∆
x = x
2
–
x
1
= –720 m.
If we set
∆
x'
= 0 in Eq. 3725, we find
0
720
500 10
6
=−
=
−−
×
−
γγ
()
(
.
∆∆
xv
t
v
ms
)
c
h
which yields
v
= –1.44
×
10
8
m/s, or
/
0.480
vc
β
=
=
.
(b) The negative sign in part (a) implies that frame
S'
must be moving in the –
x
direction.
(c) Eq. 3728 leads to
∆
tt
vx
c
'.
(.
F
H
G
I
K
J
=×
−
−×
−
×
F
H
G
I
K
J
−
2
6
8
8
144 10
2 998 10
s
m / s)( 720m)
m/s)
2
which turns out to be positive (regardless of the specific value of
γ
). Thus, the order of
the flashes is the same in the
S'
frame as it is in the
S
frame (where
∆
t
is also positive).
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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