vector
DESCRIPTION
About VectorsTRANSCRIPT
Vectors and Scalars
A vector has magnitude as
well as direction.
Some vector quantities:
displacement, velocity, force,
momentum
A scalar has only a magnitude.
Some scalar quantities: mass,
time, temperature
Distance: A Scalar Quantity
A scalar quantity:
Contains magnitude only and consists of a number and a unit.
(20 m, 40 mi/h, 10 gal)
A
B
Distance is the length of the actual path taken by an object.
s = 20 m
DisplacementβA Vector Quantity
A vector quantity:
Contains magnitude AND direction, a number, unit & angle.
(12 m, 300; 8 km/h, N)
A
B D = 12 m, 20o
β’ Displacement is the straight-line separation of two points in a specified direction.
q
More about Vectors
β’ A vector is represented on paper by an
arrow
1. the length represents magnitude
2. the arrow faces the direction of
motion
3-2 Addition of VectorsβGraphical Methods
For vectors in one
dimension, simple
addition and subtraction
are all that is needed.
You do need to be careful
about the signs, as the
figure indicates.
Applications of Vectors VECTOR ADDITION β If 2 similar vectors point in the SAME
direction, add them.
β’ Example: A man walks 54.5 meters east, then another 30 meters east. Calculate his displacement relative to where he started?
54.5 m, E 30 m, E +
84.5 m, E
Notice that the
SIZE of the arrow
conveys
MAGNITUDE and
the way it was
drawn conveys
DIRECTION.
Applications of Vectors VECTOR SUBTRACTION - If 2 vectors are going in
opposite directions, you SUBTRACT.
β’ Example: A man walks 54.5 meters east, then walks
30 meters west. Calculate his displacement relative
to where he started?
54.5 m, E
30 m, W +
24.5 m, E
Easy Addingβ¦
All these planes have the same reading on
their speedometer. (plane speed not speed
with respect to the ground (actual speed)
What
factor is
affecting
their
velocity?
A B C
Addition of VectorsβGraphical Methods
If the motion is in two dimensions, the situation is
somewhat more complicated.
Here, the actual travel paths are at right angles to
one another; we can find the displacement by
using the Pythagorean Theorem.
Perpendicular Vectors When 2 vectors are perpendicular, you may use the
Pythagorean theorem.
95 km,E
55 km, N
A man walks 95 km, East
then 55 km, north.
Calculate his
RESULTANT
DISPLACEMENT.
kmc
c
bacbac
8.10912050
5595Resultant 22
22222
Example A bear, searching for food wanders 35 meters east then 20 meters north.
Frustrated, he wanders another 12 meters west then 6 meters south. Calculate
the bear's displacement.
3.31)6087.0(
6087.23
14
93.262314
1
22
Tan
Tan
mR
q
q
35 m, E
20 m, N
12 m, W
6 m, S
- = 23 m, E
- = 14 m, N
23 m, E
14 m, N
The Final Answer:
R q
3.31,93.26
3.3193.26
m
m
Addition of VectorsβGraphical Methods
Even if the vectors are not at right
angles, they can be added graphically by
using the tail-to-tip method.
To add vectors, we put the initial point of the second
vector on the terminal point of the first vector. The
resultant vector has an initial point at the initial point
of the first vector and a terminal point at the terminal
point of the second vector (see below--better shown
than put in words).
v w
Initial point of v v
w
Move w over keeping
the magnitude and
direction the same.
Terminal
point of w
w
The negative of a vector is just a vector going the opposite
way.
v
v
A number multiplied in front of a vector is called a scalar. It
means to take the vector and add together that many times.
vv
vv3
u
vw
vu u
vw3
ww
w
Using the vectors shown,
find the following:
vu u
vvwu 32
uu w
ww
v
3-3 Subtraction of Vectors, and
Multiplication of a Vector by a Scalar
In order to subtract vectors, we
define the negative of a vector, which
has the same magnitude but points
in the opposite direction.
Then we add the negative vector.
3-4 Adding Vectors by Components
Any vector can be expressed as the sum
of two other vectors, which are called its
components. Usually the other vectors are
chosen so that they are perpendicular to
each other.
3-4 Adding Vectors by Components
If the components are
perpendicular, they can be
found using trigonometric
functions.
Remember:
soh
cah
toa
Adding Vectors by Components
The components are effectively one-dimensional,
so they can be added arithmetically.
3-4 Adding Vectors by Components
Adding vectors:
1. Draw a diagram; add the vectors graphically.
2. Choose x and y axes.
3. Resolve each vector into x and y components.
4. Calculate each component using sines and cosines.
5. Add the components in each direction.
6. To find the length and direction of the vector, use:
and .
3-4 Adding Vectors by
Components
Example 3-2: Mail carrierβs
displacement.
A rural mail carrier leaves the
post office and drives 22.0 km
in a northerly direction. She
then drives in a direction 60.0Β°
south of east for 47.0 km. What
is her displacement from the
post office?
3-4 Adding Vectors by
Components
Example 3-3: Three short trips.
