the termination of the second its velocity must have reached two feet per second.  Let the triangle A B F represent this accelerated motion, and the distance, of one foot, moved through during the first second, and let the line B F represent the velocity of two feet per second, acquired by the body at the end of it.  Now let us imagine the action of the accelerating force suddenly to cease, and the body to move on merely with the velocity it has acquired.  During the next second it will move through two feet, as represented by the square B F C I. But in fact, the action of the accelerating force does not cease.  This force continues to be exerted, and produces on the body during the next second the same effect that it did during the first second, causing it to move through an additional foot of distance, represented by the triangle F I G, and to have its velocity accelerated two additional feet per second, as represented by the line I G. So in two seconds the body has moved through four feet.  We may follow the operation of this law as far as we choose.  The figure shows it during four seconds, or any other unit, of time, and also for any unit of distance.  Thus: 

Time 1           Distance 1
"  2               "    4
"  3               "    9
"  4               "   16

So it is obvious that the distance moved through by a body whose motion is uniformly accelerated increases as the square of the time.

But, you are asking, what has all this to do with a revolving body?  As soon as your minds can be started from a state of rest, you will perceive that it has everything to do with a revolving body.  The centripetal force, which acts upon a revolving body to draw it to the center, is a constant force, and under it the revolving body must move or be deflected through distances which increase as the squares of the times, just as any body must do when acted on by a constant force.  To prove that a revolving body obeys this law, I have only to draw your attention to Fig. 2.  Let the equal arcs, A B and B C, in this figure represent now equal times, as they will do in case of a body revolving in this circle with a uniform velocity.  The versed sines of the angles, A O B and A O C, show that in the time, A C, the revolving body was deflected four times as far from the tangent to the circle at A as it was in the time, A B. So the deflection increased as the square of the time.  If on the table already given, we take the seconds of arc to represent equal times, we see the versed sine, or the amount of deflection of a revolving body, to increase, in these minute angles, absolutely so far as appears up to the fifteenth place of decimals, as the square of the time.