If a particle is in equilibrium is the initial speed always 0?

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dont know it
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#1
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#1
I get that there's no acceleration if the particle is in equilibrium and that the particle is either stationary or moving at a constant velocity.

Am I wrong in saying neither of those prove the initial speed must be 0?

Essentially the question I'm referring to the particle is originally in equilibrium before a force is removed. I've worked out the acceleration, am given that t=3, and I'm meant to find out the distance travelled by the particle, so of course I need either the initial speed, final velocity. The answer says u=0 which prompted me to ask this question.
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SYEPHEN17
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#2
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You are exactly right. Can be constant velocity or at rest. Hard to give a sure answer here for this specific question. Check the wording to look for any hints
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Notnek
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#3
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(Original post by dont know it)
I get that there's no acceleration if the particle is in equilibrium and that the particle is either stationary or moving at a constant velocity.

Am I wrong in saying neither of those prove the initial speed must be 0?

Essentially the question I'm referring to the particle is originally in equilibrium before a force is removed. I've worked out the acceleration, am given that t=3, and I'm meant to find out the distance travelled by the particle, so of course I need either the initial speed, final velocity. The answer says u=0 which prompted me to ask this question.
Please post the full question.
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RichE
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#4
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(Original post by dont know it)
I get that there's no acceleration if the particle is in equilibrium and that the particle is either stationary or moving at a constant velocity.

Am I wrong in saying neither of those prove the initial speed must be 0?

Essentially the question I'm referring to the particle is originally in equilibrium before a force is removed. I've worked out the acceleration, am given that t=3, and I'm meant to find out the distance travelled by the particle, so of course I need either the initial speed, final velocity. The answer says u=0 which prompted me to ask this question.
I would normally understand "equilibrium" to mean that the particle is in the same place for all time, so necessarily velocity and acceleration are zero.
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#5
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(Original post by SYEPHEN17)
You are exactly right. Can be constant velocity or at rest. Hard to give a sure answer here for this specific question. Check the wording to look for any hints
The question says to give the distance after the first 3 seconds of motion. That means we assume it was stationary when in equilibrium right, and so we know from that u must be 0.
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#6
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(Original post by Notnek)
Please post the full question.
Think the answer I was looking for was in fact in the question, it asks to find the distance travelled during the first 3 seconds of motion, so we know that u=0. Correct me if I'm wrong
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Notnek
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#7
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(Original post by dont know it)
Think the answer I was looking for was in fact in the question, it asks to find the distance travelled during the first 3 seconds of motion, so we know that u=0. Correct me if I'm wrong
Can you please post the question in full i.e. post the exact wording that you're seeing in e.g. your textbook.
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#8
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#8
(Original post by RichE)
I would normally understand "equilibrium" to mean that the particle is in the same place for all time, so necessarily velocity and acceleration are zero.
I think Newton's first law states the particle can also be moving at a constant velocity. By definition, acceleration is the rate of change of velocity. Even though acceleration may be 0, it doesn't necessarily mean the velocity is 0, instead the rate of change of velocity is 0. So for example, on a V-T graph, we may have a straight line showing us a particle moves at a constant velocity of 5m/s for 10 seconds. Although acceleration would be 0(because the velocity hasn't changed during this time), the velocity is actually moving at a constant 5m/s. Hope that makes sense!
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#9
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(Original post by Notnek)
Can you please post the question in full i.e. post the exact wording that you're seeing in e.g. your textbook.
Oh sorry.

In this question i and j represent the unit vectors east and north respectively.

The forces (3ai + 4bj) N, (5bi + 2aj) N and (-15i-18j)N act on a particle of mass 2kg which is in equilibrium.

a)find the values of a and b.
b) The force (-15i-18j) is removed. Work out:
i)The magnitude and direction of the resulting acceleration of the particle.
ii)The distance travelled by the particle in the first 3 seconds of its motion.
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oShahpo
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#10
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#10
Equilibrium in this context normally refers to equilibrium of forces. At which point, there is no net force. This means that the particle is at a constant velocity (which could be zero, implying particle is at rest.) So in this sense, you are right, equilibria don't necessitate rest. The question should have been clearer, good spot.
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RichE
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#11
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(Original post by Notnek)
So you're saying that for you, non-zero constant speed would not mean that a particle was in equilibrium? Or am I misunderstanding you?
Yes, that's how I'd usually use the term. I guess in line with

https://en.wikipedia.org/wiki/Equilibrium_point

but I can imagine the word may get used in different ways.

PS now reading the given question, it doesn't matter whether the particle is in motion or not.
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Notnek
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#12
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(Original post by RichE)
Yes, that's how I'd usually use the term. I guess in line with

https://en.wikipedia.org/wiki/Equilibrium_point

but I can imagine the word may get used in different ways.
This agrees with the Edexcel (well at least the Edexcel M1 textbook) definition:

Image

although I can't find this definition anywhere else outside of A Level which is a bit odd.

I'm pretty sure that an exam question would never be ambiguous and just say equilibrium if the body is at rest - normally questions say "rests in equilibrium".
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username3262012
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#13
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Im not 100% sure but if a particle is in equilibrium isnt it traveling at 10x the speed of light? (3x10>4 km>-1s^-1) ?
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oShahpo
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#14
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(Original post by RichE)
Yes, that's how I'd usually use the term. I guess in line with

https://en.wikipedia.org/wiki/Equilibrium_point

but I can imagine the word may get used in different ways.

PS now reading the given question, it doesn't matter whether the particle is in motion or not.
Surely it does? At initial speed of 1 million ms^-1 and acceleration of whatever, the particle will come cover a much longer distance in 3 seconds than at initial speed of 0.
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RichE
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#15
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#15
(Original post by oShahpo)
Surely it does? At initial speed of 1 million ms^-1 and acceleration of whatever, the particle will come cover a much longer distance in 3 seconds than at initial speed of 0.
Sorry I only read (b)(i). Yes, whether it's in motion initially or not affects (b)(ii).
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SYEPHEN17
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#16
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(Original post by dont know it)
The question says to give the distance after the first 3 seconds of motion. That means we assume it was stationary when in equilibrium right, and so we know from that u must be 0.
I would assume that it was at rest, as ‘first three seconds of motion’ implies that at t=0 it was not moving, and beginning motion.
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SYEPHEN17
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#17
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(Original post by tome123)
Im not 100% sure but if a particle is in equilibrium isnt it traveling at 10x the speed of light? (3x10>4 km>-1s^-1) ?
No that just means that there is no acceleration or ******ation
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#18
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#18
(Original post by SYEPHEN17)
I would assume that it was at rest, as ‘first three seconds of motion’ implies that at t=0 it was not moving, and beginning motion.
Agreed.
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#19
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#19
(Original post by Notnek)
This agrees with the Edexcel (well at least the Edexcel M1 textbook) definition:

Image

although I can't find this definition anywhere else outside of A Level which is a bit odd.

I'm pretty sure that an exam question would never be ambiguous and just say equilibrium if the body is at rest - normally questions say "rests in equilibrium".
It actually says in the textbook for the new specification that the particle could be at rest or moving at constant velocity so maybe they've corrected it, however yes you're right the question would never be ambiguous like that.
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