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```									     Vectors

Magnitude + Direction
I. Scalars and Vectors

A. Scalars – Number and unit. No direction.
Ex: Mass, time, temperature

B. Vectors – Quantities that have a direction associated with them.

Ex. Velocity, force, momentum

1. Each separate vector is called a COMPONENT.

2. The single vector that produces the same result as the combined components
is called the RESULTANT.

=
V1        V2                                    V1 + V2

COMPONENTS                                   RESULTANT
Each vector has two ends

TIP
Vectors can be used to represent the
increasing speed of a ball in free fall.

The distance the ball falls each second increases
because the ball is accelerating.

This is shown by drawing a longer
vector arrow for each time interval.
constant velocity (v) – motion in a straight line at a constant speed; vf = vo.

Vectors can also be used to represent a ball rolling horizontally
on a table at a constant velocity.

.

Each vector arrow is drawn the same length to represent the constant velocity.

The velocity would remain constant but friction makes it slow down
and eventually stop.
II. Vector Addition and Subtraction
A. Addition of COLINEAR Vectors – when vectors point along the same direction.

Ex: If a person walks 10.0 meters, east, stops, and continues 5.0 meters, east.
What is the resultant?
Tail to Tip Method of Drawing
Vectors:

10 m, E                5 m, E             -Tail of the second arrow
Vector A                Vector B           located At the head of the
first arrow, the two lengths
=
add to give the length of the
10 m, E + 5 m E = 15 m, E                  total displacement.
- Can only be done when the
Vector R
vectors point along the same
direction.

A+B=R
Ex: If a person walks 10.0 meters, east, stops, and continues 5.0 meters, west.
What is the resultant?
Colinear Vector Addition in
Opposite Directions:
-Draw both vector components in
10 m, E                   5 m, W         their respective direction.
Vector A                   Vector B       - Find the difference between the
two vectors and select the direction
=
of the vector with the greater
10 m, E - 5 m E = 5 m, E                  magnitude.
Vector R

A+B=R
Ex: What is the velocity of a swimmer who is swimming 10 m/s upstream if the
current downstream is 5 m/s?

5 m/s;                              10 m/s up – 5 m/s; down
10 m/s;              downstream                          = 5 m/s; up
upstream
=
B. Addition of Vectors that are Neither Colinear Nor Perpendicular

Graphical Technique – A diagram is constructed in which the arrow are drawn
tail to tip. The lengths of the vector arrows are drawn to scale, and the angles
are drawn accurately (with a protractor). Then, the length of the arrow
representing the resultant vector is measured with a ruler. This length is
converted into the magnitude of the resultant vector by using the scale
factor with which the drawing was constructed.
Ex: A car moves with a displacement A
of 275 m, due west and then with a
R                displacement B of 125 m in a direction
B                       55.0° north of west.
55°

A

The two displacement vectors A and
Vector A = 275 m; W              B are neither colinear nor perpendicular
but add to give the resultant vector R.
Vector B = 125 m; @ 55° N of W
Ex: I ran 10.0 km due east but was pushed 5.0 km, north by a crosswind.
How far did I actually run? What was my angle from the start line?

R
5.0 km; north

Θ

10.0 km; east

- The resultant, R can be found by using
a ruler to measure the magnitude, and a
protractor to measure, the angle Θ.
C. Parallelogram Method to Vector Addition
Two vectors are drawn to scale and joined at their tails. Parallel lines
are drawn to complete the parallelogram.

The resultant is drawn from where the tails touch to the opposite corner.

Ex: A boat travels 2 km north, then 4 km north of east at an angle of 30°.
Determine the magnitude and direction of the resultant.

Two ways to solve for R

- Use a ruler and a protractor

R                       -Use vector resolution
2 km; n                   4 km, e        (math: Pythagorean’s Theorem;
30°                   Trig functions)
D. Vector Component Method (Vector Resolution)

1. Resolve each vector into x and y components.

2. Assign positive or negative values to each component based on
which quadrant it falls in.

3. Reduce the problem to the sum of the two vectors.

Σ x = v1x + v2x + any others
Σ y = v1y + v2y + any others

4. Determine the magnitude and direction of the resultant. Since Σ x and Σ y
are at right angles, the Pythagorean Theorem can be used to determine the
magnitude of the resultant.

