User:Narendra Sisodiya/mechanics

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
 * $$ k= \vec U_i $$

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

constant acceleration can be found using initial and final condition

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)

$$

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

Taking $$ t = T_f $$ 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


 * $$ \vec{S_f} = \vec{S_i} + \vec U_i ( {T_f} - {T_i} ) + \frac { { \vec V_f - \vec U_i } } {2 } (T_f - T_i )  $$

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

or