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4-1.Newton's Laws of Motion
hard
A particle of mass $m$ is at rest at the origin at time $t = 0$. It is subjected to a force $F(t) = F_0e^{-bt}$ in the $x$ direction. Its speed $v (t)$ is depicted by which of the following curves?
A
B
C
D
(AIEEE-2012)
Solution
$\begin{array}{l}
Given\,that\,F\left( t \right) = {F_0}{e^{ – bt}} \Rightarrow m\frac{{dv}}{{dt}} = {F_0}{e^{ – bt}}\\
\frac{{dv}}{{dt}} = \frac{{{F_0}}}{m}{e^{ – bt}} \Rightarrow \int\limits_0^v {dv} = \frac{{{F_0}}}{m}\int\limits_0^t {{e^{ – bt}}} dt\\
v = \frac{{{F_0}}}{m}\left[ {\frac{{{e^{ – bt}}}}{{ – b}}} \right]_0^t = \frac{{{F_0}}}{{mb}}\left[ { – \left( {{e^{ – bt}} – {e^{ – 0}}} \right)} \right]\\
\Rightarrow \,v\, = \frac{{{F_0}}}{{mb}}\left[ {1 – {e^{ – bt}}} \right]
\end{array}$
Given\,that\,F\left( t \right) = {F_0}{e^{ – bt}} \Rightarrow m\frac{{dv}}{{dt}} = {F_0}{e^{ – bt}}\\
\frac{{dv}}{{dt}} = \frac{{{F_0}}}{m}{e^{ – bt}} \Rightarrow \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 = \frac{{{F_0}}}{{mb}}\left[ { – \left( {{e^{ – bt}} – {e^{ – 0}}} \right)} \right]\\
\Rightarrow \,v\, = \frac{{{F_0}}}{{mb}}\left[ {1 – {e^{ – bt}}} \right]
\end{array}$
Standard 11
Physics
