Narendra Sisodiya: Equations of Motion

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) = U_i + A (t - T_i)

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

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

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 $\vec v (T_i) = \vec U_i$

$\vec U_i = \vec A \, ({T_i} - {T_i}) + K$

$k= \vec U_i$

$\vec v (t) = \vec U_i + \vec A \, (t - {T_i})$

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) = S_i + U_i(t-T_i) + (1/2)A (t - T_i)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 $t = T f$ and putting eq (3) in eq (4) , we will get

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

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

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

eq ... (6)

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

$\vec v (t) = \vec U_i + \vec A \, (t - {T_i})$

$\vec v (t) . \vec v (t) = \big(\vec U_i + \vec A \, (t - {T_i}) \big).\big(\vec U_i + \vec A \, (t - {T_i}) \big)$

$\vec v (t) . \vec v (t) = \vec U_i . \vec U_i + \vec A . \vec A \ (t - T_i)^2 + 2 \vec A . \vec U_i (t- T_i)$

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

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