SECTION 2
Plan and Prepare
Preview Vocabulary
Scientific Meaning The word
simultaneous is used for phenomena
that occur together at the same time.
Ask students to list some simultaneous
phenomena, such as the pressure of a
confined gas decreasing while its volume
is increasing.
Teach
Teaching Tip
Review the sign conventions for
coordinate systems that were established in the chapter “Motion in One
Dimension.” Movements to the right
along the x-axis and upward along the
y-axis are considered positive, and
movements to the left along the x-axis
and downward along the y-axis are
considered negative.
SECTION 2
Objectives
Identify appropriate coordinate
systems for solving problems
with vectors.
Apply the Pythagorean theorem
and tangent function to
calculate the magnitude and
direction of a resultant vector.
Resolve vectors into
components using the sine and
cosine functions.
Add vectors that are not
perpendicular.
FIGURE 2.1
Using a Coordinate System
A gecko’s displacement while climbing
a tree can be represented by an arrow
pointing along the y-axis.
Vector Operations
Key Term
components of a vector
Coordinate Systems in Two Dimensions
In the chapter “Motion in One Dimension,” the motion of a gecko
climbing a tree was described as motion along the y-axis. The direction
of the displacement of the gecko was denoted by a positive or negative
sign. The displacement of the gecko can now be described by an arrow
pointing along the y-axis, as shown in Figure 2.1. A more versatile system
for diagramming the motion of an object, however, employs vectors and
the use of both the x- and y-axes simultaneously.
The addition of another axis helps describe motion in two dimensions
and simplifies analysis of motion in one dimension. For example, two
methods can be used to describe the motion of a jet moving at 300 m/s to
the northeast. In one approach, the coordinate system can be turned so
that the plane is depicted as moving along the y-axis, as in Figure 2.2(a).
The jet’s motion also can be depicted on a two-dimensional coordinate
system whose axes point north and east, as shown in Figure 2.2(b).
One problem with the first method is that the axis must be turned
again if the direction of the plane changes. Another problem is that the
first method provides no way to deal with a second airplane that is not
traveling in the same direction as the first airplane. Thus, axes are often
designated using fixed directions. For example, in Figure 2.2(b), the
positive y-axis points north and the positive x-axis points east. Similarly,
when you analyze the motion of objects thrown into the air, orienting the
y-axis parallel to the vertical direction simplifies problem solving.
Tips and Tricks
There are no firm rules for applying coordinate systems to situations involving vectors.
As long as you are consistent, the final answer will be correct regardless of the system
you choose. Perhaps your best choice for orienting axes is the approach that makes
solving the problem easiest for you.
TEACH FROM VISUALS
FIGURE 2.2 Point out the two very
different choices for coordinate axes.
Ask Which set of axes will give the
correct answer?
Answer: Either set of axes must give the
same result.
FIGURE 2.2
Two Different Coordinate
Systems A plane traveling northeast at
a velocity of 300 m/s can be represented
as either (a) moving along a y-axis chosen
to point to the northeast or (b) moving at
an angle of 45° to both the x- and y-axes,
which line up with west-east and southnorth, respectively.
y
N
W
E
S
v = 300 m/s at 45˚
v = 300 m/s northeast
(a)
x
(b)
84
Chapter 3
Problem
Solving
Take It Further
For Figure 2.2, ask students how the plane’s
motion would be changed if there is a
headwind of 10 m/s from the northeast. The
plane’s velocity would be reduced to 290 m/s
northeast. Invite a student to draw the vectors
for this situation on the board, and to show
classmates how to solve this problem both
visually and mathematically.
Untitled-241 84
84
Chapter 3
Then ask, how would the situation be
different with a tailwind of 10 m/s from the
southwest? The plane’s velocity would be
increased to 310 m/s northeast. What would
be the total difference in velocity between a
headwind of 10 m/s and a tailwind of 10 m/s?
20 m/s
5/4/2011 2:50:58 PM
Untitled-241 85
Determining Resultant Magnitude and Direction
Misconception Alert!
Earlier, we found the magnitude and direction of a resultant graphically.
