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18.
(a) In frame
S
, our coordinates are such that
x
1
= +1200 m for the big ﬂash, and
x
2
= 1200
−
720 =
480 m for the small ﬂash (which occurred later). Thus, ∆
x
=
x
2
−
x
1
=
−
720 m. If we set ∆
x
0
=0
in Eq. 3824, we Fnd
0=
γ
(∆
x
−
v
∆
t
)=
γ
(
−
720 m
−
v
(5
.
00
×
10
−
6
s)
)
which yields
v
=
−
1
.
44
×
10
8
m/s. Therefore, frame
S
0
must be moving in the
−
x
direction with a
speed of 0
.
480
c
.
(b) Eq. 3827 leads to
∆
t
0
=
γ
µ
∆
t
−
v
∆
x
c
2
¶
=
γ
µ
5
.
00
×
10
−
6
s
−
(
−
1
.
44
×
10
8
m
/
s)(
−
720 m)
(2
.
998
×
10
8
m
/
s)
2
¶
which turns out to be positive (regardless of the speciFc value of
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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