A particle of mass m is at rest at the origin | vedclass

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Question

$\begin{array}{l}
Given\,that\,F\left( t ight) = {F_0}{e^{ - bt}} \Rightarrow m\frac{{dv}}{{dt}} = {F_0}{e^{ - bt}}\
\frac{{dv}}{{dt}} = \frac{{{

Vedclass · Accepted Answer

Vedclass · Answer

$\begin{array}{l}
Given\,that\,F\left( t ight) = {F_0}{e^{ - bt}} \Rightarrow m\frac{{dv}}{{dt}} = {F_0}{e^{ - bt}}\
\frac{{dv}}{{dt}} = \frac{{{F_0}}}{m}{e^{ - bt}} \Rightarrow \int\limits_0^v {dv}  = \frac{{{F_0}}}{m}\int\limits_0^t {{e^{ - bt}}} dt\
v = \frac{{{F_0}}}{m}\left[ {\frac{{{e^{ - bt}}}}{{ - b}}} ight]_0^t = \frac{{{F_0}}}{{mb}}\left[ { - \left( {{e^{ - bt}} - {e^{ - 0}}} ight)} ight]\
 \Rightarrow \,v\, = \frac{{{F_0}}}{{mb}}\left[ {1 - {e^{ - bt}}} ight]
\end{array}$

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?

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