Answer
See below
Work Step by Step
The differential equation for the velocity of an object at time $t$, we have: $\frac{dv}{dt}+kv=80ke^{-kt}$
Integrating factor could be: $I(x)=e^{\int kdt}=e^{kt}$
The equation becomes: $ \frac{d}{dt}(ve^{kt})=80k$
Integrating both sides:
$\rightarrow ve^{kt}=80kt+c\\\rightarrow v(t)=\frac{80kt+c}{e^{kt}}$
Use the initial-value condition $v(0)=20$ to find $c=20$
Thus, the solution is: $v(t)=\frac{80kt+20}{e^{kt}}$
a) The velocity of the object is not changing and remaining the same $20$
b) Given: $\frac{dv}{dt}(0)=2$
then $\frac{dv}{dt}=e^{-kt}80k-ke^{-kt}80kt-ke^{-kt}20\\=20ke^{-kt}(4-4kt-1)\\=20ke^{-kt}(3-4kt)$
Substitute: $20.3.k=2\\
\rightarrow k=\frac{1}{30}$
c) The velocity of the object $t$: $v(t)=(\frac{8}{3}t+20)e^{-\frac{1}{30}t}$
d) Find a value for $t$: $(\frac{8}{3}t+20)e^{-\frac{1}{30}t}=0\\
\rightarrow \frac{8}{3}t+20=0 \lor e^{-\frac{1}{30}t}=0\\
\rightarrow t=-\frac{60}{8} \lor t=\infty$
Since time can not be a negative value, there is not a finite time $t\gt 0$ at which object is at rest.
e) When $t\rightarrow \infty$, it follows that $v(t) \rightarrow 0$
