This post shows the derivation of the model between two motion variables, acceleration and time, of an object experiencing free-fall motion with drag resistance.
WeTheStudy lets you connect ideas
Learn more

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:

\(v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)

Finding the Solution for Acceleration-Time

Want to access the remaining content?
You're a Member!
Click to expand on exclusive content
Want to access the remaining content?

Become a Member

When you sign-up and subscribe to WeTheStudy, you’ll get the following benefits:

No ads! (yey!)
Complete access to all articles
Ability to track your progress in the tree

SIGN-UP

Complete Your Checkout

When you complete your account, here are the following benefits:

No ads! (yey!)
Complete access to all articles
Ability to track your progress in the tree

PROCEED CHECKOUT

Solution 1

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

Express acceleration as the ratio of velocity and time:

\(a(t)=\frac{\partial v}{\partial t}=\frac{\partial\left(\sqrt{\frac{m g}{k}} \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\right)}{\partial t}\)

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:

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

Created On
June 5, 2023
Updated On
July 20, 2024
Contributors
Edgar Christian Dirige
Founder
References

WeTheStudy original content

Revision
1.00
Got some questions? Something wrong? Contact us