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

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(D) Given that the force is $F(t) = F_0e^{-bt}$.According to Newton's second law,$F = m \frac{dv}{dt}$.So,$m \frac{dv}{dt} = F_0e^{-bt}$.Rearranging t

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(D) Given that the force is $F(t) = F_0e^{-bt}$.According to Newton's second law,$F = m \frac{dv}{dt}$.So,$m \frac{dv}{dt} = F_0e^{-bt}$.Rearranging the terms,we get $\frac{dv}{dt} = \frac{F_0}{m} e^{-bt}$.Integrating both sides with respect to time $t$ from $0$ to $t$,and velocity $v$ from $0$ to $v(t)$:$\int_0^{v(t)} dv = \int_0^t \frac{F_0}{m} e^{-bt} dt$.$v(t) = \frac{F_0}{m} \left[ \frac{e^{-bt}}{-b} ight]_0^t$.$v(t) = \frac{F_0}{m} \left( \frac{e^{-bt}}{-b} - \frac{e^0}{-b} ight)$.$v(t) = \frac{F_0}{mb} (1 - e^{-bt})$.As $t 	o \infty$,$v(t) 	o \frac{F_0}{mb}$.This represents an exponential growth curve that approaches a horizontal asymptote at $v = \frac{F_0}{mb}$.Comparing this with the given options,curve $D$ represents this behavior.

$A$ 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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