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1. We use Eq. 22 and Eq. 23. During a time
t
c
when the velocity remains a positive
constant, speed is equivalent to velocity, and distance is equivalent to displacement, with
Δ
x
=
v t
c
.
(a) During the first part of the motion, the displacement is
Δ
x
1
= 40 km and the time
interval is
t
1
40
133
==
(
.
km)
(30 km / h)
h.
During the second part the displacement is
Δ
x
2
= 40 km and the time interval is
t
2
40
067
(
.
(60 km / h)
h.
Both displacements are in the same direction, so the total displacement is
Δ
x
=
Δ
x
1
+
Δ
x
2
= 40 km + 40 km = 80 km.
The total time for the trip is
t
=
t
1
+
t
2
= 2.00 h. Consequently, the average velocity is
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.
 Spring '06
 Stoler
 Physics

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