However, this approach is time-consuming, and the accuracy of the answer
depends on how carefully the diagram is drawn and measured. A simpler
method uses the Pythagorean theorem and the tangent function.
Use the Pythagorean theorem to find the magnitude of the resultant.
Imagine a tourist climbing a pyramid in Egypt. The tourist knows the
height and width of the pyramid and would like to know the distance
covered in a climb from the bottom to the top of the pyramid. Assume
that the tourist climbs directly up the middle of one face.
As can be seen in Figure 2.3, the magnitude of the tourist’s vertical
displacement, ∆y, is the height of the pyramid. The magnitude of the
horizontal displacement, ∆x, equals the distance from one edge of the
pyramid to the middle, or half the pyramid’s width. Notice that these
two vectors are perpendicular and form a right triangle with the displacement, d.
As shown in Figure 2.4(a), the Pythagorean theorem states that for any
right triangle, the square of the hypotenuse—the side opposite the right
angle—equals the sum of the squares of the other two sides, or legs.
FIGURE 2.3
A Triangle Inside of a Pyramid
Because the base and height of a pyramid
are perpendicular, we can find a tourist’s
total displacement, d, if we know the
height, ∆y, and width, 2∆x, of the
pyramid.
∆y
∆x
d
2∆x
Pythagorean Theorem for Right Triangles
c2 = a2 + b2
(length of hypotenuse)2 = (length of one leg)2 + (length of other leg)2
In Figure 2.4(b), the Pythagorean theorem is applied to find the tourist’s
displacement. The square of the displacement is equal to the sum of the
square of the horizontal displacement and the square of the vertical
displacement. In this way, you can find out the magnitude of the displacement, d.
Teaching Tip
Point out that finding the resultant for
the pyramid is fairly simple because the
height, half-width, and hypotenuse form
a right triangle. It is important to
mention at this point that right triangles
will also allow students to find the x and
y components that are important for
vector addition.
FIGURE 2.4
Using the Pythagorean Theorem
(a) The Pythagorean theorem can be applied to any right triangle.
(b) It can also be applied to find the magnitude of a resultant displacement.
d
c
(a)
b
c 2 = a2 + b 2
Students often try to apply the
Pythagorean theorem to triangles that
do not contain a right angle. Point out
that the Pythagorean theorem can be
used only with a right triangle. Some
students may know the Law of Cosines,
which applies to all triangles. This law
states that c2 = a2 + b2 − 2abcosθ. The
Law of Cosines can be used to calculate
one side of any triangle when the
opposite angle and the lengths of the
other two sides are known. In this
expression, c is the unknown side, θ is
the angle opposite c, and a and b are
the two known sides. Some students
may attempt to use the Law of Cosines
to add nonperpendicular vectors. This
approach will give the correct answer,
but it entails more computation and is
more prone to student error when more
than two vectors are to be added.
y
a
(b)
d2 =
x
x2 + y2
Two-Dimensional Motion and Vectors
85
Deconstructing Problems
Emphasize the importance of recognizing what
each symbol in an equation stands for. For
example, in d 2 = Δx2 + Δy2, the Δ is neither
a variable nor a coefficient. In fact, both Δx
and Δy are unique variables. That is, Δx or Δy
can be replaced with only one value. Challenge
students to investigate more about the
symbol Δ, delta, and what it stands for in
mathematics and physics.
5/4/2011 2:50:59 PM
Two-Dimensional Motion and Vectors
85
Use the tangent function to find the direction of the resultant.
FIGURE 2.5
Teach continued
Classroom Practice
Finding Resultant Magnitude and
Direction A plane travels from Houston,
Texas, to Washington, D.C., which is
1540 km east and 1160 km north of
Houston. What is the total displacement
of the plane?
Using the Tangent
Function (a) The tangent function
can be applied to any right triangle,
and (b) it can also be used to find the
direction of a resultant displacement.
Hypotenuse
tan =
For any right triangle, the tangent of an angle is defined as the ratio of
the opposite and adjacent legs with respect to a specified acute angle of a
right triangle, as shown in Figure 2.5(a).
