Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Jun 8, 2018 #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!

    kinematics.jpg
     
  2. jcsd
  3. Jun 8, 2018 #2
    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.
     
  4. Jun 8, 2018 #3
    Very helpful. Thank you so much!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted