copyright © by holt, rinehart and winston. all rights reserved....and ,,, = = )of
TRANSCRIPT
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
PHYSICS IN ACTION
Although the streams of water of the Place
de la Concorde fountain in Paris, France,
seem to flow through transparent pipes,
they are actually fired in highly concentrated
solitary streams. Each of the streams fol-
lows a parabolic path, just like the path a
baseball follows when it is thrown through
the air. In this chapter you will analyze two-
dimensional motion and solve problems
involving objects projected into the air.
• Why does all of the water in a stream followthe same path?
• How does the angle at which the water isfired affect its path?
CONCEPT REVIEW
Displacement (Section 2-1)
Velocity (Section 2-1)
Acceleration (Section 2-2)
Free fall (Section 2-3)
CHAPTER 3
Two-DimensionalMotion andVectors
83Two-Dimensional Motion and VectorsCopyright © by Holt, Rinehart and Winston. All rights reserved.
Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 384
SCALARS AND VECTORS
In Chapter 2 our discussion of motion was limited to two directions, forward
and backward. Mathematically, we described these directions of motion with
a positive or negative sign. This chapter explains a method of describing the
motion of objects that do not travel along a straight line.
Vectors indicate direction; scalars do not
Each of the physical quantities we will encounter in this book can be catego-
rized as either a scalar or a vector quantity. A scalar is a quantity that can be
completely specified by its magnitude with appropriate units; that is, a scalar
has magnitude but no direction. Examples of scalar quantities are speed, vol-
ume, and the number of pages in this textbook. A vector is a physical quantity
that has both direction and magnitude.
As we look back to Chapter 2, we can see that displacement is an example
of a vector quantity. An airline pilot planning a trip must know exactly how
far and which way to fly. Velocity is also a vector quantity. If we wish to
describe the velocity of a bird, we must specify both its speed (say, 3.5 m/s)
and the direction in which the bird is flying (say, northeast). Another example
of a vector quantity is acceleration, which was also discussed in Chapter 2.
Vectors are represented by symbols
In physics, quantities are often represented by symbols, such as t for
time. To help you keep track of which symbols represent vector quan-
tities and which are used to indicate scalar quantities, this book will
use boldface type to indicate vector quantities. Scalar quantities are
designated by the use of italics. Thus, to describe the speed of a bird
without giving its direction, you would write v = 3.5 m/s. But a veloc-
ity, which includes a direction, is written as v = 3.5 m/s to the north-
east. Handwritten, a vector can be symbolized by showing an arrow
drawn above the abbreviation for a quantity, such as v�� = 3.5 m/s to
the northeast.
One way to keep track of vectors and their directions is to use
diagrams. In diagrams, vectors are shown as arrows that point in the
direction of the vector. The length of a vector arrow in a diagram is
related to the vector’s magnitude. For example, in Figure 3-1 the
arrows represent the velocities of the two soccer players running
toward the soccer ball.
3-1Introduction to vectors
3-1 SECTION OBJECTIVES
• Distinguish between a scalarand a vector.
• Add and subtract vectorsusing the graphical method.
• Multiply and divide vectorsby scalars.
Figure 3-1The lengths of the vector arrows represent the magnitudes of these two soccer players’ velocities.
scalar
a physical quantity that has onlya magnitude but no direction
vector
a physical quantity that has botha magnitude and a direction
Copyright © by Holt, Rinehart and Winston. All rights reserved.
The word vector is also used by air-line pilots and navigators. In thiscontext, a vector is the particularpath followed or to be followed,given as a compass heading.
85Two-Dimensional Motion and Vectors
Because the player on the right of the soccer ball is moving faster, the arrow
representing that velocity is drawn longer than the arrow representing the
velocity of the player on the left.
Vectors can be added graphically
When adding vectors, you must make certain that they have the same units
and describe similar quantities. For example, it would be meaningless to add a
velocity vector to a displacement vector because they describe different physi-
cal quantities. Similarly, it would be meaningless, as well as incorrect, to add
two displacement vectors that are not expressed in the same units (in meters,
in feet, and so on).
Section 2-1 covered vector addition and subtraction in one dimension.