As shown below, the magnitude of the opposite leg divided by the
magnitude of the adjacent leg equals the tangent of the angle.
Adjacent
opp
adj
Definition of the Tangent Function for Right Triangles
d
Answer: 1930 km at 37.0° north of east
A camper travels 4.5 km northeast and
4.5 km northwest. What is the camper’s
total displacement?
Answer: 6.4 km north
Opposite
(a)
In order to completely describe the tourist’s displacement, you must also
know the direction of the tourist’s motion. Because ∆x, ∆y, and d form a
right triangle, as shown in Figure 2.5(b), the inverse tangent function can be
used to find the angle θ, which denotes the direction of the tourist’s
displacement.
(b)
x
tan
=
= tan-1
Teaching Tip
opposite leg
tangent of angle = __
adjacent leg
opp
tan θ = _
adj
y
The inverse of the tangent function, which is shown below, gives
the angle.
opp
θ = tan−1 _
adj
( )
Finding Resultant Magnitude and Direction
Explain that trigonometric functions
such as the one in the text, tan‒1, have a
different role. This is not an exponent.
Instead, it is used to represent the
inverse of a trigonometric function. For
example, the inverse of the functions
sin x = 0.5 and cos 3a = 0.92 take on
the following forms:
x = sin‒1 0.5 = 30°
cos-1 0.92
​= 7.7°
3a = cos‒1 0.92 ⇒ a = ​ _
3
Sample Problem A An archaeologist climbs the Great
Pyramid in Giza, Egypt. The pyramid’s height is 136 m and its
width is 2.30 × 102 m. What is the magnitude and the direction of
the displacement of the archaeologist after she has climbed from
the bottom of the pyramid to the top?
ANALYZE
1 (width) = 115 m
∆x = _
2
Given:
∆y = 136 m
Unknown:
d=?
Diagram:
Choose the archaeologist’s starting position
as the origin of the coordinate system.
θ=?
y
d
x = 115 m
y = 136 m
x
Continued
86
Chapter 3
Problem
Solving
Take It Further
Give students directions for a treasure hunt
with movements that are all perpendicular to
one another (i.e, move front/back and right/
left). Have a student follow the directions to
find a “treasure”. Then have the class resolve
the directions into two components and
calculate the resultant vector. Have a student
use the resultant vector to go directly from
the start position to the treasure.
Untitled-241 86
86 Chapter 3
5/4/2011 2:51:00 PM
ntitled-241 87
Finding Resultant Magnitude and Direction
PLAN
(continued)
d 2 = ∆x2 + ∆y2
∆y
tan θ = _
∆x
d=
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = Sample Problem Set II (online)
Solving for:
Tips and Tricks
Rearrange the equations to isolate the unknowns:
∆x2 + ∆y2
√�����
( )
∆y
θ = tan-1 _
∆x
SOLVE
PROBLEM guide A
Choose an equation or situation:
The Pythagorean theorem can be used to find the magnitude of the
archaeologist’s displacement. The direction of the displacement can
be found by using the tangent function.
Be sure your calculator is set
to calculate angles measured
in degrees. Some calculators
have a button labeled “DRG”
that, when pressed, toggles
between degrees, radians,
and grads.
(115 m)2 + (136 m)2
√��������
(
)
Answers
θ = 49.8°
CHECK YOUR
WORK
Sample, 1–4;
Ch. Rvw.
21–22, 23*
PW 2, 4–5, 7*
PB Sample, 1–5
*Challenging Problem
d = 178 m
136 m
θ = tan-1 _
115 m
SE
component PW Sample, 1, 3*, 6*
PB 6–10
Substitute the values into the equations and solve:
d=
resultant
Practice A
1. a. 23 km
b. 17 km to the east
2. 45.6 m at 9.5° east of north
3. 15.7 m at 22° to the side of downfield
4. 1.8 m at 49° below the horizontal
Because d is the hypotenuse, the archaeologist’s displacement should
be less than the sum of the height and half of the width. The angle is
expected to be more than 45° because the height is greater than half of
the width.
