Linear Air Resistance

Horizontal motion

If $n$ is unity then the velocity is exponentially approaching zero.

$F(v) = -bv$ : negative sign indicates a retarding force and

$b$ is a proportionality constant

$\sum \vec {F}_{ext} = -bv = m \frac{dv}{dt}$

$\Rightarrow \int_{v_i}^{v_f} \frac{dv}{v} = \int_{t_i}^{t_f} \frac{-b}{m}dt$

$\ln\frac{v_f}{v_i} = \frac{-b}{m}t$ ;

$t_i \equiv 0$

$v_f = v_i e^{-\frac{b}{m}t}$

The displacement is given by

$x = \int_0^t v_i e^{-\frac{b}{m}t} dt$

$= \left . v_i \left ( \frac {e^{-\frac{b}{m}t}}{-\frac{b}{m}} \right ) \right |_0^t$

$= \left . v_i \left ( -\frac{m}{b} e^{-\frac{b}{m}t} \right ) \right |_0^t$

$= \left . v_i \left ( \frac{m}{b} e^{-\frac{b}{m}t} \right ) \right |_t^0$

$= v_i \left ( \frac{m}{b} e^{-\frac{b}{m}0} -\frac{m}{b} e^{-\frac{b}{m}t} \right )$

$= \frac{m}{b} v_i \left ( 1-e^{-\frac{b}{m}t} \right )$

Consider a ball falling under the influence of gravity and a frictional force that is proportion to its velocity

$\sum \vec{F}_{ext} = mg -bv = m \frac{dv}{dt}$

let

$b=$ coefficient of air resistance

$v_t = \frac{mg}{b} =$ Terminal speed

$v_t -v = \frac{1}{b} \frac{dv}{dt}$

$b dt= \frac{dv}{v_t -v}$

$-b dt= \frac{dv}{v -v_t}$

$-\int_0^t b dt= \int_{v_0}^v \frac{dv}{v -v_t}$

$-bt = \ln{\left( v -v_t \right)} - \ln{\left ( v_0-v_t \right )}$

$-bt = \ln \left(\frac{ v -v_t }{v_0-v_t}\right )$

$e^{-bt} = \left(\frac{ v -v_t }{v_0-v_t}\right )$

$v -v_t = \left ( v_0-v_t\right )e^{-bt}$

$v = v_0e^{-bt} + v_t \left (1 -e^{-bt}\right )$

The posiiton as a function of time may be determined by directly integrating the above equation

$\frac{dy}{dt} = v_0e^{-bt} + v_t \left (1 -e^{-bt}\right )$

$\int_0^y = \int_0^t \left ( v_0e^{-bt} + v_t \left (1 -e^{-bt}\right ) \right ) dt$

$y = \int_0^t v_0e^{-bt}dt + \int_0^t v_t \left (1 -e^{-bt}\right ) dt$

$= \frac{v_0}{-b}\left ( e^{-bt}-e^{-b0} \right ) + v_t t + \frac{v_t}{b}\left ( e^{-bt} - e^{-b0}\right ) dt$

$= \frac{v_0}{b}\left ( 1- e^{-bt} \right ) + v_t t + \frac{v_t}{b}\left ( e^{-bt} - 1\right ) dt$

$= v_t t + \frac{1}{b}\left ( v_0 - v_t) \right ) \left ( 1- e^{-bt} \right ) dt$