Chapter 3 motion in two d 2

Chapter 3
Kinematics in Two
Dimensions; Vectors

There are two kinds of
quantities…
• Scalars are quantities that have magnitude only,
such as
– position
– speed
– time
– mass
• Vectors are quantities that have both magnitude
and direction, such as
– displacement
– velocity
– acceleration

Notating vectors

• This is how you
notate a
vector…
R R

• This is how you
draw a
vector…
R head
tail

Direction of Vectors
• Vector direction is the direction of the
arrow, given by an angle.
• This vector has an angle that is
between 0o and 90o.

A
θ x

Vector angle ranges
y
Quadrant II Quadrant I
90o < θ < 180o 0 < θ < 90o

θ θ
x
θ θ
Quadrant III Quadrant IV
180o < θ < 270o 270o < θ < 360o

Direction of Vectors

• What angle range would this vector
have?
• What would be the exact angle, and how
would you determine it?
Between 180o and 270o
θ x
θ
B or between
- 90o and -180o

Magnitude of Vectors
• The best way to determine the magnitude
(or size) of a vector is to measure its
length.
• The length of the vector is proportional to
the magnitude (or size) of the quantity it
represents.

Sample Problem
• If vector A represents a displacement of three miles to the
north, then what does vector B represent? Vector C?

B
A
C

Equal Vectors

• Equal vectors
have the same
length and
direction, and
represent the
same quantity
(such as force
or velocity).
• Draw several
equal vectors.

Inverse Vectors
• Inverse vectors
have the same A
length, but
opposite
direction.
• Draw a set of -A
inverse vectors.

The Right Triangle

hy
po
opposite

te
nu
se

θ
adjacent

Pythagorean Theorem
• hypotenuse2 = opposite2 + adjacent2
• c2 = a2 + b2

hy
po
opposite

te
nu
se

θ
adjacent

Basic Trigonometry functions
• sin θ = opposite/hypotenuse
• cos θ = adjacent/hypotenuse
• tan θ = opposite/adjacent

hy
po
opposite

te
nu
se
SOHCAHTOA

θ
adjacent

Inverse functions
• θ = sin-1(opposite/hypotenuse)
• θ = cos-1(adjacent/hypotenuse)
• θ = tan-1(opposite/adjacent)

hy
po
opposite

te
nu
se
SOHCAHTOA

θ
adjacent

Sample problem
• A surveyor stands on a riverbank directly across the river from a
tree on the opposite bank. She then walks 100 m downstream, and
determines that the angle from her new position to the tree on the
opposite bank is 50o. How wide is the river, and how far is she from
the tree in her new location?

Sample problem
• You are standing at the very top of a tower and notice that in order
to see a manhole cover on the ground 50 meters from the base of
the tower, you must look down at an angle 75o below the horizontal.
If you are 1.80 m tall, how high is the tower?

Vectors: x-component
• The x-component of a vector is the
“shadow” it casts on the x-axis.
• cos θ = adjacent ∕ hypotenuse
• cos θ = Ax ∕ A
• Ax = A cos θ
A

θ
x
Ax

Vectors: y-component
• The y-component of a vector is the
“shadow” it casts on the y-axis.
• sin θ = opposite ∕ hypotenuse
• sin θ = Ay ∕ A
• Ay = A sin θ y

Ay A
Ay
θ x

Vectors: angle
• The angle a vector
makes with the x-
axis can be y
determined by the
components.
Ry
• It is calculated by
the inverse tangent θ x
Rx
function
• θ = tan-1 (Ay/Ax)

Vectors: magnitude
• The magnitude of a
vector can be
determined by the y
components.
R
• It is calculated using
the Pythagorean Ry
Theorem.
x
• R2 = Rx2 + Ry2 Rx

Practice Problem
• You are driving up a long inclined road. After 1.5 miles you
notice that signs along the roadside indicate that your
elevation has increased by 520 feet.
a) What is the angle of the road above the horizontal?

