Chapter 3
Motion in Two Dimensions
Most interesting kinematics problems occur in two dimensions (or
three, but the extension from 2D to 3D is fairly simple). As it turns
out, our 1D equations from chapter 2 are still extraordinarily useful in
solving 2D and 3D problems. In this chapter, we will see how this is
accomplished. It will require the
use of vectors – and in
particular, the component
notation of vectors. Make
sure that you are comfortable
with the use of vectors – it’s
the most difficult
mathematical concept that
you will encounter in PC141.
PC141 Intersession 2013
Day 5 – May 14 – WBL 3.1-3.2
Slide 1
3.0 Review of Trigonometry
A trig “refresher” can be found in
Appendix 1D of the text. The most
important concepts are shown here,
and will be discussed in class.
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Slide 2
3.0 Review of Trigonometry
Angles – reference point and units
Unless otherwise specified, angles are measured relative to the
positive direction of the x-axis. They are positive if they are
measured in the counterclockwise direction and negative if
they are measured in the clockwise direction.
Angles may be measured in degrees or radians (rad). To convert
between these units, recall that a full circle is 360° or 2π rad. For
example, to convert 40° into radians, write
2𝜋 rad
40° ×
= 0.70 rad
360°
When using a calculator, always be aware of whether it is in
“degrees” mode or “radians” mode. Advanced scientific
computation software (MasteringPhysics, in particular) almost
always uses the latter by default.
PC141 Intersession 2013
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Slide 3
3.0 Review of Trigonometry
Inverse trig functions
The inverse trig functions sin-1 x, cos-1 x, and tan-1 x are not equal
to 1/(sin x), 1/(cos x), 1/(tan x), respectively. Rather, this
notation indicates that you must invert the operation of the trig
function itself; for example, y = sin x implies that x = sin-1 y.
The function sin-1 x is called the “arcsine” of x (similarly, we have
the “arccosine”, “arctangent”, “arccosecent”, etc.)
In the unlikely case that you need to need to use an inverse trig
function in MasteringPhysics, the preferred syntax is “acos”,
“asin”, etc.
PC141 Intersession 2013
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Slide 4
3.0 Review of Trigonometry
Inverse trig functions
Note however that there are an infinite
number of solutions of the inverse trig
functions, two for each 2π rotation
around a circle.
For instance, sin-1 0.5 has the solutions
30° and 150°. Your calculator will only
give you the first of these. Deciding if
the calculator’s answer makes sense is
one reason why it is strongly suggested
that you sketch a diagram that
represents the problem you are wishing
to solve. Later in this chapter, we will
come across this issue in one of the
problems that we will solve.
PC141 Intersession 2013
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0.5
30°
150°
Slide 5
3.1 Components of Motion
In chapter 2, we assumed that objects moved in straight lines
(although they could move back and forth along these lines).
These lines were labeled as the x- or y-axes (for horizontal and
vertical lines, respectively).
If an object moves in a straight line that is at an angle to these
axes, it’s still a 1D problem…we are free to reorient our axes in
any direction we wish.
For example, if an
object is sliding
downhill, it is usually
beneficial to orient
your axes as shown to
the right.
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Slide 6
3.1 Components of Motion
However, we are frequently called upon to analyze problems in
which an object undergoes curvilinear motion – that is, its
trajectory does not follow a straight line. The Angry Birds
depicted on the title slide for this chapter is an example.
On the other hand, a situation might arise where multiple
objects are moving in
straight lines that aren’t
parallel or perpendicular
to each other.
In both of these cases, we
must describe the motion
in more than one
dimension.
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Slide 7
3.1 Components of Motion
In the figure on the previous slide, the ball is moving in a
direction θ with respect to the +x-direction, and with a velocity
of magnitude v. It can be thought of as having velocity
components in both the x-direction and the y-direction. These
are labeled vx and vy, respectively.