1. A truck driver is attempting to deliver some furniture. First, he travels 8 km east,
and then he turns around and travels 3 km west. Finally, he turns again and travels
12 km east to his destination.
a. What distance has the driver traveled?
b. What is the driver’s total displacement?
2. While following the directions on a treasure map, a pirate walks 45.0 m north and
then turns and walks 7.5 m east. What single straight-line displacement could the
pirate have taken to reach the treasure?
3. Emily passes a soccer ball 6.0 m directly across the field to Kara. Kara then kicks
the ball 14.5 m directly down the field to Luisa. What is the total displacement of
the ball as it travels between Emily and Luisa?
4. A hummingbird, 3.4 m above the ground, flies 1.2 m along a straight path. Upon
spotting a flower below, the hummingbird drops directly downward 1.4 m to hover
in front of the flower. What is the hummingbird’s total displacement?
Two-Dimensional Motion and Vectors
87
Alternative Approaches
To find an angle in a right triangle, you can
apply different trigonometric functions, such
as the following.
( )
5/4/2011 2:51:01 PM
( )
∆y
136
θ = sin‒1 ​ _
​   ​ ​= sin‒1 ​ ​ _ ​ ​= 49.8°
178
d
Two-Dimensional Motion and Vectors 87
Resolving Vectors into Components
Teach continued
components of a vector the
projections of a vector along the axes
of a coordinate system
Teaching Tip
In mathematics, the components of a
vector are called projections. The x
component is the projection of the
vector along the x-axis, and the y
component is the projection of the
vector along the y-axis.
In this textbook, components of vectors are shown as outlined, open
arrows. Components have arrowheads to indicate their direction.
Components are scalars (numbers), but they are signed numbers. The
direction is important to determine their sign in a coordinate system.
You can often describe an object’s motion more conveniently by
breaking a single vector into two components, or resolving the vector.
Resolving a vector allows you to analyze the motion in each direction.
FIGURE 2.6
Diagramming a Movie Scene
Misconception Alert!
Because of the prominence of angles
measured from the x-axis, students may
develop the misconception that the x
component of a vector is always
calculated using the cosine function.
This misconception may be corrected by
using examples on the board in which
the angles are measured from the y-axis.
In the pyramid example, the horizontal and vertical parts that add up to
give the tourist’s actual displacement are called components. The x component is parallel to the x-axis. The y component is parallel to the y-axis. Any
vector can be completely described by a set of perpendicular components.
A truck carrying a film crew must be driven
at the correct velocity to enable the crew
to film the underside of a plane. The plane
flies at 95 km/h at an angle of 20° relative
to the ground.
v plane
This point is illustrated by examining a scene on the set of an action
movie. For this scene, a plane travels at 95 km/h at an angle of 20° relative
to the ground. Filming the plane from below, a camera team travels in a
truck directly beneath the plane at all times, as shown in Figure 2.6.
To find the velocity that the truck must maintain to stay beneath the
plane, we must know the horizontal component of the plane’s velocity.
Once more, the key to solving the problem is to recognize that a right
triangle can be drawn using the plane’s velocity and its x and y components. The situation can then be analyzed using trigonometry.
The sine and cosine functions are defined in terms of the lengths of
the sides of such right triangles. The sine of an angle is the ratio of the leg
opposite that angle to the hypotenuse.
20˚
vtruck
Definition of the Sine Function for Right Triangles
sin θ =
opp
_
hyp
sine of an angle =
opposite leg
__
hypotenuse
In Figure 2.7, the leg opposite the 20° angle represents the y component, vy, which describes the vertical speed of the airplane. The hypotenuse, vplane, is the resultant vector that describes the airplane’s total
velocity.
The cosine of an angle is the ratio between the leg adjacent to that
angle and the hypotenuse.
