Let's break down the problem step by step: --- ## **Given Data** - **Landing speed:** 150 mph - **Slows down to:** 110 mph at a deceleration of -10 mph/min - **Remains at:** 110 mph (constant speed) - **Accelerates back up to:** 150 mph at 10 mph/min - **Total sequence time:** 13 minutes - **Longest runway in the world:** 5 km (≈ 3 miles) --- ## **Step 1: Calculate Time Spent in Each Segment** ### **Segment 1: Deceleration (150 mph to 110 mph at -10 mph/min)** - Speed change = 150 mph - 110 mph = 40 mph - Time = Speed change / Rate \( \text{Time} = \frac{40 \text{ mph}}{10 \text{ mph/min}} = 4 \text{ min} \) **Explanation:** It takes 4 minutes to slow from 150 mph to 110 mph at a deceleration of 10 mph per minute. --- ### **Segment 2: Acceleration (110 mph to 150 mph at 10 mph/min)** - Speed change = 150 mph - 110 mph = 40 mph - Time = Speed change / Rate \( \text{Time} = \frac{40 \text{ mph}}{10 \text{ mph/min}} = 4 \text{ min} \) **Explanation:** It takes 4 minutes to accelerate from 110 mph to 150 mph at an acceleration of 10 mph per minute. --- ### **Segment 3: Constant Speed** - Total time = 13 min - Time spent decelerating and accelerating = 4 min + 4 min = 8 min - Time at constant speed = 13 min - 8 min = 5 min **Explanation:** The plane spends 5 minutes traveling at 110 mph between slowing down and speeding up. --- ## **Step 2: Calculate Distance for Each Segment** ### **Segment 1: Deceleration (Uniformly Changing Speed)** - Formula for distance under constant acceleration: \( d_1 = \frac{v_i + v_f}{2} \times t \) Where: \( v_i = 150 \) mph, \( v_f = 110 \) mph, \( t_1 = 4 \) min \( = \frac{4}{60} \) hr \( d_1 = \frac{150 + 110}{2} \times \frac{4}{60} = 130 \times .0667 = 8.67 \) miles --- ### **Segment 2: Constant Speed** - Distance = Speed × Time \( d_2 = 110 \times \frac{5}{60} = 110 \times .0833 = 9.17 \) miles --- ### **Segment 3: Acceleration (Uniformly Changing Speed)** - Again, average speed = (initial + final) / 2 = (110 + 150) / 2 = 130 mph Time = 4 min = .0667 hr \( d_3 = 130 \times .0667 = 8.67 \) miles --- ### **Total Distance** - \( \text{Total runway length} = d_1 + d_2 + d_3 = 8.67 + 9.17 + 8.67 = 26.51 \) miles --- ## **Final Answers** ### **(i) How long is the runway that the plane was on?** **Answer:** The runway would need to be approximately **26.5 miles** long. --- ### **(ii) Could the sequence happen in real life?** - **Longest real runway:** 3 miles - **Required runway:** 26.5 miles **Answer:** No, the sequence could not happen in real life—the required runway is nearly **nine times longer** than the longest runway in the world. --- **Summary Table:** | Segment | Time (min) | Distance (mi) | |----------------------|------------|--------------| | Deceleration | 4 | 8.67 | | Constant speed | 5 | 9.17 | | Acceleration | 4 | 8.67 | | **Total** | **13** | **26.51** |