Kinematics derivation: what's "ds/dv" ?

  • #1
Hello, here's a derivation for one of the equations for Uniform Accelerated Motion (UAM). I like it because it's far more concise than the algebra version. Question though: is there any kind of meaningful interpretation of ds/dv (which starts everything off in step 1)? Thank you!

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  • #2
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The derivation is usually done as follows: $$a=\frac{dv}{dt}=\frac{dv}{ds}\frac{ds}{dt}=v\frac{dv}{ds}$$where dv/ds is the rate of change of speed with respect to distance. So ds/dv is the reciprocal of dv/ds, but it doesn't have much fundamental significance.
 
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  • #3
Very helpful. Thank you so much!
 

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