A particle of mass m is at rest at the origin at time t=0 It is...

Given ($\mathrm{F}(\mathrm{t})=\mathrm{FO} \mathrme^{-\mathrm{bt}}$), acceleration ($\mathrm{a}(\mathrm{t})=\mathrm{F}(\mathrm{t})$/ $\mathrm{m}=(\mathrm{FO}$/ $\mathrm{m}) \mathrme^{-\mathrm{bt}}$), so (dv/dt=($\mathrm{FO}$/ $\mathrm{m}) \mathrme^{-\mathrm{bt}}$). Integrating with ($\mathrm{v}($0)=0) gives ($\mathrm{v}(\mathrm{t})=(\mathrm{FO}$/(\mathrm{mb}))\left[1-\mathrme^{-$\mathrm{bt}$}\right]). This rises from 0 and asymptotically approaches ($\mathrm{FO}$/($\mathrm{mb})$). Hence the correct ($\mathrm{v}-\mathrm{t}$) curve is a rising exponential saturating at ($\mathrm{FO}$/($\mathrm{mb})$), which corresponds to (b).