Science:Math Exam Resources/Courses/MATH102/December 2016/Question B 03 (c)/Solution 1

We can rewrite the equation in the form

${\frac {dv}{dt}}=g-kv=-k(v-{\frac {g}{k}})=-k(v-v_{ss})$ where $v_{ss}={\frac {g}{k}}$ (recall part (a)).

This means that if we denote $z=v-v_{ss}$ then since $v_{ss}$ is a constant we have ${\frac {dz}{dt}}={\frac {dv}{dt}}$ so

${\frac {dz}{dt}}=-kz,z_{0}=v(0)-v_{ss}=0-10^{-6}=-10^{-6}$, which has a solution $z(t)=-10^{-6}e^{-kt}\ {\text{or}}\ v-10^{-6}=-10^{-6}e^{-kt}$ .

Now, the time to go from $v(0)=0$ to $v(t)=(1-e^{-2})v_{ss}=(1-e^{-2})10^{-6}$ is equivalent to solving the following equation for t:

$(1-e^{-2})10^{-6}-10^{-6}=-10^{-6}e^{-kt}$

i.e. $e^{-kt}={\frac {(1-e^{-2})10^{-6}-10^{-6}}{-10^{-6}}}=e^{-2}$. Indeed $\color {blue}t={\frac {2}{k}}$.