Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
The motion of a body is given by equation \[\frac{d\,v(t)}{dt}=6.0-3v(t),\]where\[v(t)\]is the speed in m/s and t in sec. If body was at rest at\[t=0\] (1) the terminal speed is 2.0 m/s (2) the speed varies with the time as \[v(t)=2(1-{{e}^{-3t}})m/s\] (3) the magnitude of initial acceleration is\[6.0\text{ }m/{{s}^{2}}\] (4) the speed is 0.1 m/s when the acceleration if half the initial valueA) 1, 2 and 3 are correct.
B) 1 and 2 are correct.
C) 2 and 4 are correct.
D) 1 and 3 are correct.
Correct Answer: A
Solution :
\[\frac{dv}{dt}=6-3v\] \[\Rightarrow \] \[\frac{dv}{6-3v}=dt\] Integrating both sides\[\int{\frac{dv}{6-3v}}=\int{dt}\] \[\Rightarrow \] \[\frac{{{\log }_{e}}(6-3v)}{-3}=t+{{k}_{1}}\] \[\Rightarrow \] \[{{\log }_{e}}(6-3v)=-3t={{k}_{2}}\] At\[t=0,v=0\]\[\therefore \]\[{{\log }_{e}}6={{k}_{2}}\] Substituting the value of\[{{k}_{2}}\]in Eq. (i) \[{{\log }_{e}}(6-3v)=-3t+{{\log }_{e}}6\] \[\Rightarrow \] \[{{\log }_{e}}\left( \frac{6-3v}{6} \right)=-3t\] \[\Rightarrow \] \[{{e}^{-3t}}=\frac{6-3v}{6}\] \[\Rightarrow \] \[6-3v=6{{e}^{-3t}}\] \[\Rightarrow \] \[v=2(1-{{e}^{-3t}})\] \[\therefore \] \[{{v}_{ter\min al}}=2m/s,a=6{{e}^{-3t}}\] \[{{a}_{initial}}=6m/{{s}^{2}}\]
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