Find the distance with the relativity problem

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SUMMARY

The discussion centers on calculating the distance Popeye's boat travels north along the west bank while crossing a channel affected by a current and wind. The boat's velocity is 12 km/h at a 45° angle east of north, with a current of 4 km/h flowing south. The calculations reveal that the eastward component of the boat's velocity is approximately 8.485 km/h, which is essential for determining the time taken to cross the channel. The final distance traveled north along the west bank is derived from these velocity components and the time spent crossing.

PREREQUISITES
  • Understanding of vector addition in physics
  • Knowledge of basic trigonometry for resolving velocity components
  • Familiarity with the concept of relative velocity
  • Ability to perform unit conversions (e.g., km to hours)
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  • Calculate the time taken to cross the channel using the eastward component of velocity
  • Explore the effects of wind and current on boat navigation
  • Learn about vector decomposition in physics
  • Investigate real-world applications of relative velocity in sailing
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Students studying physics, particularly in mechanics and vector analysis, as well as anyone interested in navigation and sailing dynamics.

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Homework Statement


Popeye was sailing his boat in a straight channel one half km wide. The steady current flows south at four km/h and a wind blows from the north, parallel to the current. Popeye starts from the west shore. He sets his sail and tiller so that he moves diagonally across and up the channel. The velocity of the boat with respect to the water is twelve km/h in a direction 45° East of North. When he gets very close to the east shore, he stops readjusts the sil and tiller, and establishes an identical upstream tack back toward the west bank (his direction through the water is now 45° West of north). Due to lack of spinach, he was effectively stopped in the water for one whole minute at the turn around. Calculate the distance measured north along the west bank between his start and finish points.

I have created a little description image by myself:
http://s1323.beta.photobucket.com/user/mommadaddy2/media/hhh_zpsa050c917.jpg.html
This image has a error the current is at the same side as wind

The Attempt at a Solution


okay this is how i decided to start this off.
V of current to water = Vcurrent to boat + V boat to water
V = 4 km/h [ S ] + 12 km/h [E 45 N]
V = 4 km/h + 8.485 km/h [ E] + 8.485 km/h [N]
V = 4.485 km/h [ N] + 8.485 km/h [ E]
i do not know what to do next
 
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Can you use your eastward component of velocity to find the time to cross the channel from the west shore to the east shore?
 
TSny said:
Can you use your eastward component of velocity to find the time to cross the channel from the west shore to the east shore?

alright so the velocity is 12 km/h and the distance is 1.5 km
1.5 km/ 12 km/h = 0.125 h
 
Please Wait said:
alright so the velocity is 12 km/h and the distance is 1.5 km
1.5 km/ 12 km/h = 0.125 h

What distance is 1.5 km? I don't believe this is correct.
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Instead, try to answer the following two questions:

(a) How far east does the boat need to travel to get across the channel? The boat will also travel some in the north direction, but just think about how far the boat needs to move eastward to get across.

(b) What is the rate at which the boat moves eastward? (Hint: this is just another way of asking for the eastward component of velocity.) You've already calculated it in your first post.

Use the answers to (a) and (b) to find the time to cross the channel from the west side to the east side.
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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