | | 5 | 61 | 69 | | 6 | 68 | 74 | | 7 | 74 | 80 | | 8 | 80 | 86 | | 9 | 86 | 92 | | 10 | 89 | 98 | | | |102 | **But:** For the midpoint rule, we use the midpoint velocities between each interval. ### **Step 3: Set Up the Midpoint Rule** - The time interval is from 0 to 10 seconds (\(\Delta t = 1\) second for each interval). - We need to estimate the **distance traveled** by integrating velocity over time: \[ \text{Distance} = \int_0^{10} v(t) \, dt \approx \sum_{i=1}^n v(\text{midpoint}) \cdot \Delta t \] - Each interval: 1 second - There are 10 intervals for 0 to 10 seconds. #### **Convert seconds to hours** Since velocities are in mph, and times are in seconds, we need to convert seconds to hours: \[ 1 \text{ second} = \frac{1}{3600} \text{ hour} \] So, \[ \Delta t = 1 \text{ sec} = \frac{1}{3600} \text{ hour} \] ### **Step 4: Calculate Midpoint Velocities** For each interval, the midpoint is at: - 0.5, 1.5, 2.5, ..., 9.5 seconds The velocity at each midpoint is the average of the velocities at the two endpoints. #### **For Chris (v_C):** - (0+22)/2 = 11 - (22+33)/2 = 27.5 - (33+47)/2 = 40 - (47+55)/2 = 51 - (55+61)/2 = 58 - (61+68)/2 = 64.5 - (68+74)/2 = 71 - (74+80)/2 = 77 - (80+86)/2 = 83 - (86+89)/2 = 87.5 #### **For Kelly (v_K):** - (0+24)/2 = 12 - (24+37)/2 = 30.5 - (37+53)/2 = 45 - (53+62)/2 = 57.5 - (62+69)/2 = 65.5 - (69+74)/2 = 71.5 - (74+80)/2 = 77 - (80+86)/2 = 83 - (86+92)/2 = 89 - (92+98)/2 = 95 ### **Step 5: Calculate Total Distance Using the Midpoint Rule** \[ \text{Distance} \approx \sum_{i=1}^{10} v(\text{midpoint}) \cdot \Delta t \] #### For Chris: \[ D_C \approx (11 + 27.5 + 40 + 51 + 58 + 64.5 + 71 + 77 + 83 + 87.5) \times \frac{1}{3600} \] \[ = 570.5 \times \frac{1}{3600} \] \[ = 0.1585 \text{ miles} \] #### For Kelly: \[ D_K \approx (12 + 30.5 + 45 + 57.5 + 65.5 + 71.5 + 77 + 83 + 89 + 95) \times \frac{1}{3600} \] \[ = 626 \times \frac{1}{3600} \] \[ = 0.1739 \text{ miles} \] ### **Step 6: How Much Farther Did Kelly Go?** \[ \Delta D = D_K - D_C = 0.1739 - 0.1585 = 0.0154 \text{ miles} \] ### **Final Answer (Rounded to four decimal places):** \[ \boxed{0.0154 \text{ miles}} \] --- **Summary:** - Find midpoint velocities for each interval for both Chris and Kelly. - Apply the midpoint rule, multiplying by the interval (converted to hours). - Subtract Chris's distance from Kelly's to get the answer. **Final answer:** **0.0154 miles**