Practice Problem
• You are driving up a long inclined road. After 1.5 miles you
notice that signs along the roadside indicate that your
elevation has increased by 520 feet.
b) How far do you have to drive to gain an additional 150 feet of
elevation?

Practice Problem
• Find the x- and y-components of the
following vectors
a) R = 175 meters @ 95o

Practice Problem
following vectors
b) v = 25 m/s @ -78o

Practice Problem
following vectors
c) a = 2.23 m/s2 @ 150o

Graphical Addition of Vectors

Day 2

Graphical Addition of Vectors

1) Add vectors A and B graphically by drawing
them together in a head to tail arrangement.
2) Draw vector A first, and then draw vector B
such that its tail is on the head of vector A.
3) Then draw the sum, or resultant vector, by
drawing a vector from the tail of A to the head
of B.
4) Measure the magnitude and direction of the
resultant vector.

Practice Graphical Addition

B
B
A
R
A + B = R
R is called the resultant vector!

The Resultant and the
Equilibrant
• The sum of two or more vectors is called
the resultant vector.
• The resultant vector can replace the
vectors from which it is derived.
• The resultant is completely canceled out
by adding it to its inverse, which is called
the equilibrant.

The Equilibrant Vector

B

A -R
R A + B =
The vector -R is calledRthe
equilibrant.
If you add R and -R you get a
null (or zero) vector.

Graphical Subtraction of Vectors

1) Subtract vectors A and B graphically
by adding vector A with the inverse of
vector B (-B).
2) First draw vector A, then draw -B such
that its tail is on the head of vector A.
3) The difference is the vector drawn
from the tail of vector A to the head of
-B.

Practice Graphical Subtraction

-B B
C

A
A - B = C

Practice Problem
• Vector A points in the +x direction and
has a magnitude of 75 m. Vector B has a
magnitude of 30 m and has a direction of
30o relative to the x axis. Vector C has a
magnitude of 50 m and points in a
direction of -60o relative to the x axis.
a) Find A + B
b) Find A + B + C
c) Find A – B.

Vector Addition Lab
1.Attach spring scales to force board such that they all have different

2.Slip graph paper between scales and board and carefully trace your set
In Class

3.Record readings of all three spring scales.
4.Detach scales from board and remove graph paper.
5.On top of your tracing, draw a force diagram by constructing vectors
Homework

6.On a separate sheet of graph paper, add the three vectors together

7.Did you get a resultant? Did you expect one?
8.You must have a separate set of drawings for each member of

Component Addition of Vectors

1) Resolve each vector into its x- and y-
components.
Ax = Acosθ Ay = Asinθ
Bx = Bcosθ By = Bsinθ
Cx = Ccosθ Cy = Csinθ etc.
1) Add the x-components (Ax, Bx, etc.)
together to get Rx and the y-components
(Ay, By, etc.) to get Ry.

Component Addition of Vectors

3) Calculate the magnitude of the
resultant with the Pythagorean
Theorem (R = √Rx2 + Ry2).
4) Determine the angle with the equation
θ = tan-1 Ry/Rx.

Practice Problem
• In a daily prowl through the neighborhood, a cat makes a
displacement of 120 m due north, followed by a displacement of
72 m due west. Find the magnitude and displacement required if
the cat is to return home.

Practice Problem
• If the cat in the previous problem takes 45 minutes to complete
the first displacement and 17 minutes to complete the second
displacement, what is the magnitude and direction of its average
velocity during this 62-minute period of time?

Relative Motion
• Relative motion problems are difficult to do
unless one applies vector addition
concepts.
• Define a vector for a swimmer’s velocity
relative to the water, and another vector
for the velocity of the water relative to the
ground. Adding those two vectors will give
you the velocity of the swimmer relative to
the ground.

Relative Motion

Vs
Vw
Vw

Vt = V s + V w

Relative Motion

Vw
Vs
Vw
Vt = V s + V w

Practice Problem
• You are paddling a canoe in a river that is flowing
at 4.0 mph east. You are capable of paddling at
5.0 mph.
a) If you paddle east, what is your velocity relative to
the shore?
b) If you paddle west, what is your velocity relative to
the shore?
c) You want to paddle straight across the river, from
the south to the north.At what angle to you aim
your boat relative to the shore? Assume east is 0o.

