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anonymous

  • one year ago

Mhm

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  1. anonymous
    • one year ago
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    \[\text{Total area} = X + Y \implies \frac{ 1 }{ 2 }t(\vec v_f- \vec v_i) +\vec v_i t\]

  2. anonymous
    • one year ago
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    Where total area is displacement

  3. anonymous
    • one year ago
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    \[\vec d = \frac{ \vec v_f t }{ 2 }- \frac{ \vec v_i t }{ 2 }+v_i t\] \[\vec d = \frac{ \vec v_f t }{ 2 }+\frac{ \vec v_i t }{ 2 } = \left( \frac{ \vec v_f + \vec v_i }{ 2 } \right)t\] \[\text{Therefore} ~ \vec d = \left( \frac{ \vec v_f + \vec v_i }{ 2 } \right)t\]

  4. anonymous
    • one year ago
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    Ah, so now similarly can derive the other equations of uniform motion, using just the graphs :).

  5. anonymous
    • one year ago
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    |dw:1437888152528:dw|Using above being derived from

  6. anonymous
    • one year ago
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    Well we can derive \[\vec d = \vec v_i t + \frac{ 1 }{ 2 } \vec at^2\] and then to find \[\vec v_f ^2= \vec v_i^2+ 2a \vec d\] we can substitute \[\vec a = \frac{ \vec v_f - \vec v_i }{ t } \] into \[\vec d = \vec v_i t + \frac{ 1 }{ 2 } \vec a t^2\]

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