Tangential & normal acceleration derivations

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SUMMARY

The discussion focuses on the derivation of tangential and normal acceleration components in physics. The user initially struggles with the relationship between the sides of a triangle and the trigonometric functions involved. They specifically question the use of sine versus tangent in the context of calculating the opposite side of a triangle. Ultimately, the user resolves their confusion, confirming that the correct relationship is indeed sine equals opposite over hypotenuse.

PREREQUISITES
  • Understanding of basic trigonometric functions (sine, cosine, tangent)
  • Familiarity with the concepts of tangential and normal acceleration
  • Knowledge of triangle properties and relationships
  • Basic physics principles related to motion and acceleration
NEXT STEPS
  • Study the derivation of tangential and normal acceleration in circular motion
  • Learn about the applications of trigonometric functions in physics problems
  • Explore advanced topics in kinematics and dynamics
  • Review graphical representations of motion and their mathematical implications
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Students of physics, educators teaching kinematics, and anyone looking to deepen their understanding of motion dynamics and trigonometric applications in physics.

Markel
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I'm trying to understand how my book derives the tangential accelration. I drew a picture because it's kind of confusing to explain.

From the triangle we get the tangential and normal components of the velocity.

What my problem is (and I think, and hope, it's something simple that I just can't see) that in the second equation where they find SQ.

They times they take adjacent x sin(angle) = oposite.

but isn't sin = oposite/hypoteneuse ??

shouldn't the formula be adjacent x tan(angle) = opposite = SQ ?
 

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oops. figured this out. Sorry, this was really silly.
 

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