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81. The problem consists of two constantacceleration parts:
part 1 with
v
0
= 0,
v
= 6.0
m/s,
x
= 1.8 m, and
x
0
= 0 (if we take its original position to be the coordinate origin); and,
part 2 with
v
0
= 6.0 m/s,
v
= 0, and
a
2
= –2.5 m/s
2
(negative because we are taking the
positive direction to be the direction of motion).
(a) We can use Eq. 217 to find the time for the first part
x
–
x
0
=
1
2
(
v
0
+
v
)
t
1
¡
1.8 m – 0 =
1
2
(0 + 6.0 m/s)
t
1
so that
t
1
= 0.6 s. And Eq. 211 is used to obtain the time for the second part
02
2
vv a
t
=+
¡
0 = 6.0 m/s + (–2.5 m/s
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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, Acceleration

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