The definition of the tangent of the angle can be used to determine the
direction of the resultant.
tan Θ = vy                     Θ = tan-1
vx                              (v)
v
y
x
4 km, e                                                  4 km, e

28°                                                         28°
R                                                           R
2 km; n                                                   2 km; n

Vector 1                                4 km, e                                  Vector 2
(2 km; n)                                                  v2y               (4 km; e @ 28°)
28°
V1x = 0 km                                        v2x                          V2x = 4 km

V1y = 2 km                 V2y = sin θ = o/h           V2x = cos θ = a/h       V2y = 2 km
V2y = o = sin θ (h)         V2x = a = cos θ (h)
V2y = sin (28°) ( 4 km)     V2x = cos (28°) ( 4 km)
V2y = 2 km                  V2x = 4 km
v2x                             Σx

v2y
Σy
R                              R
v1y

Σ x = v1x + v2x = 0 km + 4 km = 4 km

Σ y = v1y + v2y = 2 km + 2 km = 4 km
Σ x = 4 km
Θ = tan-1       vy
( )     vx
Σ y = 4 km
R
θ
Θ = tan-1 Σ x
( Σy         )
Θ = tan-1 4 km
(
4 km           )
Θ = 45º; East of North

sin Θ = o                 sin Θ = Σ x
R
h
R=     Σx                  R=      4
(sin Θ)                    (sin 37)

R = 7 km
Rule for Adding Vectors
When adding vectors, it doesn’t matter which
order they are added in. The resultant will
always be the same.
Ex: A man walks 100 ft at 30º north of east; 200 ft at 40º south of east; and 400 ft
at 90º.
ΣRx = V1x + V2x + V3x
ΣRx = V1cosΘ + V2cosΘ + V3cosΘ
R
ΣRx = 100(cos30) + 200(cos40) + 0
V3         ΣRx = 240 ft
V1              40º
30º
V2
90º

ΣRy = V1y + (-V2y) + V3y
ΣRy = V1sinΘ - V2sinΘ + V3sinΘ

ΣRy = 100(sin30) - 200(sin40) + 400

ΣRy = 321 ft
Θ = tan-1 Ry
( )
Rx

R
Θ = tan-1 321 ft
Ry = 321 ft
( 240 ft   )
Θ                                 Θ = 53º; North of East
Rx = 240 ft

sin Θ = Ry
sin Θ = o
R
h
R=      321
(sin 53)

R = 402 ft
III. Relative Motion/Velocity

The concept of a frame of reference is important when solving for
these problems. It is important to always keep in mind the relationship
between each variable involved.

A. Each velocity vector is labeled by two subscripts:
First subscript refers to the object
Second subscript refers to the reference frame in which it has a velocity

Ex: Boat goes across a river
VBW = velocity of the boat with respect to the water:
how fast the boat travels in the water across the river.
vws
vBS
VWS = velocity of the water with respect to the shore:
vBW               how fast the water in the river travels when compared
to the shore.

VBS = velocity of the boat with respect to the shore:
how fast the boat (that’s in the water) is traveling
Current             when compared to the shore).
vBS = vBW + vSW

vBW = -vWB

Ex: A passenger is walking towards the front of a
moving train. The people sitting on the train see
the passenger walking with a velocity of +2.0 m/s,
where the plus sign denotes a direction to the right.
Suppose the train is moving with a velocity of
+9.0 m/s relative to an observer standing on
the ground.
What would the ground-based observer see the
passenger moving with?

vPT = velocity of the PASSENGER relative to the TRAIN = + 2.0 m/s
vTG = velocity of the TRAIN relative to the GROUND = + 9.0 m/s
vPG = vPT + vTG
vPG = velocity of the PASSENGER relative to the GROUND = + 11 m/s
=               vPG
vPT        vTG
Ex: The engine of a boat drives it across a river that is 1800 m wide. The
velocity VBW of the boat relative to the water is 4.0 m/s, directed perpendicular
to the current, as shown below. The velocity VWS of the water relative to the shore
is 2.0 m/s.
a.) What is the velocity VBS of the boat relative to the shore?
b.) How long does it take for the boat to cross?

a.) The velocity VBS of the boat relative to the shore is the vector sum of the velocity
VBW of the boat relative to the water and the velocity VWS of the water
relative to the shore: VBS = VBW + VWS.

VBS = √(VBW 2 + VWS2)                         Θ = tan-1 VBW
(
VWS   )
VBS = √( (4.0 m/s)2 + (2.0 m/s)2 )
Θ = tan-1 4.0
= 4.5 m/s                                              (2.0   )
= 63º
b.) The component of VBS that is parallel to the width of the river determines
how fast the boat is moving across the river; this parallel component is
VBS (sin Θ) = VBW = 4.0 m/s. The time for the boat to cross the river is equal
to the width of the river divided by the magnitude of this velocity component.

t=x
v
t=    width        = 1800 m     = 450 s
VBS (sin Θ)     4.0 m/s
IV. Projectile Motion
The motion of an object that is projected into the air at an angle.

