VECTORS
Honors Physics
SCALAR
A SCALAR quantity
is any quantity in
physics that has
MAGNITUDE ONLY
Number value
with units
Scalar
Example Magnitude
Speed 35 m/s
Distance 25 meters
Age 16 years
VECTOR
A VECTOR quantity
is any quantity in
physics that has
BOTH MAGNITUDE
and DIRECTION
Vector
Example
Magnitude and
Direction
Velocity 35 m/s, North
Acceleration 10 m/s2, South
Force 20 N, East
An arrow above the symbol
illustrates a vector quantity.
It indicates MAGNITUDE and
DIRECTION
VECTOR APPLICATION
ADDITION: When two (2) vectors point in the SAME direction, simply
add them together.
EXAMPLE: A man walks 46.5 m east, then another 20 m east.
Calculate his displacement relative to where he started.
66.5 m, E
MAGNITUDE relates to the
size of the arrow and
DIRECTION relates to the
way the arrow is drawn
46.5 m, E + 20 m, E
VECTOR APPLICATION
SUBTRACTION: When two (2) vectors point in the OPPOSITE direction,
simply subtract them.
EXAMPLE: A man walks 46.5 m east, then another 20 m west.
Calculate his displacement relative to where he started.
26.5 m, E
46.5 m, E
-
20 m, W
Vectors are always added head to tail
For example draw what you think it would look like if I
added these two vectors together: A+B. Then draw what
you think A-B would look like.
A B
A+B
A-B
You try on the whiteboard
A B
C
a.A+C
b.B+C
c.A-B
d.C-B
e.A+B
NON-COLLINEAR VECTORS
When two (2) vectors are PERPENDICULAR to each other, you must
use the PYTHAGOREAN THEOREM
Example: A man travels 120 km east
then 160 km north. Calculate his
resultant displacement.
120 km, E
160 km, N
the hypotenuse is
called the RESULTANT
HORIZONTAL COMPONENT
VERTICAL
COMPONENT
S
T
A
R
T
FINISH
c2
 a2
 b2
 c  a2
 b2
c  resultant  120
 
2
 160
 
2
 
c  200km
WHAT ABOUT DIRECTION?
In the example, DISPLACEMENT asked for and since it is a VECTOR quantity,
we need to report its direction.
N
S
E
W
N of E
E of N
S of W
W of S
N of W
W of N
S of E
E of S
NOTE: When drawing a right triangle that
conveys some type of motion, you MUST
draw your components HEAD TO TOE.
N of E
NEED A VALUE – ANGLE!
Just putting N of E is not good enough (how far north of east ?).
We need to find a numeric value for the direction.
N of E
160 km, N
120 km, E
To find the value of the
angle we use a Trig
function called TANGENT.

Tan 
opposite side
adjacent side

160
120
1.333
  Tan1
(1.333)  53.1o

200 km
So the COMPLETE final answer is : 200 km, 53.1 degrees North of East
What are your missing
components?
Suppose a person walked 65 m, 25 degrees East of North. What
were his horizontal and vertical components?
65 m
25
H.C. = ?
V.C = ?
The goal: ALWAYS MAKE A RIGHT
TRIANGLE!
To solve for components, we often use
the trig functions since and cosine.
E
m
C
H
opp
N
m
C
V
adj
hyp
opp
hyp
adj
hypotenuse
side
opposite
hypotenuse
side
adjacent
,
47
.
27
25
sin
65
.
.
,
91
.
58
25
cos
65
.
.
sin
cos
sine
cosine














 Draw the vector diagram
 Combine ‘like’ vectors
 Determine the resultant AND the angle of direction
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.
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.
35 m, E
20 m, N
12 m, W
6 m, S
- =
23 m, E
- =
14 m, N
23 m, E
14 m, N

3
.
31
)
6087
.
0
(
6087
.
23
14
93
.
26
23
14
1
2
2








Tan
Tan
m
R


The Final Answer: 26.93 m, 31.3 degrees NORTH or EAST
R

 Draw the vector diagram
 Combine ‘like’ vectors (if necessary)
 Determine the resultant AND the angle of direction
A 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.
Example
A 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.
15 m/s, N
8.0 m/s, W
Rv 

