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Callisto

  • 3 years ago

Deriving kinematic equations

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  1. Callisto
    • 3 years ago
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    For a = constant

  2. Callisto
    • 3 years ago
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    \[v=\int a dt=at+c\]c = initial velocity So, \[v=u+at\]

  3. gerryliyana
    • 3 years ago
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    good job

  4. gerryliyana
    • 3 years ago
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    if a constan, a =0 v = u + at v = u + 0 v = u

  5. Callisto
    • 3 years ago
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    \[x_f = \int v(t) dt=\int (\int a dt) dt = \int(at+u)dt = \frac{1}{2}at^2 + ut +c\], c = initial position So, \[x_f = ut +\frac{1}{2}at^2 + x_i\]\[x_f-x_i = ut +\frac{1}{2}at^2\]\[s = ut+\frac{1}{2}at^2\]s=displacement.

  6. gerryliyana
    • 3 years ago
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    for a = 0; s = ut + 0.5 at^2 s = ut + 0 s = ut

  7. Callisto
    • 3 years ago
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    From the third post, \[s=ut+\frac{1}{2}at^2\]\[at^2+2ut -2s = 0\]\[t^2 + \frac{2u}{a}t - \frac{2s}{a}=0\]\[(t+\frac{u}{a})^2 - \frac{u^2}{a^2}-\frac{2s}{a}=0\]\[(t+\frac{u}{a})^2 = \frac{u^2-2as}{a^2}\]\[t+\frac{u}{a} = \frac{\sqrt{u^2-2as}}{a}\]\[t = \frac{\sqrt{u^2-2as}-u}{a} -(1)\] Put (1) into v=u+at \[v =u+a(\frac{\sqrt{u^2-2as}-u}{a})\]\[v = u +\sqrt{u^2-2as}-u\]\[v^2 = u^2 +2as\]

  8. Callisto
    • 3 years ago
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    \[x_f = \int v(t) dt = v_{ave}t+C\]c = initial position \[x_f = v_{ave}t +x_i\]\[x_f = \frac{1}{2}(u+v)t +x_i \]\[s = \frac{1}{2}(u+v)t\]displacement = s = \(x_f - x_i\)

  9. Callisto
    • 3 years ago
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    Probably something wrong with the last post, which is s=(1/2) (u+v)t.

  10. experimentX
    • 3 years ago
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    that's correct ... it assumes constant acceleration. that;s all.

  11. Callisto
    • 3 years ago
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    Seriously?! I did it!?!

  12. experimentX
    • 3 years ago
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    let me see how can i put it up logically.

  13. experimentX
    • 3 years ago
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    |dw:1351833416633:dw|

  14. experimentX
    • 3 years ago
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    now you just need to show that for constant accn V_av = (u+v)/2 ... wanna try it?

  15. Callisto
    • 3 years ago
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    Huh!? I thought it's some maths.. Oh..How to start??

  16. experimentX
    • 3 years ago
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    try using MVT

  17. Callisto
    • 3 years ago
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    MVT.... again.../_\

  18. experimentX
    • 3 years ago
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    well, you can do this without MVT ,,, i was wondering if i could improve my skills with MVT. try expanding v(t) in the expression of average.

  19. experimentX
    • 3 years ago
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    or, V_av = s/t

  20. Callisto
    • 3 years ago
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    expanding v(t)?

  21. experimentX
    • 3 years ago
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    v(t) = u + at

  22. experimentX
    • 3 years ago
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    or simply put ,,, average velocity = distance/time

  23. Callisto
    • 3 years ago
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    Fail ._.

  24. Callisto
    • 3 years ago
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    \[v_f= v_i+\int_0^{t_f} a dt=v_i + at\] So, \[v_f=v_i + at\]

  25. Callisto
    • 3 years ago
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    \[x_f = x_i+\int_0^{t_f} v(t) dt= x_i+\int_0^{t_f} (v_i+at) dt = x_i+v_it+\frac{1}{2}at^2\] So, \[x_f = x_i+v_it+\frac{1}{2}at^2\] That is \[x_f - x_i = v_it+\frac{1}{2}at^2\]\[s= ut+\frac{1}{2}at^2\]

  26. experimentX
    • 3 years ago
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    |dw:1351853269560:dw|

  27. experimentX
    • 3 years ago
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    |dw:1351853406639:dw|

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