The motion of a body is given by the equation | vedclass

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Question

(d) $\frac{{dv}}{{dt}} = 6 - 3v \Rightarrow \frac{{dv}}{{6 - 3v}} = dt$

Integrating both sides, $\int {\frac{{dv}}{{6 - 3v}}} = \int {dt} $

$&rA

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Vedclass · Answer

(d) $\frac{{dv}}{{dt}} = 6 - 3v \Rightarrow \frac{{dv}}{{6 - 3v}} = dt$

Integrating both sides, $\int {\frac{{dv}}{{6 - 3v}}} = \int {dt} $

$&rArr; \frac{{{{\log }_e}(6 - 3v)}}{{ - 3}} = t + {K_1}$

$&rArr; {\log _e}(6 - 3v) = - 3t + {K_2}$&hellip;(i)

At $t = 0,\;v = 0$

$	herefore {\log _e}6 = {K_2}$

Substituting the value of ${K_2}$ in equation (i)

${\log _e}(6 - 3v) = - 3t + {\log _e}6$

$&rArr; {\log _e}\left( {\frac{{6 - 3v}}{6}} ight) = - 3\,t$ $&rArr;$ ${e^{ - 3t}} = \frac{{6 - 3v}}{6}$

$&rArr; 6 - 3v = 6{e^{ - 3\,t}}$ $&rArr;$ $3v = 6(1 - {e^{ - 3\,t}})$

$&rArr; v = 2(1 - {e^{ - 3\,t}})$

$	herefore {v_{{m{terminal}}}} = 2\;m/s$  (When $t = \infty $).

Acceleration $a = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left[ {2\left( {1 - {e^{ - 3\;t}}} ight)} ight] = 6{e^{ - 3\,t}}$

Initial acceleration =$6\;m/{s^2}$.

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 $t = 0$

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