# Acceleration Question

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#### grogberries

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##### Acceleration Question
« on: April 06, 2010, 10:38:06 PM »
Can something with a velocity of 0 ever accelerate? I ask the question because anything multiplied by 0 is 0. I mean I understand that things at rest will move if a force is applied to them. But does this create any mathematical complications?
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#### bowler

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##### Re: Acceleration Question
« Reply #1 on: April 07, 2010, 02:54:35 AM »
Wha? Acceleration is the rate of change of velocity. While your assertion that 0x0=0 is quite correct what does this have to do with acceleration?

#### Its a Sphere

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##### Re: Acceleration Question
« Reply #2 on: April 07, 2010, 04:21:21 AM »
Not only that, but something with 0 velocity can be accelerating at the same moment in time as well.
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#### SupahLovah

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##### Re: Acceleration Question
« Reply #3 on: April 09, 2010, 07:53:27 AM »
Not only that, but something with 0 velocity can be accelerating at the same moment in time as well.
Quite frequently in fact. Toss a ball (or anything) into the air. At the instant it starts coming back down, the velocity is 0 and acceleration is 9.8 m/s^2 towards the earth.
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#### Parsifal

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##### Re: Acceleration Question
« Reply #4 on: April 09, 2010, 07:57:58 AM »
I'm going to side with the white supremacists.

#### SupahLovah

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##### Re: Acceleration Question
« Reply #5 on: April 09, 2010, 08:37:04 AM »
* SupahLovah sees your pedantic point.

At the instant it stops rising, before it begins falling, its velocity is 0m/s.

EDIT:

TBH, I just started going over my old calculus work and book again, need me to draw a graph of a rising and falling object?
"Study Gravitation; It's a field with a lot of potential!"

#### WardoggKC130FE

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##### Re: Acceleration Question
« Reply #6 on: April 11, 2010, 04:05:05 PM »

#### Lorddave

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##### Re: Acceleration Question
« Reply #7 on: April 11, 2010, 06:44:52 PM »
Indeed it can.
But only for an instant.  That instant is the time in which the acceleration is applied but the velocity hasn't gone beyond 0.  This is best seen when you're going from one direction to another.

For example:
Let's say you have an object moving forward with an acceleration of 10m/s^2.  You then suddenly turn that acceleration into -10m/s^2.  Essentially you change it's direction but it still has a velocity.  The velocity will decrease until it reaches 0 (called Deceleration) then start up again as the object accelerates in the opposite direction.
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#### Parsifal

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##### Re: Acceleration Question
« Reply #8 on: April 12, 2010, 05:48:35 AM »
I'm going to side with the white supremacists.

#### WardoggKC130FE

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##### Re: Acceleration Question
« Reply #9 on: April 12, 2010, 10:18:17 AM »

#### SupahLovah

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##### Re: Acceleration Question
« Reply #10 on: April 12, 2010, 10:21:52 AM »
the velocity is 0

Wrong.

Depends on the FoR.

No it doesn't. Velocity is never 0.

Wrong.
* SupahLovah sees your pedantic point.

At the instant it stops rising, before it begins falling, its velocity is 0m/s.

EDIT:

TBH, I just started going over my old calculus work and book again, need me to draw a graph of a rising and falling object?
He's just being an asshole about units.
"Study Gravitation; It's a field with a lot of potential!"

#### Parsifal

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##### Re: Acceleration Question
« Reply #11 on: April 12, 2010, 10:24:00 AM »
Warrdog is actually correct. I was wrong in my earlier assessment.
I'm going to side with the white supremacists.

#### WardoggKC130FE

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##### Re: Acceleration Question
« Reply #12 on: April 12, 2010, 10:25:06 AM »
Warrdog is actually correct. I was wrong in my earlier assessment.
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#### parsec

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##### Re: Acceleration Question
« Reply #13 on: April 19, 2010, 10:04:32 PM »
Hey groggy,

I think what you are confused with is the different use between growth or inflation as used in Economics and in Physics.