Think back to the example of the gecko that ran up a tree from a 20 cm mark-
er to an 80 cm marker. Then the gecko reversed direction and ran back to the
50 cm marker. Because the two parts of this displacement are opposite, they
can be added together to give a total displacement of 30 cm. The answer found
by adding two vectors in this way is called the resultant.Consider a student walking to school. The student walks 1600 m to a
friend’s house, then 1600 m to the school, as shown in Figure 3-2. The stu-
dent’s total displacement during his walk to school is in a direction from his
house to the school, as shown by the dotted line. This direct path is the vector
sum of the student’s displacement from his house to his friend’s house and his
displacement from the friend’s house to school. How can this resultant dis-
placement be found?
One way to find the magnitude and direction of the
student’s total displacement is to draw the situation to
scale on paper. Use a reasonable scale, such as 50 m on
land equals 1 cm on paper. First draw the vector repre-
senting the student’s displacement from his house to his
friend’s house, giving the proper direction and scaled
magnitude. Then draw the vector representing his walk
to the school, starting with the tail at the head of the first
vector. Again give its scaled magnitude and the right
direction. The magnitude of the resultant vector can
then be determined by using a ruler to measure the
length of the vector pointing from the tail of the first
vector to the head of the second vector. The length of
that vector can then be multiplied by 50 (or whatever
scale you have chosen) to get the actual magnitude of
the student’s total displacement in meters.
The direction of the resultant vector may be deter-
mined by using a protractor to measure the angle
between the first vector and the resultant.
(b)
(a)
(c)
Figure 3-2A student walks from his house to his friend’s house (a), thenfrom his friend’s house to the school (b).The student’s resultantdisplacement (c) can be found using a ruler and a protractor.
resultant
a vector representing the sum oftwo or more vectors
Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 386
PROPERTIES OF VECTORS
Now consider a case in which two or more vectors act at the same point.
When this occurs, it is possible to find a resultant vector that has the same net
effect as the combination of the individual vectors. Imagine looking down
from the second level of an airport at a toy car moving at 0.80 m/s across a
moving walkway that moves at 1.5 m/s, as graphically represented in Figure3-3. How can you determine what the car’s resultant velocity will look like
from your vantage point?
Vectors can be moved parallel to themselves in a diagram
Note that the car’s resultant velocity while moving from one side of the walk-
way to the other will be the combination of two independent velocities. Thus,
the car traveling for t seconds can be thought of as traveling first at 0.80 m/s
across the walkway for t seconds, then down the walkway at 1.5 m/s for t sec-
onds. In this way, we can draw a given vector anywhere in the diagram as long
as the vector is parallel to its previous alignment and still points in the same
direction. Thus, you can draw one vector with its tail starting at the tip of the
other as long as the size and direction of each vector do not change.
Determining a resultant vector by drawing a vector from the tail of the first
vector to the tip of the last vector is known as the triangle method of addition.
Again, the magnitude of the resultant vector can be measured using a ruler,
and the angle can be measured with a protractor. In the next section, we will
develop a technique for adding vectors that is less time-consuming because it
involves a calculator instead of a ruler and protractor.
Vectors can be added in any order
When two or more vectors are added, the sum is independent of the order of
the addition. This idea is demonstrated by a runner practicing for a marathon
along city streets, as represented in Figure 3-4. In (a) the runner takes one
path during a run, and in (b) the runner takes another. Regardless of which
path the runner takes, the runner will have the same total displacement,
expressed as d. Similarly, the vector sum of two or more vectors is the same
regardless of the order in which the vectors are added, provided that the mag-
nitude and direction of each vector remain the same.
To subtract a vector, add its opposite
Vector subtraction makes use of the definition of the negative of a vector. The
negative of a vector is defined as a vector with the same magnitude as the origi-
nal vector but opposite in direction. For instance, the negative of the velocity
of a car traveling 30 m/s to the west is −30 m/s to the west, or 30 m/s to the
east. Thus, as shown below, adding a vector to its negative vector gives zero.
v + (−v) = 30 m/s + (−30 m/s) = 30 m/s − 30 m/s = 0 m/s
Car
(c)
(a)(b)
vwalkway = 1.5 m/s
vresultant
v car
= 0
.80
m/s
Figure 3-3The resultant velocity (a) of a toycar moving at a velocity of 0.80 m/s(b) across a moving walkway with avelocity of 1.5 m/s (c) can be foundusing a ruler and a protractor.
Figure 3-4A marathon runner’s displacement,d, will be the same regardless of whether the runner takes path(a) or (b).
