This question is a duplicate of [Does turning sharply on a bicycle conserve more energy than a wide turn?][1], but the accepted answer to that question is incorrect, so it's probably safe for now. In general, in the absence of friction, the force of turning applied to something like a bike curving through a turn is approximately perpendicular to the path, and hence does no work.

However, realistically, friction losses will probably play a decent part in it. So, if I were you I'd try setting up the simulation so that 
$$\frac{dv}{dt}=-\alpha v \left|\frac{d\theta}{dt}\right|$$
or something of the sort, where $\alpha$ is a constant of proportionality which encodes the frictional energy loss associated with turning around the car and $\frac{d\theta}{dt}=\omega$ is the angular velocity of the car. 

This is a model differential equation which states that the speed-damping coefficient is $\alpha\left|\omega(t)\right|$, ie, the sharper you turn, the more speed you lose due to friction. This way, when you take a sharp turn, you lose a lot of speed from inefficiency and friction, but when you take a slow, soft turn, you don't lose as much speed.

If you need this in 2D rectilinear coordinates, you can just use the fact that $\theta(t)=\mbox{ArcTan}[x(t),y(t)]$ in Mathematica notation, from which it follows that 

$$\frac{dx}{dt}=-\alpha x(t)\left|\frac{x(t) y'(t)-y(t) x'(t)}{x(t)^2+y(t)^2}\right|$$
and
$$\frac{dy}{dt}=-\alpha y(t)\left|\frac{x(t) y'(t)-y(t) x'(t)}{x(t)^2+y(t)^2}\right|.$$

Try various values of $\alpha$, and if possible run tests. Which ones look realistic, and which ones look goofy? Testing and debugging to find something which looks "right" is probably the most important part.


  [1]: http://physics.stackexchange.com/questions/91639/does-turning-sharply-on-a-bicycle-conserve-more-energy-than-a-wide-turn