g-force experienced by a jumper using DE

Question

The eqation $mv' = Kv^2 - mg$ describes the velocity (negative velocity points downwards of a parachute jumper of mass $m$ subject to gravity $g$, and wind resistance from the open parachute of $Kv^2$, with $K$ a constant. $m= 100 kg$, $g=10 m/s^2$, $K=10 kg/m$.

Suppose that at $t = 0$, $v(0) = -20 m/s$. Find the time at which the acceleration $v'$ is largest in absolute value. At that moment, what is the g-force experienced by the jumper?

We are given that $v' = \frac{1}{10}v^2 - g$.

I tried using implicit differentiation so that $v'' = \frac{1}{5}vv' = \frac{1}{5}v(\frac{1}{10}v^2-g)$ and set this equal to $0$. Hence we have 3 critical points, at $v= 0$, and $v = \pm \sqrt{10g}$.

Calculating $v''(0)=-120$, we know the function is concave, but I'm not really sure where to go from here. Any help would really be appreciated!

Answer 1

$$m\frac{dv}{dt}=Kv^2-mg$$ With the given numerical values $\quad m= 100$, $\quad g=10$, $\quad K=10$ : $$v'=\frac{dv}{dt}=\frac{1}{10}v^2-10$$ $$dt=\frac{dv}{\frac{1}{10}v^2-10}$$ $$t=\int\frac{dv}{\frac{1}{10}v^2-10}=\frac12\ln\left(\frac{v-10}{v+10} \right)+c$$ Initial condition : $v(0)=-20$

$0=\frac12\ln\left(\frac{-20-10}{-20+10} \right)+c\quad\implies\quad c=-\frac12\ln(3)$ $$t=\frac12\ln\left(\frac{v-10}{3(v+10)} \right)$$ $$\boxed{v=-10\frac{3e^{2t}+1}{3e^{2t}-1}}$$ $$v'=120\frac{e^{2t}}{(3e^{2t}-1)^2}$$ $v'>0\quad$ thus $v(t)$ is increasing, starting from $-20$ and tending to $-10$. $$v''=-240\frac{e^{2t}(3e^{2t}+1)}{(3e^{2t}-1)^3}$$ $t>0\quad\implies\quad 3e^{2t}-1>0$

$v''<0\quad$ thus $v'(t)$ is decreasing, starting from $30$ and tending to $0$.

Hence the maximum of $v'$ is $30$ at $t=0$.

Of course the above result is valid only for $t\geq 0$.

Answer 2

The differential equation $ \ \frac{dv}{dt} \ = \ \frac{K}{m}·v^2 - g \ \ $ has two equilibrium solutions (at which $ \ \frac{dv}{dt} \ = \ 0 \ ) \ , $ namely $ \ v \ = \ -\sqrt{\frac{mg}{K}} \ \ $ and $ \ v \ = \ + \sqrt{\frac{mg}{K}} \ \ . $ Since the "negative direction" has been identified as "downward" in this problem, the second of these solutions is never attained, since an "upward" velocity would be unphysical for "falling" through a resistive medium.

We also see that $ \ \frac{dv}{dt} \ > \ 0 \ \ $ for $ \ |v| \ > \ \sqrt{\frac{mg}{K}} \ $ and $ \ \frac{dv}{dt} \ < \ 0 \ \ $ for $ \ -\sqrt{\frac{mg}{K}} \ < \ v \ < \ +\sqrt{\frac{mg}{K}} \ \ . $ This indicates that the "downward" terminal velocity $ \ v_{t} \ = \ -\sqrt{\frac{mg}{K}} \ $ is an "attractor": larger downward speeds will be decelerated to $ \ v_{t} \ $ by the air resistance, while slower downward speeds will be increased to $ \ v_{t} \ $ by the gravitational force.

[The "upward" equilibrium velocity $ \ +\sqrt{\frac{mg}{K}} \ $ is a "repellor": lesser upward speeds will be "reduced" to downward speeds by the combination of gravity and air resistance. An upward speed faster than $ \ +\sqrt{\frac{mg}{K}} \ $ is "unphysical" because this differential equation becomes an inaccurate model, since the air resistance would then be "downward", making the proper differential equation $ \ \frac{dv}{dt} \ = \ -\frac{K}{m}·v^2 - g \ \ . \ ] $

It is not necessary then to solve the differential equation to answer the question. At the initial velocity $ \ v(0) \ = \ -20 \ $ m/sec , the acceleration $ \ \frac{dv}{dt} \ $ is the largest it ever will be, since the downward speed is subsequently decelerated to the terminal velocity $ \ v_{t} \ \approx \ -\sqrt{\frac{100·10}{10}} \ = \ -10 \ $ m/sec. At this equilibrium speed, the acceleration has become equal to zero (gravity and air resistance "in balance"). So the moment when (the absolute value of) the person's acceleration is greatest is at $ \ t \ = \ 0 \ . $

At that time, the force due to air resistance is upward with magnitude $ \ K·v^2 \ = \ 10·(-20)^2 \ = \ 4000 \ $ Newtons, while the force of gravity (the person's "weight") is downward (always) with magnitude $ \ mg \ = \ 100·10 \ = \ 1000 \ $ Newtons. The sum of these is a net upward force of $ \ 3000 \ $ Newtons, which is equal to $ \ 3·mg \ \ . $ So when the skydiver first begins their "jump", they experience an upward "g-force" of $ \ 3 \ $ g's, which declines to zero (effective "weightlessness") at terminal velocity (presuming they have enough vertical "drop" available to reach that equilibrium speed).