(a)
d
(b)
d
TOPIC: VectorsGO TO: www.scilinks.orgsciLINKS CODE: HF2031
NSTA
Copyright © by Holt, Rinehart and Winston. All rights reserved.87Two-Dimensional Motion and Vectors
When adding vectors in two dimensions, you can add a negative vector to a
positive vector that does not point along the same line by using the triangle
method of addition.
Multiplying or dividing vectors by scalars results in vectors
There are mathematical operations in which vectors can multiply other vec-
tors, but they are not necessary for the scope of this book. This book does,
however, make use of vectors multiplied by scalars, with a vector as the result.
For example, if a cab driver obeys a customer who tells him to go twice as fast,
that cab’s original velocity vector, vcab, is multiplied by the scalar number 2.
The result, written 2vcab, is a vector with a magnitude twice that of the origi-
nal vector and pointing in the same direction.
On the other hand, if another cab driver is told to go twice as fast in the
opposite direction, this is the same as multiplying by the scalar number −2.
The result is a vector with a magnitude two times the initial velocity but
pointing in the opposite direction, written as −2vcab.
Section Review
1. Which of the following quantities are scalars, and which are vectors?
a. the acceleration of a plane as it takes off
b. the number of passengers on the plane
c. the duration of the flight
d. the displacement of the flight
e. the amount of fuel required for the flight
2. A roller coaster moves 85 m horizontally, then travels 45 m at an angle of
30.0° above the horizontal. What is its displacement from its starting
point? Use graphical techniques.
3. A novice pilot sets a plane’s controls, thinking the plane will fly at
2.50 × 102 km/h to the north. If the wind blows at 75 km/h toward the
southeast, what is the plane’s resultant velocity? Use graphical techniques.
4. While flying over the Grand Canyon, the pilot slows the plane’s engines
down to one-half the velocity in item 3. If the wind’s velocity is still
75 km/h toward the southeast, what will the plane’s new resultant velo-
city be? Use graphical techniques.
5. Physics in Action The water used in many fountains is recycled.
For instance, a single water particle in a fountain travels through an 85 m
system and then returns to the same point. What is the displacement of a
water particle during one cycle?
Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 388
3-2Vector operations
COORDINATE SYSTEMS IN TWO DIMENSIONS
In Chapter 2, the motion of a gecko climbing a tree was described as motion
along the y-axis. The direction of the displacement of the gecko along the axis
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 3-5.A more versatile system for diagraming the motion of an object, however,
employs vectors and the use of both the x- and y-axes simultaneously.
The addition of another axis not only helps describe motion in two dimen-
sions but also 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 3-6(a). The jet’s
motion also can be depicted on a two-dimensional coordinate system whose
axes point south to north and west to east, as shown in Figure 3-6(b).The problem with the orientation in the first case is that the axis must be
turned again if the direction of the plane changes. Also, it will become difficult
to describe the direction of another plane that is not traveling exactly north-
east. Thus, axes are often designated with the positive y-axis pointing north
and the positive x-axis pointing east, as shown in Figure 3-6(b).Similarly, when analyzing the motion of objects thrown into the air, orient-
ing the y-axis perpendicular to the ground, and therefore parallel to the direc-
tion of free-fall acceleration, greatly simplifies things.
3-2 SECTION OBJECTIVES
• Identify appropriate coordi-nate systems for solvingproblems with vectors.
• Apply the Pythagorean theorem and tangent func-tion to calculate the magni-tude and direction of aresultant vector.
• Resolve vectors into compo-nents using the sine andcosine functions.
• Add vectors that are notperpendicular.
∆y
y
Figure 3-5A gecko’s displacement while climb-ing a tree can be represented by anarrow pointing along the y-axis.
v = 300 m/s northeast
yN
S
W E
(a)
Figure 3-6A plane traveling northeast at a velocity of 300 m/s can be repre-sented as either (a) moving along a y-axis chosen to point to thenortheast or (b) moving at an angle of 45° to both the x- and y-axes, which line up with west-east and north-south, respectively.
v = 300 m/s northeast
y
(b)
x
Copyright © by Holt, Rinehart and Winston. All rights reserved.89Two-Dimensional Motion and Vectors
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 the best choice for orienting axes
is the approach that makes solving the problem easiest.
DETERMINING RESULTANT MAGNITUDE AND DIRECTION
In Section 3-1, the magnitude and direction of a resultant were found graphi-
cally by making a drawing. There are, however, drawbacks to this approach: it
is time-consuming, and the accuracy of the answer depends on how carefully
the diagram is drawn and measured. There is a better, simpler method using
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.
