A) The terminal speed is 2.0 \[m/s\]
B) The speed varies with the time as \[v(t)=2(1-{{e}^{-3t}})m/s\]
C) The speed is \[0.1m/s\] when the acceleration is half the initial value
D) The magnitude of the initial acceleration is \[6.0m/{{s}^{2}}\]
Correct Answer: A
Solution :
\[\frac{dv}{dt}=6-3v\Rightarrow \frac{dv}{6-3v}=dt\] Integrating both sides, \[\int{\frac{dv}{6-3v}}=\int{dt}\] Þ\[\frac{{{\log }_{e}}(6-3v)}{-3}=t+{{K}_{1}}\] Þ\[{{\log }_{e}}(6-3v)=-3t+{{K}_{2}}\] ?(i) At \[t=0,\ v=0\]\[\therefore \] \[{{\log }_{e}}6={{K}_{2}}\] Substituting the value of \[{{K}_{2}}\] in equation (i) \[{{\log }_{e}}(6-3v)=-3t+{{\log }_{e}}6\] Þ \[{{\log }_{e}}\left( \frac{6-3v}{6} \right)=-3\,t\] Þ \[{{e}^{-3t}}=\frac{6-3v}{6}\] Þ \[6-3v=6{{e}^{-3\,t}}\] Þ \[3v=6(1-{{e}^{-3\,t}})\] Þ \[v=2(1-{{e}^{-3\,t}})\] \[\therefore \] \[{{v}_{\text{trminal}}}=2\ m/s\](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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