This preview shows page 1. Sign up to view the full content.
1
(a) The average velocity during any time interval is the displacement during that interval divided
by the duration of the interval:
v
avg
=
∆
x/
∆
t
, where
∆
x
is the displacement and
∆
t
is the time
interval. In this case the interval is divided into two parts. During the first part the displacement
is
∆
x
1
= 40 km and the time interval is
∆
t
1
=
(40 km)
(30 km
/
h)
=1
.
33 h
.
During the second part the displacement is
∆
x
2
= 40 km and the time interval is
∆
t
2
=
(40 km)
(60 km
/
h)
=0
.
67 h
.
Both displacements are in the same direction, so the total displacement is
∆
x
=
∆
x
1
+
∆
x
2
=
40 km + 40 km = 80 km. The total time interval is
∆
t
=
∆
t
1
+
∆
t
2
.
33 h + 0
.
67 h = 2
.
00 h.
The average velocity is
v
avg
=
(80 km)
(2
.
0h)
=40km
/
h
.
(b) The average speed is the total distance traveled divided by the time. In this case the total
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 03/20/2008 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas at Austin.
 Spring '07
 KOPP

Click to edit the document details