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Homework Help: Is this equation wrong?

  1. Aug 13, 2005 #1
    Is this equation wrong?

    This is a equation for
    "uniform acceleration directed line motion at zero initial speed" in our textbook.

    S=1/2 at^2

    Here is the list to compare the answer:
    t: 1 2 3
    v: 2 4 6

    If we use the equation to calculate, the answer is different to the list.

    For example,

    1/2(2)(3*3) = 9, s=9

    but in the list, it shows the answer is 12!

    Is my wrong? Or the equation is really wrong?
     
    Last edited: Aug 13, 2005
  2. jcsd
  3. Aug 13, 2005 #2

    ZapperZ

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    It is wrong because the kinematical equation that is used to derived that expression is for constant acceleration only. If the acceleration varies, that equation isn't valid.

    Zz.
     
  4. Aug 13, 2005 #3
    But in my trial calculation, the a is already constant. The a is 2, it is still not equal to the list result. Is me there is any wrong?
     
  5. Aug 13, 2005 #4

    ZapperZ

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    What list? All you did was plug in t and a, and calculate s. YOu got an value for s. Why is the eqn. wrong?

    Zz.

    edit: I see you EDITED your list to now list v. This is no longer relevant and appropriate for the equation you are using. How could you plug "a" for "v"? This is getting VERY confusing.
     
    Last edited: Aug 13, 2005
  6. Aug 13, 2005 #5
    The reason of a comes from v is because the v is a speed with initial zero speed, so the first second, 1/s' speed is the a. But anyway I calculate it, I really can't get the true result.
     
  7. Aug 13, 2005 #6

    ZapperZ

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    And you don't see ANYTHING wrong with plugging in the values of "v" into "a" in

    [tex] s = \frac{1}{2} at^2[/tex]?

    If there is an INITIAL velocity at t=0, then your question is MISSING a term. The full equation for the displacement is

    [tex]s= s_0 + v_0t + \frac{1}{2} at^2[/tex]

    It is ONLY for s(t=0)=0 and v(t=0)=0 do you get the displacement equation that you quoted in the first place.

    Zz.
     
    Last edited: Aug 13, 2005
  8. Aug 13, 2005 #7
    More details,

    t: 1 2 3
    v: 2 4 6

    The v is a speed with zero initial speed, front mentioned. So the first time unit is the a, thus the a is 2.
    If we put 2 into a for calculating, we will get:

    1/2 a t^2
    =1/2 (2) 3^2
    =9

    but the list shows that 3 seconds (t) = 12 (v). This is really making me doubt, whether the equation is wrong.
     
  9. Aug 13, 2005 #8

    ZapperZ

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    What the......?

    I give up. Maybe someone else can translate what's going on.

    Zz.
     
  10. Aug 13, 2005 #9
    How come is the s_0 and v_0?
    The initial velocity is already the a, and the v is variable, and the s_0 where can find?

    Let me trial calculate it, ignore the s_0,

    v_0t+1/2 at^2
    =2*3 + 1/2 (2) (3^2)
    =15

    only "v_0t+1/2 at^2" result is ready far higher than this equation "S=1/2 at^2".
     
  11. Aug 13, 2005 #10

    LeonhardEuler

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    The equation s=1/2at^2 goves the total distance traveled. After 3 seconds this is 1/2*2*3^2=9. The list says that after 3 seconds the velocity is 6. Where is the problem with this?
     
  12. Aug 13, 2005 #11

    LeonhardEuler

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    Why are you taking v_0 = 2? If the accelration is constant then you can extrapolate v_0 to be 0.
     
  13. Aug 13, 2005 #12
    If you're starting at postition s = 0 and time t = 0, the equation is just

    [tex]s = v_{0}t + \frac{1}{2}at^2[/tex]

    If you're starting with velocity v = 0 at time t = 0, the equation reduces further

    [tex]s = \frac{1}{2}at^2[/tex]

    The [itex]s_{0}[/itex] and [itex]v_{0}[/itex] are just the displacement and velocity at time t = 0, respectively. What's hard to understand? ZZ's explanation seemed pretty clear to me :/.
     
  14. Aug 13, 2005 #13
    This is an accelerated motion, the a is 2 , so the volecity is not wrong.

    v=at
    v=2*3
    v=6
     
  15. Aug 13, 2005 #14

    ZapperZ

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    I'm sorry, but INITIAL VELOCITY is "a"? Since when can an acceleration becomes a velocity? Where did you learn this?

    Zz.
     
  16. Aug 13, 2005 #15
    Oh, sorry. I missed understanding to it. In my textbook there is no this mention.
     
  17. Aug 13, 2005 #16
    Oh, I missed understanding to it. In my textbook there is no mention of this. But the result is still wrong.
     
  18. Aug 13, 2005 #17

    LeonhardEuler

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    He seems to have gotten this from the fact that it is a uniform acceleration, so:
    [tex]a = \frac{\Delta v}{\Delta t} = \frac{v_1-v_0}{1-0}=2[/tex]
     
  19. Aug 13, 2005 #18

    LeonhardEuler

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    Is it possible that the initial position is not zero? Could you explain to me exactly how you know that the result is wrong. I don't seem to see any contradiction within the data you gave.
     
  20. Aug 13, 2005 #19
    The initial speed is at zero, that is not wrong. It has many motive examples, from still to accelerate. It is also used for calculating certain time's distance. But I found it is not correct.
     
  21. Aug 13, 2005 #20

    ZapperZ

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    Yes, but he's going to cause himself a lot of grief if he keeps insisting that a=v. a=dv/dt is certainly not a = v.

    Zz.
     
  22. Aug 13, 2005 #21
    Not right that. Many motions have the feature from still to accelerate, also the acceleration is average, for example the rockets or trains, in theoretical, it is possible. But often we will ignore the variable volecity for easy to calculate.
     
  23. Aug 13, 2005 #22
    You mistook my meaning... Also this is the second paper of me in my scheme. Today now I found me is right, I can be brave to write it. Thank you.
     
  24. Aug 13, 2005 #23

    ZapperZ

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    You are right about what, that the equation is "wrong"? Or did you not realize you were USING it wrongly?

    I hate to think this is the "paper" you have been touting about in the other part of PF.

    Zz.
     
  25. Aug 13, 2005 #24

    BobG

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    Your answer is correct!

    Even without the acceleration being given, you can look at the increase in 'v' each second and see that the velocity is increasing by 2 each second, which gives you an acceleration of 2 (which you correctly used in your equation).

    Or you could differentiate the position equation and solve for acceleration (the acceleration is a little too obvious to bother with that, in this case).
     
  26. Aug 13, 2005 #25

    HallsofIvy

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    You repeatedly said "the list shows the answer is 12".

    What list?? You didn't show us any list that include "12".
     
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