PowerPoint Presentation - VECTORS VECTORS
Level 1 Physics
Objectives and Essential
Questions
 Objectives
 Distinguish between basic
trigonometric functions (SOH
CAH TOA)
 Distinguish between vector
and scalar quantities
and analytical methods
 Essential Questions
 What is a vector quantity?
What is a scalar quantity?
Give examples of each.
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
r r r r
x, v , a, F
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
EXAMPLE: A man walks 46.5 m east, then another 20 m east.
Calculate his displacement relative to where he started.
46.5 m, E
+
66.5 m, E
20 m, E
MAGNITUDE relates to the
size of the arrow and
DIRECTION relates to the
way the arrow is drawn
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.
46.5 m, E
20 m, W
26.5 m, E
NON-COLLINEAR VECTORS
When two (2) vectors are PERPENDICULAR to each other, you must
use the PYTHAGOREAN THEOREM
FINISH
Example: A man travels 120 km east
then 160 km north. Calculate his
resultant displacement.
the hypotenuse is
called the RESULTANT
160 km, N
c 2  a2  b2  c  a2  b2
c  resultant 
c  200km
VERTICAL
COMPONENT
120  160 
2
2
S
R
T
T
A
120 km, E
HORIZONTAL COMPONENT
In the example, DISPLACEMENT asked for and since it is a VECTOR quantity,
we need to report its direction.
N
W of N
E of N
N of E
N of E
N of W
E
W
S of W
NOTE: When drawing a right triangle that
conveys some type of motion, you MUST
S of E
W of S
E of S
S
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.
To find the value of the
angle we use a Trig
function called TANGENT.
200 km
160 km, N
oppositeside 160
Tan 

 1.333
 N of E
  Tan1 (1.333)  53.1o
120 km, E
So the COMPLETE final answer is : 200 km, 53.1 degrees North of East

components?
Suppose a person walked 65 m, 25 degrees East of North. What
were his horizontal and vertical components?
H.C. = ?
V.C = ?
25
65 m
The goal: ALWAYS MAKE A RIGHT
TRIANGLE!
To solve for components, we often use
the trig functions since and cosine.
opposite side
sine  
hypotenuse
hypotenuse
opp  hyp sin 
cosine 
adj  V .C.  65 cos 25  58.91m, N
opp  H .C.  65 sin 25  27.47m, E
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.
-
12 m, W
-
=
6 m, S
20 m, N
35 m, E
14 m, N
R

23 m, E
=
14 m, N
R  14 2  232  26.93m
14
Tan 
 .6087
23
  Tan 1 (0.6087)  31.3
23 m, E
The Final Answer: 26.93 m, 31.3 degrees NORTH or EAST
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.
Rv  82  152  17 m / s
8.0 m/s, W
15 m/s, N
Rv

8
Tan   0.5333
15
  Tan 1 (0.5333)  28.1
The Final Answer : 17 m/s, @ 28.1 degrees West of North
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.
opposite side
cosine 
sine  
hypotenuse
hypotenuse
opp  hyp sin 
H.C. =?
32
63.5 m/s
V.C. = ?
adj  H .C.  63.5 cos 32  53.85 m / s, E
opp  V .C.  63.5 sin 32  33.64 m / s, S
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.
1500 km
opposite side
sine  
hypotenuse
hypotenuse
V.C.
opp  hyp sin 
cosine 
40
5000 km, E
H.C.
adj  H .C.  1500 cos 40  1149.1 km, E
opp  V .C.  1500 sin 40  964.2 km, N
5000 km + 1149.1 km = 6149.1 km
R  6149.12  964.2 2  6224.2 km
964.2
 0.157
6149.1
  Tan1 (0.157)  8.92o
Tan 
R
964.2 km

The Final Answer: 6224.2 km @ 8.92
degrees, North of East
6149.1 km
