2.11.I
DENTIFY
:
The average velocity is given by
av
x
x
v
t
∆
=
∆
. We can find the displacement
t
∆
for each constant velocity
time interval. The average speed is the distance traveled divided by the time.
S
ET
U
P
:
For
0
t
=
to
2.0 s
t
=
,
2.0 m/s
x
v
=
. For
2.0 s
t
=
to
3.0 s
t
=
,
3.0 m/s
x
v
=
. In part (b),
3.0 m/s
x
v
= −
for
2.0 s
t
=
to
3.0 s
t
=
. When the velocity is constant,
x
x
v
t
∆
=
∆
.
E
XECUTE
:
(a)
For
0
t
=
to
2.0 s
t
=
,
(2.0 m/s)(2.0 s)
4.0 m
x
∆
=
=
. For
2.0 s
t
=
to
3.0 s
t
=
,
(3.0 m/s)(1.0 s)
3.0 m
x
∆
=
=
. For the first 3.0 s,
4.0 m
3.0 m
7.0 m
x
∆
=
+
=
. The distance traveled is also 7.0
m. The average velocity is
av
7.0 m
2.33 m/s
3.0 s
x
x
v
t
∆
=
=
=
∆
. The average speed is also 2.33 m/s.
(b)
For
2.0 s
t
=
to 3.0 s,
(
3.0 m/s)(1.0 s)
3.0 m
x
∆
= −
= −
. For the first 3.0 s,
4.0 m
(
3.0 m)
1.0 m
x
∆
=
+ −
= +
.
The dog runs 4.0 m in the
x
+
direction and then 3.0 m in the
x
−
direction, so the distance traveled is still 7.0
m.
av
1.0 m
0.33 m/s
3.0 s
x
x
v
t
∆
=
=
=
∆
. The average speed is
7.00 m
2.33 m/s
3.00 s
=
.
E
VALUATE
:
When the motion is always in the same direction, the displacement and the distance traveled are
equal and the average velocity has the same magnitude as the average speed. When the motion changes direction
during the time interval, those quantities are different.
2.15.I
DENTIFY
and
S
ET
U
P
:
Use
x
dx
v
dt
=
and
x
x
dv
a
dt
=
to calculate
( )
x
v
t
and
( ).
x
a
t
E
XECUTE
:
2
2.00 cm/s
(0.125 cm/s
)
x
dx
v
t
dt
=
=
−
2
0.125 cm/s
x
x
dv
a
dt
=
= −
(a)
At
0,
t
=
50.0 cm,
x
=
2.00 cm/s,
x
v
=
2
0.125 cm/s
.
x
a
= −
(b)
Set
0
x
v
=
and solve for
t
:
16.0 s.
t
=
(c)
Set
50.0 cm
x
=
and solve for
t
. This gives
0
t
=
and
32.0 s.
t
=
The turtle returns to the starting point after
32.0 s.
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 Fall '07
 Evrard
 Velocity, m/s, ax

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