(d) The direction of
avg
v
G
is
1
tan [( 0.22 km/h)/(0.67 km/h)]
18
θ
−
=−
=
−
°
, or 18
°
south
of east.
(e) The average speed is distinguished from the magnitude of average velocity in that it
depends on the total distance as opposed to the net displacement. Since the camel travels
140 km, we obtain (140 km)/(90 h) = 1.56 km/h
1.6 km/h
≈
.
(f) The net displacement is required to be the 90 km East from
A
to
B
. The displacement
from the resting place to
B
is denoted
3
.
r
Δ
G
Thus, we must have
123
ˆ
+
+
= (90 km)i
rrr
ΔΔΔ
GGG
which produces
3
ˆˆ
(30 km)i
(20 km)j
r
Δ=
+
G
in unitvector notation, or (36
33 )
∠°
in
magnitudeangle notation.
Therefore, using Eq. 48 we obtain
avg
36km

 =
= 1.2km/h.
(120 90) h
v
−
G
(g) The direction of
v
G
is the same as
G
r
3
(that is, 33° north of east).
121. On the one hand, we could perform the vector addition of the displacements with a
vectorcapable calculator in polar mode ((75
37 ) + (65
90 ) = (63
18 )),
∠
−
°
∠
−°
but in keeping with Eq. 35 and Eq. 36 we will show the details in unitvector notation.
This is the end of the preview. Sign up
to
access the rest of the document.
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

Click to edit the document details