3.1.
I
DENTIFY
and
S
ET
U
P
:
Use Eq.(3.2), in component form.
E
XECUTE
:
(
)
2
1
av
2
1
5.3 m
1.1 m
1.4 m/s
3.0 s
0
x
x
x
x
v
t
t
t
∆
−
−
=
=
=
=
∆
−
−
(
)
2
1
av
2
1
0.5 m
3.4 m
1.3 m/s
3.0 s
0
y
y
y
y
v
t
t
t
∆
−
−
−
=
=
=
= −
∆
−
−
E
VALUATE
:
Our calculation gives that
av
v
r
is in the 4th quadrant. This corresponds to increasing
x
and
decreasing
y
.
3.6.
I
DENTIFY
:
Use Eq.(3.8), written in component form.
S
ET
U
P
:
2
2
2
2
(0.45m s )cos31.0
0.39m s ,
(0.45m s )sin31.0
0.23m s
x
y
a
a
=
° =
=
° =
E
XECUTE
:
(a)
av
x
x
v
a
t
∆
=
∆
and
2
2.6 m s
(0.39 m s )(10.0 s)
6.5 m s
x
v
=
+
=
.
av
y
y
v
a
t
∆
=
∆
and
2
1.8 m s
(0.23 m s )(10.0 s)
0.52 m s
y
v
= −
+
=
.
(b)
2
2
(6.5m s)
(0.52m s)
6.48m s
v
=
+
=
, at an angle of
0.52
arctan
4.6
6.5
⎛
⎞
=
°
⎜
⎟
⎝
⎠
above the horizontal.
(c)
The velocity vectors
1
v
r
and
2
v
r
are sketched in Figure 3.6. The two velocity vectors differ in magnitude and
direction.
E
VALUATE
:
1
v
r
is at an angle of
35
°
below the
axis
x
+
and has magnitude
1
3.2 m/s
v
=
, so
2
1
v
v
>
and the
direction of
2
v
r
is rotated counterclockwise from the direction of
1
v
r
.
Figure 3.6
(
)
(
)
av
av
1.3 m/s
tan
0.9286
1.4 m/s
y
x
v
v
α
−
=
=
= −
360
42.9
317
α
=
° −
° =
°
(
)
(
)
2
2
av
av
av
x
y
v
v
v
=
+
2
2
av
(1.4 m/s)
(
1.3 m/s)
1.9 m/s
v
=
+ −
=
Figure 3.1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
3.9.
I
DENTIFY
:
The book moves in projectile motion once it leaves the table top. Its initial velocity is horizontal.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 Evrard
 Acceleration, Velocity, m/s, 0.5 m, 0.23 m, 0.23M

Click to edit the document details