Question

A differential equation for the velocity v of a falling mass m subjected to air resistance

proportional to the square of the instantaneous velocity is:

𝑚 (𝑑𝑣/𝑑𝑡) = mg-kv²

Where k >0 is a constant of proportionality. The positive direction is downward.

(a) Solve the equation subject to the initial condition v(0)=v₀.

(b) Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass.

(c) If the distance s, measured from the point where the mass was released above ground, is

related to velocity v by ds/dt=v(t), find an explicit expression for s(t) if s(0)= 0.

Answer 1

$𝑚(𝑑𝑣/𝑑𝑡) = mg-kv^2$

Put $mg=ku_0^2$ and $u_0=gT$. Then

$𝑚 (𝑑𝑣/𝑑𝑡) = k(u_0^2-v^2)$

1. Consider the case $u_0>v_0$

$\frac{dv}{u_0^2-v^2}=\frac{k}{m}dt$

$\int\limits_{v(0)}^{v(t)}\frac{dv}{u_0^2-v^2}=\int\limits_{0}^{t}\frac{k}{m}dt=\frac{kt}{m}$

$\int\limits_{v(0)}^{v(t)}\frac{dv}{u_0^2-v^2}=\int\limits_{v_0}^{v(t)}\left(\frac{1}{u_0-v}+\frac{1}{u_0+v}\right)\frac{dv}{2u_0}=$

$=\frac{1}{2u_0} \log\frac{u_0+v(t)}{u_0-v(t)}-\frac{1}{2u_0} \log\frac{u_0+v_0}{u_0-v_0}$

$\log\frac{u_0+v(t)}{u_0-v(t)}=\log\frac{u_0+v_0}{u_0-v_0}+2t\frac{ku_0}{m}$

But $\frac{ku_0}{m}=\frac{gku_0}{gm}=\frac{gku_0}{ku_0^2}=\frac{g}{u_0}=\frac{1}{T}$

$\frac{u_0+v(t)}{u_0-v(t)}=\frac{u_0+v_0}{u_0-v_0}e^{\frac{2kv_0t}{m}}=\frac{u_0+v_0}{u_0-v_0}e^{\frac{2t}{T}}=e^{\frac{2(t-t_0)}{T}}$ ,

where $t_0=-\frac{1}{2}\log\frac{u_0+v_0}{u_0-v_0}$ is the time when $v(t)=0$. Then

$\frac{2u_0}{u_0-v(t)}=1+\frac{u_0+v(t)}{u_0-v(t)}=1+e^{\frac{2(t-t_0)}{T}}$

$u_0-v(t)=\frac{2u_0}{1+e^{\frac{2(t-t_0)}{T}}}$

$v(t)=u_0-\frac{2u_0}{1+e^{\frac{2(t-t_0)}{T}}}=u_0\frac{e^{\frac{2(t-t_0)}{T}}-1}{e^{\frac{2(t-t_0)}{T}}+1}=u_0\tanh\frac{t-t_0}{T}$

where $u_0=\sqrt\frac{mg}{k}$, $T=\frac{u_0}{g}=\sqrt\frac{m}{kg}$ and $t_0=-\frac{1}{2}\log\frac{u_0+v_0}{u_0-v_0}$.

(b) $\lim\limits_{t\to+\infty}v(t)=\lim\limits_{t\to+\infty}u_0\tanh\frac{t-t_0}{T}=u_0=\sqrt\frac{mg}{k}$

(c) $s(t)=s(0)+\int\limits_{0}^{t}v(t)dt=\int\limits_{0}^{t}u_0\tanh\frac{t-t_0}{T}dt=$

$=u_0T\int\limits_{0}^{t}\frac{d\cosh\frac{t-t_0}{T}}{\cosh\frac{t-t_0}{T}}=u_0T(\log\cosh\frac{t-t_0}{T}-\log\cosh\frac{t_0}{T})$

$=u_0T\log\frac{\cosh\frac{t-t_0}{T}}{\cosh\frac{t_0}{T}}$

2. Consider the case $u_0<v_0$

$\frac{dv}{u_0^2-v^2}=\frac{k}{m}dt$

$\int\limits_{v(0)}^{v(t)}\frac{dv}{u_0^2-v^2}=\int\limits_{0}^{t}\frac{k}{m}dt=\frac{kt}{m}$

$\int\limits_{v(0)}^{v(t)}\frac{dv}{u_0^2-v^2}=\int\limits_{v_0}^{v(t)}\left(\frac{1}{u_0-v}+\frac{1}{u_0+v}\right)\frac{dv}{2u_0}=$

$=\frac{1}{2u_0} \log\frac{v(t)+u_0}{v(t)-u_0}-\frac{1}{2u_0} \log\frac{v_0+u_0}{v_0-u_0}$

$\log\frac{v(t)+u_0}{v(t)-u_0}=\log\frac{v_0+u_0}{v_0-u_0}+2t\frac{ku_0}{m}=\log\frac{v_0+u_0}{v_0-u_0}+\frac{2t}{T}$