Since the x-axis and y-axis are perpendicular, simple
trigonometry allows us to resolve the velocity vector into its
scalar components:
𝑣𝑥 = 𝑣 cos 𝜃
𝑣𝑦 = 𝑣 sin 𝜃
On the other hand, if we know vx and vy, we can find the
magnitude and angle of the velocity using Pythagoras’s theorem
and a bit more trigonometry:
𝑣=
PC141 Intersession 2013
𝑣𝑥2
+ 𝑣𝑦2 ,
𝜃=
𝑣𝑦
−1
tan
𝑣𝑥
Day 5 – May 14 – WBL 3.1-3.2
Slide 8
3.1 Components of Motion
The benefit of using component notation for vectors is that a 2D
kinematics problem can be converted into two 1D kinematics
problems. These problems are solved individually using the
methods of the last chapter, then combined at the end if
necessary. For instance, in the case of a constant acceleration a
with components 𝑎𝑥 = 𝑎 cos 𝜃 and 𝑎𝑦 = 𝑎 sin 𝜃, we can write
1
𝑥 = 𝑥0 + 𝑣𝑥0 𝑡 + 𝑎𝑥 𝑡 2
2
motion in x
𝑣𝑥 = 𝑣𝑥0 + 𝑎𝑥 𝑡
1
𝑦 = 𝑦0 + 𝑣𝑦0 𝑡 + 𝑎𝑦 𝑡 2
2
motion in y
𝑣𝑦 = 𝑣𝑦0 + 𝑎𝑦 𝑡
Here, 𝑣𝑥0 and 𝑣𝑦0 are the velocity components at t = 0.
PC141 Intersession 2013
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Slide 9
Problem #1: Curvilinear Motion
WBL LP 3.3
For an object in curvilinear motion (that is, it’s not
traveling in a straight line):
PC141 Intersession 2013
A
The object’s velocity components are constant
B
The y-velocity component is necessarily greater than
the x-velocity component
C
There is an acceleration nonparallel to the object’s path
D
The velocity and acceleration vectors must be at right
angles (90°)
Day 5 – May 14 – WBL 3.1-3.2
Slide 10
Problem #1: Curvilinear Motion
WBL LP 3.3
For an object in curvilinear motion (that is, it’s not
traveling in a straight line):
A
The object’s velocity components are constant
B
The y-velocity component is necessarily greater than
the x-velocity component
C
There is an acceleration nonparallel to the object’s path
D
The velocity and acceleration vectors must be at right
angles (90°)
(extra page…the solution and discussion is rather lengthy)
PC141 Intersession 2013
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Slide 11
Problem #2: Rolling Ball
WBL EX 3.9
A ball rolling on a table has a velocity with rectangular components vx = 0.60 m/s
and vy = 0.80 m/s. What is the displacement of the ball in an interval of 2.5 s?
Solution: In class
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Slide 12
Problem #3: Particle Acceleration
WBL EX 3.13
A particle moves at a speed of 3.0 m/s in the +x-direction. Upon reaching the
origin, the particle receives a continuous constant acceleration of 0.75 m/s2 in the
– y-direction. What is the position of the particle 4.0 s later?
Solution: In class
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Slide 13
3.2 Vector Addition and Subtraction
We will use boldface and over-arrows (A) to indicate vectors.
On pp. 73-74 of the text, the geometrical method of vector
addition is explained.
This method is sometimes
useful for checking the
“reasonableness” of your
answer to 2D problems.
However, it is almost always
better to perform vector
addition and subtraction using
component notation.
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Slide 14
3.2 Vector Addition and Subtraction
To perform vector addition and subtraction using component
notation, we must introduce the unit vector.
A unit vector has a particular direction, but it always has a
magnitude of one (in whatever units are appropriate). In our
text, it is notated using lower-case boldface and a hat. That is,
the unit vector corresponding to A is a.
Any vector can then be expressed as the product of its
magnitude and its unit vector: A = 𝐴a.
If we define x and y as unit vectors in
the x- and y-directions, then any vector
C in the xy-plane can be written as C =
𝐶𝑥 x + 𝐶𝑦 y.
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Slide 15
3.2 Vector Addition and Subtraction
To add vectors using component notation,
we simply resolve the vectors into their
components and add the components for
each axis separately.
For example, consider two vectors: F1 =
𝐹𝑥1 x + 𝐹𝑦1 y and F2 = 𝐹𝑥2 x + 𝐹𝑦2 y.
Their sum is F = F1 + F2 .
Since each component of F must equal the
corresponding component of F1 + F2 , we
can write
𝐹𝑥 = 𝐹𝑥1 + 𝐹𝑥2
𝐹𝑦 = 𝐹𝑦1 + 𝐹𝑦2
PC141 Intersession 2013
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Slide 16
3.2 Vector Addition and Subtraction
Vector subtraction works the same way
as scalar subtraction. If you have two
scalars, a and b, their difference is
a – b = a + (-b)
Likewise, the difference of two vectors
A and B is
A − B = A + −B ,
where −B is a vector with the same
magnitude as B, but with the opposite
direction.
