Solving for Displacement & Avg Velocity of Ocean Swimmer

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SUMMARY

The discussion focuses on calculating the displacement and average velocity of an ocean swimmer based on their position relative to a lighthouse. The swimmer starts 5 km from the coastline with a bearing of N70E and ends up 5 km due west of the lighthouse after 30 minutes. The calculated displacement is 1.82 km using trigonometric functions, specifically tan(70) = 5/x, leading to x = 1.82. The average velocity can be derived from this displacement over the time interval of 30 minutes.

PREREQUISITES
  • Understanding of basic trigonometry, specifically tangent functions.
  • Familiarity with vector displacement concepts.
  • Knowledge of average velocity calculations.
  • Ability to interpret bearings and directional movement.
NEXT STEPS
  • Study trigonometric functions and their applications in physics.
  • Learn about vector addition and displacement in two dimensions.
  • Explore average velocity calculations in different contexts.
  • Review nautical navigation principles, including bearings and distances.
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This discussion is beneficial for physics students, educators teaching kinematics, and anyone interested in applying trigonometry to real-world scenarios, particularly in navigation and movement analysis.

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Homework Statement


At a particular time, a long distance ocean swimmer is 5km from a coast line that stretches in a north south direction. One of the passengers on the boat accompanying the swimmer determines that a lighthouse on the coastline has a bearing of N70E. After 30 minutes the swimmer is 5km due west of the lighthouse.
a) What is the displacement of the swimmer between the two measurements?
b) What is the average velocity of the swimmer?


Homework Equations




The Attempt at a Solution


I drew a diagram with this question but if going according to the scenario, it would mean the swimmer ends up on the other side of the shore. But how thick is the shore? I don't think my diagram is correct. Should the last sentence contain the word east instead of west?
 
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No, I think it is correct. I think it means the swimmer is 5 km west from the point of view of the lighthouse, not the other way around.
 
In that case, the swimmer hasn't changed position longitudinally (i.e. still 5km from the coastline). Simple trig shows that the displacement is 1.82km with tan(70)=5/x => x=1.82 which is what the answers suggested. So that's that figured out.
 

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