written 6.2 years ago by | • modified 4.0 years ago |
Show that the distance traveled by the particle at any time t is $x=\left(\frac{g}{k^2}+\frac{u}{k}\right)(1-e^{-kt})-\frac{g}{k}t$
written 6.2 years ago by | • modified 4.0 years ago |
Show that the distance traveled by the particle at any time t is $x=\left(\frac{g}{k^2}+\frac{u}{k}\right)(1-e^{-kt})-\frac{g}{k}t$
written 6.2 years ago by | • modified 6.0 years ago |
$ \frac{dv}{dt}=-g-kv\\ \frac{dv}{g+kv}=-dt$
Integrating the above equation we get,
$\frac{1}{k}\log (g+kv)=-t+c$
$Initially\ when\ t=0,\ v=u$
$\frac{1}{k}(g+ku)=c$
$\frac{1}{k}\log (g+kv)=-t+\frac{1}{k}\log(g+ku)$
$ t=\frac{1}{k}\log\frac{(g+ku)}{(g+kv)}$
$kt=\log\frac{(g+ku)}{(g+kv)} $
$ e^{kt}=\frac{(g+ku)}{(g+kv)}$
$e^{-kt}(g+ku)=(e+gv)$
$-g+(g+ku)e^{-kt}=(kv)$
$v=-\frac{g}{k}+\frac{1}{k}(g+ku)e^{-kt}$
$ \frac{dx}{dt}=\frac{-g}{k}+\frac{1}{k}(g+ku)e^{-kt} $
Integrating the above equation we get,
$x=-\frac{g}{k}t-\frac{1}{k^2}(g+ku)e^{-kt}+c $
Initially when $t=0,x=0$ we get $c=\frac{(g+ku)}{k^2}$
$ x=-\frac{g}{k}tt-\frac{1}{k^2}(g+ku)e^{-kt}+\frac{(g+ku)}{k^2}$
$x= \frac{(g+ku)}{k^2}(1-e^{-kt})-\frac{g}{k}t$
$x=\left( \frac{g}{k^2}+\frac{u}{k}\right)(1-e^{-kt})-\frac{g}{k}t$