If s is distance and v is velocity, then: $$v=\frac{ds}{dt}$$and$$a=\frac{dv}{dt}$$ $$\frac{a}{v}=\frac{\frac{dv}{dt}}{\frac{ds}{dt}}$$ $$\frac{a}{v}=\frac{\frac{d}{dt}(\frac{ds}{dt})}{\frac{ds}{dt}}=\frac{dv}{ds}$$

I don't understand the last step .Is it possible? Numerator is first derivative whereas denominator is second derivative. Please explain it . Thanks.


Acceleration is rate of change of velocity.
Velocity is rate of change of displacement.
Acceleration by velocity would be rate of change of velocity by rate of change of displacement.

Let velocity in some small time period t, change by v.
And in the same time period displacement changes by d.
Then a/v = v/d, since the time period is the same.
This is essentially what the last step is trying to convey.

  • $\begingroup$ well,i understand what you say.if we consider dv/dt as small change in v with respect to small change in t,then you are right .i agree. BUT if we consider acceleration as second derivative and velocity as first derivative then how you eliminate dt term??? $\endgroup$ Sep 3 '16 at 0:52
  • $\begingroup$ That dt term is essentially an infinitely small change in time. What I am trying to make you understand is that both the changes in time, correspond to the same value so they do in fact cancel out. $\endgroup$
    – novice
    Sep 3 '16 at 0:53
  • $\begingroup$ well ,i try to understand using the definition of derivative and with the explaination yours i understand it.thanks .i need more help from you when i have doubts $\endgroup$ Sep 3 '16 at 1:06

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