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The total displacement is
123
ˆˆˆ
ˆ
(40.0 km)i (15.3 km)i (12.9 km) j
(50.0 km)i
ˆˆ
(5.30 km) i (12.9 km) j.
rrrr
Δ=Δ+Δ+Δ=
+
+
−
=+
GGGG
The time for the trip is (40.0 + 20.0 + 50.0) min = 110 min, which is equivalent to 1.83 h.
Eq. 48 then yields
avg
5.30 km
12.9 km
ˆ
ˆ
i
j = (2.90 km/h)i + (7.01 km/h) j.
1.83 h
1.83 h
v
§·
¨¸
©¹
G
The magnitude is
22
avg


(2.90 km/h)
(7.01 km/h)
7.59 km/h.
v
=
G
(b) The angle is given by
1
7.01 km/h
tan
67.5
(north of east),
2.90 km/h
θ
−
==
°
or 22.5
°
east of due north.
7. The average velocity is given by Eq. 48. The total displacement
Δ
G
r
is the sum of
three displacements, each result of a (constant) velocity during a given time. We use a
coordinate system with +
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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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