Paradox of dV/dS*v=a at Highest Point

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The discussion centers on the equation dV/dS*v=a and its implications at the highest point of a thrown ball, where velocity (v) equals zero. Participants clarify that v=0 does not imply acceleration (a) is also zero, as acceleration can still exist even when velocity is momentarily at rest. The conversation highlights that substituting v=0 into the equation leads to a situation where acceleration appears undefined or infinite, raising questions about the applicability of the formula under constant acceleration conditions. Graphical representations of velocity against time and displacement are suggested to illustrate these concepts more clearly. Ultimately, the participants agree that while the formula can be applied, its interpretation at specific points, like when v=0, can lead to complexities.
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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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welcome to pf!

hi carhah! welcome to pf! :smile:
carhah said:
v=0
which implies a=0

no, v= 0 does not imply dv/dt = 0 :wink:

(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...
 


tiny-tim said:
hi carhah! welcome to pf! :smile:


no, v= 0 does not imply dv/dt = 0 :wink:

(draw a graph of v against t)


but if we put v=0 in equation-dV/dS*v=a
 
hi carhah! :smile:

(you don't need to send a pm … everyone gets automatic email notification of any new post in any thread they've posted in :wink:)
carhah said:
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) = ∞ :wink:
 


tiny-tim said:
hi carhah! :smile:

(you don't need to send a pm … everyone gets automatic email notification of any new post in any thread they've posted in :wink:)


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) = ∞ :wink:





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


can u draw graph of dv/ds ? and prove it is infinity?
 
carhah said:
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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