## dV/dS*v = a

dV/dS*v=a
now at the higest point when we throw a ball.
v=0
which implies a=0

but that is npot true....any expanations?
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hi carhah! welcome to pf!
 Quote by carhah v=0 which implies a=0
no, v= 0 does not imply dv/dt = 0

(draw a graph of v against t)
 but if we put v=0 in equation-dV/dS*v=a we get a=0? i am in doubt...

## dV/dS*v = a

 Quote by tiny-tim hi carhah! welcome to pf! no, v= 0 does not imply dv/dt = 0 (draw a graph of v against t)

but if we put v=0 in equation-dV/dS*v=a

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hi carhah!

(you don't need to send a pm … everyone gets automatic email notification of any new post in any thread they've posted in )
 Quote by carhah but if we put v=0 in equation-dV/dS*v=a
but dv/ds = ∞ (draw the graph of v against s)

eg, if a is constant, and if so = vo = 0, then s = 1/2 at2 and v = at,

so v = √(2as), so dv/ds|t=0 = √(a/so)= √(a/0) = ∞

 Quote by tiny-tim hi carhah! (you don't need to send a pm … everyone gets automatic email notification of any new post in any thread they've posted in ) but dv/ds = ∞ (draw the graph of v against s) eg, if a is constant, and if so = vo = 0, then s = 1/2 at2 and v = at, so v = √(2as), so dv/ds|t=0 = √(a/so)= √(a/0) = ∞

thanks tim :D
BUT DOES that mean that we cannot apply that formula a=dv/ds*v in case of constant acceleration. :)

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 Quote by carhah thanks tim :D BUT DOES that mean that we cannot apply that formula a=dv/ds*v in case of constant acceleration. :)
no, you can always apply that formula …
but if it says a = ∞*0, that doesn't help very much!
 can u draw graph of dv/ds ? and prove it is infinity?

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 Quote by carhah can u draw graph of dv/ds ?
you can draw a graph of v against s (as i've already asked you to) …
where it's vertical, dv/ds = ∞

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