An airplane trip involves three legs,
with two stopovers. The first leg is
due east for 620 km; the second leg
is 45Β° south of east (315) for 440
km; and the third leg is at 53Β°
south of west (233Β°), for 550 km, as
shown. What is the planeβs total
displacement?
Analytical Method
β’ Polar Form of Vectors
β’ π΄ = π΄ , ππ΄
β π΄ = ππππππ‘π’ππ ππ π£πππ‘ππ π΄
β ππ΄ = ππππππ‘πππ ππ π£πππ‘ππ π΄
β’ Example
β π΄ = 10 ππ, 450
β π΄ = 10 ππ
β ππ΄ = 450
Caution
β’ Addition of vectors in polar form cannot be done algebraically
Ex. A = 5 km, 45 deg
B = 4 km, 135 deg
C = 3 km, 270 deg
R = 12 km, 450 deg
Vectors can only be added algebraically if they are parallel or antiparallel
Component Form
β’ π΄ = π΄ , ππ΄
β’ π΄ = π΄ cos ππ΄ π₯ + π΄ sin ππ΄ π¦
β π¦ and π₯ πππ ππππππ π’πππ‘ π£πππ‘πππ
β π΄ = 10 ππ, 450
β π΄ = 10 πππππ (45)π₯ + 10ππ sin 45 π¦
β π΄ = 7.07 ππ π₯ + 7.07 ππ π¦
π΄ = 10 ππ, 450
π΄ = 10 πππππ (45)π₯ + 10ππ sin 45 π¦
π΄ = 7.07 ππ π₯ + 7.07 ππ π¦
7.07 ππ
7.07 ππ 10 ππ
β π΅ = 20 ππ, 1200
β π΅ = 20 πππππ (120)π₯ + 20ππ sin 120 π¦
β π΅ = β10 ππ π₯ + 17.32 ππ π¦
17.32 ππ
20 ππ
β10 ππ
β’ π΄ = 7.07 ππ π₯ + 7.07 ππ π¦
β’ π΅ = β10 ππ π₯ + 17.32 ππ π¦
β’ R = -2.93 km x + 24.39 km y
17.32 ππ
β10 ππ
7.07 ππ
7.07 ππ
Graphical Representation of the
Analytical Method
β’ π΄ = 12 km, 30 deg
β’ π΅ = 6 km, 60 deg
β’ π΄ + π΅ = πΆ
π΄
π΅
π΄
π΅
π΄π₯
π΄π¦
π΅π₯
π΅π¦
π
ππ₯ = ππ₯ +ππ₯
ππ¦ = ππ¦ +ππ¦
Component Method
x y
D1 620km, 0 deg 620km (cos 0) 620km (sin 0)
D2 440km, 315 deg 440km (cos 315) 440km (sin 315)
D3 550 km, 233 deg 550km (cos 233) 550km (sin 233)
x y
D1 620km, 0 deg 620km 0km
D2 440km, 315 deg 311km -311 km
D3 550 km, 233 deg -331 km -439 km
R 600 km -750 km
β’ π·1 +π·2+ π·3 = π
Adding vectors:
1. Draw a diagram; add the vectors graphically.
2. Choose x and y axes.
3. Resolve each vector into x and y components.
4. Calculate each component using sines and cosines.
5. Add the components in each direction.
6. To find the magnitude of the vector, use:
π = π₯2 + π¦2
Direction:
1. Resolve what quadrant is the vector pointing at?
2. Get the Reference Angle π½πππ = πππβππ
π
3. if Quadrant 1 π½ = π½πππ
Quadrant 2 π½ = πππ β π½πππ
Quadrant 3 π½ = πππ + π½πππ
Quadrant 4 π½ = πππ β π½πππ
+x
+y
βx
+y
β x
β y +x
β y
Q1 Q2
Q3 Q4
From Component to Polar
β’ Magnitude of R = 6002 + β750 2 =960 ππ
β’ Angle of R = π‘ππβ1β750
600=51 deg at Q4
β’ 360-51 = 309 degrees
π = 600 ππ π₯ β750πππ¦
β750 ππ
600 ππ
Scaling Vectors
x y
3*D1 = 1860 km, 0 deg
β2*D2 = β 880 km, 315 deg
4*D3 = 2200 km, 233 deg
R
3π·1 β 2π·2 + 4π·3 = π»
D1 620km, 0 deg 3*D1 = 1860 km, 0 deg
D2 440km, 315 deg β2*D2 = β 880 km, 315 deg
D3 550 km, 233 deg +4*D3 = 2200 km, 233 deg
Problem Set
β’ π΄ = 4 ππ, 30
β’ π΅ = 13 ππ, 120
β’ πΆ = 10 ππ, 240
β’ π· = 7 ππ, 310 β’ Find the magnitude and direction of the following
1. π΄ + π΅ + πΆ + π· = πΈ
2. π΄ β π΅ + πΆ + π· = πΉ
3. π΄ + π΅ β πΆ + π· = πΊ
4. π΄ + π΅ + πΆ β π· = π»
5. π΄ + π΅ β πΆ β π· = π½
6. π΄ β π΅ β πΆ + π· = πΎ