Vectors & ScalarsVectors are measurements which have both

magnitude (size) and a directional component.

Scalars are measurements which have only magnitude (size) and no directional component

EXAMPLES OF VECTOR VALUES:Displacement

VelocityAcceleration

ForceDirection counts in all of these measurements.

EXAMPLES OF SCALAR VALUES:Distance

SpeedTemperature

Comparing Vector & Scalar ValuesDisplacement (a vector) versus distance (a scalar)

LAKE TRANQUILITY

A

B

We want to get from point A to point B. If we follow the road around the lake our direction is always changing. There is no specific

direction. The distance traveled on the road is a scalar quantity.

A straight line between A and B is the displacement. It has a specificdirection and is therefore a vector.

Speed & VelocitySpeed and velocity are not the same.

Velocity requires a directional component and is therefore a vector quantity.

Speed tells us how fast we are going but not which way.Speed is a scalar (direction doesn’t count!)

9080

706050

403020

10

SPEEDOMETER

N

S

EW

COMPASS

Vectors are typically illustrated by drawing an ARROW

The direction of the physical quantity is given by the direction of the arrow.

The magnitude of the quantity is given by the length of the arrow and it is denoted by |F|.

It is a good habit to label your vectors in a diagram

The direction of the physical quantity is given by the direction of the arrow.

The magnitude of the quantity is given by the length of the arrow and it is denoted by |F|.

It is a good habit to label your vectors in a diagram

F=10N

VECTOR

MAGNITUDE & DIRECTION

MAGNITUDE DISPLACEMENT - METERS, FEET, MILES ETC. FOR VELOCITY - METERS PER SECOND, FEET PER

MINUTE, etc FORCE - NEWTONS, DYNES OR POUNDS.

DIRECTION

- DEGREES- RADIANS - GEOGRAPHIC INDICATORS SUCH AS NORTH,

EAST, SOUTH, NORTHEAST, ETC.

- KNOWLEDGE OF COORDINATE GEOMETRY

(X-Y PLANE)

Up = + Down = - Right = + Left = -

y

x

+

+

-

-

Quadrant IQuadrant II

Quadrant III Quadrant IV

0 o East

90 o North

West 180 o

270 o South

360 o

Rectangular Coordinates

0O East

90O North

West 180O

270O South

360O

+x

+y

- x

- y

120O

-240O

30O West of North30O Left of +y

60O North of West60O Above - x

MEASURING THESAME DIRECTION

IN DIFFERENT WAYS

VECTOR NOTATIONSVECTOR NOTATION MAY TAKE SEVERAL DIFFERENT

FORMS:• POLAR /CIRCULAR FORM INDICATES A MAGNITUDE VALUE AND A DIRECTIONAL

VALUE. THE DIRECTION VALUE MAY BE IN DEGREES, RADIANS OR GEOGRAPHIC TERMS.

• RECTANGULAR FORM IDENTIFIES THE X-Y COORDINATES OF THE VECTOR. THE VECTOR ITSELF EXTENDS FROM ORIGIN TO THE X-Y POINT.– EXAMPLES: (X = +10, Y = -10) or (10, -10) – THE MAGNITUDE OF THE VECTOR CAN BE FOUND USING THE

PYTHAGOREAN THEOREM (102 + (-102))1/2 = 14.1 - THE DIRECTION CAN BE FOUND USING AN INVERSE TANGENT

FUNCTION TAN-1 (10/10) = TAN-1 (1.0) = 45O SINCE X IS POSITIVE AND Y IS NEGATIVE THE ANGLE IS -45O AND IS IN QUADRANT IV OR 315O

Circular NotationCircular notation defines a vector by

designating the vector’s magnitude |A| and angle θ relative to the +x axis. Using that notation the vector is written:

315O

0O East

90O North

West 180O

270O South

360O

+x

+y

- x

- y

-45O or45O SOUTH OF EAST

•POLAR COORDINATES14 METERS , θ=315O

14 METERS, θ= -45O

14 METERS, 45O SOUTH OF EAST

CIRCULAR NOTATIONS

A=14 m

Circular NotationExample:Force vector with magnitude 12 Newtons oriented at 210 degrees with the + x-axis.

