()
21
12
2
2 160 m
4.0 s .
30 m/s
50 m/s
xx
tt
vv
−
−=
=
=
++
(c) Since the train is at rest (
v
0
= 0) when the clock starts, we find the value of
t
1
from Eq.
211:
10
1
1
2
30 m/s
6.0 s .
5.0 m/s
a
t
t
=+
¡
==
(d) The coordinate origin is taken to be the location at which the train was initially at rest
(so
x
0
= 0).
Thus, we are asked to find the value of
x
1
. Although any of several
equations could be used, we choose Eq. 217:
(
)
1
1
11
30 m/s 6.0 s
90 m .
22
xv
v
t
=
=
(e) The graphs are shown below, with SI units assumed.
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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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