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Vectors
Pearland ISD Physics
Mrs. Akibola
Scalars and Vectors
• A scalar quantity is one that can be
described by a single number:
– Examples:
• temperature, speed, mass
• A vector quantity deals inherently
with both magnitude and direction:
– Examples:
• Acceleration, velocity, force, displacement
Identify the following as
scalars or vectors – not in your notes 
• The acceleration of an airplane as it
takes off vector
• The number of passengers and
crew on the airplane scalar
• The duration of the flight scalar
• The displacement of the flight vector
• The amount of fuel required for the
flight scalar
Representing vectors
• Vectors can be represented graphically.
– The direction of the arrow is the direction of
the vector.
– The length of the arrow tells the magnitude
4N
8N
• Vectors can be moved parallel to
themselves and still be the same vector.
– Vectors only tell magnitude (amount) and
direction, so a vector can starts anywhere
Adding vectors
• The sum of two vectors is called the
resultant.
• To add vectors graphically, draw each
vector to scale.
• Place the tail of the second vector at the
tip of the first vector. (tip-to-tail method)
• Vectors can be added in any order.
• To subtract a vector, add its opposite.
Vector Addition and Subtraction
• Often it is necessary to add one vector
to another.
Vector Addition and Subtraction
3m
5m
8m
Adding Vector Problem
A parachutist jumps from a plane. He has
not pulled is parachute yet. His weight or
force is 800 N downward. The wind is
applying a small drag force of 50 N
upward. What is the vector sum of the
forces acting on him?
750 N downward
Adding perpendicular vectors
• Perpendicular vectors can be easily
added and we use the Pythagorean
theorem to find the magnitude of the
resultant.
• Use the tangent function to find the
direction of the resultant.
Vector Addition and Subtraction
1.6 Vector Addition and Subtraction
2.00 m
6.00 m
Vector Addition and Subtraction
R  2.00 m   6.00 m 
2
2
R
2.00 m
2
2
 6.00 m  6.32m
2
 2.0 

  tan 
  18.4
 6.0 
R
1
2.00 m
6.00 m
Adding perpendicular
Vectors – this slide is not in your notes
Note: in this example, In order to use the tip-to-tail
Method, Vector B must be moved.
Vector Addition and Subtraction
This slide is not in your notes
YOU MAY WANT TO
MAKE A NOTE OF
THIS!!
When a vector is multiplied
by -1, the magnitude of the
vector remains the same, but
the direction of the vector is
reversed.
Multiplying Vectors
• A vector multiplied or divided by a
scalar results in a new vector.
– Multiplying by a positive number changes
the magnitude of the vector but not the
direction.
– Multiplying by a negative number changes
the magnitude and reverses the direction.
Resolving vectors into
components.
• Any vector can be resolved, that is,
broken up, into two vectors, one that
lies on the x-axis and one on the y-axis.
1.7 The Components of a Vector


x and y are called the x vector component

and the y vector component of r.
1.7 The Components of a Vector

The vector components of A are two perpendicular


vectors A x and A y that are parallel to the x and y axes,
 

and add together vectorially so that A  A x  A y .
Resolving Vectors - Practice
Does 0.5 + 0.7 = 1 ?
• Consider two forces acting on an object,
one force with 0.7 N of force at 600 above
the horizontal and to the right, and the
other pulling with 0.5 N of force at 450
above the horizontal to the left.
• How much total force is produced by the
two forces?
Adding non-perpendicular
vectors
• Resolve each vector into x and y
components, using sin and cos.
• Add the x components together to get the
total x component. Add the y component
together to get the total y component.
• Find the magnitude of the resultant using
Pythagorean theorem.
• Find the direction of the resultant using the
inverse tan function.
Adding non-perpendicular
vectors
• Resolving Vector 1:
F1Y
60o
F1X
Adding non-perpendicular
vectors
• Resolving Vector 1:
F1 X
cos(60 ) 
F1

F1 cos(60 )  F1 X  0.7  cos(60 )  0.35  0.4
F1Y
F1Y
sin(60 ) 
F1

60o
F1X
F1 sin(60 )  F1Y  .7  sin(60 )  .6


Adding non-perpendicular
vectors
• Resolving Vector 1:
F1
F2
FR
X
Y
.4
.6
Adding non-perpendicular
vectors
• Resolving Vector 2:
F2Y
45o
F2X
Adding non-perpendicular
vectors
• Resolving Vector 2
F2 X
cos(45 ) 
F2