F = 12 N, θ = 210°F = 12 N, θ = 210°

F = 12 N, θ = -150°F = 12 N, θ = -150°

F = 12 N, 30° South of WestF = 12 N, 30° South of West

Circular Notation

In this picture, we have a force vector of 4 Newtons oriented along the -x axis.

F = 4 N, θ = 180°

Circular Notation

In the picture above we have 2 vectors C and D. How do we characterize the two vectors?

C = 2, θ = 30°

D = 4 , θ = -50° or D = 4, θ = 310°

Circular Notation

Exercises: Draw the vectors graphically.

1. A= 6, 30° North of West

2. B= 3, θ = 330 °

3. C= 20, 45° West of South

4. D= 5, θ = -240 °

5. E= 8, 60° SE

ADDITION/ SUBTRACTION OF VECTORS

• TWO METHODS1. GRAPHICAL METHOD (DRAWING) 2. COMPONENT METHOD (MATHEMATICAL).

• GRAPHICAL ADDITION AND SUBTRACTION REQUIRES THAT EACH VECTOR BE REPRESENTED AS AN ARROW WITH A LENGTH PROPORTIONAL TO THE MAGNITUDE VALUE AND POINTED IN THE PROPER DIRECTION ASSIGNED TO THE VECTOR.

(Note: We need ruler and protractor)

SCALE

= 10 METERS

50 METERS ,θ= 0O

30 METERSθ= 90O

30 METERS ,θ= 45O

VECTOR ARROWS MAY BE DRAWN ANYWHERE ON THE PAGE AS

LONG AS THE PROPER LENGTH AND DIRECTION ARE MAINTAINED

• VECTORS ARE ADDED GRAPHICALLY BY DRAWING EACH VECTOR TO SCALE AND ORIENTED IN THE PROPER DIRECTION. THE VECTOR ARROWS ARE PLACED HEAD TO TAIL. THE ORDER OF PLACEMENT DOES NOT AFFECT THE RESULT

(A + B = B + A)• THE RESULT OF THE VECTOR ADDITION IS CALLED

THE RESULTANT VECTOR. IT IS MEASURED FROM THE TAIL OF THE FIRST VECTOR ARROW TO THE HEAD OF THE LAST ADDED VECTOR ARROW.

• THE LENGTH OF THE RESULTANT VECTOR ARROW CAN THEN BE MEASURED AND USING THE SCALE FACTOR CONVERTED TO THE CORRECT MAGNITUDE VALUE. THE DIRECTIONAL COMPONENT CAN BE MEASURED USING A PROTRACTOR.

GRAPHICAL METHOD

A

B

C

D

A

BC

DR

A + + + =B C D R

ALL VECTORS MUST BE DRAWN TO

SCALE & POINTED INTHE PROPER DIRECTION

SCALE

= 10 METERS

Vector B50 METERS @ 0O

Vector C30 METERS

@ 90O

Vector A30 METERS @ 45O

A

B

C

Resultant = 9 x 10 = 90 meters

Angle is measured at 40o

To add the vectorsPlace them head to tail

• IN ALGEBRA, A – B = A + (-B) OR IN OTHER WORDS, ADDING A NEGATIVE VALUE IS ACTUALLY SUBTRACTION. THIS IS ALSO TRUE IN VECTOR SUBTRACTION. IF WE ADD A NEGATIVE VECTOR B TO VECTOR A THIS IS REALLY SUBTRACTING VECTOR B FROM VECTOR A.

• VECTOR VALUES CAN BE MADE NEGATIVE BY REVERSING THE VECTOR’S DIRECTION BY 180 DEGREES. IF VECTOR A IS 30 METERS DIRECTED AT 45 DEGREES (QUADRANT I), NEGATIVE VECTOR A IS 30 METERS AT 225 DEGREES (QUADRANT II).

WORKING WITH VECTORS GRAPHIC SUBTRACTION

Vector A30 METERS @ 45O

Vector - A30 METERS @ 225O

A

B

C

D A + - - =B C D R

A + + ( - ) + ( - ) =B C D R

-C

=

-D=

A

-D

R

B

-C

Graphical Method - Examples

1. 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+

R= 84.5 m, E

Notice that the SIZE of the arrow conveys MAGNITUDE and the way it was drawn conveys DIRECTION.