In Economics, one usually reports the relative growth of a certain quantity (GDP, overall prices, population, etc.) between two consecutive periods. Let the quantity we measure in the nth period be denoted with Pn. Then, the increase in the quantity between the nth period and the (n+1)-st period is simply given as the difference:

$\Delta&space;P_{n}&space;\equiv&space;P_{n&space;+&space;1}&space;-&space;P_{n}$.

The relative increase is given by:

$r_{n}&space;\equiv&space;\frac{\Delta&space;P_{n}}{P_{n}}&space;=&space;\frac{P_{n&space;+&space;1}&space;-&space;P_{n}}{P_{n}}&space;=&space;\frac{P_{n+1}}{P_{n}}&space;-&space;1$.

Sometimes we define the growth quotient:

$q_{n}&space;\equiv&space;1&space;+&space;r_{n}$.

Let us give a specific example. If someone says that the growth rate is 5%, it means that rn = 0.05 and qn = 1.05. Sometimes the growth can be negative (like in a recession). Then, rn < 0 and qn < 1.

If we know the values of qn for every n and the initial value P0, then we can calculate Pn:

$q_{n}&space;=&space;\frac{P_{n+1}}{P_{n}}&space;\Rightarrow&space;P_{n&space;+&space;1}&space;=&space;P_{n}&space;q_{n},&space;\&space;n&space;\ge&space;0$

$\begin{array}{rcl}&space;P_{1}&space;&&space;=&space;&&space;P_{0}&space;\,&space;q_{0}&space;\\&space;P_{2}&space;&&space;=&space;&&space;P_{1}&space;\,&space;q_{1}&space;=&space;P_{0}&space;\,&space;q_{0}&space;\,&space;q_{1}&space;\\&space;P_{3}&space;&&space;=&space;&&space;P_{2}&space;\,&space;q_{2}&space;=&space;P_{0}&space;\,&space;q_{0}&space;\,&space;q_{1}&space;\,&space;q_{2}&space;\\&space;&&space;\cdots&space;&&space;\end{array}$

$P_{n}&space;=&space;P_{0}&space;\,&space;q_{0}&space;\,&space;q_{1}&space;\,&space;\cdots&space;\,&space;q_{n&space;-&space;1}&space;\equiv&space;P_{0}&space;\,&space;\prod_{k&space;=&space;0}^{n&space;-&space;1}&space;{q_{k}},&space;n&space;\ge&space;1$
So, if you had this in mind for the rate of change, then you were right:

$P_{0}&space;=&space;0&space;\Rightarrow&space;P_{n}&space;=&space;0,&space;\forall&space;n&space;\ge&space;0$

However, this is not what is meant as a rate of change in Physics. To see this, we will take the continuum limit in the definition for rn in the following manner. Let T be the time between two periods. Then, if the initial instant t0 = 0, we can take the instant when the nth period begins to be tn = n*T. Then, we may consider any discrete sequence {xn} (such as Pn or rn) simply as the value of a continuous function x(t) at discrete time intervals. Then, if we formally let $\inline&space;T&space;\rightarrow&space;0$, we will have the values of the function for all instants of time. Notice that:

$\inline&space;P_{n+1}&space;\equiv&space;P(t_{n+1})&space;=&space;P(t_{n}&space;+&space;T)$

so

$\frac{r_{n}}{T}&space;=&space;\frac{\frac{P(t_n&space;+&space;T)&space;-&space;P(t_{n})}{T}}{P(t_{n})}&space;\rightarrow&space;\frac{\frac{dP(t)}{dt}}{P(t)}&space;=&space;\frac{d(\ln(P(t)))}{dt},&space;\&space;T&space;\rightarrow&space;0$

So, the growth rate is really the logarithmic derivative of a quantity (albeit taken at a discrete set of instants in time).

The acceleration is simply defined as the ordinary derivative of the velocity:

$a(t)&space;=&space;\frac{dv(t)}{dt}$

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#### grogberries

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##### Re: Acceleration Question
« Reply #14 on: April 19, 2010, 11:46:42 PM »
Bah. For some reason I had the idea in my head that the equation for acceleration was something different. Art students shouldn't meddle with math. I have learned my lesson.
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