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4.59.
Model:
Let the ground frame be S and the car frame be S.
′
S
′
moves relative to S with a velocity
V
along the
x
direction.
Solve:
The Galilean transformation of velocity is
vvV
′
=
+
G
G
G
where
and
vv
′
G
G
are the velocities of the raindrops
in frames S and S
′
. While driving north,
( )
ˆ
25 m/s
Vi
=
G
and
RR
ˆˆ
cos
sin
.
j
v
i
θ
=−
−
Thus,
′
G
GG
( )
sin
25 m/s
cos
vi
v
j
−
−
Since the observer in the car finds the raindrops making an angle of 38
°
with the vertical, we have
R
R
sin
25 m/s
tan38
cos
v
v
+
=
°
While driving south,
()
ˆ
25 m/s
,
G
and
cos
sin
.
j
v
i
−
G
Thus,
( )
sin
cos
i
v
j
′=−
+
−
G
Since the observer in the car finds the raindrops falling vertically straight, we have
R
R
sin
tan0
0
cos
v
v
−+
=
°=
R
sin
25 m/s
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This note was uploaded on 12/10/2011 for the course PHYS 1211 taught by Professor Geller during the Fall '09 term at University of Georgia Athens.
 Fall '09
 geller

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