A force F = Be^{-Ct} acts on a particle of ma | vedclass

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(B) Given force $F = Be^{-Ct}$.According to Newton's second law,$F = ma$,so $ma = Be^{-Ct}$.Acceleration $a = \frac{dv}{dt} = \frac{B}{m} e^{-Ct}$.Int

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$\frac{B}{mC}$

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(B) Given force $F = Be^{-Ct}$.According to Newton's second law,$F = ma$,so $ma = Be^{-Ct}$.Acceleration $a = \frac{dv}{dt} = \frac{B}{m} e^{-Ct}$.Integrating with respect to time $t$ from $0$ to $t$:$\int_{0}^{v} dv = \int_{0}^{t} \frac{B}{m} e^{-Ct} dt$.$v = \frac{B}{m} \left[ \frac{e^{-Ct}}{-C} ight]_{0}^{t} = \frac{B}{mC} (1 - e^{-Ct})$.Terminal velocity is the velocity as $t 	o \infty$.As $t 	o \infty$,$e^{-Ct} 	o 0$.Therefore,$v_{terminal} = \frac{B}{mC} (1 - 0) = \frac{B}{mC}$.

$A$ force $F = Be^{-Ct}$ acts on a particle of mass $m$,which is at rest at $t = 0$. Its terminal velocity is:

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