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 
$$\frac{dv}{dt}=-\alpha v \frac{d\theta}{dt}$$
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 $\frac{d\theta}{dt}$ is the angular velocity of the car. 

This is a model differential equation which states that the speed-damping coefficient is $\alpha\frac{d\theta}{dt}$, 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.

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