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A ship travels 200 miles due west, then adjusts its course 30° north of west. The ship continues on this course for 30 miles. Approximately how far is the ship from where it began?

[tex] \approx 1031 \, \text{mi} [/tex]

Sagot :

To solve this problem, we need to determine the total displacement of the ship from its starting point after traveling in two different legs.

First leg of the journey:
1. The ship travels 200 miles due west.

Second leg of the journey:
1. The ship then adjusts its course and travels 30 miles on a bearing of 30° north of west.

To find the ship’s final position, we need to break down the second leg of the journey into its component vectors. This involves computing the x (westward) and y (northward) components of its travel.

1. Angle Conversion: The angle given is 30°. We should convert this angle into radians:
- Angle in radians = 0.5235987755982988

2. Calculate the components of the north-west vector:
- The x-component of the movement can be found using [tex]\( \cos \)[/tex] of the angle:
[tex]\[ x \text{ (westward component)} = 30 \times \cos(30^\circ) \approx 25.98076211353316 \text{ miles} \][/tex]
- The y-component of the movement can be found using [tex]\( \sin \)[/tex] of the angle:
[tex]\[ y \text{ (northward component)} = 30 \times \sin(30^\circ) \approx 15 \text{ miles} \][/tex]

3. Total movement in the x-direction:
- The total x-distance is the sum of the initial westward travel and the westward component of the north-west travel:
[tex]\[ 200 \text{ miles (initial westward travel)} + 25.98076211353316 \text{ miles (x-component)} \approx 225.98076211353316 \text{ miles} \][/tex]

4. Total movement in the y-direction:
- Since the initial y-distance is zero (as the ship only traveled west), the total y-distance is:
[tex]\[ y \text{ (northward component)} \approx 15 \text{ miles} \][/tex]

5. Combine these to find the resultant distance:
- Using the Pythagorean theorem to find the total distance from the starting point:
[tex]\[ \text{Total distance} = \sqrt{ (225.98076211353316)^2 + (15)^2} \approx 226.4780449522939 \text{ miles} \][/tex]

Thus, the ship is approximately 226.48 miles from where it began.