Transcript 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
 Add vectors using graphical
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
add them together.
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
WHAT ABOUT DIRECTION?
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
draw your components HEAD TO TOE.
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
adjacentside 120
 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

What are your missing
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.
adjacent side
opposite side
sine  
hypotenuse
hypotenuse
adj  hyp cos
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.
adjacent side
opposite side
cosine 
sine  
hypotenuse
hypotenuse
adj  hyp cos
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
adjacent side
opposite side
sine  
hypotenuse
hypotenuse
V.C.
adj  hyp cos
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
