This post shows the solution for the acceleration-time (a-t) model for free-falling bodies with fluid resistance \(F_f = kv^2\). Before delving in, we recommend reading first how the velocity-time model was derived:
\(v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)
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This solution involves using the chain rule (derivation made by Rigel Melaan).
Express acceleration as the ratio of velocity and time:
\(a(t)=\frac{\partial v}{\partial t}=\frac{\partial\left(\sqrt{\frac{m g}{k}} \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\right)}{\partial t}\)
Factor out constants:
\(a(t)=\sqrt{\frac{m g}{k}} \times \frac{\partial\left(\tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\right)}{\partial t}\)
\(a(t)=\sqrt{\frac{m g}{k}} \times \frac{\partial\left(\tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\right)}{\partial t}\)
Consider the partial derivatives as a chain:
\(a(t)=\sqrt{\frac{m g}{k}} \times \frac{d \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}{d\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}\)
\(\times \frac{\partial\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}{\partial t}\)
Using trigonometric identity for each derivative term in the chain rule:
\(\frac{d \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}{d\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}=\)
\(\operatorname{sech}^2\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)
\(\frac{\partial\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}{\partial t}=\)
\(\frac{\partial\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)\right)}{\partial t}-\frac{\partial\left(t \sqrt{\frac{k g}{m}}\right)}{\partial t}\)
Simplify. Note that the derivative of a constant is zero:
\(\frac{\partial\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)}{\partial t}=\)
\(\frac{\partial\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)\right)}{\partial t}-\frac{\partial\left(t \sqrt{\frac{k g}{m}}\right)}{\partial t}\)
\(-\sqrt{\frac{k g}{m}} \frac{d(t)}{d t}=-\sqrt{\frac{k g}{m}}\)
Apply the chain rule: the derivatives of the original function is the product of individual derivatives:
\(a(t)=\sqrt{\frac{m g}{k}} \times \operatorname{sech}^2\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right) \times-\sqrt{\frac{k g}{m}}\)
Simplify to get the acceleration-time model equation for free-falling bodies with air resistance:
\(a(t)=-g \operatorname{sech}^2\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)
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