1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
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