topic 7 vectors and applications ii two dimensional vectors and their algebraic and geometric...
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Topic 7
Vectors and Applications II
• two dimensional vectors and their algebraic and geometric
representation• unit vectors• resolution of vectors into components acting at right
angles to each other• operations on vectors including addition and multiplication
by a scalar• application of vectors in both life-related and purely
mathematical situations – velocity and relative velocity
Topic 7
Vectors and Applications II
Investigation
At 10:05 am, Wal Taylor took off in his helicopter from Dogwood Homestead.
He and his son were making a routine inspection of part of the property and hoping to locate a mob of cattle they had not seen for a while.
They were carrying enough fuel to fly for about three hours.
Soon after take-off, Wal called his wife on the radio.
At 10:22, Liz lost contact. This concerned her because they could normally call from anywhere on the 350 000 hectare property.
At 2 pm, Wal had not returned. Liz contacted the police to commence search and rescue operations.
To make things difficult, Wal had not told Liz where he was heading, but she knew that he normally flew at about 120 km/h
Liz got out a map of the property.
She marked in Wal’s possible locations, assuming that he flew in a straight line and crashed as soon as radio contact was lost.
Liz got out a map of the property.
She marked in Wal’s possible locations, assuming that he flew in a straight line and crashed as soon as radio contact was lost.
Mark these locations on your copy of the map.
Unknown to Liz, a jilleroo named Kate, working in a paddock near the house, had seen Wal fly over in a north-easterly direction just after 10 am. When she got back to her quarters, she heard that he had crashed but not at what time.
Kate got her map of the property and marked the possible locations of the crash site assuming that the helicopter flew in a straight line until going down.
Liz had one bit of information – the estimated distance of the crash from the house.
Kate had one bit of information – the direction of the crash from the house.
Later that day, Kate went over to the house to see if she could help. When they put the two maps together, they knew where to search.
Liz contacted Search and Rescue who then found the helicopter before dark. Both men were injured but alive.
On another day, Kate rode out to a point whose displacement from the homestead was
(59 km, 337oT)
Later in the week, Kate rode from the homestead to Marcoo No. 1 Bore. What was her displacement from the homestead ?
The magnitude of her displacement was 47 km and the direction was 157oT
Suppose a visitor landed at the homestead and asked where Garrad Bore was.
One way to tell him would be to give the displacement vector from the homestead to the bore i.e. (25 km, 007oT)
Alternatively, you could say “Go east for 3 km and then go north for 24.8 km” i.e. 3i + 24.8j
Model: Show the following displacements on the same set of axes: (a) 3m east (b) 2m north (c) 6m south-west
x
y
-6 -4 -2 0 2 4 6
-4
-2
0
2
4
(a)
(b)
(c)
Model: Show the following displacements on the same set of axes: (a) 3m, 30o
(b) 2m,150o
(c) 5m, 300o
x
y
-6 -4 -2 0 2 4 6
-4
-2
0
2
4
(a)(b)
(c)
Model: Convert (6, 30o) to component form.
X
Y
-2 -1 1 2 3 4 5 6
-2
-1
1
2
3
4
5
6
0
6
30
Model: Convert (6, 30o) to component form.
X
Y
-2 -1 1 2 3 4 5 6
-2
-1
1
2
3
4
5
6
030
6
a
b
a = 6 cos 30
= 33
b = 6 sin 30
= 3
(6, 30o) = (33 , 3)
X
Y
-8 -6 -4 -2
-8
-6
-4
-2
0
formpolartoConvertModel
5
3:
θ
tan θ = 5/3
θ = tan-1 (5/3)
= 59o
r = (32 + 52)
= 34
∴ (34, -121o)
r
∴ = 180 – 59 = 121
