Part 1In searching for a boat lost at sea, a Coast Guard cutter leaves a port and travels 9595.0 mi due east. It then turns 3535degrees° north of east and travels another 9595.0 mi, and finally turns another 3535degrees° toward the west and travels another 9595.0 mi. What is its displacement from the port? Part 1The distance from the port is [input]enter your response here miles.(Round to the nearest mile as needed.)

Question

Part 1In searching for a boat lost at​ sea, a Coast Guard cutter leaves a port and travels 9595.0 mi due east. It then turns 3535degrees° north of east and travels another 9595.0 ​mi, and finally turns another 3535degrees° toward the west and travels another 9595.0 mi. What is its displacement from the​ port? Part 1The distance from the port is [input]enter your response here miles.​(Round to the nearest mile as​ needed.)

BlackTom AI · Accepted Answer

First, restate the problem in my own words to ensure understanding: the Coast Guard travels three straight legs of equal length, each 9595.0 miles, in directions east, then 35° north of east, then 35° toward the west. We are asked for the straight-line displacement from the starting port to the final position, i.e., the magnitude of the resultant vector after summing all three legs, and to round to the nearest mile.

Now analyze each leg as a vector in components (x is east/west, y is north/south):
- Leg 1: 9595.0 mi due east. This contributes x1 = +9595.0, y1 = 0.
- Leg 2: 9595.0 mi at 35° north of east. This has x2 = 9595 cos(35°) and y2 = 9595 sin(35°).
  Using approximate values cos(35°) ≈ 0.819152 and sin(35°) ≈ 0.573576, we get x2 ≈ 9595 × 0.819152 ≈ 7859.8 and y2 ≈ 9595 × 0.573576 ≈ 5500 (more precisely around 5500–5505).
- Leg 3: 9595.0 mi turned 35° toward the west from the previous direction, which places this leg at 145° from the +x axis (east). Its components are x3 = 9595 cos(145°) and y3 = 9595 sin(145°). Since cos(145°) = −cos(35°) and sin(145°) = sin(35°), we have x3 ≈ −9595 cos(35°) ≈ −7859.8 and y3 ≈ 9595 sin(35°) ≈ +5500.

Sum the components to get the overall displacement:
- Total x: x1 + x2 + x3 ≈ 9595.0 + 7859.8 − 7859.8 ≈ 9595.0 miles (the x components from legs 2 and 3 cancel each other, as expected from the symmetric directions).
- Total y: y1 + y2 + y3 ≈ 0 + 5500 + 5500 ≈ 11000 miles.

Thus the displacement vector has components (approximately) (9595, 11000).

Compute the magnitude of the displacement (the straight-line distance from the port):
Distance = sqrt( (9595)^2 + (11000)^2 ).

Calculate the squares: 9595^2 ≈ 92,064,025 and 11,000^2 = 121,000,000. Sum ≈ 213,064,025. The square root of this is approximately 14,600.7 miles.

Rounding to the nearest mile gives about 14,601 miles. The result indicates the final position is roughly 14.6 thousand miles away from the starting port, in a direction that has a large northerly component and an easterly component equal to the first leg (due to the cancellations from legs 2 and 3).

If you were entering a numeric answer, you would input approximately 14601 miles. Note that depending on how you carry intermediate rounding, you may see values very close to this, such as 14600 or 14602, but 14601 is the closest integer.

Homework:Module 11 Homework (Vectors)

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