1
.
28
)
5333
.
0
(
5333
.
0
15
8
/
17
15
8
1
2
2








Tan
Tan
s
m
Rv


The Final Answer : 17 m/s, @ 28.1 degrees West of North
A plane moves with a velocity of 63.5 m/s at 32
degrees South of East. Calculate the plane's
horizontal and vertical velocity components.
 Draw the vector diagram
 Combine ‘like’ vectors (if necessary)
 Determine the resultant AND the angle of direction
Example
A plane moves with a velocity of 63.5 m/s at 32 degrees South of East. Calculate
the plane's horizontal and vertical velocity components.
63.5 m/s
32
H.C. =?
V.C. = ?
S
s
m
C
V
opp
E
s
m
C
H
adj
hyp
opp
hyp
adj
hypotenuse
side
opposite
hypotenuse
side
adjacent
,
/
64
.
33
32
sin
5
.
63
.
.
,
/
85
.
53
32
cos
5
.
63
.
.
sin
cos
sine
cosine














A storm system moves 5000 km due east, then shifts course at
40 degrees North of East for 1500 km. Calculate the storm's
resultant displacement.
 Draw the vector diagram
 Combine ‘like’ vectors (if necessary)
 Determine the resultant AND the angle of direction
Example
A storm system moves 5000 km due east, then shifts course at 40
degrees North of East for 1500 km. Calculate the storm's
resultant displacement.
N
km
C
V
opp
E
km
C
H
adj
hyp
opp
hyp
adj
hypotenuse
side
opposite
hypotenuse
side
adjacent
,
2
.
964
40
sin
1500
.
.
,
1
.
1149
40
cos
1500
.
.
sin
cos
sine
cosine














5000 km, E
40
1500 km
H.C.
V.C.
5000 km + 1149.1 km = 6149.1 km
6149.1 km
964.2 km
R


R  6149.12
 964.22
 6224.2km
Tan 
964.2
6149.1
 0.157
  Tan1
(0.157)  8.92o
The Final Answer: 6224.2 km @ 8.92
degrees, North of East

