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

1. Jul 26, 2008

### 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

2. Jul 26, 2008

### rcgldr

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
3. Jul 27, 2008

### Hootenanny

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

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