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Derivation of S = ut + 1/2at^2

  1. Jul 26, 2008 #1
    Hi

    I tried to derive the distance traveled by a body at contact acceleration from the definition of acceleration (increase in speed every sec), but the ended with a different result. Can you see what I am doing wrong.

    u = initial speed
    t = time taken

    S = {distance in 1st sec} + {distance in 2nd sec } + {distance in 3rd sec) + ............... + {distance in t sec}

    S = {u } + {u + a} + {u + 2a } + ................... + {u + (t - 1)a}

    S = u * t + {a + 2a + .........+ (t - 1)a }

    S = ut + a( 1 + 2 + 3 ........+ (t-1))

    S = ut + a * t * (t -1) / 2

    Vijay
     
  2. jcsd
  3. Jul 26, 2008 #2

    rcgldr

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    Homework Helper

    The distance moved is based on average velocity during each period, not the final velocity at the end of each time period:

    S = {u + (1/2)a } + {u + (3/2)a} + {u + (5/2)a } + ... + {u + ((2t-1)/2)a}

    To calculate the sum of the coefficients for a:

    Code (Text):

    c  = (   1)/2 + (   3)/2 + (   5)/2 +    ...   + (2t-5)/2 + (2t-3)/2 + (2t-1)/2
     
    2c = (   1)/2 + (   3)/2 + (   5)/2 +    ...   + (2t-5)/2 + (2t-3)/2 + (2t-1)/2
       + (2t-1)/2 + (2t-3)/2 + (2t-5)/2 +    ...   + (   5)/2 + (   3)/2 + (   1)/2
         -------------------------------------------------
         (2t  )/2 + (2t  )/2 + (2t  )/2 + (2t  )/2 + (2t  )/2 + (2t  )/2 + (2t  )/2

         = t^2

    c    = 1/2 t^2
     
    S = u * t + 1/2 * a * t^2
     
    Last edited: Jul 26, 2008
  4. Jul 27, 2008 #3

    Hootenanny

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    Staff Emeritus
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    Gold Member

    A much more straight forward method would be to solve the second order ODE:

    [tex]\frac{d^2s}{dt^2} = \text{const.}[/tex]
     
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