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4-1.Newton's Laws of Motion
hard
$m$ દળનો એક કણ $t = 0$ સમયે સ્થિર સ્થિતિમાં છે. તેને $x$ $-$ દિશામાં $F(t) = F_0e^{-bt}$ બળ લાગુ પાડવામાં આવે છે. નીચેનામાંથી કયા દ્વારા તેની ઝડપ $v(t)$ દર્શાવાય છે?
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