As can be seen in Figure 3-7, the magnitude of the tourist’s vertical dis-
placement, ∆y, is the height of the pyramid, and the magnitude of the hori-
zontal 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 per-
pendicular and form a right triangle with the displacement, d.As shown in Figure 3-8(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.
In Figure 3-8(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 displace-
ment. In this way, you can find out the magnitude of the displacement, d.
PYTHAGOREAN THEOREM FOR RIGHT TRIANGLES
c2 = a2 + b2
(length of hypotenuse)2 = (length of one leg)2 + (length of other leg)2
d
2∆x
∆y
∆x
Figure 3-7Because the base and height of apyramid are perpendicular, we canfind a tourist’s total displacement,d, if we know the height, ∆y, andwidth, 2∆x, of the pyramid.
∆x
∆ydc
c2 = a2 + b2 d2 = ∆x2 + ∆y2b
a
(a) (b)
Figure 3-8(a) The Pythagorean theoremcan be applied to any right tri-angle. (b) It can also be appliedto find the magnitude of a resul-tant displacement.
Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 390
Use the tangent function to find the direction of the resultant
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 trian-
gle, as shown in Figure 3-9(b), the tangent function can be used to find the
angle q, which denotes the direction of the tourist’s displacement.
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 tri-
angle, as shown in Figure 3-9(a).As shown below, the quantity of the opposite leg divided by the magnitude
of the adjacent leg equals the tangent of the angle.
The inverse of the tangent function, which is shown below, indicates the
angle.
q = tan−1�o
a
p
d
p
j�
DEFINITION OF THE TANGENT FUNCTION FOR RIGHT TRIANGLES
tan q = o
a
p
d
p
j tangent of angle =
o
ad
p
j
p
a
o
c
s
e
i
n
te
t l
l
e
e
g
g
θ
HypotenuseOpposite
Adjacent(a)
(b)
θtan = opp
adj
θtan = ∆y∆x
θ tan−1= ∆y
∆x( )
θ
∆x
∆yd
Figure 3-9(a) The tangent function can beapplied to any right triangle, and (b) it can also be used to find thedirection of a resultant displacement.
SAMPLE PROBLEM 3A
Finding resultant magnitude and direction
P R O B L E MAn archaeologist climbs the Great Pyramid in Giza,Egypt. If the pyramid’s height is 136 m and its width is2.30 � 102 m, what is the magnitude and the directionof the archaeologist’s displacement while climbingfrom the bottom of the pyramid to the top?
S O L U T I O NGiven: ∆y = 136 m ∆x = 1
2(width) = 115 m
Unknown: d = ? q = ?
Diagram: Choose the archaeologist’s starting
position as the origin of the coordinate system.
Choose an equation(s) 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.
d2 = ∆x2 + ∆y2
tan q = ∆∆
x
y
1. DEFINE
2. PLAN
∆y =136 m
∆x = 115 m
d
x
y
θ
Copyright © by Holt, Rinehart and Winston. All rights reserved.91Two-Dimensional Motion and Vectors
Rearrange the equation(s) to isolate the unknown(s):
d = �∆�x�2�+� ∆�y�2�
q = tan−1�∆∆
x
y�
Substitute the values into the equation(s) and solve:
d = �(1�15� m�)2� +� (�13�6�m�)2�
q = tan−1�113
1
6
5
m
m�
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.
q = 49.8°
d = 178 m
3. CALCULATE
4. EVALUATE
CALCULATOR SOLUTION
Be sure your calculator is set to cal-culate angles measured in degrees.Most calculators have a buttonlabeled “DRG” that, when pressed,toggles between degrees, radians,and grads.
1. A truck driver attempting to deliver some furniture travels 8 km east,
turns around and travels 3 km west, and then 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, then turns and walks 7.5 m east. What single straight-line dis-
placement could the pirate have taken to reach the treasure?
3. Emily passes a soccer ball 6.0 m directly across the field to Kara, who
then kicks the ball 14.5 m directly down the field to Luisa. What is the
ball’s total displacement as it travels between Emily and Luisa?
4. A hummingbird flies 1.2 m along a straight path at a height of 3.4 m
above the ground. 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?
PRACTICE 3A
Finding resultant magnitude and direction
Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 392
RESOLVING VECTORS INTO COMPONENTS
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. These compo-
nents can be either positive or negative numbers with units.