FIGURE 2.7
Using Vector Components
To stay beneath the biplane, the truck
must be driven with a velocity equal to the
x component (vx ) of the biplane’s velocity.
plane
= 95 km/h
20˚
Definition of the Cosine Function for Right Triangles
cos θ =
adj
_
hyp
cosine of an angle =
adjacent leg
__
hypotenuse
In Figure 2.7, the adjacent leg represents the x component, vx, which
describes the airplane’s horizontal speed. This x component equals the
speed required of the truck to remain beneath the plane. Thus, the truck
must maintain a speed of vx = (cos 20°)(95 km/h) = 90 km/h.
88
Chapter 3
Differentiated
Instruction
Below level
Students who need a refresher on trigonometry
can be directed to Appendix A, which includes
a more detailed discussion of the sine, cosine,
and tangent functions, as well as the
Pythagorean theorem.
Untitled-241 88
88
Chapter 3
5/4/2011 2:51:02 PM
ntitled-241 89
PREMIUM CONTENT
Interactive Demo
Resolving Vectors
Classroom Practice
HMDScience.com
Sample Problem B Find the components of the velocity of a
helicopter traveling 95 km/h at an angle of 35° to the ground.
ANALYZE
Given:
v = 95 km/h
Resolving Vectors An arrow is shot
from a bow at an angle of 25° above
the horizontal with an initial speed of
45 m/s. Find the horizontal and vertical
components of the arrow’s initial
velocity.
θ = 35°
Unknown:
vx = ?
Diagram:
The most convenient coordinate system
is one with the x-axis directed along the
ground and the y-axis directed vertically.
vy = ?
= 95 km/h
35˚
Answer: 41 m/s, 19 m/s
PLAN
Choose an equation or situation:
Because the axes are perpendicular, the sine and cosine functions can
be used to find the components.
vy
Tips and Tricks
sin θ = _
v
vx
cos θ = _
v
Rearrange the equations to isolate the unknowns:
vy = v sin θ
vx = v cos θ
SOLVE
Don’t assume that the cosine
function can always be used for
the x-component and the sine
function can always be used for
the y-component. The correct
choice of function depends on
where the given angle is located.
Instead, always check to see
which component is adjacent and
which component is opposite to
the given angle.
PROBLEM guide B
vy = 54 km/h
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = Sample Problem Set II (online)
Solving for
vx = (95 km/h) (cos 35°)
one
component
Substitute the values into the equations and solve:
vy = (95 km/h) (sin 35°)
vx = 78 km/h
CHECK YOUR
ANSWER
The arrow strikes the target with a
speed of 45 m/s at an angle of −25°
with respect to the horizontal. Calculate
the horizontal and vertical components
of the arrow’s final velocity.
Answer: 41 m/s, −19 m/s
SE 1–2
PW Sample, 1, 3
PB 1–4
v 2 = vx2 + vy2
SE
both
components
PW
PB
(95)2 = (78)2 + (54)2
*Challenging Problem
Because the components of the velocity form a right triangle with the
helicopter’s actual velocity, the components must satisfy the
Pythagorean theorem.
Sample, 3–4;
Ch. Rvw. 24–25
2–8
Sample, 5–10
9025 ≈ 9000
The slight difference is due to rounding.
Continued
Problem Solving
Two-Dimensional Motion and Vectors
89
Alternative Approach
For Step 4 (Check Your Answer) students can
apply the sine and cosine ratios as shown:
5/4/2011 2:51:02 PM
vy
54
sin θ = _
​ v ​⇒ sin 35° = _
​   ​⇒ 0.573 ≈ 0.568
95
v
78
​   ​⇒ 0.819 ≈ 0.821
cos θ = _
​ vx ​⇒ cos 35° = _
95
The slight difference is due to rounding.
Two-Dimensional Motion and Vectors 89
Resolving Vectors
Teach continued
(continued)
1. How fast must a truck travel to stay beneath an airplane that is moving 105 km/h
at an angle of 25° to the ground?
Answers
2. What is the magnitude of the vertical component of the velocity of the plane in item 1?
Practice B
1. 95 km/h
2. 44 km/h
3. 21 m/s, 5.7 m/s
4. 0 m, 5 m
3. A truck drives up a hill with a 15° incline. If the truck has a constant speed of
22 m/s, what are the horizontal and vertical components of the truck’s velocity?
