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Home » At 30 m: \( 1.21 \times 1.1 = 1.331 \) atm - Veritas Home Health

At 30 m: \( 1.21 \times 1.1 = 1.331 \) atm - Veritas Home Health

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Thus, the bird reaches its maximum altitude at $ \boxed{3} $ minutes after takeoff.Question: A precision agriculture drone programmer needs to optimize the route for monitoring crops across a rectangular field measuring 120 meters by 160 meters. The drone can fly in straight lines and covers a swath width of 20 meters per pass. To minimize turn-around time, it must align each parallel pass with the shorter side of the rectangle. What is the shortest total distance the drone must fly to fully scan the field?

📌 Solution: The field is 120 meters wide (short side) and 160 meters long (long side). To ensure full coverage, the drone flies parallel passes along the 120-meter width, with each pass covering 20 meters in the 160-meter direction. The number of passes required is $\frac{120}{20} = 6$ passes. Each pass spans 160 meters in length. Since the drone turns at the end of each pass and flies back along the return path, each pass contributes $160 + 160 = 320$ meters of travel—except possibly the last one if it doesn’t need to return, but since every pass must be fully flown and aligned, the drone must complete all 6 forward and 6 reverse segments. However, the problem states it aligns passes to scan fully, implying the drone flies each pass and returns, so 6 forward and 6 backward segments. But optimally, the return can be integrated into flight planning; however, since no overlap or efficiency gain is mentioned, assume each pass is a continuous straight flight, and the return is part of the route. But standard interpretation: for full coverage with back-and-forth, there are 6 forward passes and 5 returns? No—problem says to fully scan with aligned parallel passes, suggesting each pass is flown once in 20m width, and the drone flies each 160m segment, and the turn-around is inherent. But to minimize total distance, assume the drone flies each 160m segment once in each direction per pass? That would be inefficient. But in precision agriculture standard, for 120m width, 6 passes at 20m width, the drone flies 6 successive 160m lines, and at the end turns and flies back along the return path—typically, the return is not part of the scan, but the drone must complete the loop. However, in such problems, it's standard to assume each parallel pass is flown once in each direction? Unlikely. Better interpretation: the drone flies 6 passes of 160m each, aligned with the 120m width, and the return from the far end is not counted as flight since it’s typical in grid scanning. But problem says shortest total distance, so we assume the drone must make 6 forward passes and must return to start for safety or data sync, so 6 forward and 6 return segments. Each 160m. So total distance: $6 \times 160 \times 2 = 1920$ meters. But is the return 160m? Yes, if flying parallel. But after each pass, it returns along a straight line parallel, so 160m. So total: $6 \times 160 \times 2 = 1920$. But wait—could it fly return at angles? No, efficient is straight back. But another optimization: after finishing a pass, it doesn’t need to turn 180 — it can resume along the adjacent 160m segment? No, because each 160m segment is a new parallel line, aligned perpendicular to the width. So after flying north on the first pass, it turns west (180°) to fly south (return), but that’s still 160m. So each full cycle (pass + return) is 320m. But 6 passes require 6 returns? Only if each turn-around is a complete 180° and 160m straight line. But after the last pass, it may not need to return—it finishes. But problem says to fully scan the field, and aligned parallel passes, so likely it plans all 6 passes, each 160m, and must complete them, but does it imply a return? The problem doesn’t specify a landing or reset, so perhaps the drone only flies the 6 passes, each 160m, and the return flight is avoided since it’s already at the far end. But to be safe, assume the drone must complete the scanning path with back-and-forth turns between passes, so 6 upward passes (160m each), and 5 downward returns (160m each), totaling $6 \times 160 + 5 \times 160 = 11 \times 160 = 1760$ meters. But standard in robotics: for grid coverage, total distance is number of passes times width times 2 (forward and backward), but only if returning to start. However, in most such problems, unless stated otherwise, the return is not counted beyond the scanning legs. But here, it says shortest total distance, so efficiency matters. But no turn cost given, so assume only flight distance matters, and the drone flies each 160m segment once per pass, and the turn between is instant—so total flight is the sum of the 6 passes and 6 returns only if full loop. But that would be 12 segments of 160m? No—each pass is 160m, and there are 6 passes, and between each, a return? That would be 6 passes and 11 returns? No. Clarify: the drone starts, flies 160m for pass 1 (east). Then turns west (180°), flies 160m return (back). Then turns north (90°), flies 160m (pass 2), etc. But each return is not along the next pass—each new pass is a new 160m segment in a perpendicular direction. But after pass 1 (east), to fly pass 2 (north), it must turn 90° left, but the flight path is now 160m north—so it’s a corner. The total path consists of 6 segments of 160m, each in consecutive perpendicular directions, forming a spiral-like outer loop, but actually orthogonal. The path is: 160m east, 160m north, 160m west, 160m south, etc., forming a rectangular path with 6 sides? No—6 parallel lines, alternating directions. But each line is 160m, and there are 6 such lines (3 pairs of opposite directions). The return between lines is instantaneous in 2D—so only the 6 flight segments of 160m matter? But that’s not realistic. In reality, moving from the end of a 160m east flight to a 160m north flight requires a 90° turn, but the distance flown is still the 160m of each leg. So total flight distance is $6 \times 160 = 960$ meters for forward, plus no return—since after each pass, it flies the next pass directly. But to position for the next pass, it turns, but that turn doesn't add distance. So total directed flight is 6 passes × 160m = 960m. But is that sufficient? The problem says to fully scan, so each 120m-wide strip must be covered, and with 6 passes of 20m width, it’s done. And aligned with shorter side. So minimal path is 6 × 160 = 960 meters. But wait—after the first pass (east), it is at the far west of the 120m strip, then flies north for 160m—this covers the north end of the strip. Then to fly south to restart westward, it turns and flies 160m south (return), covering the south end. Then east, etc. So yes, each 160m segment aligns with a new 120m-wide parallel, and the 160m length covers the entire 160m span of that direction. So total scanned distance is $6 \times 160 = 960$ meters. But is there a return? The problem doesn’t say the drone must return to start—just to fully scan. So 960 meters might suffice. But typically, in such drone coverage, a full scan requires returning to begin the next strip, but here no indication. Moreover, 6 passes of 160m each, aligned with 120m width, fully cover the area. So total flight: $6 \times 160 = 960$ meters. But earlier thought with returns was incorrect—no separate returnline; the flight is continuous with turns. So total distance is 960 meters. But let’s confirm dimensions: field 120m (W) × 160m (N). Each pass: 160m N or S, covering a 120m-wide band. 6 passes every 20m: covers 0–120m W, each at 20m intervals: 0–20, 20–40, ..., 100–120. Each pass covers one 120m-wide strip. The length of each pass is 160m (the length of the field). So yes, 6 × 160 = 960m. But is there overlap? In dense grid, usually offset, but here no mention of offset, so possibly overlapping, but for minimum distance, we assume no redundancy—optimize path. But the problem doesn’t say it can skip turns—so we assume the optimal path is 6 straight segments of 160m, each in a new

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Subtract (1) from (2):
\[ (8a + 4b + 2c + d) - (a + b + c + d) = 15 - 4 \]
\[ 7a + 3b + c = 11 \quad \text{(E1)} \]
Subtract (2) from (3):
\[ (27a + 9b + 3c + d) - (8a + 4b + 2c + d) = 40 - 15 \]
\[ 19a + 5b + c = 25 \quad \text{(E2)} \]
Subtract (3) from (4):
\[ (64a + 16b + 4c + d) - (27a + 9b + 3c + d) = 85 - 40 \]
\[ 37a + 7b + c = 45 \quad \text{(E3)} \]
Subtract E1 from E2:

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