Graphical Method - Examples

2. You walk 50 meters 60 degrees north of west, then another 85 meters in opposite direction. Calculate your displacement relative to where you started?

N

E W

S

60°

60°

R=35, θ=300°

COMPONENT METHOD

• PYTHAGOREAN THEOREM• TRIGONOMETRIC FUNCTIONS• HORIZONTAL AND VERTICAL

COMPONENTS OF THE VECTOR

A

B

C

Sin = A / C

Cos = B / C

Tan = A / B

AC

B

A

BA RIGHT TRIANGLE

C

= Arctan (A / B)

y

x

+

+

-

- 0 radians radians

3/2 radians

2 radians

Quadrant III

Quadrant IV

Quadrant I

Quadrant II

Sin Cos Tan + + +

+ - -- - +

- + -

/2 radians

90 o

0 o

180 o

270 o

360 o

A

X COMPONENT

Y COMPONENT

X COMPONENT

Y COMPONENTB

Y COMPONENT

X COMPONENT

C

X

Y

A

Ax

Ay

Ax A = COS

A= SIN Ay

is the angle between the +x-axis and the vector.

• THE SIGNS OF THE X AND Y COMPONENTS DEPEND ON WHICH QUADRANT THE VECTOR LIES.

• VECTORS IN QUADRANT I (0 TO 90 DEGREES) HAVE POSITIVE X AND POSITIVE Y VALUES

• VECTORS IN QUADRANT II (90 TO 180 DEGREES) HAVE NEGATIVE X VALUES AND POSITIVE Y VALUES.

• VECTORS IN QUADRANT III (180 TO 270 DEGREES) HAVE NEGATIVE X VALUES AND NEGATIVE Y VALUES.

• VECTORS IN QUADRANT IV (270 TO 360 DEGREES) HAVE POSITIVE X VALUES AND NEGATIVE Y VALUES.

VECTOR COMPONENTS

WHAT ARE THE X AND Y COMPONENTS OF THE FOLLOWING VECTORS?

1. A =40 m, = 60O ?

2. V= 60 m/s, = 2450 ?

VECTOR COMPONENTS

Ax A = COS A= SIN Ay

VECTOR COMPONENTS

Exercises: Find the x and y components of the following vectors.

1. A= 100 N, 30° North of East

2. B= 30 km, θ = 330 °

3. C= 20 m, 45° West of South

4. D= 50 m/s, θ = -240 °

5. E= 80 N, 60° Southeast

Analytical Method of Vector AdditionAnalytical Method of Vector Addition1. Find the x- and y-components of each vector.

