A)
B)
C)
D)
Correct Answer: C
Solution :
[c] \[F=ma={{F}_{0}}{{e}^{-bt}}\] |
\[\frac{dv}{dt}=\frac{{{F}_{0}}}{m}{{e}^{-bt}}\] |
\[\int\limits_{0}^{v}{dv}=\frac{{{F}_{0}}}{m}\int\limits_{0}^{t}{{{e}^{-bt}}dt}\] |
\[v=\frac{{{F}_{0}}}{m}\left[ \frac{{{e}^{-bt}}}{-b} \right]_{0}^{t}\] |
\[v=\frac{{{F}_{0}}}{mb}\left( 1-{{e}^{-bt}} \right)\] |
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