Question

For positive constants k and g, the velocity, v, of a particle of mass m at time t is given by v=mgk(1−e−kt/m). At what rate is the velocity is changing at time 0? At t=2? What do your answers tell you about the motion? At what rate is the velocity changing at time 0?

Answer 1

I assume you mean
\[\large v=mgk\left(1-e^{-kt/m}\right)~~?\]

Answer 2

yes

Answer 3

\[\large\begin{align*}\frac{dv}{dt}&=\frac{d}{dt}\left[mgk\left(1-e^{-kt/m}\right)\right]\\\\
&=\color{gray}{mgk\frac{d}{dt}\left[\left(1-e^{-kt/m}\right)\right]}\\\\
&=\color{gray}{mgk\left(\frac{d}{dt}[1]-\frac{d}{dt}\left[e^{-kt/m}\right]\right)}\\\\
&=\color{lightgray}{mgk\left(0-e^{-kt/m}\frac{d}{dt}\left[-\frac{kt}{m}\right]\right)}\\\\
&=\color{lightgray}{mgk\left(\frac{k}{m}e^{-kt/m}\right)}\\\\
&=\color{white}{gk^2e^{-kt/m}}\end{align*}\]
Plug in \(t=0\).