A Meaningful Physics Model - Simple car on 2D plane

Very often people try to describe how things work to students using very simple examples.  One of my disliked versions is the simple nonholonomic ( don't worry about this I'll explain later) car model. When someone tries to explain how a car works, they make examples so simple, these examples never work like anything real.  This causes students - including me - real problems understanding harder models.

Imagine you are trying to understand how mechanics - simple Newtonian physics - works.  Imagine a card like object, something that drives forward along it's front wheels direction of travel.  OK, seems simple enough.  We all know that when you push an object with wheels it moves forward (perpendicular to the wheel axis) over ground.

Let's assume some simple things that don't change Newton's Laws of Motion.  Let's assume the entire world is flat.  If so, we can ignore gravity so this makes a 2D force equation.  Let's just move in X and Y and forget Z.  Imagine an x-y plane. The car drives over it.  These are huge real world assumptions that don't apply but to start off they are critical.

Now how we push anything is to hit it with a force. Punch your brother, he moves. How, you transfered momentum to him but that started as accelerating a mass.
 Here are my variables:  P represents the position of the car model.  Let's say that $$p$$ is the point midway through the wheel axle along the travel direction. P consists of an x and y component.
$$\vec{p} = (x, y)$$
The angle of wheels when measured from my imaginary origin at (0,0) is theta
$$\theta$$
 Angle can change independently of the velocity applied to the system.
Now let's call the force I apply to the vehicle as
$$F(t)$$
where t is time. The force applied can vary over time.  But the force does not depend on the vehicle it is pushing.  When we add friction, that will change.
 Let the car have a mass of $$m$$ or it's weight at the Earth's surface (another assumption).

Most people start with a kinematic model, one that ignores how mechanical systems work by ignoring Newton's Laws.

Here is Newton's Second Law of Motion in terms I used:

$$ m \otimes  \ddot{\vec{p}} = F(t)   $$

Forces applied at all times to an object equals that mass cross product acceleration. Let's apply force at the center of the mass, so this simplifies to

$$ m  \cdot \ddot{\vec{p}} = F(t)   $$

Another way to say this is:
$$  \ddot{\vec{p}} = \frac{F(t)}{m }   $$
This puts it into a second order ordinary differential equation.


Now, instead of just claiming that the car has a velocity in the x and y directions let's derive how you get that.  Since the mass is a constant, we can ignore a change in mass.  We will make one more assumption, that during one tiny infinitessimal period the amount of force applied to the mass does not change.  This is wrong, because it's possible that Force varies instantaneously but in reality this is a small error term for the area under the integral curve.  This is a big departure from other people's interpretation.  On the other hand, I don't make the small angle assumption which estimates $$ \theta \approxeq \sin \theta $$
to make the equations linear.  You can work with nonlinear systems, they don't bite.

Now how do you figure out how the position changes over time based on a force applied to the car?  We can solve the acceleration differential equation twice, integrate once to get the velocity equation and twice to get the position equation.

$$ \int \ddot{\vec{p}} dt = \int \frac{F(t)}{m } dt  $$

Force and mass constant:


$$ \int \ddot{\vec{p}} dt = \frac{F(t)}{m } \int dt  $$


$$  \dot{\vec{p}} = ( \frac{F(t)}{m } ) * t + C $$

When you integrate, this multiplies the constants by t and adds a constant term.  You need initial conditions to know what C should be.  For example, if your wheeled car was already accelerating there would be a constant velocity applied.  This is the velocity equation.



 $$ \int \dot{\vec{p}} dt  = ( \frac{F(t)}{m } ) \int ( t + C_{0}) dt $$

 $$ {\vec{p}}  = ( \frac{F(t)}{m } )  ( t^{2} +  C_{0}t +  C_{1}) $$

This is the position equation.  If the car's angle is used to determine the velocity in the x and y directions, then the velocity is the projection of vector velocity along the axes.

$$ (\dot{x} = \dot{p}\cos (\theta) ,  \dot{y} = \dot{p}\sin (\theta) )$$


$$\dot{\vec{p}} = (\dot{x},\dot{y}) = (   )$$

Ok this is complicated, but now you can apply a force to the wheeled vehicle and it will move like a real object would.