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anonymous

  • one year ago

Guy riding a bike accelerates for 8 seconds. He covers 60 meters. Using the formula d=vt+1/2at^2 calculate his acceleration.

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  1. anonymous
    • one year ago
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    Alright so basically plugging in the numbers you'd think right

  2. anonymous
    • one year ago
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    Look 60m=1/2(a)64s^2

  3. anonymous
    • one year ago
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    Units don't cancel out

  4. anonymous
    • one year ago
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    60m=32s^2a

  5. anonymous
    • one year ago
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    I mean wtf

  6. anonymous
    • one year ago
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    a=(m/s^2)

  7. anonymous
    • one year ago
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    This is such a paradox

  8. anonymous
    • one year ago
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    No way to derive units for acceleration with the given kinematics equation which holds true

  9. anonymous
    • one year ago
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    wtf

  10. anonymous
    • one year ago
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    Hey dude what do you think

  11. anonymous
    • one year ago
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    lol you stuck with this paradox too

  12. anonymous
    • one year ago
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    don't worry you are not the only one

  13. anonymous
    • one year ago
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    The only way to do this is to assume "a" doesn't bear any value.

  14. anonymous
    • one year ago
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    Nope.

  15. anonymous
    • one year ago
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    Merely a means to represent imaginary number which follows from the function of this equation

  16. anonymous
    • one year ago
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    \[v=m/s \]

  17. anonymous
    • one year ago
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    But from the principles of physics which assigns unit to every different letter this is a paradox

  18. anonymous
    • one year ago
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    eh?

  19. anonymous
    • one year ago
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    \[d=m\] \[a=m/s^2\]

  20. anonymous
    • one year ago
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    Then the prospect of using a is misconceived because a in this equation doesn't assume the units m/s^2 No

  21. anonymous
    • one year ago
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    Instead I am left with the presupposition that a will have the units and must

  22. anonymous
    • one year ago
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    60m/32s^2=a

  23. anonymous
    • one year ago
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    total distance traveled=initial distance+(velocity)(time)+0.50(acceleration)(time)^2

  24. anonymous
    • one year ago
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    1.875m/s^2

  25. anonymous
    • one year ago
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    In this case the initial distance and initial velocity is ignored as the cyclist starts from the state of being at rest

  26. anonymous
    • one year ago
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    \[m=m+(\frac{ m }{ s }*s)+0.50*\frac{ m }{ s^2 }*s^2\]

  27. anonymous
    • one year ago
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    total m=m+m+0.50*m

  28. anonymous
    • one year ago
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    you are wrong

  29. anonymous
    • one year ago
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    you can't assume a to have (m/s^2) internally.

  30. anonymous
    • one year ago
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    the units of a are assigned subsequently from the calculation. Assigning such units without solid foundation only serves to confuse the user.

  31. anonymous
    • one year ago
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    acceleration is m/s^2

  32. anonymous
    • one year ago
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    Look 60m=32s^2a a=1.875m/s^2

  33. anonymous
    • one year ago
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    This is correct.

  34. anonymous
    • one year ago
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    But if you assume a=(s/m^2) that gets you nowhere.

  35. anonymous
    • one year ago
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    Try it for yourself with units to be accorded

  36. anonymous
    • one year ago
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    This is where the teaching of physics fails.

  37. anonymous
    • one year ago
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    Each representative is not to be assumed to have the units but rather guidelines for what they stand for and what the user should expect to get following the calculations.

  38. anonymous
    • one year ago
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    acceleration is velocity over time. Acceleration is how fast velocity is changing with time since velocity/time is acceleration. and velocity is m/s and time is s (m/s)*(1/s)=m/s^2

  39. anonymous
    • one year ago
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    By no means assign the units prior to calculation as it throws of the equation.

  40. anonymous
    • one year ago
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    But it's interesting.

  41. anonymous
    • one year ago
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    so we know acceleration is m/s^2 because acceleration is known from how fast velocity is changing. So they plug a into your equation and the units cancels.

  42. anonymous
    • one year ago
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    No

  43. anonymous
    • one year ago
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    a=v/t came first before d=d1+vt+1/2)at^2

  44. anonymous
    • one year ago
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    When a is imaginary you can't even assign m/s^2

  45. anonymous
    • one year ago
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    However when a is known you can naturally assign the units however "naturally" as in being circumstantial and not at the hands of calculator

  46. anonymous
    • one year ago
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    there is no imaginary. there is an estimated A.

  47. anonymous
    • one year ago
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    hypothetical A

  48. anonymous
    • one year ago
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    so that hypothetical a doesn't assume units in itself

  49. anonymous
    • one year ago
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    You know what I mean? So many amateurs like me are presupposing upon seeing the hypothetical a or v or even d that they have units in themselves but this assumption is totally off

  50. anonymous
    • one year ago
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    Yes, it can if velocity and time can too?

  51. anonymous
    • one year ago
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    and acceleration is derived from them

  52. anonymous
    • one year ago
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    This is quite contradictory from what physics teachers force you to assume.

  53. anonymous
    • one year ago
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    we use seconds for time and units for speed why not use acceleration by dividing the units out

  54. anonymous
    • one year ago
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    Nothing imaginary about that. It is not abstract like math.

  55. anonymous
    • one year ago
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    I think my best bet is whenever you see a letter that makes a reference to set of units you need to assume that they in themselves do not possess anything of the sort however it's a guideline instructing the user of that equation to assume that the final work will be concordant with the standard form.

  56. anonymous
    • one year ago
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    Yeah physics is very paradoxical sometimes but that is our passion right? XD

  57. anonymous
    • one year ago
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    Yes, but it is also inaccurate in a lot of ways.

  58. anonymous
    • one year ago
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    PV=nRT for ideal gas law n does not make sense for me also.

  59. anonymous
    • one year ago
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    that is my opinion on some part of physics, but I think the acceleration actually does make sense.

  60. anonymous
    • one year ago
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    Yeah. Sometimes physics ignores the principles of math and heavily relies on intuitive reasoning which often fails. But this is the reality.

  61. anonymous
    • one year ago
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    The prospect of entirely relying on single solution upon a problem of kind shows it all. Unfortunate but yet convenient.

  62. anonymous
    • one year ago
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    Yep, that is the difference between experimental and theoretical physicists. They try to prove each other wrong. Well I got to sleep now. See you another day.

  63. anonymous
    • one year ago
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    Yeah Interesting talk my friend;) Good to see you.

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