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Analysis at Constant Acceleration

We have at                              t =0;S = S0 , v =v0.

(1.2) =>                               dv = a dt

or Integrating        

or                                 v – v0 =at

v =v0 +at

Integrating ,      

1/2 [ v2 –v02] = a (S –S0)

v2 =v02 +2a (S –S0)

 

From (1.1) =>                         dS =v  dt

dS = (v0+at) dt

S –S0 = v0 t +1/2 at2

S =S0 + v0 t +1/2 at2

Note that equation (1.5), (1.6) and (1.7) should be applied to a case of constant acceleration only.