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Slide 17
3.2 Vector Addition and Subtraction
For 3D problems, almost everything we’ve learned in two
dimensions can still be applied. For example, if an object is
moving in three dimensions, its velocity has x-, y-, and zcomponents: v = 𝑣𝑥 x + 𝑣𝑦 y + 𝑣𝑧 z. Its magnitude is
𝑣=
𝑣𝑥2 + 𝑣𝑦2 + 𝑣𝑧2
(Pythagoras’ equation works just fine in higher dimensions).
The concept of angles becomes much more complicated in 3D –
it’s ignored in PC141.
PC141 Intersession 2013
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Slide 18
3.2 Vector Addition and Subtraction
Three final notes:
1. The process of decomposing 2D and 3D problems into separate 1D
problems will be utilized repeatedly during PC141, so I strongly
suggest that you get comfortable with it now.
2. We have seen that for 2D problems occurring in a vertical plane,
we use x to label the horizontal axis and y to label the vertical axis.
When dealing with 3D problems, it is customary to use x and y to
represent the two horizontal axes and z to represent the vertical
axis.
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Slide 19
3.2 Vector Addition and Subtraction
3. For problems referenced to a map, it is customary for the +x-axis
to point east, the +y-axis to point north, and the +z-axis to point
upward. The term “northeast” means exactly northeast…that is, at
45 degrees to both x and y (and similarly for “northwest”, etc.).
The term “north of east” is a more vague term referring to the first
quadrant (where the x- and y-coordinates are both positive)
y
y
“northeast”
z
x
z
x
“south of
east”
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Slide 20
Problem #4: Vector Magnitude & Direction
If two vectors A and B are given, such that A + B = 0, what can be
said about the magnitude and direction of vectors A and B?
A
They must have the same magnitude, and can be in any
direction
B
They must have the same magnitude, but must be in the same
direction
C
They can have different magnitudes, but must have the same
direction
D
They must have the same magnitude, and must have the
opposite direction
PC141 Intersession 2013
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Slide 21
Problem #5: Doppler Radar
WBL EX 3.39
A meteorologist tracks the movement of a thunderstorm with
Doppler radar. At 8:00 PM, the storm was 60 miles northeast of
her station. At 10:00 PM, the storm is 75 miles north of the
station. The general direction of the storm’s velocity is…
A
South of East
B
North of West
C
D
North of East
PC141 Intersession 2013
South of West
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Slide 22
Problem #5: Doppler Radar
WBL EX 3.39
A meteorologist tracks the movement of a thunderstorm with
Doppler radar. At 8:00 PM, the storm was 60 miles northeast of
her station. At 10:00 PM, the storm is 75 miles north of the
station. The general direction of the storm’s velocity is…
A
South of East
B
North of West
C
D
North of East
South of West
(extra page…the solution and discussion is rather lengthy)
PC141 Intersession 2013
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Slide 23
Problem #6: Terrible Golfer
A golfer must make just one putt to win the Masters. The hole is 10.0 m away
(assume that the green is perfectly flat).
The putt is terrible…it’s off by 20° (to the right of the hole) and it travels 14.0 m.
If the 2nd putt is to be accurate, how far should it travel? In what direction should
it travel, relative to the true alignment of the initial putt?
Solution: In class
PC141 Intersession 2013
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Slide 24
Problem #7: Zero-Sum Vectors
WBL EX 3.43
Two vectors are shown in the figure, with magnitudes F1 = 100 and F2 = 150. A
third vector (not shown) must be found such that the sum of all three vectors is
the zero vector. What is this third vector?
Solution: In class
Note: This concept of “vector balancing” is very important in mechanics, when the
vectors are forces. We’ll get to that in Chapter 4.
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Slide 25
Problem #8: Mountain Climber
A mountaineer uses a global positioning system (GPS) receiver to measure his
displacement from base camp to the summit of Mt. McKinley. The coordinates of
the base camp are x = 0 m, y = 0 m, and z = 4300 m, where z represents altitude.
The coordinates of the summit are x = 1600 m, y = 4200 m, and z = 6200 m. What
is the magnitude of the displacement in going from the base camp to the
summit?
Solution: In class
PC141 Intersession 2013
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Slide 26