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jaersyn

  • 3 years ago

Can someone explain to me why the slope on a distance vs. time^2 graph represents 1/2 the acceleration of that object? And also how to figure out the spring constant given a force vs. displacement graph? Thanks

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  1. imron07
    • 3 years ago
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    1) Maybe this gives you a clue: \[\frac{dv}{dt^2}=\frac{dv}{2tdt}=\frac{a}{2t}\] 2) Find the slope,

  2. jaersyn
    • 3 years ago
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    explain like i'm five? ^^"

  3. jaersyn
    • 3 years ago
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    y = 1/2 * gt^2 the slope of the line is y = ax^2 so my thinking is x^2 corresponds to t^2 while a corresponds to 1/2*gt but i don't know how to express that algebraically

  4. imron07
    • 3 years ago
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    The slope of distance vs time^2: \[\frac{dy}{dt^2}= \frac{dy}{2tdt}=\frac{v}{2t}=\frac{a}{2} \] Is this clear enough?

  5. inkyvoyd
    • 3 years ago
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    \(\Large\frac{d^2s}{dt^2}=a\) This is acceleration by definition. \(\Large\frac{ds}{dt}=at+C_1=v\) This is velocity by definition. \(\Large s=\frac{1}{2}at^2+C_1t+C_2\) This is distance. Now we figure out what the constants are. C1 here is the initial velocity, while C2 here is the starting distance. Thus we have \(\Large\frac{ds}{dt}=at+v_0=v\) \(\Large s=\frac{1}{2}at^2+v_0t+s_0\)

  6. inkyvoyd
    • 3 years ago
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    So, where'd the 1/2 come from? well take the derivative of at^2. You'll get 2at, which is twice off of what we want, which is at. In fact, we fix this issue by dividing at^2 by 2. Try the derivative of (1/2)at^2, you should get at.

  7. jaersyn
    • 3 years ago
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    so I have slope y = 4.8711x

  8. jaersyn
    • 3 years ago
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    how do you find a

  9. inkyvoyd
    • 3 years ago
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    Can you like give me what your problem that you have to solve is? :S

  10. imron07
    • 3 years ago
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    x here t^2? If yes, then a=1/2*4.8711 :D

  11. jaersyn
    • 3 years ago
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    it's to compute the experimental value of g (gravity) from the slope value which is y = 4.8711x

  12. inkyvoyd
    • 3 years ago
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    You're giving me an equation, but I have no idea what this equation means. Can you clarify?

  13. jaersyn
    • 3 years ago
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    imron, how'd you know to substitute that way though?

  14. jaersyn
    • 3 years ago
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    so i have this graph where the slope of the distance vs time^2 is y=4.8711x

  15. inkyvoyd
    • 3 years ago
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    you could just draw the graph, but nvm I got it now.

  16. jaersyn
    • 3 years ago
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    i did

  17. imron07
    • 3 years ago
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    |dw:1347849013566:dw| If this graph is what you get?

  18. inkyvoyd
    • 3 years ago
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    Kk. you have \(\huge s(x)=\frac{1}{2}at^2\) s(x) is the distance, a is acceleration, and t is time.

  19. jaersyn
    • 3 years ago
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    yezzir

  20. imron07
    • 3 years ago
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    From the derivation before, you can see that the slope is 1/2*a. a=2*4.8711 (I was mistaken back then :) )

  21. inkyvoyd
    • 3 years ago
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    Well you have a 1/2 in that equation that comes from integrating that I just explained earlier. a is assumed to be constant (gravity), and there you go it's the equation.

  22. jaersyn
    • 3 years ago
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    assume we don't know a

  23. jaersyn
    • 3 years ago
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    or g rather

  24. inkyvoyd
    • 3 years ago
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    We don't know a, but it's a constant.

  25. jaersyn
    • 3 years ago
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    4.8711x = (1/2)at^2

  26. jaersyn
    • 3 years ago
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    do the x and t^2 cancel out somehow?

  27. jaersyn
    • 3 years ago
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    since x represents t^2 in the graph?

  28. inkyvoyd
    • 3 years ago
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    They are the same thing.

  29. inkyvoyd
    • 3 years ago
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    So there you go. Cancel them both out, multiply both sides by 2, and here you have a.

  30. jaersyn
    • 3 years ago
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    hmm

  31. inkyvoyd
    • 3 years ago
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    So, where did the magical formula come from? Well I explained it way back there. \(\Huge s(x)=\frac{1}{2}at^2+v_0t+s_0\) But, we know that the initial velocity and the initial distance are both 0, we we effectively get \(\Huge s(x)=\frac{1}{2}at^2\)

  32. jaersyn
    • 3 years ago
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    wtf who are you and tracks?

  33. jaersyn
    • 3 years ago
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    sorry. proceed

  34. inkyvoyd
    • 3 years ago
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    I finished already. ;)

  35. jaersyn
    • 3 years ago
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    |dw:1347849594808:dw|

  36. jaersyn
    • 3 years ago
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    it feels not right canceling x and t^2

  37. inkyvoyd
    • 3 years ago
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    Well, you defined x as t^2. So too bad.

  38. inkyvoyd
    • 3 years ago
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    The x axis IS t^2.

  39. jaersyn
    • 3 years ago
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    yeahhhh but

  40. jaersyn
    • 3 years ago
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    flutterit i'll see what happens

  41. jaersyn
    • 3 years ago
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    thanks for all your help, the derivations were cool

  42. inkyvoyd
    • 3 years ago
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    Yeah, I likes dem too :D

  43. jaersyn
    • 3 years ago
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    awwwwwww yeahhhh math.

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