1.
VECTORS
Pre-AP Physics

2.
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

3.
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

4.
Vector quantities can be
identified by bold type
with an arrow above the
symbol.
V = 23 m/s NE

5.
Vectors are represented
by drawing arrows

6.
The length and direction
of a vector should be
drawn to a reasonable
scale size and show its
magnitude
20 km
10 km

7.
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.

8.
A man walks 46.5 m east, then another 20 m east.
Calculate his displacement relative to where he started.
•Let’s Practice
66.5 m, E
46.5 m, E
+ 20 m, E

9.
•VECTOR APPLICATION
SUBTRACTION: When
two (2) vectors point in
the OPPOSITE
direction,
simply subtract them.

10.
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.
26.5 m, E
46.5 m, E
-
20 m, W

11.
Graphical Method
Aligning vectors head to tail
and then drawing the
resultant from the tail
of the first to the
head of the last.

12.
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.

13.
NON CO-LINEAR VECTORS
When two (2) vectors are
PERPENDICULAR to each
other, you must use the
PYTHAGOREAN
THEOREM

14.
Let’s Practice
A man travels 120 km
east then 160 km
north. Calculate his
resultant
displacement.
c2
 a2
 b2
 c  a2
 b2
c  resultant  120
 
2
 160
 
2
 
c  200km
VERTICAL
COMPONENT
FINISH
120 km, E
160 km, N
the hypotenuse is
called the RESULTANT
HORIZONTAL COMPONENT
S
T
A
R
T

15.
•WHAT ABOUT DIRECTION?
In the example, DISPLACEMENT is 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

16.
•Directions
• There is a difference between Northwest and West of North

17.
•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.
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
N of E

200 km
So the COMPLETE final answer is : 200 km, 53.1 degrees North of East

18.
•What are your missing
components?
Suppose a person walked 65 m, 25 degrees East of North. What were
his horizontal and vertical components?
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














65 m
25˚
H.C. = ?
V.C = ?
The goal: ALWAYS MAKE A RIGHT
TRIANGLE!
To solve for components, we often use the
trig functions sine and cosine.

19.
•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.

3
.
31
)
6087
.
0
(
6087
.
23
14
93
.
26
23
14
1
2
2








Tan
Tan
m
R


35 m, E
20 m, N
12 m, W
6 m, S
- =
23 m, E
- =
14 m, N
23 m, E
14 m, N
The Final Answer: 26.93 m, 31.3 degrees NORTH of EAST
R


20.
•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.

1
.
28
)
5333
.
0
(
5333
.
0
15
8
/
17
15
8
1
2
2








Tan
Tan
s
m
Rv


15 m/s, N
8.0 m/s, W
Rv 
The Final Answer : 17 m/s, @ 28.1 degrees West of North

21.
•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.
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














63.5 m/s
32˚
H.C. =?
V.C. = ?

22.
•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















R  6149.12
 964.22
 6224.2km
Tan 
964.2
6149.1
 0.157
  Tan1
(0.157)  8.92o
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

The Final Answer: 6224.2 km @ 8.92
degrees, North of East