we know that

$$a = \dfrac{dv}{dt}$$

dividing numerator and denominator by $dx$, we get $$a=v\dfrac{dv}{dx}$$ provided that $dx$ is not equal to zero or instantaneous velocity not equal to zero

when I questioned my teacher that this formula implies instantaneous velocity should not be zero or their should be no turnaround points then why we use this formula for deriving equations of particle's position , velocity which is performing SHM, but got no satisfactory answer.

what is wrong in my argument and what are the conditions under which above mentioned equation is not true?

  • $\begingroup$ so if we have v=dx/dt and by your saying that dx is not equal to zero , dt not equal to zero then v=dx/dt can never be zero which os completely wrong where is the flaw $\endgroup$ May 2 '21 at 21:39

What is wrong is assuming that dv/dx is finite when v=0. Try this for motion with uniform acceleration. For example, for object thrown upwards with some initial velocity. At the top turning point the expression fir dv/dx tends to infinity.

  • $\begingroup$ Please use MathJax to write all math. $\endgroup$
    – G. Smith
    May 2 '21 at 21:45
  • $\begingroup$ so why do we use this equation in such cases $\endgroup$ May 2 '21 at 23:31
  • $\begingroup$ @lalittolani Who is we? $\endgroup$ May 3 '21 at 2:28
  • $\begingroup$ @lalittolani i dont think we use the a = v dv/dx equation $\endgroup$ May 3 '21 at 5:19
  • $\begingroup$ hmm i think coorect equation is a/v=dv/dx $\endgroup$ May 3 '21 at 6:34

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