Objectives and essential questions

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

ρr 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
Example: A man travels 120 km east
then 160 km north. Calculate his
resultant displacement.

FINIS
H

the hypotenuse is
called the RESULTANT

160 km, N

c 2 = a2 + b2 → c = a2 + b2
2

VERTICAL
COMPONENT
2

c = resultan t = [(120) + (160) ]
c = 200 km

120 km, E

HORIZONTAL COMPONENT


WHAT ABOUT DIRECTION?

In the example, DISPLACEMENT asked for and since it is a VECTOR quanti
we need to report its direction.

N
W of N
N of E

E of N
N of E

N of W
W

NOTE: When drawing a right triangle
that conveys some type of motion, you
MUST draw your components HEAD
TO TOE.

E
S of W

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.

200 km

θ N of E
120 km, E

160 km, N

To find the value of the
angle we use a Trig
function called
TANGENT.

opposite side 160
Tanθ =
=
= 1.333
adjacent side 120
θ = Tan −1 (1.333) = 53.1o

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. = ?

The goal: ALWAYS MAKE A RIGHT
TRIANGLE!

25

To solve for components, we
often use the trig functions since
andadjacent
cosine. side
opposite side

V.C = ?

65 m

cosineθ =

hypotenuse
adj = hyp cos θ

sineθ =

hypotenuse
opp = hyp sin θ

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

20 m, N

35 m, E

R = 14 2 + 232 = 26.93m
14 m, N

R
θ

23 m, E

=

23 m, E

14
= .6087
23
θ = Tan −1 (0.6087) = 31.3
Tanθ =

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

θ

8
Tanθ = = 0.5333
15
θ = Tan −1 (0.5333) = 28.1

The Final Answer : 17 m/s, @ 28.1 degrees West of No

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

adjacent side
cosineθ =
hypotenuse
adj = hyp cos θ

V.C. = ?

opposite side
sineθ =
hypotenuse
opp = hyp sin θ

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 θ
cosineθ =

opposite side
hypotenuse
opp = hyp sin θ

sineθ =

adj = H .C. = 1500 cos 40 = 1149.1 km, E
opp = V .C. = 1500 sin 40 = 964.2 km, N

2
2
R
=
6149.1
+
964.2
= 6224.2 km
5000 km + 1149.1 km = 6149.1 km

964.2
Tanθ =
= 0.157
6149.1
R
964.2 km
θ = Tan −1 (0.157) = 8.92 o

θ

6149.1 km

The Final Answer: 6224.2 km @
8.92 degrees, North of East