The motion of a body is given by the equation $\frac{{dv(t)}}{{dt}} = 6.0 - 3v(t)$. where $v(t)$ is speed in $m/s$ and $t$ in $\sec $. If body was at rest at $t = 0$
The motion of a body is given by the equation $\frac{{dv(t)}}{{dt}} = 6.0 - 3v(t)$. where $v(t)$ is speed in $m/s$ and $t$ in $\sec $. If body was at rest at $t = 0$
The terminal speed is $2.0 \,m/s$
The speed varies with the time as $v(t) = 2(1 - {e^{ - 3t}})\,m/s$
The magnitude of the initial acceleration is $6.0\,m/{s^2}$
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The motion of a body is given by the equation $\frac{{dv(t)}}{{dt}} = 6.0 - 3v(t)$. where $v(t)$ is speed in $m/s$ and $t$ in $\sec $. If body was at rest at $t = 0$
$\frac{{dv}}{{dt}} = 6 - 3v \Rightarrow \frac{{dv}}{{6 - 3v}} = dt$
Integrating both sides, $\int {\frac{{dv}}{{6 - 3v}}} = \int {dt} $
$⇒ \frac{{{{\log }_e}(6 - 3v)}}{{ - 3}} = t + {K_1}$
$⇒ {\log _e}(6 - 3v) = - 3t + {K_2}$…(i)
At $t = 0,\;v = 0$
$\therefore {\log _e}6 = {K_2}$
Substituting the value of ${K_2}$ in equation (i)
${\log _e}(6 - 3v) = - 3t + {\log _e}6$
$⇒ {\log _e}\left( {\frac{{6 - 3v}}{6}} \right) = - 3\,t$ $⇒$ ${e^{ - 3t}} = \frac{{6 - 3v}}{6}$
$⇒ 6 - 3v = 6{e^{ - 3\,t}}$ $⇒$ $3v = 6(1 - {e^{ - 3\,t}})$
$⇒ v = 2(1 - {e^{ - 3\,t}})$
$\therefore {v_{{\rm{terminal}}}} = 2\;m/s$ (When $t = \infty $).
Acceleration $a = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left[ {2\left( {1 - {e^{ - 3\;t}}} \right)} \right] = 6{e^{ - 3\,t}}$
Initial acceleration =$6\;m/{s^2}$.
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