VECTORS
PRE-AP 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
Displacement
20 m, East
Vector quantities can be
identified by bold type
with an arrow above the
symbol.
V = 23 m/s NE
Vectors are represented
by drawing arrows
The length and direction
of a vector should be
drawn to a reasonable
scale size and show its
magnitude
20 km
10 km
VECTOR APPLICATION
•ADDITION: When two (2) vectors point
in the SAME direction, simply add them
together.
•When vectors are added together they
should be drawn head to tail to determine
the resultant or sum vector.
•The resultant goes from tail of A to head
of B.
LET’S
PRACTICE
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
VECTOR APPLICATION
SUBTRACTION: When two (2)
vectors point in the OPPOSITE
direction,
simply subtract them.
LET’S PRACTICE SOME MORE….
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
GRAPHICAL METHOD
Aligning vectors head to tail
and then drawing the
resultant from the tail
of the first to the
head of the last.
GRAPHICAL VECTOR ADDITION A + B
Step 1 – Draw a start point
Step 2 – Decide on a scale
Step 3 – Draw Vector A to scale
Step 4 – Vector B’s tail begin at Vector A’s head. Draw Vector B
to scale.
Step 5 – Draw a line connecting the initial start point to the
head of B. This is the resultant.
NON CO-LINEAR VECTORS
When two (2) vectors are
PERPENDICULAR to
each other, you must use
the PYTHAGOREAN
THEOREM
LET’S PRACTICE
A man travels 120 km
east then 160 km
north. Calculate his
resultant
displacement.
FINISH
the hypotenuse is
called the RESULTANT
160 km, N
VERTICAL
COMPONENT
c 2  a2  b2  c  a2  b2
c  resul tan t 
c  200km
120  160 
2
2
S
R
T
T
A
120 km, E
HORIZONTAL COMPONENT
WHAT ABOUT DIRECTION?
In the example, DISPLACEMENT is
asked for and since it is a VECTOR
quantity, we need to report its direction.
N
W of N
N of E
NOTE: When drawing a right triangle that
conveys some type of motion, you MUST
draw your components HEAD TO TOE.
E of N
N of E
N of W
W
E
S of W
S of E
W of S
E of S
S
DIRECTIONS
• There is a difference between Northwest and West of
North
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
 N of E
Tan 
opposite side 160

 1.333
adjacent side 120
  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 sine and cosine.
adjacent side
hypotenuse
adj  hyp cos 
cosine 
opposite side
hypotenuse
opp  hyp sin 
sine 
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
=
14 m, N
R  14 2  232  26.93m
20 m, N
35 m, E
23 m, E
14
 .6087
23
  Tan 1 (0.6087)  31.3
Tan 
14 m, N
R

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.
H.C. =?
32˚
63.5 m/s
V.C. = ?
adjacent side
cosine 
hypotenuse
adj  hyp cos 
opposite side
sine 
hypotenuse
opp  hyp sin 
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
40
5000 km, E
H.C.
adjacent side
hypotenuse
V.C.
adj  hyp cos 
cosine 
opposite side
hypotenuse
opp  hyp sin 
sine 
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)  8.92 o
Tan 
R

964.2 km
The Final Answer: 6224.2 km @ 8.92
degrees, North of East
6149.1 km
