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Scientific American Supplement, No. 531, March 6, 1886 by Various
page 38 of 142 (26%)
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The explanation of this method may be a little more abstruse than the
explanations already given, but it is very simple and elegant when you
see it, and I fancy I can make it quite clear. I shall have to preface
it by the explanation of two simple laws. The first of these is, that a
body acted on by a constant force, so as to have its motion uniformly
accelerated, suppose in a straight line, moves through distances which
increase as the square of the time that the accelerating force continues
to be exerted.

The necessary nature of this law, or rather the action of which this law
is the expression, is shown in Fig. 3.

[Illustration: Fig. 3]

Let the distances A B, B C, C D, and D E in this figure represent four
successive seconds of time. They may just as well be conceived to
represent any other equal units, however small. Seconds are taken only
for convenience. At the commencement of the first second, let a body
start from a state of rest at A, under the action of a constant force,
sufficient to move it in one second through a distance of one foot. This
distance also is taken only for convenience. At the end of this second,
the body will have acquired a velocity of two feet per second. This is
obvious because, in order to move through one foot in this second, the
body must have had during the second an average velocity of one foot per
second. But at the commencement of the second it had no velocity. Its
motion increased uniformly. Therefore, at 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,
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