Physics - Kinematics Question with Solution | TestHub

For (A) : At $\mathrm{t}\to \infty$, x will be maximum so ${\mathrm{x}}_{\text{max }}={\mathrm{x}}_0\left(1-{\mathrm{e}}^{-\infty }\right)={\mathrm{x}}_0$

For (B) : $\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}={\mathrm{x}}_0\left[0-\left({\mathrm{e}}^{-\mathrm{at}}\right)(-\mathrm{a})\right]={\mathrm{ax}}_0{\mathrm{e}}^{-\mathrm{at}}\Rightarrow {\mathrm{v}}_{\text{max }}={\mathrm{ax}}_0$

For (C) : $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}={\mathrm{ax}}_0\left({\mathrm{e}}^{-\mathrm{at}}\right)(-\mathrm{a})=-{\mathrm{a}}^2{\mathrm{x}}_0{\mathrm{e}}^{-\mathrm{at}}<0$

For (D) : a (-ve) \& v (+ve) ⇒ speeding down