F2 cos(45 )  F2 X  0.5  cos(45 )  .35  .4
F2X  .4
F2Y
45o
F2X
F2Y
sin( 45 ) 
F2

F2 sin(45 )  F2Y  0.5  sin(45 )  0.4
Adding non-perpendicular
vectors
• Resolving Vector 2:
X
Y
F1
.4
.6
F2
-.4
.4
FR
Adding non-perpendicular
vectors
• Add the x
components to get
the resultant x
component.
X
Y
F1
.4
.6
F2
-.4
.4
FR
0
Adding non-perpendicular
vectors
• Add the y
components to get
the resultant y
component.
X
Y
F1
.4
.6
F2
-.4
.4
FR
0
1.0
Adding non-perpendicular
vectors
• Use the
Pythagorean
Theorem to find the
magnitude of the
resultant.
• Use the inverse tan
function to find the
direction of the
resultant.
X
Y
F1
.4
.6
F2
-.4
.4
FR
0
1.0
Adding non-perpendicular
vectors
• Use the
2
2
2
Pythagorean
FR  FRX  FRY  0  1  1
Theorem to find the
FR  1 Newton
magnitude of the
resultant.
• Use the inverse tan
function to find the
direction of the
-1 opposite
1 0
0


tan

tan

0
resultant.
adjacent
1
What does this tell us about the direction
of motion of the object?
Calculate the following:
Practice!!
• A roller coaster moves 85 meters horizontally,
then travels 45 meters at an angle of 30.0°
above the horizontal. What is its
displacement from its starting point?
• A pilot sets a plane’s controls, thinking the
plane will fly at 2.50 X 102 km/hr to the north.
If the wind blows at 75 km/hr toward the
southeast, what is the plane’s resultant
velocity?
Addition of Vectors by Means of
Components
  
C AB

A  Ax xˆ  Ay yˆ

B  Bx xˆ  B y yˆ
I will not accept late work
CLASS ASSIGNMENT /
HOMEWORK
SECTION: 4
RELATIVE VELOCITY
Relative Motion
• The motion of an object
can be expressed from
different points of view
• These points of view
are known as “frames
of reference”
• Depending on the
frame of reference
used, the description of
the motion of the object
may change
Relative Velocity
(motion of
objects is independent of each other)
• Velocity of A relative to B: using the ground as a reference frame
• The Language
 vAG : v of A with respect to G (ground) = 40 m/s
 vBG : v of B with respect to G = 30 m/s
 vAB : v of A with respect to B:



v AB  v AG  v BG


v AB  40m / s  30m / s

v AB  10m / s
Example 1
• The white speed boat has
a velocity of 30km/h,N,
and the yellow boat a
velocity of 25km/h, N,
both with respect to the
ground. What is the
relative velocity of the
white boat with respect to
the yellow boat?



vWY  vWG  vYG

vWY  30km / h  25km / h

vWY  5km / h
Example 2- The Bus Ride
Lets do this together
A passenger is seated on a bus that is
traveling with a velocity of 5 m/s, North.
If the passenger remains in her seat,
what is her velocity:
a) with respect to the ground? 5m / sNorth
b) with respect to the bus? 0m / sNorth
Example 2 –continued
Lets do this together
The passenger decides to approach the driver
with a velocity of 1 m/s, N, with respect to the
bus, while the bus is moving at 5m/s, N.
What is the velocity of the passenger with
respect to the ground?
6m / sNorth
Resultant Velocity (motion of
objects is dependent on each other)
The resultant velocity is the net velocity
of an object with respect to a reference
frame.
Example 3- Airplane and Wind relative to the ground
An airplane has a velocity of 40 m/s, N, in still air.
It is facing a headwind of 5m/s with respect to the
ground.
What is the resultant velocity of the airplane?



v PG  v PG  vWG

v PG  40m / s  5m / s

v PG  35m / s
Frame of reference
• What is this guy’s velocity?
2m / s
• What about now?
2m / s
or
11m / s
Frame of reference
• What about now?
 2m / s or 7 m / s
Relative Velocity:
Example 4: Crossing a River
The engine of a boat drives it across a river that is 1800m wide.
The velocity of the boat relative to the water is 4.0m/s directed
perpendicular to the current. The velocity of the water relative
to the shore is 2.0m/s.
A) What is the velocity of the boat
relative to the shore?
B) How long does it take for the boat
to cross the river?
Relative Velocity

v BW  4.0m / s 

v WS  2.0m / s 
• G:

• U: v BS  ?



• E: v BS  v BW  v WS
• S:
• S:
2
2
vBS  vBW
 vWS

(Pythagorean Theorem)
4.0 m s2  2.0 m s2
 4.5 m s
 4.0 

  tan 
  63
 2.0 
1
Relative Velocity
• G:
• U:
t ?
• E:
v
x
t
(Kinematics)
x
• S: t 
v
1800 m
 450 s
• S: t 
4.0 m s
I will not accept late work
CLASS ASSIGNMENT /
HOMEWORK