vectorshonors.ppt

  • 1.
  • 2.
    SCALAR A SCALAR quantity isany quantity in physics that has MAGNITUDE ONLY Number value with units Scalar Example Magnitude Speed 35 m/s Distance 25 meters Age 16 years
  • 3.
    VECTOR A VECTOR quantity isany quantity in physics that has BOTH MAGNITUDE and DIRECTION Vector Example Magnitude and Direction Velocity 35 m/s, North Acceleration 10 m/s2, South Force 20 N, East An arrow above the symbol illustrates a vector quantity. It indicates MAGNITUDE and DIRECTION
  • 4.
    VECTOR APPLICATION ADDITION: Whentwo (2) vectors point in the SAME direction, simply add them together. EXAMPLE: A man walks 46.5 m east, then another 20 m east. Calculate his displacement relative to where he started. 66.5 m, E MAGNITUDE relates to the size of the arrow and DIRECTION relates to the way the arrow is drawn 46.5 m, E + 20 m, E
  • 5.
    VECTOR APPLICATION SUBTRACTION: Whentwo (2) vectors point in the OPPOSITE direction, simply subtract them. EXAMPLE: A man walks 46.5 m east, then another 20 m west. Calculate his displacement relative to where he started. 26.5 m, E 46.5 m, E - 20 m, W
  • 6.
    Vectors are alwaysadded head to tail For example draw what you think it would look like if I added these two vectors together: A+B. Then draw what you think A-B would look like. A B A+B A-B
  • 7.
    You try onthe whiteboard A B C a.A+C b.B+C c.A-B d.C-B e.A+B
  • 8.
    NON-COLLINEAR VECTORS When two(2) vectors are PERPENDICULAR to each other, you must use the PYTHAGOREAN THEOREM Example: A man travels 120 km east then 160 km north. Calculate his resultant displacement. 120 km, E 160 km, N the hypotenuse is called the RESULTANT HORIZONTAL COMPONENT VERTICAL COMPONENT S T A R T FINISH c2  a2  b2  c  a2  b2 c  resultant  120   2  160   2   c  200km
  • 9.
    WHAT ABOUT DIRECTION? Inthe example, DISPLACEMENT asked for and since it is a VECTOR quantity, we need to report its direction. N S E W N of E E of N S of W W of S N of W W of N S of E E of S NOTE: When drawing a right triangle that conveys some type of motion, you MUST draw your components HEAD TO TOE. N of E
  • 11.
    NEED A VALUE– ANGLE! Just putting N of E is not good enough (how far north of east ?). We need to find a numeric value for the direction. N of E 160 km, N 120 km, E To find the value of the angle we use a Trig function called TANGENT.  Tan  opposite side adjacent side  160 120 1.333   Tan1 (1.333)  53.1o  200 km So the COMPLETE final answer is : 200 km, 53.1 degrees North of East
  • 12.
    What are yourmissing components? Suppose a person walked 65 m, 25 degrees East of North. What were his horizontal and vertical components? 65 m 25 H.C. = ? V.C = ? The goal: ALWAYS MAKE A RIGHT TRIANGLE! To solve for components, we often use the trig functions since and cosine. E m C H opp N m C V adj hyp opp hyp adj hypotenuse side opposite hypotenuse side adjacent , 47 . 27 25 sin 65 . . , 91 . 58 25 cos 65 . . sin cos sine cosine              
  • 13.
     Draw thevector diagram  Combine ‘like’ vectors  Determine the resultant AND the angle of direction 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.
  • 14.
    Example A bear, searchingfor 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. 35 m, E 20 m, N 12 m, W 6 m, S - = 23 m, E - = 14 m, N 23 m, E 14 m, N  3 . 31 ) 6087 . 0 ( 6087 . 23 14 93 . 26 23 14 1 2 2         Tan Tan m R   The Final Answer: 26.93 m, 31.3 degrees NORTH or EAST R 
  • 15.
     Draw thevector diagram  Combine ‘like’ vectors (if necessary)  Determine the resultant AND the angle of direction A 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.
  • 16.
    Example A boat moveswith 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. 15 m/s, N 8.0 m/s, W Rv   1 . 28 ) 5333 . 0 ( 5333 . 0 15 8 / 17 15 8 1 2 2         Tan Tan s m Rv   The Final Answer : 17 m/s, @ 28.1 degrees West of North
  • 17.
    A plane moveswith a velocity of 63.5 m/s at 32 degrees South of East. Calculate the plane's horizontal and vertical velocity components.  Draw the vector diagram  Combine ‘like’ vectors (if necessary)  Determine the resultant AND the angle of direction
  • 18.
    Example A plane moveswith a velocity of 63.5 m/s at 32 degrees South of East. Calculate the plane's horizontal and vertical velocity components. 63.5 m/s 32 H.C. =? V.C. = ? S s m C V opp E s m C H adj hyp opp hyp adj hypotenuse side opposite hypotenuse side adjacent , / 64 . 33 32 sin 5 . 63 . . , / 85 . 53 32 cos 5 . 63 . . sin cos sine cosine              
  • 19.
    A storm systemmoves 5000 km due east, then shifts course at 40 degrees North of East for 1500 km. Calculate the storm's resultant displacement.  Draw the vector diagram  Combine ‘like’ vectors (if necessary)  Determine the resultant AND the angle of direction
  • 20.
    Example A storm systemmoves 5000 km due east, then shifts course at 40 degrees North of East for 1500 km. Calculate the storm's resultant displacement. N km C V opp E km C H adj hyp opp hyp adj hypotenuse side opposite hypotenuse side adjacent , 2 . 964 40 sin 1500 . . , 1 . 1149 40 cos 1500 . . sin cos sine cosine               5000 km, E 40 1500 km H.C. V.C. 5000 km + 1149.1 km = 6149.1 km 6149.1 km 964.2 km R   R  6149.12  964.22  6224.2km Tan  964.2 6149.1  0.157   Tan1 (0.157)  8.92o The Final Answer: 6224.2 km @ 8.92 degrees, North of East