Practice Problem
• You are flying a plane with an airspeed of 400
mph. If you are flying in a region with a 80 mph
west wind, what must your heading be to fly due
north?

Solving 2-D Problems
• Resolve all vectors into components
– x-component
– Y-component
• Work the problem as two one-dimensional
problems.
– Each dimension can obey different equations of
motion.
• Re-combine the results for the two components
at the end of the problem.

Sample Problem
• You run in a straight line at a speed of 5.0 m/s in a direction
that is 40o south of west.
a) How far west have you traveled in 2.5 minutes?
b) How far south have you traveled in 2.5 minutes?

Sample Problem
• A roller coaster rolls down a 20o incline with an
acceleration of 5.0 m/s2.
a) How far horizontally has the coaster traveled in 10 seconds?
b) How far vertically has the coaster traveled in 10 seconds?

Sample Problem
A particle passes through the origin with a speed of 6.2 m/s
traveling along the y axis. If the particle accelerates in the
negative x direction at 4.4 m/s2.
a) What are the x and y positions at 5.0 seconds?

Sample Problem
A particle passes through the origin with a speed of 6.2 m/s
traveling along the y axis. If the particle accelerates in the
negative x direction at 4.4 m/s2.
b) What are the x and y components of velocity at this time?

3-5 Projectile Motion

A projectile is an object
moving in two
dimensions under the
influence of Earth's
gravity; its path is a
parabola.

Projectile Motion
• Something is fired, thrown, shot, or hurled
near the earth’s surface.
• Horizontal velocity is constant.
• Vertical velocity is accelerated.
• Air resistance is ignored.

1-Dimensional Projectile
• Definition: A projectile that moves in a vertical
direction only, subject to acceleration by gravity.
• Examples:
– Drop something off a cliff.
– Throw something straight up and catch it.
• You calculate vertical motion only.
• The motion has no horizontal component.

2-Dimensional Projectile
• Definition: A projectile that moves both
horizontally and vertically, subject to
acceleration by gravity in vertical direction.
• Examples:
– Throw a softball to someone else.
– Fire a cannon horizontally off a cliff.
– Shoot a monkey with a blowgun.
• You calculate vertical and horizontal motion.

Horizontal Component of
Velocity
• Is constant
• Not accelerated
• Not influence by gravity
• Follows equation:
• x = Vo,xt

Horizontal Component of
Velocity

Vertical Component of Velocity
• Undergoes accelerated motion
• Accelerated by gravity (9.8 m/s2 down)
• Vy = Vo,y - gt
• y = yo + Vo,yt - 1/2gt2
• Vy2 = Vo,y2 - 2g(y – yo)

Launch angle
• Definition: The angle at which a projectile
is launched.
• The launch angle determines what the
trajectory of the projectile will be.
• Launch angles can range from -90 o
(throwing something straight down) to +90 o
(throwing something straight up) and
everything in between.

Zero Launch angle
vo

• A zero launch angle implies a perfectly
horizontal launch.

Sample Problem
• The Zambezi River flows over Victoria Falls in Africa. The falls are
approximately 108 m high. If the river is flowing horizontally at 3.6 m/s just
before going over the falls, what is the speed of the water when it hits the
bottom? Assume the water is in freefall as it drops.

Sample Problem
• An astronaut on the planet Zircon tosses a rock horizontally with a
speed of 6.75 m/s. The rock falls a distance of 1.20 m and lands a
horizontal distance of 8.95 m from the astronaut. What is the
acceleration due to gravity on Zircon?

Sample Problem
• Playing shortstop, you throw a ball horizontally to the second baseman
with a speed of 22 m/s. The ball is caught by the second baseman 0.45
s later.
a) How far were you from the second baseman?
b) What is the distance of the vertical drop?

