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

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Discussion Overview

The discussion revolves around the equation dV/dS*v=a in the context of projectile motion, specifically examining the implications at the highest point of a thrown ball where the velocity (v) is zero. Participants explore the relationship between velocity, acceleration, and the mathematical interpretation of the equation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that at the highest point of a thrown ball, v=0, which leads to the conclusion that a=0, questioning the validity of this implication.
  • Another participant counters that v=0 does not imply dv/dt=0, suggesting that a graph of velocity against time would clarify this point.
  • Concerns are raised about substituting v=0 into the equation dV/dS*v=a, leading to confusion about the resulting acceleration being zero.
  • Participants discuss the concept of dv/ds approaching infinity when v=0, referencing the behavior of the graph of velocity against displacement.
  • There is mention of constant acceleration and the potential application of the formula a=dv/ds*v, with uncertainty about its validity under these conditions.
  • Questions arise regarding the graphical representation of dv/ds and its implications for understanding the behavior of the system at the highest point.

Areas of Agreement / Disagreement

Participants express differing views on the implications of v=0 in the context of the equation, with no consensus reached on the validity of applying the formula under conditions of constant acceleration or the interpretation of infinity in the context of dv/ds.

Contextual Notes

There are unresolved assumptions regarding the behavior of acceleration and velocity at the highest point of projectile motion, particularly in relation to the mathematical limits and the definitions used in the discussion.

carhah
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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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