What Does Rest Mean? | Acceleration Explained

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In summary, "rest" means "not moving in the chosen frame of reference" and is represented by v=0, regardless of any other higher derivatives such as acceleration or jerk. However, there may be some ambiguity in the usage of the term, as it may also refer to an object remaining at rest for a non-zero time period or "super rest" where both v and a are 0. It is important to understand calculus in order to fully comprehend the concept of rest.
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tony873004
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What does "rest" mean?

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.
 
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  • #2


tony873004 said:
If I throw a ball straight up, at its highest point, is it momentarily at rest?

Yes, "at rest" means "not moving in the chosen frame of reference". In the frame of reference of a spot on the moon, for example, it is never at rest throughout your example, but I'm assuming you mean relative to YOU as the frame of reference.
 
  • #3


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


I always took it to mean not moving, so the object can still be accelerating.
 
  • #5


russ_watters said:
I always took it to mean not moving, so the object can still be accelerating.

Agreed. This is not even limited to the given example of a ball at the top of it's trajectory. In that example the ball is at rest for an instant of zero duration as it reverses direction. If I am in a free falling elevator and have a ball in my hand the ball is at rest in my frame of reference for period of non-zero duration (until we hit the ground). An observer on the ground sees the ball and I accelerating for the duration of the experiment.
 
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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 v0 = 0 mean not moving?

[tex]\Delta y = v_{0}t + 0.5 at^{2}[/tex]

Any non-0 number for t will result in an accelerating object to have a non-zero Δy, even if v0 = 0.

My curiosity is from a problem where 5 free-body diagrams are shown: A, B, C, D, E.
It asks "Which of the following could represent an object at rest?"
If we take v0 = 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


Isn't v0[\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


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 v0 = 0 and t is actually Δt, some non-zero amount of time after the instant of the highest point.
 
  • #9


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 non-zero 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


tony873004 said:
My curiosity is from a problem where 5 free-body diagrams are shown: A, B, C, D, E.
It asks "Which of the following could represent an object at rest?"
If we take v0 = 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.
Yes, in problem statements like this the term "being at rest" is sometimes used for "remaining at rest for a non-zero time period". It's a bit ambiguous and you have to get from the context what is meant.
 
  • #11


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.
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


Naty1 said:
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'...
dv/dt is non-zero. There is absolutely nothing ambiguous about it. The object undergoes constant acceleration. That means dv/dt=-g at all times, regardless of whether v is zero or not.
 
  • #13


"At rest" means that, at least momentarily, the velocity is 0. That has nothing at all to do with the acceleration.
 
  • #14


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


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


azizlwl said:
Very good information.
Since there is no icon that I click that I can follow this tread, then i have to make a reply.

There is, at the top there is a thread tools drop down, then in that there is the option to 'subscribe to thread' :biggrin:
 
  • #17


genericusrnme said:
There is, at the top there is a thread tools drop down, then in that there is the option to 'subscribe to thread' :biggrin:

Thanks so much.
 

1. What is the definition of rest in terms of physics?

In physics, rest refers to the state of an object when it is not moving or experiencing any acceleration. This means that the object's velocity is zero and there is no change in its position over time.

2. How is rest related to acceleration?

Acceleration is the rate of change of an object's velocity over time. When an object is at rest, there is no change in its velocity, and therefore, no acceleration. This means that the object's acceleration is zero when it is at rest.

3. Can an object be at rest while still moving?

No, an object cannot be at rest while still moving. In order for an object to be considered at rest, it must have a velocity of zero. If an object is moving, it has a non-zero velocity and is not at rest.

4. How is rest different from motionlessness?

Rest and motionlessness are often used interchangeably, but they have slightly different meanings in physics. Rest refers to an object's state of having zero velocity, while motionlessness refers to an object's lack of movement. An object can be at rest while still being in motion, such as a ball thrown up in the air at its highest point, but it cannot be motionless while in motion.

5. What are some examples of rest in everyday life?

Some examples of rest in everyday life include a book sitting on a table, a car parked in a parking lot, or a person standing still. In each of these situations, the object or person is not moving and is at rest. However, it's important to note that even when an object appears to be at rest, it is still moving at the atomic level due to thermal motion.

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