8
   

Can an object be accelerating and yet -not- moving?

 
 
Reply Tue 12 Feb, 2013 05:13 pm
This seems like a dumb question, I know, and it really might be. I'm working through some physics problems, and one states: "An object is accelerating in a northern direction. In that moment, which of the following applies..."
One of the correct answers is "The object may or may not be moving," and it's really confusing me. If the object is accelerating, its velocity is changing, and it is in motion. How can it not be? My resource cryptically adds, "an object can accelerate, and at a specific time, not be moving." Fine. But if "in that moment" the object is accelerating, how is there a chance it isn't moving?

-Thanks; a confused student
 
dalehileman
 
  1  
Reply Tue 12 Feb, 2013 06:29 pm
@becquerel,
Quote:
If the object is accelerating, its velocity is changing, and it is in motion. How can it not be?
Could it be Bec there's a large object in motion nearby

I's suppose the fella would assume he's accelerating but actually he's just compensating for the passing field--a sort of relativity
becquerel
 
  1  
Reply Tue 12 Feb, 2013 06:55 pm
@dalehileman,
I don't think so - the question is from the point of view of an (I assume) stationary observer who is observing an object that is actually accelerating. If that's what you meant; please correct me if I'm wrong.

From the same set of questions, I also found this: "You see an object that is not moving. At that moment, which of these applies..." And here, "the object may or may not be accelerating" is also a correct answer. The explanation is, "the object may be accelerating, just momentarily at rest. Thus, a force may be acting on the object, just not producing motion at that instant."

I understand that forces can act on an object without producing motion (and no net force = no acceleration), but I still don't really understand how an object could be accelerating and yet (potentially) simultaneously at rest, or definitely at rest but (potentially) simultaneously accelerating. Humph. But thanks for the reply!
roger
 
  1  
Reply Tue 12 Feb, 2013 07:29 pm
@becquerel,
I'm glad I claim no knowledge of physics, because this strikes me as nonsencical. If they are saying this object has some kind of average acceleration, including stops and starts, I'm still left with nothing.

If it accelerates, stops, accelerates, and stops again, it would seem that acceleration would be canceled by the accompanying decceleration, with and average acceleration being zero.

I am grateful it is you that is required to provide an answer, and not me.
engineer
 
  5  
Reply Tue 12 Feb, 2013 07:34 pm
@becquerel,
If you throw a ball into the air, when it reaches its peak height and is about to fall back to earth, it is accelerating and motionless. If the velocity (upward from your throw) and the acceleration (downward due to gravity) are in different directions, there will be a moment when the velocity drops to zero before changing directions.
rosborne979
 
  1  
Reply Tue 12 Feb, 2013 07:52 pm
@engineer,
Are you sure the definition of acceleration accurately describes the object right at the moment in time when it's velocity becomes zero.

Acceleration is the rate at which the velocity of a body changes with time. So if you're talking about a tiny moment of time when velocity is essentially zero, then both time and velocity equal zero in that definition, and there can be no "rate of change" as the moment of time approaches zero.
becquerel
 
  1  
Reply Tue 12 Feb, 2013 07:54 pm
@roger,
I can hardly claim any knowledge of physics either, hah.

I did poke around online, and found this: "Acceleration is nothing but the rate of change of velocity. Now consider a car that is at rest but is about to start moving. You press the accelerator and the car starts moving. As the velocity of car has changed from zero to a nonzero value, the car must have had acceleration even though it was at rest."

I guess this is what my resource is talking about, but it still seems counter-intuitive. Of course an object can accelerate -from- rest and decelerate -to- rest , but I would argue that it can't be accelerating and actually at rest (not moving at all) simultaneously, unlike my questions seem to be suggesting. But apparently it's a valid view that an object accelerating from rest can still be considered "at rest" during a point in its actual acceleration.
0 Replies
 
becquerel
 
  1  
Reply Tue 12 Feb, 2013 08:13 pm
@engineer,
Hmm. I believe you're correct there - the acceleration of a thrown object at maximum height, although its instantaneous velocity equals zero, is still the downward acceleration of gravity.

