Understanding Acceleration Formulas: Instantaneous vs Average

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Average acceleration can be calculated using the formula aav = Δs/Δt, but it is also valid as aav = 0.5(afi + ain) only when acceleration varies linearly with time. In the equations s = v0t + 0.5at², vfi = vin + at, and (vfi)² - (vin)² = 2as, the acceleration "a" represents both instantaneous and average acceleration, assuming constant acceleration. If acceleration is not constant, these equations will yield incorrect results. The discussion also highlights that when acceleration is a function of time, it can be referred to as "jerk," which describes the rate of change of acceleration.
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Homework Statement


This is not a homework, just my understanding problem.

Homework Equations


We know that average acceleration can be calculated by: aav=Δs/Δt
Is this formula also correct: aav=0.5(afi+ain) ??

One more question is: in the formulas: s = v0t + 0.5at2 and vfi = vin + at and (vfi)2 - (vin)2 = 2as

a in these equations is instantaneous acceleration or average acceleration ?

The Attempt at a Solution

 
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stephenranger said:

Homework Statement


This is not a homework, just my understanding problem.

Homework Equations


We know that average acceleration can be calculated by: aav=Δs/Δt
Is this formula also correct: aav=0.5(afi+ain) ??
The first equation is never correct, since it doesn't even have the right units. The second equation is correct only if the acceleration is varying linearly with time.
One more question is: in the formulas: s = v0t + 0.5at2 and vfi = vin + at and (vfi)2 - (vin)2 = 2as

a in these equations is instantaneous acceleration or average acceleration ?
Both. These equations assume that the acceleration is constant, so it is both the instantaneous acceleration and the average acceleration.

Chet
 
stephenranger said:
...

We know that average acceleration can be calculated by: aav=Δs/Δt
A correct statement is: aav=Δv/Δt .
Perhaps you had a typo.

Also true is vav=Δs/Δt .
 
SammyS said:
A correct statement is: aav=Δv/Δt .
Perhaps you had a typo.

Also true is vav=Δs/Δt .
Yes. I have a typo, sorry.

What's about this:
stephenranger said:
s = v0t + 0.5at2 and vfi = vin + at and (vfi)2 - (vin)2 = 2as

the acceleration "a" in these equations is instantaneous or average acceleration or both like Chestermiller said ?
 
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Chestermiller said:
The second equation is correct only if the acceleration is varying linearly with time.
Yes sir. if the acceleration is varying linearly with time, that means that it's no longer constant, right ?

the changing acceleration is not taught in high school physics, so aav = Δv/Δt is enough for me.
 
stephenranger said:
the acceleration "a" in these equations is instantaneous or average acceleration or both like Chestermiller said ?
Don't you believe me?

Chet
 
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Chestermiller said:
Don't you believe me?

Chet
Come on sir. I need multiple viewpoints to have a good comparison.
Your phrase:
Chestermiller said:
The second equation is correct only if the acceleration is varying linearly with time.
gives me an idea that when acceleration is an equation of time it becomes instantaneous acceleration. Is that correct ?
 
stephenranger said:
Come on sir. I need multiple viewpoints to have a good comparison.
Your phrase:

gives me an idea that when acceleration is an equation of time it becomes instantaneous acceleration. Is that correct ?
The following equations that you presented, s = v0t + 0.5at2 and vfi = vin + at and (vfi)2 - (vin)2 = 2as, are valid only if the acceleration is constant. Otherwise they will give the wrong answer. If the acceleration is constant, then the instantaneous acceleration is equal to the average acceleration.

Chet
 
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Chestermiller said:
The following equations that you presented, s = v0t + 0.5at2 and vfi = vin + at and (vfi)2 - (vin)2 = 2as, are valid only if the acceleration is constant. Otherwise they will give the wrong answer. If the acceleration is constant, then the instantaneous acceleration is equal to the average acceleration.

Chet
Yes. Could you tell me in what phenomenon will the acceleration be an equation of time, a = a(t) ? and when I take its derivative da(t)/dt, what does da(t)/dt stand for ?
I ask that out of my curiosity.
 
  • #10
stephenranger said:
Yes. Could you tell me in what phenomenon will the acceleration be an equation of time, a = a(t) ? and when I take its derivative da(t)/dt, what does da(t)/dt stand for ?
I ask that out of my curiosity.
The acceleration of a body is a function of time if the net force acting on the body is varying with time.

As far as the derivative of the acceleration with respect to time is concerned, I believe this quantity is called the "jerk."

http://en.wikipedia.org/wiki/Jerk_(physics)

Chet
 
  • #11
Chestermiller said:
The acceleration of a body is a function of time if the net force acting on the body is varying with time.

As far as the derivative of the acceleration with respect to time is concerned, I believe this quantity is called the "jerk."

http://en.wikipedia.org/wiki/Jerk_(physics)

Chet

Thanks very much.
 
  • #12
Chet has answered well, but I'm going to add my answer, if you're interested. In the general case, the acceleration is a function of time. It is only in special cases that the acceleration is constant. So, when the acceleration is not constant, as you say, we have: a(t) a function of time. For example, you are in a car, and the car is accelerating forward, due to the driver pushing their foot down on the gas. You have to use your neck muscles to hold your head forward, due to the acceleration. But then, suppose the driver decides to put their foot down on the gas even more, so acceleration increases, and your head would be pushed backwards, unless you strained your neck muscles even more. So this intuitively explains why the rate of change of acceleration is called the jerk.
 
  • #13
BruceW said:
Chet has answered well, but I'm going to add my answer, if you're interested. In the general case, the acceleration is a function of time. It is only in special cases that the acceleration is constant. So, when the acceleration is not constant, as you say, we have: a(t) a function of time. For example, you are in a car, and the car is accelerating forward, due to the driver pushing their foot down on the gas. You have to use your neck muscles to hold your head forward, due to the acceleration. But then, suppose the driver decides to put their foot down on the gas even more, so acceleration increases, and your head would be pushed backwards, unless you strained your neck muscles even more. So this intuitively explains why the rate of change of acceleration is called the jerk.

Thanks. Now I know the origin of its name.
 
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