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13. Since
v
=
dx
dt
(Eq. 2-4), then ∆
x
=
R
vdt
, which corresponds to the area under the
v
vs
t
graph. Dividing
the total area
A
into rectangular (base
×
height) and triangular (
1
2
base
×
height) areas, we have
A
=
A
0
<t<

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