Understanding Maxium Displacement Problem in Serway 8th Edition Chapter 3

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SUMMARY

The Maximum Displacement Problem in Serway's 8th Edition, Chapter 3, involves solving a triangle using trigonometric principles. The hypotenuse is given as 1.4 m, and the problem can be approached using the Law of Sines, where a/(sin A) = b/(sin B). Additionally, the Pythagorean theorem can be applied to find the other side of the triangle. The perpendicular and parallel components can be calculated using the formulas y = r cos θ and x = r sin θ, respectively.

PREREQUISITES
  • Understanding of trigonometry, specifically the Law of Sines
  • Familiarity with the Pythagorean theorem
  • Basic knowledge of angles and their components in triangles
  • Ability to interpret problems from Serway's Physics textbook
NEXT STEPS
  • Review the Law of Sines and its applications in triangle problems
  • Practice using the Pythagorean theorem in various contexts
  • Study trigonometric functions and their graphical representations
  • Explore additional examples from Serway's Physics textbook for deeper understanding
USEFUL FOR

Students studying physics, particularly those tackling problems in Serway's 8th Edition, as well as anyone needing to strengthen their understanding of trigonometry and its applications in physics problems.

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I understand that i should post all relevant work that I've done, but i don't even know where to start, in the serway 8th edition chapter 3 this problem is listed, i read all of chapter 3 yet still can't even attempt this problem. It's for my online homework which is due monday morning and was assigned last night, so i don't even have the chance to goto the tutors for help. Any and all help would be greatly appreciated.

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This is just trigonometry. They are telling you that your hypotenuse is 1.4 m and then giving you an angle. I believe that you can use the law of sines to solve this problem where a/(sin A) = b/(sin B). Look up law of sines on wikipedia if you don't remember because it will have an illustration that helps.
After you do that once you can just use the Pythagorean to get the other side.

Basically you are just solving for this triangle.
 
I was just thinking about it and I'm pretty sure you can just take y = r cos theta to get the perpendicular component and x = r sin theta to get the parallel component.

Both ways should work. Sorry if I over complicated it.
 

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