Find the displacement vector from A(-2,-1) to B(1,3)
x
y
-6 -4 -2 0 2 4 6
-4
-2
0
2
4
4
3AB
A
B
Find the displacement vector from B(1,3) to A(-2,-1)
x
y
-6 -4 -2 0 2 4 6
-4
-2
0
2
4
4
3BA
A
B
Exercise
NewQ P 281
Exercise 11.1
22
11
2
1
2
1
2
1
2
1
ba
ba
b
b
a
aba
then
b
bband
a
aaIf
a
ba+b
22
11
2
1
2
1
2
1
2
1
ba
ba
b
b
a
aba
then
b
bband
a
aaIf
a
ba+b
Vector addition and Scalar Mulitiplication
Find the resultant of displacements of 3m at 30o and 6m at 60o
X
Y
2 4 6 8
2
4
6
8
0
X
Y
2 4 6 8
2
4
6
8
0
X
Y
2 4 6 8
2
4
6
8
0
X
Y
2 4 6 8
2
4
6
8
0
X
Y
2 4 6 8
2
4
6
8
0
(3cos30, 3sin30)
(6cos60, 6sin60)
X
Y
2 4 6 8
2
4
6
8
0
(2.598, 1.5)
(3, 5.196)
X
Y
2 4 6 8
2
4
6
8
0
(2.598, 1.5)
(3, 5.196)
(5.598, 6.696)
X
Y
2 4 6 8
2
4
6
8
0
(2.598, 1.5)
(3, 5.196)
(5.598, 6.696)
r
r = (5.5982 + 6.6962)
= 8.728
tan θ = 6.696/5.598
θ = tan-1(6.696/5.598)
= 50.1o
θ
X
Y
2 4 6 8
2
4
6
8
0
3
6
60
30
120
X
Y
2 4 6 8
2
4
6
8
0
3
6
150
r
1.50
1.2030
1.20
728.8
150sin6sin
sin
6
150sin
sin
)3(728.8
)2(18.76
150cos63263
cos222
r
ruletheUsing
dpr
dp
r
ruletheUsing
θ
Model: If a = , show the vector (a) 2a(b) ½ a(c) -a
(a) 2a = 2
=
(b) ½ a = ½
=
(c) –a =
3
1
3
1
6
2
3
1
5.1
5.0
3
1
X
Y
-2 2 4 6 8
-2
2
4
6
8
0
Model: The velocity of a boat changes from 20 knots south-east to 10 knots at N 60o E. Find the change in velocity.
X
Y
-20 -10 10 20 30
-20
-10
10
20
30
0
Model: The velocity of a boat changes from 20 knots south-east to 10 knots at N 60o E. Find the change in velocity.
X
Y
-20 -10 10 20 30
-20
-10
10
20
30
0v1
v2
Change in velocity = v2-v1
-v1
v1 = =
v2 = =
)45sin(20
)45cos(20
210
210
30sin10
30cos10
5
35
v2-v1 = -
=
=
5
35
210
210
2105
21035
14.19
48.5
X
Y
-20 -10 10 20 30
-20
-10
10
20
30
0v1
v2
-v1
WN16atknots91.19
74
)(tan
tan
91.19
14.19)48.5(
14.19
48.5
48.514.191
48.514.19
22
v
v
v
Exercise
NewQ P 290
Exercise 11.2
Model: If a = , find the unit vector
4
3
54
53
51
22
4
3ˆ
5
43
a
a
X
Y
-2 2 4 6 8
-2
2
4
6
8
0
a
Model: Write (-1,5) as a linear combination of (3,-1) and (1,2)
2
12
1
31
5
1
1
2147)3()1(
)3....(15633)2(
)2....(52
)1....(13
2
3
2
3
2
1
1
3
5
1
a
bb
ba
ba
ba
ba
ba
b
b
a
a
ba
Model: (P295 Ex 11.3 No. 11)The weight of a block of ice on a slope of 22o is a force of 500N acting vertically downwards. Resolve the force into components parallel and perpendicular to the slope to find the force acting down the slope.
22
22
68
500
500 cos 22
500 sin 22
Force down the plane = 500 sin 22 N
Exercise
NewQ P 293
Exercise 11.3
2-3, 9-12
A plane is flying at constant velocity.
Four forces are acting upon it as shown in the diagram:
weight
lift
thrust
drag
The weight is produced by gravity and is 93 724 N
The thrust is produced by the jets and is 152 940 N acting left 4o up
Drag is produced by air resistance and is 147 721 N acting to the right
Lift is produced by the flow of air over the wings.
weight
lift
thrust
drag
Find the magnitude and direction of the lift vector
The weight is produced by gravity and is 93 724 N
The thrust is produced by the jets and is 152 940 N acting left 4o up
Drag is produced by air resistance and is 147 721 N acting to the right
Lift is produced by the flow of air over the wings.
93 724
F
152 940147 721
The weight is produced by gravity and is 93 724 N
The thrust is produced by the jets and is 152 940 N acting left 4o up
Drag is produced by air resistance and is 147 721 N acting to the right
Lift is produced by the flow of air over the wings.
93 724
F
152 940147 721
152 940 cos 4o
152 940 sin 4o
F cos
F sin
Since velocity is constant,
Fx = 0
147721 + Fsin - 152940cos 4 = 0
Fsin = 4846.45 ….(1)
Since plane is flying horizontally,
Fy = 0
Fcos +152940sin 4 – 93724 = 0
Fcos = 83055.44 …..(2)
(1) = Fsin = 4846.45
(2) Fcos 83055.44
= 3.34o
Subs in (1) to find F
etc