General Launch Angle Projectiles

Day 5

General launch angle

vo
θ

• Projectile motion is more complicated when the
launch angle is not straight up or down (90 o or –
90o), or perfectly horizontal (0o).

General launch angle

vo
θ

• You must begin problems like this by resolving the
velocity vector into its components.

Resolving the velocity

• Use speed and the launch angle to find
horizontal and vertical velocity components

Vo Vo,y = Vo sin θ

θ
Vo,x = Vo cos θ

Resolving the velocity

• Then proceed to work problems just like you
did with the zero launch angle problems.

Vo Vo,y = Vo sin θ

θ
Vo,x = Vo cos θ

Sample problem
• A soccer ball is kicked with a speed of 9.50 m/s at an angle of
25o above the horizontal. If the ball lands at the same level from
which is was kicked, how long was it in the air?

Sample problem
• Snowballs are thrown with a speed of 13 m/s from a roof 7.0 m
above the ground. Snowball A is thrown straight downward;
snowball B is thrown in a direction 25o above the horizontal.
When the snowballs land, is the speed of A greater than, less
than, or the same speed of B? Verify your answer by
calculation of the landing speed of both snowballs.

Projectiles launched over level
ground
• These projectiles have highly symmetric
characteristics of motion.
• It is handy to know these characteristics,
since a knowledge of the symmetry can
help in working problems and predicting
the motion.
• Lets take a look at projectiles launched
over level ground.

Trajectory of a 2-D Projectile

y

x
• Definition: The trajectory is the path
traveled by any projectile. It is plotted on
an x-y graph.


y

x
• Mathematically, the path is defined by a
parabola.


y

x
• For a projectile launched over level
ground, the symmetry is apparent.

Range of a 2-D Projectile

y

x
Range

• Definition: The RANGE of the projectile is
how far it travels horizontally.

Maximum height of a projectile

y

Maximum
Height

x
Range
• The MAXIMUM HEIGHT of the projectile
occurs when it stops moving upward.


y

Maximum
Height

x
Range
• The vertical velocity component is zero at
maximum height.


y

Maximum
Height

x
Range
• For a projectile launched over level ground, the
maximum height occurs halfway through the flight of
the projectile.

Acceleration of a projectile

y
g
g g

g g
x

• Acceleration points down at 9.8 m/s 2 for the
entire trajectory of all projectiles.

Velocity of a projectile

y
v

v v

vo
vf x

• Velocity is tangent to the path for the entire
trajectory.


y
vx

vy vx
vx vy

vy vx
vx vy x

• The velocity can be resolved into
components all along its path.


y
vx

vy vx
vx vy

vy vx
vx vy x

• Notice how the vertical velocity changes while
the horizontal velocity remains constant.


y
vx

vy vx
vx vy

vy vx
vx vy x

• Maximum speed is attained at the beginning, and
again at the end, of the trajectory if the projectile
is launched over level ground.


vo
θ θ-
vo
• Launch angle is symmetric with landing angle for
a projectile launched over level ground.

Time of flight for a projectile

t

to = 0

• The projectile spends half its time
traveling upward…

Time of flight for a projectile

t

to = 0 2t

• … and the other half traveling down.

Position graphs for 2-D
projectiles
y y x

x t t

Velocity graphs for 2-D
projectiles
Vy Vx

t t

Acceleration graphs for 2-D
projectiles
ay ax

t t

Projectile Lab
The purpose is to collect data to plot a trajectory for a
projectile launched horizontally, and to calculate the
launch velocity of the projectile. Equipment is provided,
you figure out how to use it.
• What you turn in:
1. a table of data
2. a graph of the trajectory
3. a calculation of the launch velocity of the ball
obtained from the data
• Hints and tips:
1. The thin paper strip is pressure sensitive. Striking
the paper produces a mark.
2. You might like to hang a sheet of your own graph
paper on the brown board.

The Range Equation
• Derivation is an important part of physics.
• Your book has many more equations than
your formula sheet.
• The Range Equation is in your textbook,
but not on your formula sheet. You can
use it if you can memorize it or derive it!

