Vf^2=Vi^2+2ad not proportional?

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<Moderator's note: Moved from a technical forum and thus no template.>

Hi, I am a junior in highschool and future physicist. I have a problem (I am very curious) and would like for it to be solved as I CAN NOT figure this out!

I was thinking about an object being dropped from rest from my aprox. height (2m) and its acceleration.
Using the formula in the title:
Vf^2=0^2+2*9.8*2=39.2
Sqrt(39.2)=6.26
Vf ~= 6.26m/s

Now I thought about the same object dropped from half the height. You would think, or at least I did, mathmatically you would do 6.26/2 which would be 3.13.
However, if you use the same formula the answer isn't the same.
Vf^2 = 0^2+2*9.8*1=19.6
Sqrt(19.6)=4.42
Vf ~= 4.42m/s

Why does it do this?
 

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  • #2
jtbell
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Why?
Because algebraically nothing changed but an independent variable.

Thinking about this though, if I imagine a graph, I would imagine the line to be one similar to a quadratic equation graphed. I'm not sure if that plays into this however.
 
  • #4
kuruman
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What is the implicit assumption when you believe that the object will be traveling at half the velocity over half the distance? Is that assumption valid in this case?
 
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What is the implicit assumption when you believe that the object will be traveling at half the velocity over half the distance? Is that assumption valid in this case?
The assumption was the rate is constant and therefor directly proportional. It is invalid, but why? It would makes sense for it to be in my head from what I have learned previously in math.
 
  • #6
kuruman
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The assumption was the rate is constant and therefor directly proportional.
Which rate? Can you be more specific as to what is changing with respect to what?
 
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Which rate? Can you be more specific as to what is changing with respect to what?
The rate of acceleration.
If the Vi stays 0, and the acceleration is assuned as 9.8m/s^2 because I am on Earth, why would the Vf for 2meters not be divisible by 2 for the Vf for 1m?
 
  • #8
kuruman
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The rate of acceleration.
You did not answer my question. Yes, the acceleration is a rate and yes this rate is constant. However, a rate is how fast something is changing with respect to something else. In the case of acceleration what is the "something" and what is the "something else"? If you can answer the question, you should be able to see why your approach is incorrect.
 
  • #9
sophiecentaur
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The rate of acceleration.
If the Vi stays 0, and the acceleration is assuned as 9.8m/s^2 because I am on Earth, why would the Vf for 2meters not be divisible by 2 for the Vf for 1m?
Acceleration is the increase in velocity in a given time. It is not the increase in velocity in a given distance.
There is no point in using 'words' in this argument because Maths is a much better way of describing what goes on. You have to believe that the Gravitational Potential Energy is transferred to Kinetic Energy at the end of the drop. So:
mgh = mv2/2
So, if you double the height of the drop, the Kinetic Energy doubles but the velocity increases by a factor of √2
 
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  • #10
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Acceleration is the increase in velocity in a given time. It is not the increase in velocity in a given distance.
There is no point in using 'words' in this argument because Maths is a much better way of describing what goes on. You have to believe that the Gravitational Potential Energy is transferred to Kinetic Energy at the end of the drop. So:
mgh = mv2/2
So, if you double the height of the drop, the Kinetic Energy doubles but the velocity increases by a factor of √2
I see. Thank you. We just started learning Ep and Ek this week actually so that didn't even occur to me. I want to be a physicist when I graduate and I like thinking of these types of questions that really make me think.
 
  • #11
lekh2003
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What you are thinking of is a linear transformation. If the kinematic equations were linear, then sure you would see this, but that is not the case. They are parabolic and significantly different.
I want to be a physicist when I graduate and I like thinking of these types of questions that really make me think.
Me too, physics is what I want in my future. I often find that the key to all of these questions is looking carefully at the language of physics, math. And of course practicing so much that the concept is absolutely imprinted on your mind.
 
  • #12
Ray Vickson
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I see. Thank you. We just started learning Ep and Ek this week actually so that didn't even occur to me. I want to be a physicist when I graduate and I like thinking of these types of questions that really make me think.
A very practical use of this is to note that some "driver's ed." courses emphasize the fact that when you double your speed it requires four times the distance to come to a stop. That is one good reason to not "tailgate", especially at highway speeds.
 
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  • #13
kuruman
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A very practical use of this is to note that some "driver's ed." courses emphasize the fact that when you double your speed it requires four times the distance to come to a stop. That is one good reason to not "tailgate", especially at highway speeds.
Very good point.
 
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A very practical use of this is to note that some "driver's ed." courses emphasize the fact that when you double your speed it requires four times the distance to come to a stop. That is one good reason to not "tailgate", especially at highway speeds.
I'm glad yours did, my drivers ed course was a joke. My friend didn't even know how to turn her highbeams on after she got her license. I was very dissapointed.
 

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