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

• carhah
In summary: , where it's horizontal, dv/ds = 0 …and if you want to prove it's ∞, use the definition of acceleration: a = dv/dt = dv/ds * ds/dt = dv/ds * v …if v = 0 and a = 0, then dv/ds = 0 …but if v = 0 and a = ∞, then dv/ds = ∞ …so you can't say anything about dv/ds without knowing both v and a …but you still need to think about what is happening when you throw a ball, and what happens at the top of the throw …without that, you won't understand this … :smile
carhah
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?

welcome to pf!

hi carhah! welcome to pf!
carhah said:
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...

tiny-tim said:
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

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 )
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) = ∞

tiny-tim said:
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. :)

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!

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 = ∞​

1. What is the "Paradox of dV/dS*v=a at Highest Point"?

The "Paradox of dV/dS*v=a at Highest Point" is a mathematical concept that describes a situation where the derivative of velocity (dV/dS) multiplied by the velocity (v) is equal to acceleration (a) at the highest point of a trajectory. This paradox highlights the fact that a particle can have a constant velocity while still experiencing acceleration due to the changing direction of its motion.

This paradox contradicts our intuition about motion because we typically think of acceleration as causing a change in velocity, but in this case, the particle has a constant velocity even though it is experiencing acceleration. This challenges our traditional understanding of how motion works.

3. What are some real-life examples of this paradox?

One real-life example of this paradox is a satellite in orbit around the Earth. The satellite is constantly accelerating due to the Earth's gravitational pull, but it maintains a constant velocity as it orbits. Another example is a pendulum at its highest point, where it has a momentary velocity of 0 but is still experiencing acceleration due to gravity.

4. How can this paradox be resolved?

This paradox can be resolved by understanding that acceleration is not solely dependent on changes in velocity, but also on changes in direction. In the case of the satellite, even though its speed remains constant, its direction is constantly changing, resulting in acceleration. Similarly, the pendulum experiences acceleration due to the change in direction of its motion at its highest point.

5. What is the significance of this paradox in science?

This paradox highlights the complexity of motion and challenges our traditional understanding of it. It also serves as a reminder that our intuitions and assumptions may not always align with scientific principles and that further investigation and analysis are necessary to fully understand and explain phenomena.

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