VECTORS
Honors Section 1.5 – 1.9
Regular Physics Chapter 2
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
For tonight, find the following vertical and
horizontal components
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 of 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

Homework #1
 Remember YOU CANNOT SOLVE A VECTOR
PROBLEM WITHOUT DRAWING A PICTURE!!!!! Use
lots of space.
 Honors Physics: P23 Q21 – 26
 Regular Physics: P91 Q1,2,3 and P94 Q1,2,3,4,5,6,7
 Adding vectors algebraically that are not at right angles:
 So far the vectors have been “tidy” and at right angles to
each other. In real life that is rarely the case.
 Consider this problem:
 A 200N force is placed on a box 15° W of N. Another
150N force is placed on the box 25° E of N. Find the
net horizontal force on the box.
Strategy for dealing with adding vectors:
 Draw an accurate picture of all vectors.
 Add and /or subtract collinear vectors to reduce the
problem.
 Find x and y components of each remaining vector. Be
consistent with + and – signs.
 Add all x components for all vectors.
 Add all y components for all vectors
 Reconstruct the resultant vector from x and y component
sums, STATING MAGNITUDE AND ANGLE IN ANSWER.
.
 A 200N force is placed on a box 15° W of N. Another 150N
force is placed on the box 25° E of N. Find the net horizontal
force on the box.
Projectiles – anything shot or thrown
into the air that have only gravity as a
net external force on them
Projectiles
Any object that is shot or thrown into the air.
 Examples: Baseball, bullet, basketball, tennis
ball, etc. can you think of others.
 The following are not projectiles: Rockets,
aircraft, birds, helicopters etc. can you think
why?
 The path of a projectile throught the air is called
the TRAJECTORY.
 Projectiles may be horizontal or launched at an
angle.

Horizontal Projectiles
Examples: a ball rolled of the edge of a table, a
car driving off a cliff etc. Can you think of any
more?
 The time it takes a horizontal projectile to reach
the ground is the same time taken for an object
to fall straight down from the same height.
 The range (how far it lands from the base of cliff
etc) depends on its speed at time of launch and
the height of the drop.
 Let’s do the following problem...

Projectiles launched at an angle
Examples would be throwing a ball far, throwing
a javelin. Can you think of more?
 These projectiles have a horizontal and vertical
speed at the launch.
 The range depends on speed of launch and
angle alone.
 The angle for maximum range is 45 degrees
 Angles > 45 make it stay in the air longer.
 Angles < 45 make it land sooner.
