Mathematics - Application of Derivative Question with Solution | TestHub

The differential equation of the motion is $$\begin{array}{l}\frac{\mathrm{dv}}{\mathrm{dt}}=30-3\mathrm{v}\\ \Rightarrow \frac{\mathrm{dv}}{30-3\mathrm{v}}=\mathrm{dt}\end{array}$$ Integrating we get $$\begin{array}{l}-\frac{1}{3}\log (30-3\mathrm{v})=\mathrm{t}+\mathrm{C}\\ \Rightarrow \log (30-3\mathrm{v})=-3(\mathrm{t}+\mathrm{c})\\ \Rightarrow 30-3\mathrm{v}={\mathrm{e}}^{-3\mathrm{t}-3\mathrm{c}}=\mathrm{A}{\mathrm{e}}^{-3\mathrm{t}},\mathrm{A}={\mathrm{e}}^{-3\mathrm{c}}\\ \Rightarrow 3\mathrm{v}=30-{\mathrm{Ae}}^{-3\mathrm{t}}\\ \text{ For maximum velocity }\frac{\mathrm{dv}}{\mathrm{dt}}=0\\ \Rightarrow 30-3\mathrm{v}=0\text{ from (i) }\\ \therefore \mathrm{v}=\frac{30}{3}=10\mathrm{cm}/\mathrm{s}\end{array}$$ which is the maximum velocity However from (ii) $\frac{\mathrm{dv}}{\mathrm{dt}}=3{\mathrm{Ae}}^{-3\mathrm{t}}$ Clearly $\frac{\mathrm{dv}}{\mathrm{dt}}=0$ if $\mathrm{t}\to \infty$ $\therefore$ Themaximum velocity will be achieved after infinite time in other words, the maximum velocity will hever be reached.