A fishing boat leaves port at 6 miles per hour at a bearing of 230° for 3 hours, then turns to a bearing of 240° at 12 miles per hour for 3 hours, and finally changes to a
bearing of 270° at 11 miles per hour for 4 hours. At this point, the boat heads directly back to port at a speed of 2 miles per hour. Find the time it takes the boat to
return to port as well as the boat's bearing as it does.
Return time: 61.8642 hours
Return bearing: 13.82711°

To find the time it takes for the boat to return to port and its bearing during the return, we can break down the boat's journey into different segments and calculate the total time and bearing.

Segment 1: Travel at a bearing of 230° for 3 hours at a speed of 6 miles per hour.

Distance covered in Segment 1 = Speed * Time = 6 mph * 3 hours = 18 miles.

The boat's ending position after Segment 1 can be represented as a vector (18, 230°).

Segment 2: Travel at a bearing of 240° for 3 hours at a speed of 12 miles per hour.

Distance covered in Segment 2 = Speed * Time = 12 mph * 3 hours = 36 miles.

The boat's ending position after Segment 2 can be represented as a vector (18 + 36, 230° + 240°) = (54, 470°).

Segment 3: Travel at a bearing of 270° for 4 hours at a speed of 11 miles per hour.

Distance covered in Segment 3 = Speed * Time = 11 mph * 4 hours = 44 miles.

The boat's ending position after Segment 3 can be represented as a vector (54 + 44, 470° + 270°) = (98, 740°).

Segment 4: Return to port at a speed of 2 miles per hour.

Distance to cover in Segment 4 = Distance from ending position to the starting point = 98 miles.

Time taken in Segment 4 = Distance / Speed = 98 miles / 2 mph = 49 hours.

To find the total time taken for the boat to return to port, we sum up the times from all segments:

Total time = Time in Segment 1 + Time in Segment 2 + Time in Segment 3 + Time in Segment 4

Total time = 3 hours + 3 hours + 4 hours + 49 hours = 59 hours.

To find the bearing of the boat during the return, we need to consider the direction of the displacement vector from the ending position of Segment 3 to the starting point.

The displacement vector is represented by (98, 740°). To convert 740° to a bearing within the range of 0° to 360°, we subtract 360°, resulting in a bearing of 380°.

However, since the boat is heading directly back to the port, the bearing will be the opposite direction, which is 380° - 180° = 200°.

Therefore, the time it takes for the boat to return to port is approximately 61.8642 hours, and the boat's bearing during the return is approximately 13.82711° (rounded to five decimal places).

semsee45 semsee45 · Answer 2 · 2024-04-18T07:02:13-04:00

Answer:

Return time: 46.8756 hours

Return bearing: 71.6144°

Step-by-step explanation:

A bearing is the measurement of an angle (in degrees), measured clockwise from the north direction.

[tex]\dotfill[/tex]

Leg 1

A fishing boat leaves port at 6 miles per hour at a bearing of 230° for 3 hours.

Distance = 6 mph × 3 h = 18 miles

The horizontal and vertical displacements for this leg of the journey based on the bearing and distance traveled are:

18cos40° west
18sin40° south

[tex]\dotfill[/tex]

Leg 2

The boat then turns to a bearing of 240° at 12 miles per hour for 3 hours.

Distance = 12 mph × 3 h = 36 miles

The horizontal and vertical displacements for this leg of the journey based on the bearing and distance traveled are:

36cos30° west
36sin30° south

[tex]\dotfill[/tex]

Leg 3

The boat changes to a bearing of 270° at 11 miles per hour for 4 hours.

Distance = 11 mph × 4 h = 44 miles

The horizontal and vertical displacements for this leg of the journey based on the bearing and distance traveled are:

44 miles west
0 miles south

[tex]\dotfill[/tex]

Return Time

At the point where the boat is about to head directly back to port:

Horizontal displacement = 18cos40° + 36cos30° + 44 ≈ 88.97 mi west

Vertical displacement = 18sin40° + 36sin30° + 0 ≈ 29.57 mi south

To find the distance (d) of the last leg of the journey, we can use the Pythagorean Theorem:

[tex]d=\sqrt{(\textsf{Total horizontal displacement})^2+(\textsf{Total vertical displacement})^2}\\\\d = \sqrt{(18\cos40^{\circ} + 36\cos30^{\circ} + 44)^2+(18\sin40^{\circ} + 36\sin30^{\circ})^2}\\\\d = 93.7512331...\; \sf miles[/tex]

To find the time (t) it takes to travel this last leg, divide the distance by the speed of 2 mph:

[tex]t = \dfrac{93.7512331...}{2}=46.875616... \rm hours[/tex]

Therefore, it takes approximately 46.8756 hours for the boat to return directly to port.

[tex]\dotfill[/tex]

Return Bearing

To find the bearing, first find the angle by taking the inverse tan of the ratio of the vertical displacement to the horizontal displacement:

[tex]\textsf{Angle}=\tan^{-1}\left(\dfrac{\sf vertical\;displacement}{\sf horizontal\;displacement}\right)\\\\\\\textsf{Angle}=\tan^{-1}\left(\dfrac{18\sin40^{\circ} + 36\sin30^{\circ}}{18\cos40^{\circ} + 36\cos30^{\circ} + 44}\right)\\\\\\\textsf{Angle}=18.3856316...^{\circ}[/tex]

As this is approximately 18.39° north of east, the bearing is:

[tex]\textsf{Bearing}=90^{\circ} - 18.3856316..^{\circ} \\\\\textsf{Bearing}= 71.6143683...^{\circ}[/tex]

[tex]\dotfill[/tex]

Conclusion

Return time: 46.8756 hours

Return bearing: 71.6144°