Any vector can be completely described by a set of perpendicular compo-
nents. When a vector points along a single axis, as do the quantities in Chapter
2, the second component of the vector is equal to zero.
By breaking a single vector into two components, or resolving it into its
components, an object’s motion can sometimes be described more conve-
niently in terms of directions, such as north to south or east to west.
To illustrate this point, let’s examine a scene on the set of a new action movie.
For this scene a biplane travels at 95 km/h at an angle of 20° relative to the
ground. Attempting to film the plane from below, a camera team travels in a
truck, keeping the truck beneath the plane at all times, as shown in Figure 3-10.How fast must the truck travel to remain directly below the plane?
To find out 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 oppo-
site that angle to the hypotenuse.
In Figure 3-11, 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 motion.
The cosine of an angle is the ratio between the leg adjacent to that angle
and the hypotenuse.
In Figure 3-11, the leg adjacent to the 20° angle represents the x component,
vx, which describes the horizontal speed of the airplane. This x component
equals the speed that the truck must maintain to stay beneath the plane.
DEFINITION OF THE COSINE FUNCTION FOR RIGHT TRIANGLES
cos q = h
a
y
d
p
j cosine of an angle =
a
h
d
y
j
p
a
o
c
t
e
e
n
n
t
u
le
se
g
DEFINITION OF THE SINE FUNCTION FOR RIGHT TRIANGLES
sin q = o
h
p
yp
p sine of an angle =
o
h
p
y
p
p
o
o
s
te
it
n
e
u
le
se
g
20˚
vplane
v truck
Figure 3-10A truck carrying a film crew mustbe driven at the correct velocity toenable the crew to film the under-side of a biplane flying at an angle of 20° to the ground at a speed of95 km/h.
vplane = 95 km/h
vx
vy20°
Figure 3-11To stay beneath the biplane, thetruck must be driven with a velocityequal to the x component (vx) ofthe biplane’s velocity.
components of a vector
the projections of a vector alongthe axes of a coordinate system
Copyright © by Holt, Rinehart and Winston. All rights reserved.93Two-Dimensional Motion and Vectors
SAMPLE PROBLEM 3B
Resolving vectors
P R O B L E MFind the component velocities of a helicopter traveling 95 km/h at anangle of 35° to the ground.
S O L U T I O NGiven: v = 95 km/h q = 35°
Unknown: vx = ? vy = ?
Diagram: The most convenient coordinate system is one
with the x-axis directed along the ground and
the y-axis directed vertically.
Choose an equation(s) or situation: Because the axes are
perpendicular, the sine and cosine functions can be used to find
the components.
sin q = v
v
y
cos q = v
v
x
Rearrange the equation(s) to isolate the unknown(s):
vy = v(sin q)
vx = v(cos q)
Substitute the values into the equation(s) and solve:
vy = (95 km/h)(sin 35°)
vx = (95 km/h)(cos 35°)
Because the component velocities form a right tri-
angle with the helicopter’s actual velocity, the
Pythagorean theorem can be used to check whether
the magnitudes of the components are correct.
v2 = vx2 + vy
2
(95)2 ≈ (78)2 + (54)2
9025 ≈ 9000
The slight difference is due to rounding.
vx = 78 km/h
vy = 54 km/h
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
x
y v = 95 km/h
vx
vy
35°
CALCULATOR SOLUTION
When using your calculator to solve aproblem, perform trigonometric func-tions such as sin, cos, and tan first,before multiplication. This approachwill help you maintain the proper num-ber of significant digits.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ADDING VECTORS THAT ARE NOTPERPENDICULAR
Until this point, the vector-addition problems concerned vectors that are per-
pendicular to one another. However, many objects move in one direction, and
then turn at an acute angle before continuing their motion.
Suppose that a plane initially travels 50 km at an angle of 35° to the
ground, then climbs at only 10° to the ground for 220 km. How can you deter-
mine the magnitude and direction for the vector denoting the total displace-
ment of the plane?
Because the original displacement vectors do not form a right triangle, it is
not possible to directly apply the tangent function or the Pythagorean theo-
rem 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 their x and
Chapter 394
PRACTICE 3B
Resolving vectors
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?
2. What is the magnitude of the vertical component of the velocity of the
plane in item 1?
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 it has climbed 5 m directly up a tree?
5. Find the horizontal and vertical components of the 125 m displacement
of a superhero who flies down from the top of a tall building at an angle
of 25° below the horizontal.
