The figure below shows how the PRT system can schematically be represented as the free-body diagram. We have separated the elements to make it easier to understand the forces applied to each body, but notice they are still connected. The trolley moves forward due to some traction force, $F_{traction}$, which is counteracted by the air drag $F_{drag}$. The resulting force on the trolley is therefore the sum of both individual forces.

We get: $$\begin{cases} F_{res} &=& F_{traction} + F_{drag}\\ F_{drag} &=& -\dfrac{1}{2}\cdot p\cdot v^2\cdot C_D\cdot A \end{cases}$$ where $p$ is the air density, $v$ the speed of the trolley (relative to the air), $A$ the cross-sectional area and $C_D$ the drag coefficient. The drag force is negative, because we assume that any motion forwards (to the right in the figure) is positive and any motion backwards (i.e., to the left) is negative.

Newton's Second Law of Motion states the resulting force equals the product of the mass and the acceleration ($F_{res} = m\cdot a$). We know the acceleration $a$ is the change of rate (i.e., the derivative) of the velocity $v$. The equations can now be written as: $$a = \dfrac{dv}{dt} = \dfrac{F_{traction} - \dfrac{1}{2}\cdot p\cdot v^2\cdot C_D\cdot A}{m_{trolley}}$$

The human passenger in the trolley can be represented by a mass-spring-damper system. It has a mass (the rectangle named "Passenger" in the figure), a spring (the zig-zag line) and a damper (the structure under the zig-zag). The force for which a spring opposes the compression is proportional to the spring's compression. If, in the figure, an external force (i.e., $F_{inertia}$) pushes the passenger to the left, displacing them by $-x$, the spring will exert a force in the opposite direction: $F_{spring} = k\cdot(-x)$, where $k$ is a constant.

The damper opposes the compression quite differently: a slow compression will not be opposed with significant force, but a fast compression will. If, in the figure, an external force pushes the passenger to the left, displacing them with a speed of $-v$, the damper will excert an opposite force: $F_{damper} = c\cdot (-v)$, where $c$ is a constant.

Applying Newton's Second Law, under the influence of an external force $F_{inertia}$, the following set of equations can be obtained: $$\begin{cases} F_{spring} &=& k(-x)\\ F_{damper} &=& c(-v)\\ m_{psgr}\cdot a &=& F_{inertia} + F_{spring} + F_{damper}\\ \dfrac{dv}{dt} &=& a\\ \dfrac{dx}{dt} &=& v\\ \end{cases}$$ which can be converted to the following first-order ODE: $$\begin{cases} \dfrac{dv}{dt} &=& \dfrac{- F_{inertia} + k(-x) + c(-v)}{m_{psgr}}\\ \dfrac{dx}{dt} &=& v\\ \end{cases}$$

If an external force $F_{inertia}$ pushes the train forwards, the train would start moving forwards and the passenger would feel the acceleration of the train. For a small passenger (think: an ant or a spider), this force would be really weak, but for a large passenger (i.e., a human), it would be stronger. This is a fictitious force, which can be described with Newton's Second Law: $F_{inertia} = -m_{psgr}\cdot a$. Combining both ODE from above now yields the following equations. Notice that the total mass of the trolley also includes the mass of the passenger. $$\begin{cases} \dfrac{dv_{trolley}}{dt} &=& \dfrac{F_{traction} - \dfrac{1}{2}\cdot p\cdot v^2\cdot C_D\cdot A}{m_{trolley} + m_{psgr}}\\ \dfrac{dx_{trolley}}{dt} &=& v_{trolley}\\ \dfrac{dv_{psgr}}{dt} &=& \dfrac{k(-x_{psgr}) + c(-v_{psgr}) - (m_{psgr}\cdot a_{inertia})}{m_{psgr}}\\ \dfrac{dx_{psgr}}{dt} &=& v_{psgr} \end{cases}$$

Here, $a_{inertia}$ is still unknown, yet this is the acceleration felt by the passenger when the PRT starts moving. It is therefore the same intensity as the acceleration of the trolley: $a_{inertia} = a_{trolley}$. The ODE of this full system is: $$\begin{cases} \dfrac{dv_{psgr}}{dt} &=& \dfrac{k\cdot (-x_{psgr}) + c\cdot (-v_{psgr}) - m_{psgr}\cdot\dfrac{F_{traction} - \dfrac{1}{2}\cdot p\cdot v^2\cdot C_D\cdot A}{m_{trolley} + m_{psgr}}}{m_{psgr}}\\ \dfrac{dv_{trolley}}{dt} &=& \dfrac{F_{traction} - \dfrac{1}{2}\cdot p\cdot v_{trolley}^2\cdot C_D\cdot A}{m_{trolley} + m_{psgr}}\\ \dfrac{dx_{psgr}}{dt} &=& v_{psgr}\\ \dfrac{dx_{trolley}}{dt} &=& v_{trolley} \end{cases}$$