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Since the desired angle is in the second quadrant, we pick 135
°
(45
°
north of due west).
Note that the displacement can be written as
()
o
56.6 135
rrr
Δ=− =
∠
°
GGG
in terms of the
magnitudeangle notation.
(c) The magnitude of
G
v
avg
is simply the magnitude of the displacement divided by the
time (
Δ
t
= 30.0 s). Thus, the average velocity has magnitude (56.6 m)/(30.0 s) = 1.89 m/s.
(d) Eq. 48 shows that
G
v
avg
points in the same direction as
Δ
G
r
, i.e, 135
°
( 45
°
north of
due west).
(e) Using Eq. 415, we have
22
o
avg
ˆˆ
(0.333 m/s )i
(0.333 m/s )j.
vv
a
t
−
==
+
Δ
GG
G
The magnitude of the average acceleration vector is therefore equal to
2
avg


(0.333 m/s )
(0.333 m/s )
0.471m/s
a
=+
=
G
.
(f) The direction of
avg
a
G
is
2
1
2
0.333 m/s
tan
45 or
135 .
θ
−
§·
°
−
°
¨¸
©¹
Since the desired angle is now in the first quadrant, we choose 45
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This note was uploaded on 05/17/2011 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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