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(b) Using the result from part (a), we find
v
bg
=
v
bw
cos
θ
= 5.5 km/h. Thus, traveling a
distance of
A
= 6.4 km requires a time of (6.4 km)/(5.5 km/h) = 1.15 h or 69 min.
(c) If her motion is completely along the
y
axis (as the problem implies) then with
v
wg
=
3.2 km/h (the water speed) we have
total
=
+
= 1.33 h
+
bw
wg
bw
wg
DD
t
vv
v
v
−
where
D
= 3.2 km. This is equivalent to 80 min.
(d) Since
+
bw
wg
bw
wg
bw
wg
bw
wg
D
D
v
v
v
v
+=
+
−−
+
the answer is the same as in the previous part, i.e.,
total
= 80 min
t
.
(e) The shortesttime path should have
0.
=°
This can also be shown by noting that the
case of general
leads to
ˆˆ
cos i
(
sin
+
) j
bg
bw
wg
bw
bw
wg
vvvv
v
v
θθ
=+=
+
GGG
where the
x
component of
G
v
bg
must equal
l
/
t
. Thus,
=
cos
bw
l
t
v
which can be minimized using
dt
/
d
= 0.
(f) The above expression leads to
t
= (6.4 km)/(6.4 km/h) = 1.0 h, or 60 min.
89. We establish coordinates with
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 Spring '08
 Any
 Physics, Current

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