What is the direction and magnitude of the third leg of the sailor's journey?

In summary, the sailor's journey consisted of sailing 2.00 km east, 3.50 km southeast, and an unknown distance in an unknown direction. The final position was 5.80 km directly east of the starting point. To find the magnitude of the third leg, one would need to find the arctan of the ratio of the y_component/x_component. To find the direction of the third leg, one could either subtract the angle found with that ratio from 90 degrees, or find the arctan of the inverse ratio (x/y).
  • #1
Soaring Crane
469
0
A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of the starting point.

Find the magnitude of the third leg of the journey.
Find the direction of the third leg of the journey.

I found the magnitude of the third leg. My question is the third leg's direction. The direction in degrees must be expressed in degrees east of north. I do not exactly understand how I am to express it like that. All I know is that if one takes the arctan of the ratio of the third leg's y_component/x_component the answer is wrong.

Thanks.
 
Physics news on Phys.org
  • #2
Your almost there. Have you drawn a diagram? Take a closer look at the ratio y_component/x_component . Which angle with this give you? Which angle do you need?
 
  • #3
The third leg is a right triangle's hypotenuse. The angle that was found with the mentioned ratio is horizontal from the rt. angle. Does that mean I subtract the angle found with that ratio from 90 degrees to find the third angle of the triangle?
 
  • #4
Soaring Crane said:
The third leg is a right triangle's hypotenuse. The angle that was found with the mentioned ratio is horizontal from the rt. angle. Does that mean I subtract the angle found with that ratio from 90 degrees to find the third angle of the triangle?
Yes, you can do that or you can simply find the arctan of the inverse ratio (x/y).
 

1. How do you determine the direction of a ship at sea?

The direction of a ship at sea is determined by using a compass, which measures the ship's orientation relative to the Earth's magnetic field. The compass needle will always point to magnetic north, allowing sailors to determine their direction of travel.

2. What is the difference between true north and magnetic north?

True north and magnetic north are two different points on the Earth's surface. True north is the geographic North Pole, while magnetic north is the point where the Earth's magnetic field lines converge. These points are not always aligned, so a compass will point towards magnetic north instead of true north.

3. How do sailors navigate using the stars?

Sailors use the stars to navigate by using a technique called celestial navigation. They use a sextant to measure the angle between a known star and the horizon, which allows them to determine their latitude. By combining this information with their compass heading, they can determine their position at sea.

4. What is dead reckoning and how is it used in navigation?

Dead reckoning is a navigation technique that uses a ship's speed, direction, and time to estimate its current position. It is used as a backup method when other navigation tools, such as a compass or GPS, are not available. Dead reckoning is also known as "deduced reckoning" or "DR."

5. What is the difference between true course and heading?

True course and heading are two different measurements used in navigation. True course is the direction of travel relative to true north, while heading is the direction the ship is pointing relative to magnetic north. To determine their true course, sailors must adjust their heading based on the difference between true north and magnetic north.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Replies
2
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
5K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
6K
Back
Top