35. (a) From the figure, we see that
x
0
= –2.0 m. From Table 21, we can apply
x
–
x
0
=
v
0
t
+
1
2
at
2
with
t
= 1.0 s, and then again with
t
= 2.0 s. This yields two equations for the
two unknowns,
v
0
and
a
:
()
(
)
(
)
2
0
2
0
1
0.0
2.0 m
1.0 s
1.0 s
2
1
6.0 m
2.0 m
2.0 s
2.0 s .
2
va
−−
=
+
=
+
Solving these simultaneous equations yields the results
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Physics

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