Narendra Sisodiya: July 2008

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)

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

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

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

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

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

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

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)

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

eq ... (7)

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

15 July 2008

Command line PNR Enquiry Script

Hi, I have developed a command line PNR Enquiry Script, though it is a small but useful and ofcourse my first perl script.

Dependency :
you need to have Perl on your system with commnad line browser w3m, inshort install w3m and perl bofore you download the script.
Note : you can view in lynx also, in case of lynx , replace w3m with lynx

Install It

$ cd
$ wget http://narendra.sisodiya.googlepages.com/pnrenq.pl
$ chmod +x pnrenq.pl

How to use it

./pnrenq.pl 8322181194

The first argument is 10 digit PNR number

#######################################
# pnrenq - PNR Enquiry Script version 1 #
# Author - Narendra Sisodiya #
# Mail - narendra.sisodiya@gmail.com #
# (C) Narendra Sisodiya - 2008 #
# This Script is released under GPLv2 #
#######################################

Narendra Sisodiya

Menu

27 July 2008

5th LUG@IITD Workshop

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

Deriving v(t) = U_i + A (t - T_i)

Deriving s(t) = S_i + U_i(t-T_i) + (1/2)A (t - T_i)²

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

15 July 2008

Command line PNR Enquiry Script

Followers

Contributors

My Blog List

Menu

27 July 2008

5th LUG@IITD Workshop

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

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

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

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

15 July 2008

Command line PNR Enquiry Script

Deriving v(t) = U_i + A (t - T_i)

Deriving s(t) = S_i + U_i(t-T_i) + (1/2)A (t - T_i)²

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