
#1
Aug2612, 02:53 PM

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If I throw a ball straight up, at its highest point, is it momentarily at rest? Its velocity is momentarily 0, but it is still accelerating.




#2
Aug2612, 03:02 PM

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#3
Aug2612, 03:02 PM

P: 852

I would consider rest to be v=0, regardless of what a or any other higher derivative is. I think this is pretty standard.




#4
Aug2612, 03:02 PM

Mentor
P: 22,001

What does "rest" mean?
I always took it to mean not moving, so the object can still be accelerating.




#5
Aug2612, 03:53 PM

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#6
Aug2612, 04:09 PM

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But does "not moving" mean v=0? Rather, could moving mean that you are not where you were some amount of time ago? In other words, in the following formula, does Δy = 0 represent not moving, or does v_{0} = 0 mean not moving?
[tex]\Delta y = v_{0}t + 0.5 at^{2}[/tex] Any non0 number for t will result in an accelerating object to have a nonzero Δy, even if v_{0} = 0. My curiosity is from a problem where 5 freebody diagrams are shown: A, B, C, D, E. It asks "Which of the following could represent an object at rest?" If we take v_{0} = 0 to be rest, then the answer is "all of them". If we take Δy = 0 to be rest, then the answer is only the ones with balanced forces. 



#7
Aug2612, 04:32 PM

P: 253

Isn't v[sub]0[\sub] the initial velocity, and (at) from the .5at^2 is the v(current), since the velocity only starts at 0 of course there is a change in y. In what situation can you have Δy!=0 and v = 0? I would consider v = Δy/t (if we are talking that one dimension only)




#8
Aug2612, 04:39 PM

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Rather than considering t being the time I threw the ball straight up, I was thinking of initial time being the moment of its highest point. So v_{0} = 0 and t is actually Δt, some nonzero amount of time after the instant of the highest point.




#9
Aug2612, 06:38 PM

P: 253

Meh, I am confused...say t=0 is the time it's at the top, at that instant v0 = 0. So if we are talking a nonzero Δt, then for that time it accelerated a, so v = a(Δt)... Moving means exactly not being where you were some time ago, which in a 1d world, means exactly having a Δy in some Δt.... v = Δy/Δt... so far as I know having a velocity is the only way to accomplish a change in y...
To the main point, at rest means v = 0, it means that the distance between the object at rest and the reference point is constant 



#10
Aug2712, 05:57 AM

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#11
Aug2712, 06:41 AM

P: 5,634

yes,no, wait......., no...er...... who knows?? One could argue the average acceleration Δv/Δt is not zero while dv/dt is zero....oh well, let's just say 'words are often ambiguous'... My old Halliday and Resnick does not even have 'rest' or 'at rest' in the index. 



#12
Aug2712, 07:57 AM

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#13
Aug2712, 08:01 AM

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"At rest" means that, at least momentarily, the velocity is 0. That has nothing at all to do with the acceleration.




#14
Aug2712, 08:06 AM

P: 615

Rest means v=0 in some chosen frame of reference, regardless of acceleration, jerk or whatever other higher derivatives you want to concider.
You could define 'super rest' or something, if you wanted, which would mean v=0 and a=0 but I'm guessing from it's lack of useage that it's not that useful of a concept. I think the problem here is that you are doing physics with no calculus, go look up some khan academy videos and learn aboud calculus and then stationary points. 



#15
Aug2712, 08:32 PM

P: 961

Very good information.
Since there is no icon that I click that I can follow this tread, then i have to make a reply. Since there is no Δy in Δt for v=0 but for acceleration Δy is not zero in (Δt)^{2} at the top the flight. 



#16
Aug2812, 07:01 AM

P: 615





#17
Aug2812, 07:49 AM

P: 961




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