4. What are the horizontal and vertical components of a cat’s displacement when the
cat has climbed 5 m directly up a tree?
Adding Vectors That Are Not Perpendicular
Teaching Tip
Until this point, the vector-addition problems concerned vectors that are
perpendicular to one another. However, many objects move in one
direction and then turn at an angle before continuing their motion.
Problems involving vectors that are not
perpendicular use both vector addition
and vector resolution. Because they act
as a nice summary to the concepts of
this section, you may want to do several
examples involving this type of problem.
These problems require a methodical
approach to problem solving, which
should prove helpful to students while
studying more difficult subjects, such as
inclined-plane problems and equilibrium
problems.
Suppose that a plane initially travels 5 km at an angle of 35° to the
ground, then climbs at only 10° relative to the ground for 22 km. How can
you determine the magnitude and direction for the vector denoting the
total displacement of the plane?
Because the original displacement vectors do not form a right triangle,
you can not apply the tangent function or the Pythagorean theorem when
adding the original two vectors.
Determining the magnitude and the direction of the resultant can be
achieved by resolving each of the plane’s displacement vectors into its x
and y components. Then the components along each axis can be added
together. As shown in Figure 2.8, these sums will be the two perpendicular
components of the resultant, d. The resultant’s magnitude can then be
found by using the Pythagorean theorem, and its direction can be found
by using the inverse tangent function.
FIGURE 2.8
Adding Vectors That Are Not Perpendicular Add the
components of the original displacement vectors to find two components
that form a right triangle with the resultant vector.
d2
d1
d
90
Chapter 3
Differentiated
Instruction
Below Level
Be sure students distinguish between the
components of d1 and d2 in Figure 2.8. Ask
students to draw each vector and its components separately. Then, have them sum the
components in each direction and add the
summed components together to find d.
Repeat with additional examples to help
prepare students for Sample Problem C,
Adding Vectors Algebraically.
Untitled-241 90
90 Chapter 3
5/4/2011 2:51:03 PM
ntitled-241 91
PREMIUM CONTENT
Interactive Demo
Adding Vectors Algebraically
Classroom Practice
HMDScience.com
Sample Problem C A hiker walks 27.0 km from her base camp
at 35° south of east. The next day, she walks 41.0 km in a direction
65° north of east and discovers a forest ranger’s tower. Find the
magnitude and direction of her resultant displacement between
the base camp and the tower.
ANALYZE
Tips and Tricks
θ1 is negative, because
clockwise angles from
the positive x-axis are
conventionally considered
to be negative.
PLAN
Select a coordinate system. Then sketch and label
each vector.
d1 = 27.0 km
θ1 = −35°
Given:
d2 = 41.0 km
Unknown:
d= ?
θ2 = 65°
Ranger’s tower
A plane flies 118 km at 15.0° south of east
and then flies 118 km at 35.0° west of
north. Find the magnitude and direction
of the total displacement of the plane.
Answer: 81 km at 55° north of east
Base
camp
θ= ?
Find the x and y components of all vectors.
Make a separate sketch of the displacements for each
day. Use the cosine and sine functions to find the
displacement components.
∆x
cos θ = _
d
∆y
sin θ = _
d
PROBLEM guide C
= -35°
= 27.0 km
(a)
(a) For day 1:
∆x1 = d1 cos θ1 = (27.0 km) [cos (−35°)] = 22 km
∆y1 = d1 sin θ1 = (27.0 km) [sin (−35°)] = −15 km
km
1.0
∆y2 = d2 sin θ2 = (41.0 km) (sin 65°) = 37 km
(b)
=4
(b) For day 2:
∆x2 = d2 cos θ2 = (41.0 km) (cos 65°) = 17 km
Find the x and y components of the total
displacement.
∆xtot = ∆x1 + ∆x2 = 22 km + 17 km = 39 km
Adding Vectors Algebraically A camper
walks 4.5 km at 45° north of east then
4.5 km due south. Find the camper’s total
displacement.
