P112 Spring 2007
Homework #3
Solutions
1. (not boxed)
Let the
x
+
direction be east (E), and the
y
+
direction be north
(N). Subscript definitions: p stands for plane, g stands for ground, and a stands
for air.
i
v
ˆ
)
km/h
220
(
a
p,
!
=
r
(i.e., 220 km/h west (W)). At
h
500
.
0
=
!
t
,
j
i
r
ˆ
)
km
20
(
ˆ
)
km
120
(
g
p,
!
+
!
=
"
r
. Therefore,
j
i
j
i
r
v
ˆ
)
km/h
40
(
ˆ
)
km/h
240
(
h
500
.
0
ˆ
)
km
20
(
ˆ
)
km
120
(
g
p,
g
p,
!
+
!
=
!
+
!
=
"
"
=
t
r
r
.
(a) This part is asking for the velocity of the air (wind) with respect to the ground,
g
a,
v
r
.
j
i
i
v
v
v
v
v
ˆ
)
km/h
40
(
ˆ
)
km/h
240
(
ˆ
)
km/h
220
(
g
p,
a
p,
g
p,
p
a,
g
a,
!
+
!
+
!
!
=
+
!
=
+
=
r
r
r
r
r
.
j
i
v
ˆ
)
km/h
40
(
ˆ
)
km/h
20
(
g
a,
!
+
!
=
r
.
km/h
45
)
km/h
40
(
)
km/h
20
(


2
2
g
a,
=
!
+
!
=
v
r
.
°
=
!
"
#
$
%
’
’
=
’
63
20
40
tan
1
g
a,
v
r
(
.
Since both
x
v
g,
a,
and
y
v
g,
a,
are negative,
g
a,
v
r
points 63
°
below the
x
!
direction or 63
°
south of west.
b) For this part of the problem,
j
v
ˆ
)
km/h
40
(
g
a,
!
=
r
(i.e., 40 km/h due south).
Again,
km/h
220
a
p,
=
v
. Here we want to know what direction of
a
p,
v
r
will
result in the plane traveling due west.
g
a,
g
p,
a
g,
g
p,
a
p,
v
v
v
v
v
r
r
r
r
r
!
=
+
=
. In order
for the plane to travel due west, require
0
g,
p,
=
y
v
. So, working with the
y
component of the velocity equation yields
km/h
40
)
km/h
40
(
0
g,
a,
g,
p,
,
a
p,
=
!
!
=
!
=
y
y
y
v
v
v
.
°
=
!
"
#
$
%
=
!
!
"
#
$
$
%
=
’
’
10
220
40
sin
sin
1
a
p,
a,
p,
1
a
p,
v
v
y
v
r
. Also notice that since
0
g,
a,
=
x
v
,
0
g,
p,
a,
p,
<
=
x
x
v
v
(negative because the plane should travel due west). Thus,
a
p,
v
r
points 10
°
above the
x
!
direction or 10
°
north of west.