Solve these problems in the same way as you would velocity problems.
Use the kinematic formulas for constant acceleration to get formulas for
projectile motion.

Horizontal Motion                              Vertical Motion

vfx = vix + axΔt                                vfy = viy + ayΔt

Δx = xi + vixt + ½ axΔt2                        Δy = yi + viyt + ½ ayΔt2
vfx2 = vix2 + 2axΔx                             vfy2 = viy2 + 2ayΔy
Idea behind trajectory/projectile motion:
Observe the motion of the ball as it falls focusing on the path followed (trajectory).
In projectile motion, we are combining the motion of the ball in free fall
with the motion of the ball rolling on the table at a constant velocity.

This is seen when rolling the ball off of the table.
The ball rolling on the table would continue forever in a straight line if
gravity is ignored. The ball
in free fall would continue to
increase its speed if air
resistance is ignored.
Since the ball is moving at a constant velocity and in free fall at the same
time, the horizontal and vertical vectors are added together during
equal time intervals. This is done for each time interval until the ball
hits the ground.
The path the ball follows can be seen by connecting the
intermittent resultant vectors.
The two-dimensional motion of the spacecraft can be viewed as the
combination of the separate x and y motions.
Kinematic Variables

y, x, ax, ay, vy, viy, vx, vix
Ex: In the x direction, a spacecraft has an initial velocity component of vix = +22 m/s
and an acceleration component ax = +24 m/s2. In the y direction, the analogous
quantities are viy = +14 m/s and ay = +12 m/s2. The directions to the right and
upward have been chosen as the positive directions. After a time of 7.0 s, find the:
a.) x and vx
b.) y and vy
c.) final velocity (magnitude and direction of the aircraft)

Treat the motion in the x and y direction separately

a.) data for motion in the x direction
ax = +24 m/s2            x = vixt + ½ axt2
vix = +22 m/s            x = (22 m/s)(7.0 s) + ½ (24 m/s2)(7.0 s) 2
t = 7.0 s
x=?                      x = +740 m
vx = ?
vx = vix + axt
vx = (22 m/s) + (24 m/s2)(7.0s)
vx = +190 m/s
b.) data for motion in the y direction
ay = +12 m/s2            y = viyt + ½ ayt2
viy = +14 m/s            y = (14 m/s)(7.0 s) + ½ (12 m/s2)(7.0 s) 2
t = 7.0 s
y=?                      y = +390 m
vy = ?
vy = viy + ayt
vy = (14 m/s) + (12 m/s2)(7.0s)
vy = +98 m/s

c.) final velocity
vfy = 98 m/s                             Vf = √(Vfx2 + Vfy2)
Vf = √( (190 m/s)2 + (98 m/s)2 )
vf
Vf = 210 m/s
θ
vfx = 190 m/s       Θ = tan-1 vy
( )vx
Θ = tan 98
-1
(190)          = 27° above the positive
x axis
Ex: The figure shows an airplane moving horizontally with a constant velocity of
+115 m/s at an altitude of 1050 m. The directions to the right and upward have been
chosen as the positive directions. The plane releases a “care package” that falls to
the ground along a curved trajectory. Ignoring air resistance, determine the time
required for the package to hit the ground.
The time required for
the package to hit the
ground is the time it
takes for it to move
1050 m. It moves both
downward and to the
right.

These two parts of motion
occur independently.

The package is not
moving in the y direction
at the instant of release,
so viy = 0 m/s.
When the package hits the ground, the y component of its displacement is
y = -1050 m as the drawing shows. The acceleration is due to gravity,
so g = - 9.80 m/s2.

(Note, if you make y = +, then make g = +. I left it as a negative number to
demonstrate that its just a frame of reference)

What we know:                       y = viyt + ½ ayt2
y = -1050 m
ay = -9.80 m/s2
viy = 0 m/s                          t = √(2y / ay)
t=?                                  t = √( 2(-1050 m) / -9.80 m/s2) )
t = 14.6 s

Note: the vertical speed of the package picks up over time, but the horizontal
component stays the same, v0x = vx = 115 cm. This is because the package
is traveling at the same speed as the plane in the x direction. Therefore, the
package does not accelerate in the x direction.
Two packages are dropped from the plane. Package A drops vertically and
package B drops at an angle. They both hit the ground at the same time because
they have the same y components of their velocities. However, package B will
hit the ground with a greater speed due to the magnitude and direction of its x
component.
Ex: Find the speed of the package and the direction of the velocity vector just
before the package in the previous problem hits the ground.