The Range Equation
• R = vo2sin(2θ)/g.
– R: range of projectile fired over level ground
– vo: initial velocity
– g: acceleration due to gravity
θ: launch angle

Sample problem
• A golfer tees off on level ground, giving the ball an initial
speed of 42.0 m/s and an initial direction of 35o above the
horizontal.
a) How far from the golfer does the ball land?

Sample problem
• A golfer tees off on level ground, giving the ball an initial
speed of 42.0 m/s and an initial direction of 35o above the
horizontal.
b) The next golfer hits a ball with the same initial speed, but at a greater
angle than 45o. The ball travels the same horizontal distance. What
was the initial direction of motion?

It can be understood by
analyzing the horizontal and
vertical motions separately.

4
monkey
problem

The speed in the x-direction
is constant; in the y-
direction the object moves
with constant acceleration g.
This photograph shows two balls
that start to fall at the same time.
The one on the right has an initial
speed in the x-direction. It can be
seen that vertical positions of the
two balls are identical at identical
times, while the horizontal
position of the yellow ball
increases linearly.

If an object is launched at an initial angle of θ0
with the horizontal, the analysis is similar except
that the initial velocity has a vertical component.

3-6 Solving Problems Involving
Projectile Motion
Projectile motion is motion with constant
acceleration in two dimensions, where the
acceleration is g and is down.

Projectile Motion
1. Read the problem carefully, and choose the
object(s) you are going to analyze.
2. Draw a diagram.
3. Choose an origin and a coordinate system.
4. Decide on the time interval; this is the same in
both directions, and includes only the time the
object is moving with constant acceleration g.
5. Examine the x and y motions separately.

Projectile Motion

6. List known and unknown quantities.
Remember that vx never changes, and that
vy = 0 at the highest point.
7. Plan how you will proceed. Use the
appropriate equations; you may have to
combine some of them.

3-7 Projectile Motion Is Parabolic
In order to demonstrate that
projectile motion is parabolic,
we need to write y as a function
of x. When we do, we find that it
has the form:

This is
indeed the
equation for
a parabola.

3-8 Relative Velocity
We already considered relative speed in one
dimension; it is similar in two dimensions
except that we must add and subtract velocities
as vectors.

Each velocity is labeled first with the object, and
second with the reference frame in which it has
this velocity. Therefore, vWS is the velocity of the
water in the shore frame, vBS is the velocity of the
boat in the shore frame, and vBW is the velocity of
the boat in the water frame.

3-8 Relative Velocity
In this case, the relationship between the
three velocities is:

(3-6)

Summary of Chapter 3

• A quantity with magnitude and direction is a
vector.
• A quantity with magnitude but no direction is
a scalar.
• Vector addition can be done either graphically
or using components.
• The sum is called the resultant vector.
• Projectile motion is the motion of an object
near the Earth’s surface under the influence of
gravity.

3-1 Vectors and Scalars

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
1

3-2 Addition of Vectors – Graphical Methods

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.

If the motion is in two dimensions, the situation is
somewhat more complicated.
Here, the actual travel paths are at right angles to
one another; we can find the displacement by
using the Pythagorean Theorem.

Adding the vectors in the opposite order gives the
same result:

Even if the vectors are not at right
angles, they can be added graphically by
using the “tail-to-tip” method.

The parallelogram method may also be used;
here again the vectors must be “tail-to-tip.”

3-3 Subtraction of Vectors, and
Multiplication of a Vector by a Scalar

In order to subtract vectors, we
define the negative of a vector, which
has the same magnitude but points
in the opposite direction.

Then we add the negative vector:

3-3 Subtraction of Vectors, and
Multiplication of a Vector by a Scalar
A vector V can be multiplied by a scalar c; the
result is a vector cV that has the same direction
but a magnitude cV. If c is negative, the resultant
vector points in the opposite direction.

3-4 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.


If the components are
perpendicular, they can be found
using trigonometric functions.


The components are effectively one-dimensional,
so they can be added arithmetically:


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:

Chapter 3 motion in two d 2

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