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HomeWork 2 solutions
Physics 203
1.
Cutnell & Johnson 8
th
Edition, Chapter 2, Problem 67
REASONING
The slope of a straightline segment in a positionversustime graph is the average velocity. The
algebraic sign of the average velocity, therefore, corresponds to the sign of the slope.
SOLUTION
a.
The slope, and hence the average velocity, is
positive
for segments
A
and
C
,
negative
for segment
B
, and
zero
for segment
D
.
b.
In the given positionversustime graph, we find the slopes of the four straightline segments to be
1.25 km 0 km
6.3 km/h
0.20 h 0 h
0.50 km 1.25 km
3.8 km/h
0.40 h 0.20 h
0.75 km 0.50 km
0.63 km/h
0.80 h 0.40 h
0.75 km 0.75 km
0 km/h
1.00 h 0.80 h
A
B
C
D
v
v
v
v
−
=
=
+
−
−
=
=
−
−
−
=
=
+
−
−
=
=
−
2.
Problem 39. Add “during this time” to the end of the last sentence.
REASONING
Because the car is traveling in the +
x
direction and decelerating, its acceleration is negative:
a
=
−2.70 m/s
2
. The final velocity for the interval is given (
v
= +4.50 m/s), as well as the elapsed time (
t
= 3.00
s). Both the car’s displacement
x
and its initial velocity
v
0
at the instant braking begins are unknown.
Compare the list of known kinematic quantities (
v
,
a
,
t
) to the equations of kinematics for constant
acceleration:
0
v
v
at
=
+
(Equation 2.4),
( )
1
0
2
x
v vt
=
(Equation 2.7),
2
1
0
2
x
v t
at
=
+
(Equation 2.8), and
22
0
2
v
v
ax
=
+
(Equation 2.9). None of these four equations contains all three known quantities and the
desired displacement
x
, and each of them contains the initial velocity
v
0
. Since the initial velocity is neither
known nor requested, we can combine two kinematic equations to eliminate it, leaving an equation in
which
x
is the only unknown quantity.
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This note was uploaded on 06/22/2009 for the course PHYS 1101 taught by Professor Richardson, b during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
 RICHARDSON, B
 Physics, Work

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