A particle of mass $$m$$ is at rest at the origin at time $$t=0$$. It is subjected to a force $$F(t) = F_0 e^{-bt}$$ in the $$x$$ direction. Its speed $$v(t)$$ is depicted by which of the following curves?

The applied force varies with time as $$F(t)=F_0 e^{-bt}$$ along the $$+x$$-direction.

Newton’s second law gives instantaneous acceleration $$a(t)=\frac{F(t)}{m}=\frac{F_0}{m}\,e^{-bt}$$

The particle starts from rest, so $$v(0)=0$$. Speed at any later time is the time-integral of acceleration:

$$\begin{aligned} v(t)&=\int_{0}^{t} a(\tau)\,d\tau \\ &=\frac{F_0}{m}\int_{0}^{t} e^{-b\tau}\,d\tau \\ &=\frac{F_0}{m}\left[\,-\frac{1}{b}e^{-b\tau}\right]_{0}^{t}\\ &=\frac{F_0}{mb}\left(1-e^{-bt}\right) \end{aligned}$$

Key features of $$v(t)=\dfrac{F_0}{mb}\left(1-e^{-bt}\right)$$:

• At $$t=0$$, $$v(0)=0$$.
• $$v(t)$$ increases monotonically because $$dv/dt = (F_0/m)e^{-bt}\gt 0$$ for all $$t\gt0$$.
• As $$t\rightarrow\infty$$, the exponential term vanishes and the speed saturates at a finite limit $$v_{\text{max}}=\dfrac{F_0}{mb}$$. Thus the curve rises quickly at first and then levels off asymptotically.

Among the given sketches, the only one that starts from the origin, rises monotonically, and approaches a horizontal asymptote is the third plot.

Hence, Option C which is: the curve that climbs from $$0$$ to a constant terminal value asymptotically.