Understanding Dimensional Analysis in Basic Linear Motion Formulas

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SUMMARY

This discussion focuses on dimensional analysis in basic linear motion formulas, specifically examining the relationships between units in equations such as \( S = \frac{1}{2}at^2 \) and \( v^2 = 2as \). The participant explores how manipulating these formulas in terms of units leads to apparent contradictions, such as deriving \( m = \frac{1}{2}m \). The conversation highlights the importance of understanding dimensional consistency and the role of dimensionless constants in physics equations.

PREREQUISITES
  • Basic understanding of physics concepts, particularly kinematics.
  • Familiarity with fundamental motion formulas like \( S = \frac{1}{2}at^2 \) and \( v = at \).
  • Knowledge of dimensional analysis techniques.
  • Ability to manipulate algebraic expressions involving units.
NEXT STEPS
  • Study the principles of dimensional analysis in physics.
  • Learn how to apply dimensional analysis to verify the consistency of complex equations.
  • Explore the significance of dimensionless constants in physics formulas.
  • Investigate advanced kinematic equations and their derivations.
USEFUL FOR

This discussion is beneficial for new physics students, educators teaching kinematics, and anyone interested in the application of dimensional analysis in understanding motion equations.

lozzajp
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One thing I enjoy as a new physics student is using formulas. I have been playing with the fundamental formulas in terms of units rather than their names.

That is instead of a=v/t
m/s/s=(m/s)/s

or v=at
m/s=(m/s/s) x s
m/s=m/s

Simple maths yea, but when I am getting to the bigger formulas I am having trouble, perhaps they just don't work the way i am thinking but if someone could enlighten me that would be great..

S=(1/2)at^2
m=(1/2)x(m/s^2)x(s^2)
m=(1/2)x(m)

why do i get m = half m?

v^2 = 2as
m^2/s^2 = 2(m/s/s) x m
m^2/s^2 = 2 x m^2/s^2
m/s = 2 x m/s

again, there is a x2 factor there.

am i just thinking about this too much?
 
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If you are only looking at the units you are throwing away any "dimensionless" constants in the formulas. From that point of view, there is not difference between "m" and "m/2"- both have dimensions of "meters" which is all you are looking at.
 

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