How to Derive Free Fall Model with Drag Resistance (Acceleration-Time)?

Solution 1

This solution involves using the chain rule (derivation made by Rigel Melaan).

Express acceleration as the ratio of velocity and time:

$a(t)=\frac{\partial v}{\partial t}=\frac{\partial (\sqrt{\frac{mg}{k}}\tanh (\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}}))}{\partial t}$

Factor out constants:

$a(t)=\sqrt{\frac{mg}{k}}\times \frac{\partial (\tanh (\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}}))}{\partial t}$

$a(t)=\sqrt{\frac{mg}{k}}\times \frac{\partial (\tanh (\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}}))}{\partial t}$

Consider the partial derivatives as a chain:

$a(t)=\sqrt{\frac{mg}{k}}\times \frac{d\tanh (\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}})}{d(\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}})}$

$\times \frac{\partial (\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}})}{\partial t}$

Using trigonometric identity for each derivative term in the chain rule:

$\frac{d\tanh (\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}})}{d(\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}})}=$

${\mathrm{sech}}^2(\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}})$

$\frac{\partial (\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}})}{\partial t}=$

$\frac{\partial \left(\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)\right)}{\partial t}-\frac{\partial \left(t\sqrt{\frac{kg}{m}}\right)}{\partial t}$

Simplify. Note that the derivative of a constant is zero:

$\frac{\partial (\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}})}{\partial t}=$

$\frac{\partial \left(\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)\right)}{\partial t}-\frac{\partial \left(t\sqrt{\frac{kg}{m}}\right)}{\partial t}$

$-\sqrt{\frac{kg}{m}}\frac{d(t)}{dt}=-\sqrt{\frac{kg}{m}}$

Apply the chain rule: the derivatives of the original function is the product of individual derivatives:

$a(t)=\sqrt{\frac{mg}{k}}\times {\mathrm{sech}}^2(\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}})\times -\sqrt{\frac{kg}{m}}$

Simplify to get the acceleration-time model equation for free-falling bodies with air resistance:

$a(t)=-g{\mathrm{sech}}^2(\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}})$