Answer
(a)
$v_{term}=\sqrt{2000gD}$
(b)
$v_{term}$ of the drops with a diameter of $10^{-3}$ m is $\sqrt{19.6}\approx4.4271 \hspace{4px} m/s$.
$v_{term}$ of the drops with a diameter of $10^{-4}$ m is $1.4 \hspace{4px} m/s$.
(c)
The raindrops accelerate more rapidly at lower velocities.
Work Step by Step
(a)
Since the terminal velocity of the drop is at zero acceleration.
$\dfrac{d^{2}s}{dt^{2}}=0$
Now substitute $\dfrac{d^{2}s}{dt^{2}}=0$ in $\dfrac{d^{2}s}{dt^{2}}=g-\dfrac{0.0005}{D}\left(\dfrac{ds}{dt}\right)^{2}$.
$0=g-\dfrac{0.0005}{D}\left(\dfrac{ds}{dt}\right)^{2}$
Now solve for $\dfrac{ds}{dt}$.
$\dfrac{0.0005}{D}\left(\dfrac{ds}{dt}\right)^{2}=g$
$\implies\left(\dfrac{ds}{dt}\right)^{2}=\dfrac{D}{0.0005}\times g$
$\implies\dfrac{ds}{dt}=\sqrt{\dfrac{D}{0.0005}\times g}$
Since, $\dfrac{1}{0.0005}=2000$
$\implies\dfrac{ds}{dt}=\sqrt{2000gD}$
Since, $\dfrac{ds}{dt}$ is the terminal velocity.
$v_{term}=\sqrt{2000gD}$
(b)
To find $v_{term}$ of the drops with a diameter of $10^{-3}$ m.
Substitute $D=10^{-3}\hspace{4px} m$ and $g=9.8\hspace{4px} m/s^{2}$ in $v_{term}=\sqrt{2000gD}$.
$v_{term}=\sqrt{2000\times9.8\times10^{-3}}$
$\implies v_{term}=\sqrt{19.6}\approx4.4271 \hspace{4px} m/s$
To find $v_{term}$ of the drops with a diameter of $10^{-4}$ m.
Substitute $D=10^{-4}\hspace{4px} m$ and $g=9.8\hspace{4px} m/s^{2}$ in $v_{term}=\sqrt{2000gD}$.
$v_{term}=\sqrt{2000\times9.8\times10^{-4}}$
$\implies v_{term}=\sqrt{1.96}=1.4 \hspace{4px} m/s$
(c)
The given equation in the model is $\dfrac{d^{2}s}{dt^{2}}=g-\dfrac{0.0005}{D}\left(\dfrac{ds}{dt}\right)^{2}$.
There is a negative sign in the velocity term.
So, as the velocity of the raindrops increases the value of acceleration decreases.
Thus, the acceleration and velocity of the raindrops are inversely related.
So, the raindrops accelerate more rapidly at lower velocities.
