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16 July 2008

Equations of Motion

Hi, most of the time I found incorrect equation of motion for classical mechanics with constant acceleration. So here are the correct equation in vector format with removing all assumptions.

Deriving the equations in vectors

  • All constant are taken in capital letters
  • All variable are given in small letter

Deriving v(t) = Ui + A (t - Ti)

\vec{a} = \frac{d \vec{v}}{dt} = \frac{d^2 \ \vec{s}}{dt^2}

so

 \vec v (t) = \int _{T_i} ^t \vec a dt

for zero or constant acceleration A we have

 \vec v (t) = \vec A  \int _{T_i} ^t  dt
 \vec v (t) = \vec A \,  [t] _{T_i} ^t
 \vec v (t) = \vec A \,  (t - {T_i}) + K

we have one unknown K , we need to consider initial or final condition

  • Lets take initially we have  \vec v (T_i) = \vec U_i
 \vec U_i = \vec A \,  ({T_i} - {T_i}) + K

ie

 k= \vec U_i
 \vec v (t) = \vec U_i + \vec A \,  (t - {T_i})  eq ... (1)

If we take final condition in consideration then ie :  \vec v (T_f) = \vec V_f

 \vec v (t) = \vec V_f + \vec A \,  ({T_f} -t )  eq ... (2)

constant acceleration can be found using initial and final condition

 \vec A =  \frac { \vec V_f - \vec U_i }{ T_f - T_i} eq ... (3)

Deriving s(t) = Si + Ui(t-Ti) + (1/2)A (t - Ti)2

now we have

 \vec v (t) = \frac{d \vec{s}}{dt} = \vec U_i + \vec A \,  (t - {T_i})
 \vec{s}(t)= \int _{T_i} ^{t} \Big(\vec U_i + \vec A \,  (t - {T_i}) \Big) {dt}
 \vec{s}(t) = (\vec U_i - \vec A \, T_i ) \int _{T_i} ^{t} {dt} + \vec A \int _{T_i} ^{t} t \ {dt}
 \vec{s}(t) = (\vec U_i - \vec A \, T_i ) \ \Big[t\Big]  ^{t} _{T_i} + \vec A \Bigg[\frac { t^2} {2} \Bigg] _{T_i} ^{t} + K


 \vec{s}(t) = (\vec U_i - \vec A \, T_i ) ( {t} - {T_i} ) + \frac {\vec A } {2 } (t^2-{T_i}^2) + K

for eliminating K we need either final or initial condition ie  \vec{s}(T_i) = \vec S_i

 \vec{S_i} = (\vec U_i - \vec A \, T_i ) ( {T_i} - {T_i} ) + \frac {\vec A } {2 } ({T_i}^2-{T_i}^2) + K
 K= \vec{S_i}
 \vec{s}(t) = \vec{S_i}  + (\vec U_i - \vec A \, T_i ) ( {t} - {T_i} ) + \frac {\vec A } {2 } (t^2-{T_i}^2)
 \vec{s}(t) = \vec{S_i}  + \vec U_i ( {t} - {T_i} ) - \vec A  ( {t}T_i - {T_i}^2 ) + \frac {\vec A } {2 } (t^2-{T_i}^2)
 \vec{s}(t) = \vec{S_i}  + \vec U_i ( {t} - {T_i} ) + \frac {\vec A } {2 } (t^2 - 2 {t}T_i +  {T_i}^2)
 \vec{s}(t) = \vec{S_i}  + \vec U_i ( {t} - {T_i} ) + \frac {\vec A } {2 } (t - T_i ) ^2  eq ... (4)

If we would have taken final condition ie  \vec{s}(T_f) = \vec S_f  then

 \vec{s}(t) = \vec{S_f}  + \vec V_f ( {T_f} - {t} ) - \frac {\vec A } {2 } (T_f - t ) ^2  eq ... (5)

Taking t = Tf and putting eq (3) in eq (4) , we will get

 \vec{s}(T_f) = \vec{S_i}  + \vec U_i ( {T_f} - {T_i} ) + \frac {\frac { \vec V_f - \vec U_i }{ T_f - T_i} } {2 } (T_f - T_i ) ^2

Or

 \vec{S_f} = \vec{S_i}  + \vec U_i ( {T_f} - {T_i} ) + \frac { { \vec V_f - \vec U_i } } {2 } (T_f - T_i )
 \vec{S_f} = \vec{S_i}  +  \frac { ( \vec V_f + \vec U_i ) (T_f - T_i ) } {2 }   eq ... (6)


Deriving |v(t)|2 = |Ui|2 + 2 A . (s(t)-S_i)

 \vec v (t) = \vec U_i + \vec A \,  (t - {T_i})
 \vec v (t) . \vec v (t) = \big(\vec U_i + \vec A \,  (t - {T_i}) \big).\big(\vec U_i + \vec A \,  (t - {T_i}) \big)
 \vec v (t) . \vec v (t) =   \vec U_i . \vec U_i + \vec A . \vec A \ (t - T_i)^2 + 2 \vec A . \vec U_i (t- T_i)
 \vec v (t) . \vec v (t) =   \vec U_i . \vec U_i + \vec A . \vec A \ (t - T_i)^2 + 2 \vec A . \Bigg( \vec s(t) - \vec S_i - \frac {\vec A } {2 } (t - T_i ) ^2 \Bigg)
 \vec v (t) . \vec v (t) =   \vec U_i . \vec U_i + \vec A . \vec A \ (t - T_i)^2 + 2 \vec A . ( \vec s(t) - \vec S_i) - 2 \vec A . ( \frac {\vec A } {2 } (t - T_i ) ^2 )


 \vec v (t) . \vec v (t) =   \vec U_i . \vec U_i + 2 \vec A . ( \vec s(t) - \vec S_i)

taking case for final velocity

 \vec V_f . \vec V_f =   \vec U_i . \vec U_i + 2 \vec A . ( \vec S_f - \vec S_i)  eq ... (7)

or

 \left | \vec V_f \right | ^ 2 =   \left | \vec U_i \right | + 2 \vec A . ( \vec S_f - \vec S_i)  eq ... (7)




Source Page : http://en.wikipedia.org/wiki/User:Narendra_Sisodiya/mechanics
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