Section 5.1
Pages 118 to 125

Evaluate the sum of two or more vectors in two
dimensions graphically.

Determine the components of vectors.

Solve the sum of two or more vectors algebraically by
adding the components of the vectors.
Vocabulary
Components
Vector Resolution
Resultant




A vector has magnitude
as well as direction.
Some vector quantities:
displacement, velocity,
force, momentum
A scalar has only a
magnitude.
Some scalar quantities:
mass, time, temperature
SCALAR
VECTOR
Any quantity that has
MAGNITUDE ONLY!!
Any quantity that has
BOTH MAGNITUDE and
DIRECTION!!
Number value with units!!
Scalar
Example
Magnitude
Vector
Example
Magnitude and
Direction
Speed
35 m/s
Velocity
35 m/s, North
Distance
25 meters
Acceleration
10 m/s2, South
Age
16 years
Force
20 N, East
One Dimensional Vectors
• For vectors in one
dimension, simple addition
and subtraction are all that
is needed.
• You do need to be careful
about the signs, as the
figure indicates.
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
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
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.
FINISH
the hypotenuse is
called the RESULTANT
160 km, N
c 2 a 2 b 2 c a 2 b 2

cresul tan t 120 160
c200km
2
2
VERTICAL
COMPONENT

START
120 km, E
HORIZONTAL COMPONENT
In the example, DISPLACEMENT is 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 TAIL.
S of E
W of S
E of S
S
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
Tan 
 N of E
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

Suppose a person walked 65 m, 25 degrees East of North.
What were his horizontal and vertical components?
65 m
VY = ?
25
The goal: ALWAYS MAKE A RIGHT
TRIANGLE!
To solve for components, we often use
the trig functions sine and cosine.
65
VX = ?
Make sure the angle is
with the x-axis (E or W)
adjacent side
opposite side
sine 
hypotenuse
hypotenuse
adj
opp
cos  
sin  
hyp
hyp
cosine 
adj  VX  65 cos 65  27.47 m, E
opp  VY  65 sin 65  58.91m, N
Adding Vectors by Components
Any vector can be expressed as the sum of two other vectors,
which are called its components. Usually the other vectors are
chosen so that they are perpendicular to each other.
Adding Vectors by Components
If the components
are perpendicular,
they can be found
using trigonometric
functions.
Adding Vectors by Components
Adding vectors:
1. Draw a diagram; add the vectors graphically.
2. Choose x and y axes.
3. Resolve each vector into x and y components.
4. Calculate each component using sines and cosines.
5. Add the components in each direction.
6. To find the length and direction of the vector, use:
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  14 2  232  26.93m
14 m, N
R
23 m, E
14
 .6087
23
  Tan 1 (0.6087)  31.3
Tan 
23 m, E
The Final Answer: 26.93 m, 31.3 degrees NORTH or EAST
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
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
hypotenuse
adj  hyp cos 
cosine 
H.C. =?
32
63.5 m/s
opposite side
hypotenuse
opp  hyp sin 
sine 
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
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
hypotenuse
V.C.
adj  hyp cos 
cosine 
opposite side
hypotenuse
opp  hyp sin 
sine 
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)  8.92 o
Tan 
R
964.2 km

The Final Answer: 6224.2 km @ 8.92
degrees, North of East
6149.1 km