Image upon impact

vfx = 115 m/s                                vfx = 115 m/s

vf       ground                               vf      ground
vfy = ?                               vfy = -143 m/s

What we know to solve for vy :           So,   vf = √( (115 m/s)2 + (-143 m/s) 2 )
ay = -9.80 m/s2                                vf = 184 m/s
viy = 0 m/s
t = 14.6 s                                Θ = cos-1 vfx
( ) vf
vfy = viy + ayt
Θ = cos 115
-1
vfy = 0m/s + (-9.80 m/s2)(14.6 s)                ( 184 )
vfy = -143 m/s                           Θ = 51.3° below the x axis
Angled projectile motion:
A projectile launched at an angle would continue in a straight line at
a constant velocity if gravity is ignored. However, gravity makes the
projectile accelerate to Earth. This shows us why a projectile launched
at an angle follows a parabolic trajectory.
Since the projectile is launched at an angle, it now has both horizontal
and vertical velocities.

When dealing with angled projectile problems, we neglect air resistance and
say that gravity is g = - 9.80 m/s2 (downward) and also say that the vertical
velocity of any object moving upwards will have a + velocity. This makes
solving the problems a lot “cleaner.”
The horizontal component of the velocity remains constant.

The vertical component of the velocity changes as the projectile
moves up or down. This is shown when looking at the vector components
of the trajectory at regular time intervals.
The up and right vectors represent the velocity given to the projectile when
launched. The vertical vectors decrease in magnitude due to gravity.
Eventually, the effects of gravity will reduce the upward velocity to zero.
This occurs at the top of the parabolic trajectory where there is only horizontal
motion.

After gravity reduces the upward (vertical)
speed to zero it begins to add a downward
velocity. This velocity increases until the

When looking at each half of the
trajectory (up and down) you can
determine that the speed of the
projectile going up is equal to the
speed of the projectile coming down.
(Provided air resistance is ignored.)
The only difference is the direction
of the motion.
Problem Solving Projectile Motion:

Notes
Since the horizontal and vertical components of a projectile launched at any angle
are still independent of one other, the equations pertaining to linear motion and
free fall can still be used in problem solving.
(As long as we look at them separately.)

Before the projectile is launched, t = 0, the position is zero, y = 0.
The other half of the equation is the part that helps us find time;
when y at the end of the trajectory is zero all the time needed to complete
the trajectory has passed.

y = viyt + ½ gt2                      x = vixt + ½ at2
When solving problems, the following procedures will be used:

1. Make a sketch of the situation and label all given and unknown quantities.
2. Determine what type of projectile motion is being observed – angled.
3. List all of the equations that apply to the problem.
4. Determine what variables are known and needed.
5. Substitute the given quantities into the equation using proper units.
6. Solve the equation, carrying units throughout the problem.
7. Check your answer to make sure it makes sense.
8. Box or circle your final answer(s).
Ex: A baseball player hits a home run, and the ball lands in the left-field seats,
7.5 m above the point at which the ball was hit. The ball lands with a velocity
of 36 m/s at an angle of 28º below the horizontal. Ignoring air resistance, find the
initial velocity with which the ball leaves the bat.

What we want to find: vi and θ
To solve for these, we need to find viy and vix

Vi = √(Vix2 + Viy2)      and          Θ = tan-1 Viy
( )
Vix
Since air resistance is ignored, the horizontal component of the velocity, vx
is constant.
vix = vfx = (36 m/s) cos 28° = 32 m/s

What we know about the y component:
y = + 7.5 m
ay = -9.80 m/s2
vfy = (-36 sin 28°) m/s
viy = ?

vfy2 = viy + 2ayy
viy = √(vfy2 – 2ayy)

viy = √( (-36 sin 28°) m/s)2 – 2(-9.80 m/s2)(7.5 m) )
viy = 21 m/s
Therefore…

Vi = √(Vix2 + Viy2)

Vi = √( (32 m/s)2 + (21 m/s)2 )

Vi = 38 m/s

Θ = tan-1 Viy
( )
Vix

Θ = tan-1 21
( )
32

Θ = 33° above the x axis

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