g. So the critical acceleration and velocity are a crit =  v crit 2 /r  =  g. Rearranging this equation gives the critical speed: v crit =  ?(rg)  =  17 m.s ?1 =  62 k.p.h. Note that, while in contact with the road, the car is rotating in the vertical plane with angular velocity ? = v/r. At the point near where it loses contact with the road, the forces exerted on the wheels are small, and so too are the torques (which we shall meet when we come to Rotation). Consequently, the car, whilst airborne, continues to rotate at this rate. This rotation adds extra danger to the manoeuvre, which is just one of the reasons for the advice, rarely given in physics, that one should trust the calculation here, rather than perform the experiment. * Racing cars often have a large foil whose effect, at high speed, is to provide a downwards force and thus allow greater frictional forces for cornering, braking and even acceleration. Some street cars have a miniature version of this foil. These have several possible functions. First, they identify the style of the driver, which may provide a warning to other drivers. Second, they increase drag and petrol consumption, which benefits oil companies (in the short term, at least). Do they have a significant effect in producing a downwards force? Taking the weight of the car as 20 kN, a 10% increase in downwards force would require the foil to exert 2 kN downwards.