The questions I'm working on aren't really about objects in free-fall though, so I think that's what threw me off.
Hulk
 
  3  
Reply Tue 12 Feb, 2013 08:17 pm
@becquerel,
There can be instances where "an object be accelerating and yet -not- moving" even if it's not a free fall problem.
Think of it this way, there can a force that is accelerating it even though it has not yet took off. It's momentary though.
If you're into cars, think of rolling backwards on a hill, so velocity is negative. Then gun the gas. As the car accelerates and velocity climbs from below zero, and soon it will pass the velocity of zero to a positive value and climb the hill. But going back, at the time the car hits zero velocity, the car is still accelerating. So when it seems like the car is still for a moment, there is acceleration but no motion.
becquerel
 
  1  
Reply Tue 12 Feb, 2013 08:33 pm
@Hulk,
That makes sense; thank you. And everyone else too.

I've been long out of practice with physics and the idea just seemed nonsensical to me, but not nearly as much now.
0 Replies
 
raprap
 
  3  
Reply Tue 12 Feb, 2013 09:27 pm
Its a boundary condition. Acceleration is the second deritive of displacement. Velocity is the first deritive.

If you are to solve a differential for displacement or velocity of a mass that is accelerated, you must have a boundary condition.

In this case this boundary a mass is being accelerated and at zero time the velocity is zero.

V=ds/dt
a=dV/dt=d^2s/dt^2
@t=0 V=0

Rap
MattDavis
 
  0  
Reply Tue 12 Feb, 2013 10:02 pm
I disagree with Engineer and Hulk.
I agree with raprap
(so long is this question is being answered within the realm of Classical Mechanics)
Allow me to show another way to demonstrate this,
which also is a way that does not require any understanding of calculus.

Velocity is a change in position over time. v = d/t
Acceleration is a change in velocity over time. a = v/t

In your situation: [ a = v/t ]
lets set velocity = to 0, since the object is not moving.
a=0/t
thus
a=0

Which tells us that at any time(t) when velocity(v) is zero, acceleration(a) is also zero.
aspvenom
 
  1  
Reply Tue 12 Feb, 2013 10:11 pm
@MattDavis,
You can't disagree with hulk and engineer and agree with rap. rap is in agreement with what hulk and engineer wrote. He worded it more elegantly perhaps, but still in agreement.
0 Replies
 
Cuterthanpaul
 
  1  
Reply Tue 12 Feb, 2013 10:16 pm
@becquerel,
I may be wrong, but could it be relativity and whether the object or the area around the object is moving?
aspvenom
 
  1  
Reply Tue 12 Feb, 2013 10:17 pm
@Cuterthanpaul,
I think you're thinking of frame of reference.
0 Replies
 
MattDavis
 
  0  
Reply Tue 12 Feb, 2013 10:49 pm
@MattDavis,
On closer inspection this is in error!!!!
Quote:
Velocity is a change in position over time. v = d/t
Acceleration is a change in velocity over time. a = v/t

In your situation: [ a = v/t ]
lets set velocity = to 0, since the object is not moving.
a=0/t
thus
a=0


In your situation: [ a = v/t ]
lets set velocity = to 0, since the object is not moving.
a = 0/t
a = 0 (except when t = 0 )

At zero time the acceleration is indeterminate (due to division by 0)
0 Replies
 
MattDavis
 
  1  
Reply Tue 12 Feb, 2013 10:52 pm
@raprap,
What is asked for however is not what velocity is at zero time, but what is acceleration is at zero time (specifically whether it is zero or not-zero).
raprap
 
  2  
Reply Wed 13 Feb, 2013 12:13 am
@MattDavis,
Not if it is a limit (or in this case) a boundary condition.

See the fundamental theorem of the Calculus.

http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

Rap

0 Replies
 
maxdancona
 
  3  
Reply Wed 13 Feb, 2013 12:27 am
@MattDavis,
Matt, Engineer is correct. You are making a basic mistake

You are correct when you say

Quote:
Acceleration is a change in velocity over time


You are incorrect when you say a = v/t. This would be Acceleration being velocity over time (which it isn't). This is not Acceleration being change in velocity over time.

The real equation for average acceleration (and instantaneous acceleration involves a bit more complicated math but works the same) is a = (Vend -Vstart)/t

Either Vend or Vstart can be zero, and still have a non-zero acceleration.
engineer
 
  2  
Reply Wed 13 Feb, 2013 07:22 am
@maxdancona,
Let's take it from the other side. Instead of saying acceleration = dV/dt let's use a = F/m or acceleration equals net force applied divided by mass. Back to the ball example, without any support, the net force applied to the ball is the force of gravity, so the ball is always accelerating until it hits the ground (and the ground provides an upward force to cancel out the force of gravity pulling downward.) If it makes it easier, think of it on the moon so you don't have to worry about air resistance. The velocity slows, goes to zero and changes direction, but the acceleration is constant so there is a time when the velocity is zero and the acceleration isn't.

In your original question, the acceleration is in the northern direction. If the initial velocity was in the southern direction, there will come a time when the object will stop going south and turn north. At that moment, the velocity is zero but it is still accelerating.
 

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