$\frac{v(t)+u_0}{v(t)-u_0}=\frac{v_0+u_0}{v_0-u_0}e^{\frac{2t}{T}}=e^{\frac{2(t-t_0)}{T}}$ ,

where $t_0=-\frac{1}{2}\log\frac{v_0+u_0}{v_0-u_0}$ is the time when $v(t)=\infty$. Then

$\frac{2u_0}{v(t)-u_0}=\frac{v(t)+u_0}{v(t)-u_0}-1=e^{\frac{2(t-t_0)}{T}}-1$

$v(t)-u_0=\frac{2u_0}{e^{\frac{2(t-t_0)}{T}}-1}$

$v(t)=u_0+\frac{2u_0}{e^{\frac{2(t-t_0)}{T}}-1}=u_0\frac{e^{\frac{2(t-t_0)}{T}}+1}{e^{\frac{2(t-t_0)}{T}}-1}=u_0\coth\frac{t-t_0}{T}$

where $u_0=\sqrt\frac{mg}{k}$, $T=\frac{u_0}{g}=\sqrt\frac{m}{kg}$ and $t_0=-\frac{1}{2}\log\frac{v_0+u_0}{v_0-u_0}$.

3. Consider the case $v_0=u_0$. It is evident that in this case $v(t)=u_0$ is the unique solution.

(b) $\lim\limits_{t\to+\infty}v(t)=\lim\limits_{t\to+\infty}u_0\tanh\frac{t-t_0}{T}=u_0$ , if $u_0>v_0$ ,

$\lim\limits_{t\to+\infty}v(t)=\lim\limits_{t\to+\infty}u_0\coth\frac{t-t_0}{T}=u_0$ , if $u_0<v_0$ ,

$\lim\limits_{t\to+\infty}v(t)=\lim\limits_{t\to+\infty}u_0=u_0$ , if $u_0=v_0$ .

Therefore, in any case $\lim\limits_{t\to+\infty}v(t)=u_0=\sqrt\frac{mg}{k}$.

(c) $s(t)=s(0)+\int\limits_{0}^{t}v(t)dt=\int\limits_{0}^{t}u_0\tanh\frac{t-t_0}{T}dt=$

$=u_0T\int\limits_{0}^{t}\frac{d\cosh\frac{t-t_0}{T}}{\cosh\frac{t-t_0}{T}}=u_0T(\log\cosh\frac{t-t_0}{T}-\log\cosh\frac{t_0}{T})$

$=u_0T\log\frac{\cosh\frac{t-t_0}{T}}{\cosh\frac{t_0}{T}}$ , if $u_0>v_0$

If $u_0<v_0$ then $s(t)=s(0)+\int\limits_{0}^{t}v(t)dt=\int\limits_{0}^{t}u_0\coth\frac{t-t_0}{T}dt=$

$=u_0T\int\limits_{0}^{t}\frac{d\sinh\frac{t-t_0}{T}}{\sinh\frac{t-t_0}{T}}=u_0T(\log\sinh\frac{t-t_0}{T}-\log\sinh\frac{t_0}{T})$

$=u_0T\log\frac{\sinh\frac{t-t_0}{T}}{\sinh\frac{t_0}{T}}$

If $u_0=v_0$ then $s(t)=s(0)+\int\limits_{0}^{t}v(t)dt=\int\limits_{0}^{t}u_0dt=u_0t$

Answer.

if $u_0>v_0$ then

(a) $v(t)=u_0\tanh\frac{t-t_0}{T}$, where $u_0=\sqrt\frac{mg}{k}$, $T=\frac{u_0}{g}=\sqrt\frac{m}{kg}$ and $t_0=-\frac{1}{2}\log\frac{u_0+v_0}{u_0-v_0}$

(b) $\lim\limits_{t\to+\infty}v(t)=\sqrt\frac{mg}{k}$

(c) $s(t)=u_0T\log\frac{\cosh\frac{t-t_0}{T}}{\cosh\frac{t_0}{T}}$

if $u_0<v_0$ then

(a) $v(t)=u_0\coth\frac{t-t_0}{T}$, where $u_0=\sqrt\frac{mg}{k}$, $T=\frac{u_0}{g}=\sqrt\frac{m}{kg}$ and $t_0=-\frac{1}{2}\log\frac{v_0+u_0}{v_0-u_0}$

(b) $\lim\limits_{t\to+\infty}v(t)=\sqrt\frac{mg}{k}$

(c) $s(t)=u_0T\log\frac{\sinh\frac{t-t_0}{T}}{\sinh\frac{t_0}{T}}$

if $u_0=v_0$ then

(a) $v(t)=v_0=u_0$

(b) $\lim\limits_{t\to+\infty}v(t)=\sqrt\frac{mg}{k}$ ,

(c) $s(t)=u_0t$