# Puzzle #7 – Resistive Forces

A particle of mass $m$ is acted on by a force

$F=\lambda e^{-\frac{x}{\gamma}}$

where $\lambda$ and $\gamma$ are positive constants, and $x$ is the particle’s displacement from the origin.

Question: By using the chain rule, show that this equation of motion is exactly equivalent to one of the form

$F=\lambda-\rho v^2$

and deduce an equivalence for $\rho$. How might you describe the two forces at work?

Extension: Given that the particle starts from rest at the origin, show that the particle’s position $x$ after time $t$ is given by

$x=2\gamma\ln\bigg(\cosh\Big[t\sqrt{\frac{\lambda}{2m\gamma}}\Big]\bigg)$

or an equivalent, expressed in terms of $\rho$.

Hint: The chain rule provides the following relation:

$\displaystyle \frac{dv}{dt}=\frac{dv}{dx}\cdot\frac{dx}{dt}$