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the force pushing itself is changing or accumulating, but rather that the
continued application of a net force always results in a buildup of momen-
tum (or a reduction of momentum, when the directions of the force and
momentum are opposed).
In fact, if we generalize the equation
P
=
f
∆t, we can discover an
even deeper relationship between force and momentum. What if, instead
of pushing the box with a constant force, Moe pushed it with a force that
varied over time? Then we can express the acceleration at any given time
t as
a
(t) =
f
(t)/m,
(12.18)
which is just Newton's second law to which we've added the notation “(t)”
to be more explicit that
a
and
f
vary with time. We learned in Chapter 11
that if we integrate acceleration over time, we get the velocity as a function
of time:
Velocity is the time
integral of acceleration,
remember?
v
(t) =
a
(t) dt.
(12.19)
Substituting Equation (12.18) into Equation (12.19), assuming that the
mass does not vary over time, we have
v
(t) =
f
(t)/m dt,
m
v
(t) =
f
(t) dt.
Finally, if we let
P
(t) be the momentum of a body as a function of time,
then by substituting
P
(t) = m
v
(t), we arrive at the important relation
Momentum as force
accumulated over time
P
(t) =
f
(t) dt.
(12.20)
Since integration is a “summing up” process, Equation (12.20) confirms our
interpretation of momentum as the result of continued application of force
over time. (
Note:
in the preceding integrals we omitted the constant of
integration, essentially assuming the initial velocity was zero.)
Remember that integration and differentiation are inverse operations.
By taking the derivative of both sides with respect to t, we state the flip
side of the relationship between momentum and force.
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