Answer: 3.4 km at 22° south of east
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = Sample Problem Set II (online)
Solving for
vector
sum
SE
Sample, 1–3, 4*;
Ch. Rvw. 26, 53*
PW Sample, 1–5
PB Sample, 1–10
*Challenging Problem
= 65°
∆ytot = ∆y1 + ∆y2 = –15 km + 37 km = 22 km
SOLVE
Use the Pythagorean theorem to find the magnitude of the
resultant vector.
d 2 = (∆xtot)2 + (∆ytot)2
�������
d = √(∆x
)2 + (∆y )2 =
tot
tot
(39 km)2 + (22 km)2 = 45 km
√��������
Use a suitable trigonometric function to find the angle.
(
)
(
)
∆ytot
22 km = 29° north of east
θ = tan–1 _
= tan–1 _
∆xtot
39 km
Continued
Problem Solving
Two-Dimensional Motion and Vectors
91
Alternative Approach
In Step 2(a), remind students that they can use
the following trigonometric formulas as an
alternative approach:
sin (-a) = -sin a
cos (-a) = cos a
5/4/2011 2:51:04 PM
Two-Dimensional Motion and Vectors
91
Teach continued
Answers
Practice C
1. 49 m at 7.3° to the right of downfield
2. 7.5 km at 26° above the horizontal
3. 13.0 m at 57° north of east
4. 171 km at 34° east of north
Assess and Reteach
Adding Vectors Algebraically
(continued)
1 A football player runs directly down the field for 35 m before turning to the right at
an angle of 25° from his original direction and running an additional 15 m before
getting tackled. What is the magnitude and direction of the runner’s total
displacement?
2. A plane travels 2.5 km at an angle of 35° to the ground and then changes direction
and travels 5.2 km at an angle of 22° to the ground. What is the magnitude and
direction of the plane’s total displacement?
3. During a rodeo, a clown runs 8.0 m north, turns 55° north of east, and runs
3.5 m. Then, after waiting for the bull to come near, the clown turns due east and
runs 5.0 m to exit the arena. What is the clown’s total displacement?
4. An airplane flying parallel to the ground undergoes two consecutive
displacements. The first is 75 km 30.0° west of north, and the second is 155 km
60.0° east of north. What is the total displacement of the airplane?
Assess Use the Formative Assessment
on this page to evaluate student
mastery of the section.
Reteach For students who need
additional instruction, download the
Section Study Guide.
Response to Intervention To reassess
students’ mastery, use the Section Quiz,
available to print or to take directly
online at HMDScience.com.
SECTION 2 FORMATIVE ASSESSMENT
Reviewing Main Ideas
1. Identify a convenient coordinate system for analyzing each of the
following situations:
a. a dog walking along a sidewalk
b. an acrobat walking along a high wire
c. a submarine submerging at an angle of 30° to the horizontal
2. Find the magnitude and direction of the resultant velocity vector for the
following perpendicular velocities:
a. a fish swimming at 3.0 m/s relative to the water across a river that
moves at 5.0 m/s
b. a surfer traveling at 1.0 m/s relative to the water across a wave that is
traveling at 6.0 m/s
3. Find the vector components along the directions noted in parentheses.
a. a car displaced 45° north of east by 10.0 km (north and east)
b. a duck accelerating away from a hunter at 2.0 m/s2 at an angle of 35° to
the ground (horizontal and vertical)
Critical Thinking
4. Why do nonperpendicular vectors need to be resolved into components
before you can add the vectors together?
92
Chapter 3
Answers
to Section Assessment
1. a. x-axis: forward and backward on
Untitled-241 92 sidewalk
y-axis: left and right on sidewalk
b. x-axis: forward and backward on rope
y-axis: up and down
c. x-axis: horizontal at water level
y-axis: up and down
2. a. 5.8 m/s at 59° downriver from its
intended path
b. 6.1 m/s at 9.5° from the direction the
wave is traveling
92
Chapter 3
3. a. 7.07 km north, 7.07 km east
b. 1.6 m/s2 horizontal, 1.1 m/s2 vertical
4. because the Pythagorean theorem and the
tangent function can be applied only to
right triangles
5/4/2011 2:51:05 PM