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

[vijay_singh]

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

 

[rcgldr]

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

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:

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

 

[Hootenanny]

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

\frac{d^2s}{dt^2} = \text{const.}