I have to work through an example from physics to strengthen my understanding of Newton's second law of motion through second-order ODEs.
I've been provided with the following settings: that the skydiver is descending under a parachute at a steady rate, where t is the time; and I have to assume zero displacement and velocity at time zero.
$$\frac{d^2y}{dt^2} + \frac{k}m^{}\frac{dy}{dt} = g$$
I've come to the following general solution y(t):
$$y(t) = A + Be^{-\frac{kt}{m}} + g\frac{m}{k}t$$
If I set the displacement and velocity at time zero, I got:
$$A=-B=-g\frac{m^2}{k^2}$$
From that point, how can I calculate:
- Time-varying vertical velocity?
- Final speed corresponding to the speed of the skydiver when descending at a steady rate?
Best Answer
The vertical velocity is given by
$$V=\frac{dy}{dt}=...$$
and the final velocity is
$$\lim_{t\to+\infty}\frac{dy}{dt}=g\frac mk$$ The exponential term goes to zero.