Ax = A cos = Ay = A sin = Bx = B cos = By = B sin =Cx = C cos = Cy = C sin =

2. Sum the x-components. This is the x-component of the resultant.

Rx =

3. Sum the y-components. This is the y-component of the resultant.

Ry =

4. Use the Pythagorean Theorem to find the magnitude of the resultant vector.Rx

2 + Ry2 = R2

ADDING & SUBTRACTING VECTORS USING COMPONENTS

Vector A30 METERS @ 45O

Vector B50 METERS @ 0O

Vector C30 METERS

@ 90O

ADD THE FOLLOWINGTHREE VECTORS USING

COMPONENTS

(1) RESOLVE EACH INTO X AND Y COMPONENTS

V= SIN Vy

Vx V = COS

COMPONENT METHODCOMPONENT METHOD

•CX = 30 COS 900 = 0 CY = 30 SIN 900 = 30

• AX = 30 COS 450 = 21.2 AY = 30 SIN 450 = 21.2

•BX = 50 COS 00 = 50 BY = 50 SIN 00 = 0

(1) SOLVE EACH INTO X AND Y COMPONENTS(1) SOLVE EACH INTO X AND Y COMPONENTS

(2) ADD THE X COMPONENTS OF EACH VECTOR ADD THE Y COMPONENTS OF EACH VECTOR

(2) ADD THE X COMPONENTS OF EACH VECTOR ADD THE Y COMPONENTS OF EACH VECTOR

Rx = 21.2 + 50 + 0 = +71.2Ry= 21.2 + 0 + 30 = +51.2

(3) CONSTUCT A NEW RIGHT TRIANGLE USING THERx AS THE BASE AND Ry AS THE OPPOSITE SIDE

(3) CONSTUCT A NEW RIGHT TRIANGLE USING THERx AS THE BASE AND Ry AS THE OPPOSITE SIDE

Rx = +71.2

Ry = +51.2

THE HYPOTENUSE IS THE RESULTANT VECTOR

R

(4) USE THE PYTHAGOREAN THEOREM TO THE LENGTH(MAGNITUDE) OF THE RESULTANT VECTOR

(4) USE THE PYTHAGOREAN THEOREM TO THE LENGTH(MAGNITUDE) OF THE RESULTANT VECTOR

Rx = +71.2

Ry = +51.2

(+71.2)2 + (+51.2)2 = 87.7

(5) FIND THE ANGLE (DIRECTION) USING INVERSETANGENT OF THE OPPOSITE SIDE OVER THE

ADJACENT SIDE

(5) FIND THE ANGLE (DIRECTION) USING INVERSETANGENT OF THE OPPOSITE SIDE OVER THE

ADJACENT SIDE

θ=TAN-1 (51.2/71.2)ANGLE = 35.7 O

QUADRANT I

R = 87.7 m , θ = 35.7 OR = 87.7 m , θ = 35.7 O

SUBTRACTING VECTORS USING COMPONENTS

Vector A30 m @ 45O

Vector C30 m

@ 90O

Vector B50 m @ 0O

A - + =B C R

A + (- ) + =B C R

Vector A

30 METERS @ 45O

- B=

50 METERS @ 180O

Vector C

30 METERS @ 90O

1. SOLVE THE X AND Y COMPONENTS

• AX = 30COS 450 = 21.2 METERS

• AY = 30 SIN 450 = 21.2 METERS

•BX = 50 COS 1800 = - 50 METERS

•BY = 50 SIN 1800 = 0 METERS

•CX = 30 COS 900 = 0 METERS

•CY = 30 SIN 900 = 30 METERS

(2) ADD THE X COMPONENTS OF EACH VECTOR ADD THE Y COMPONENTS OF EACH VECTOR

Rx = SUM OF THE Xs = 21.2 + (-50) + 0 = -28.8RY =SUM OF THE Ys = 21.2 + 0 + 30 = +51.2

(3) CONSTUCT A NEW RIGHT TRIANGLE USING THERx AS THE BASE AND RY AS THE OPPOSITE SIDE

Rx = -28.8

RY = +51.2

THE HYPOTENUSE IS THE RESULTANT VECTOR

R

(4) USE THE PYTHAGOREAN THEOREM TO THE FIND THE LENGTH (MAGNITUDE) OF THE RESULTANT VECTOR

Rx = -28.8

Ry = +51.2

ANGLE TAN-1 (51.2/-28.8)ANGLE = -60.6 0

(1800 –60.60 ) = 119.40

QUADRANT II

(-28.8)2 + (+51.2)2 = 58.7

(5) FIND THE ANGLE (DIRECTION) USING INVERSETANGENT OF THE OPPOSITE SIDE OVER THE

ADJACENT SIDE

RESULTANT = 58.7 METERS @ 119.4O

R =

ExampleA 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

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

ExampleA boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of

8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north.

1.28)5333.0(

5333.015

8

/17158

1

22

Tan

Tan

smRv

15 m/s, N

8.0 m/s, W

Rv

The Final Answer :

1.Ehra begins a three-day hiking trip by first walking 25 km, 30° south of east from her base camp. On the second day, she walks 40 km in a direction 60° north of east. On the third day, she walks 35 km, 50° south of west. What is her resultant displacement from base camp?

2.ICADEMY commuter jet departs NAIA and flies 175 km 30° East of North, turns and flies 150 km , 20° North of West, and finally flies 190 km west to its destination. What is its resultant displacement?

3. Find the resultant velocity vector for the following: V1 = 89 km/h , 55° East of SouthV2 = 46 km/h , 28° South of West V3 = 37 km/h , South