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9. Since this involves constantacceleration motion, we can apply the equations of Table
21, such as
xv
t a
t
=+
0
1
2
2
(where
x
0
0
=
). We choose to analyze the third and fifth
points, obtaining
2
0
2
0
1
0.2m
(1.0 s)
(1.0 s)
2
1
0.8m
(2.0 s)
(2.0 s)
2
va
Simultaneous solution of the equations leads to
0
0
v
=
and
a
=
040
.m
s
2
. We now have
two ways to finish the problem. One is to compute force from
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Acceleration

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