If a car travelling in a 20 metre circle at 10 kilometres per hour increases its speed to 20 kilometres per hour does the force of the tyre on the road double or square ?

Hint: the magnitude of centripetal acceleration for an object moving in a circle at constant speed is a_r = \omega^2 R = \frac { v^2} R . Memorize it! Note the v (velocity) term, and recall that the force needed to keep the object moving in the circle is detemined from \vec F = m \vec a_r . So now what do you think the answer is?

I haven't a clue. I'm an electrical engineer and that's why I'm asking the question.

Think about the formulas I gave you. Centripetal acceleration goes as the square of velocity, and the force between the tires and the road to keep the car moving in a circle is proportional to the centripetal acceleration.

Hi ebaines. So from that, I take it that when the speed doubles the force between the tyre and the road squares ! I know this happens to the wind resistance of a motor vehicle when its speed doubles the force of the air acting on it actually squares, but the laws of physiscs are something I don't use on a daily basis. Thank you for your assistance.

Correct! Another pheneomenon that goes as velocity squared is kinetic energy: KE = (1/2)mv^2.