This preview shows page 1. Sign up to view the full content.
avg
ˆˆ
(483 km)i (966 km)j
(215 km/h)i (429 km/h)j.
2.25 h
v
−
==
−
G
with a magnitude
22


(215 km/h)
( 429 km/h)
480 km/h.
v
=+
−
=
G
(d) The direction of
v
G
is 26.6° east of south, same as in part (b). In magnitudeangle
notation, we would have
(480 km/h
63.4 ).
v
=∠
−
°
G
(e) Assuming the
AB
trip was a straight one, and similarly for the
BC
trip, then

G
r
AB
is the
distance traveled during the
AB
trip, and
G
r
BC
is the distance traveled during the
BC
trip.
Since the average speed is the total distance divided by the total time, it equals
483 km
966 km
644 km/h.
2.25 h
+
=
8. Our coordinate system has
#
i
pointed east and
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Any
 Physics

Click to edit the document details