Question

In (14) of Section 1.3 we saw that a differential equation describing the velocity v of a falling mass subject to air resistance proportional to the instantaneous velocity is

$\mathit{m}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{m}\mathit{g}\mathbf{-}\mathit{k}\mathit{v}$

where $\mathit{k}\mathbf{>}\mathbf{0}$is a constant of proportionality. The positive direction is downward.

(a) Solve the equation subject to the initial condition $\mathit{v}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{v}}_{\mathbf{0}}$.

(b) Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass. We saw how to determine the terminal velocity without solving the DE in Problem 40 in Exercises 2.1.

(c) If the distance s, measured from the point where the mass was released above ground, is related to velocity v by $\mathit{d}\mathit{s}\mathbf{/}\mathit{d}\mathit{t}\mathbf{=}\mathit{v}\mathbf{\left(}\mathit{t}\mathbf{\right)}$, nd an explicit expression for s(t) if $\mathit{s}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

Short answer

$\text{a)}v\left(t\right)=\frac{mg}{k}+\left(v_0-\frac{mg}{k}\right)e^{-\frac{k}{m}t}$

$b)v=\frac{mg}{k}$

$c\left)s\right(t)=\frac{mg}{k}t+\frac{m}{k}\left(v_0-\frac{mg}{k}\right)\left(1-e^{-\frac{k}{m}t}\right)$

Step-by-step solution

Definition of Series circuits

The linear differential equation for the current i(t),

$\mathit{L}\frac{\mathbf{d}\mathbf{i}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{R}\mathit{i}\mathbf{=}\mathit{E}\mathbf{\left(}\mathit{t}\mathbf{\right)}$

where L and R are known as the inductance and the resistance, respectively. The current i(t) is also called the response of the system.

Differentiate the equation

a) We have a falling mass with a velocity is subjected to air resistance proportional to the velocity, has the differential equation

$m\frac{dv}{dt}=mg-kv$

which is

$\frac{dv}{dt}=g-\frac{k}{m}v$

and we have have to solve it as the following technique:

This equation is a first order linear and separable differential equation, then we can separate variables and then do integration as

Separate the variables and integrate

$\begin{array}{rcl}\frac{dv}{g-\frac{k}{m}v}& =& dt\\ \int \frac{1}{g-\frac{k}{m}v}dv& =& \int dt\\ -\frac{m}{k}\int \frac{-\frac{k}{m}}{g-\frac{k}{m}v}dv& =& \int dt\end{array}$

$\begin{array}{rcl}-\frac{m}{k}\ln \left(g-\frac{k}{m}v\right)& =& t+c_1\\ \ln \left(g-\frac{k}{m}v\right)& =& -\frac{k}{m}t+c_2\\ e^{\ln \left(g-\frac{k}{m}v\right)}& =& e^{\left(\frac{k}{m}t+c_2\right)}\end{array}$

$\begin{array}{rcl}g-\frac{k}{m}v& =& e^{c_2}{e}^{\frac{k}{m}t}\\ g-\frac{k}{m}v& =& c{e}^{\frac{k}{m}t}\\ \frac{k}{m}v& =& g-c{e}^{\frac{k}{m}t}\end{array}$

Then we have

$v\left(t\right)=\frac{mg}{k}-k{e}^{-\frac{k}{m}t}$

After that, to find the value of constant k, we have to apply the point of condition$(v,t)=\left(v_0,0\right)$ into equation (2), then we have

$v_0=\frac{mg}{k}-k{e}^0$

Then we have

$k=\frac{mg}{k}-v_0$

Substitution

After that, substitute with the value of constant into equation (2), then we have

$\begin{array}{rcl}v\left(t\right)& =& \frac{mg}{k}-\left(\frac{mg}{k}-v_0\right)e^{\frac{k}{m}t}\\ & =& \frac{mg}{k}+\left(v_0-\frac{mg}{k}\right)e^{-\frac{k}{m}t}\end{array}$

is the velocity of the falling mass at time t.

Find the velocity

b)

Now, to find the limiting velocity for the falling mass, we have to substitute with$t=\infty$ into equation (3), then we obtain

$\begin{array}{rcl}v& =& \frac{mg}{k}+\left(v_0-\frac{mg}{k}\right)e^{-\infty }\\ & =& \frac{mg}{k}+\left(v_0-\frac{mg}{k}\right)\left(0\right)\\ & =& \frac{mg}{k}\end{array}$

Find distance

c)

Now, we have the distance measured from releasing point to the ground below is related to the velocity of the falling mass as

$\frac{ds}{dt}=v$

with the condition $s(t=0)=0$

This differential equation (5) is a linear and separable, then we can obtain the distance at time t as

$\int ds=\int vdt$

Then we have

$\begin{array}{rcl}s& =& \int \left[\frac{mg}{k}+\left(v_0-\frac{mg}{k}\right)e^{-\frac{k}{m}t}\right]dt\\ & =& \frac{mg}{k}\int dt+\left(v_0-\frac{mg}{k}\right)\int e^{\frac{k}{m}t}dt\\ & =& \frac{mg}{k}t+\left(v_0-\frac{mg}{k}\right)\left(-\frac{m}{k}{e}^{\frac{k}{m}t}\right)+h\\ & =& \frac{mg}{k}t-\frac{m}{k}\left(v_0-\frac{mg}{k}\right)e^{\frac{k}{m}t}+h\end{array}$

After that, to find the value of constant we have to apply the point of condition$(s,t)=(0,0)$ into equation (7), then we have

$\begin{array}{rcl}0& =& \frac{mg}{k}\left(0\right)-\frac{m}{k}\left(v_0-\frac{mg}{k}\right)e^0+h\\ 0& =& 0-\frac{m}{k}\left(v_0-\frac{mg}{k}\right)+h\end{array}$

After that, substitute with the value of constantinto equation (7) , then we have

$\begin{array}{rcl}s\left(t\right)& =& \frac{mg}{k}t-\frac{m}{k}\left(v_0-\frac{mg}{k}\right)e^{-\frac{k}{m}t}+\frac{m}{k}\left(v_0-\frac{mg}{k}\right)\\ & =& \frac{mg}{k}t+\frac{m}{k}\left(v_0-\frac{mg}{k}\right)\left(1-e^{\frac{k}{m}t}\right)\\ & & \end{array}$

is the distance from position of falling mass above the ground at time

$$\begin{gathered} \text{a)}v\left(t\right)=\frac{mg}{k}+\left(v_0-\frac{mg}{k}\right)e^{-\frac{k}{m}t} \\ b)v=\frac{mg}{k} \\ c)s\left(t\right)=\frac{mg}{k}t+\frac{m}{k}\left(v_0-\frac{mg}{k}\right)\left(1-e^{-\frac{k}{m}t}\right) \end{gathered}$$