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t
c
b
==
=
2
3
230
320
10
(.
)
)
.
m/s
s.
2
3
For
t
= 0,
x
=
x
0
= 0 and for
t
= 1.0 s,
x
= 1.0 m >
x
0
. Since we seek the maximum, we
reject the first root (
t
= 0) and accept the second (
t
= 1s).
(d) In the first 4 s the particle moves from the origin to
x
= 1.0 m, turns around, and goes
back to
x
((
.
)
(
.
(
.
)
(
.
43
0
4
0
2
0
4
0
8
0
2
s)
m.
23
3
=−=
−
The total path length it travels is 1.0 m + 1.0 m + 80 m = 82 m.
(e) Its displacement is
Δ
x
=
x
2
–
x
1
, where
x
1
= 0 and
x
2
= –80 m. Thus,
80 m
x
Δ=−
.
The velocity is given by
v
= 2
ct
– 3
bt
2
= (6.0 m/s
2
)
t
– (6.0 m/s
3
)
t
2
.
(f) Plugging in
t
= 1 s, we obtain
2
(1 s)
(6.0 m/s )(1.0 s)
(6.0 m/s )(1.0 s)
0.
v
=−
=
(g) Similarly,
2
(2 s)
(6.0 m/s )(2.0 s) (6.0 m/s )(2.0 s)
12m/s .
v
=
−
(h)
2
(3 s)
(6.0 m/s )(3.0 s) (6.0 m/s )(3.0 s)
36 m/s .
v
=
−
(i)
2
(4 s)
(6.0 m/s )(4.0 s) (6.0 m/s )(4.0 s)
72 m/s .
v
=
−
The acceleration is given by
a
=
dv
/
dt
= 2
c
– 6
b
= 6.0 m/s
2
– (12.0 m/s
3
)
t
.
(j) Plugging in
t
= 1 s, we obtain
2
6.0 m/s
(12.0 m/s )(1.0 s)
6.0 m/s .
a
=
−
(k)
2
(2 s)
6.0 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

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