6. A child rides a toboggan down a hill that descends at an angle of 30.5° to
the horizontal. If the hill is 23.0 m long, what are the horizontal and ver-
tical components of the child’s displacement?
7. A skier squats low and races down an 18° ski slope. During a 5 s interval,
the skier accelerates at 2.5 m/s2. What are the horizontal (perpendicular
to the direction of free-fall acceleration) and vertical components of the
skier’s acceleration during this time interval?
Copyright © by Holt, Rinehart and Winston. All rights reserved.95Two-Dimensional Motion and Vectors
y components. Then the components along each axis can be added together.
As shown in Figure 3-12, these vector sums will be the two perpendicular
components of the resultant, d. The magnitude of the resultant can then be
found using the Pythagorean theorem, and its direction can be found using
the tangent function.
∆x2
d2
d1 d
∆x1
∆y1
∆y2 Figure 3-12Add the components of the original dis-placement vectors to find two componentsthat form a right triangle with the resul-tant vector.
SAMPLE PROBLEM 3C
Adding vectors algebraically
P R O B L E MA hiker walks 25.5 km from her base camp at 35° south of east. On the sec-ond day, she walks 41.0 km in a direction 65° north of east, at which pointshe discovers a forest ranger’s tower. Determine the magnitude and direc-tion of her resultant displacement between the base camp and theranger’s tower.
S O L U T I O NSelect a coordinate system, draw a sketch of the vectors tobe added, and label each vector.
Figure 3-13 depicts the situation drawn on a coordinate sys-
tem. The positive y-axis points north and the positive x-axis
points east. The origin of the axes is the base camp. In the
chosen coordinate system, the hiker’s direction q1 during
the first day is signified by a negative angle because clock-
wise movement from the positive x-axis is conventionally
considered to be a negative angle.
Given: q1 = −35° q2 = 65° d1 = 25.5 km d2 = 41.0 km
Unknown: d = ? q = ?
Find the x and y components of all vectors.Make a separate sketch of the displacements for each day. The values for
each of the displacement components can be determined by using the sine
and cosine functions. Because the hiker’s angle on the first day is negative,
the y component of her displacement during that day is negative.
sin q = ∆d
y
cos q = ∆d
x
1.
2.
continued onnext page
d1
d2
d
Ranger’s tower
Basecamp
x
y
2θ1θθ
Figure 3-13IN
TERACTIVE
•
T U T O RPHYSICSPHYSICS
Module 2 “Vector Additionand Resolution” provides aninteractive lesson with guidedproblem-solving practice toteach you how to add differentvectors, especially those thatare not at right angles.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d1 = 25.5 km
∆x1
∆y1
x
y
1 = −35°θ
∆y2
∆x2x
y
2 = 65°θ
d 2 =
41.
0 km
Figure 3-15
Figure 3-14
For day 1: ∆x1 = d1 (cos q1) = (25.5 km) [cos (−35°)](Figure 3-14)
∆x1 = 21 km
∆y1 = d1 (sin q1) = (25.5 km) [sin (−35°)]
∆y1 = −15 km
For day 2: ∆x2 = d2 (cos q2) = (41.0 km) (cos 65°)(Figure 3-15)
∆x2 = 17 km
∆y2 = d2 (sin q2) = (41.0 km) (sin 65°)
∆y2 = 37 km
Find the x and y components of the total displacement.First add together the x components to find the total displace-
ment in the x direction. Then perform the same operation for
the y direction.
∆xtot = ∆x1 + ∆x2 = 21 km + 17 km = 38 km
∆ytot = ∆y1 + ∆y2 = −15 km + 37 km = 22 km
Use the Pythagorean theorem to find the magnitude of the resultant vector.Because the components ∆xtot and ∆ytot are perpendicular, the Pythagore-
an theorem can be used to find the magnitude of the resultant vector.
d2 = (∆xtot)2 + (∆ytot)
2
d = �(∆�xt�ot�)2� +� (�∆�yt�ot�)2� = �(3�8�km�)2� +� (�22� k�m�)2�
Use a suitable trigonometric function to find the angle the resultant vec-tor makes with the x-axis.
The direction of the resultant can be found using the tangent function.
q = tan−1�∆∆
x
y�
q = tan−1�∆∆x
yt
t
o
o
t
t� = tan−1�232
8
k
k
m
m�
Evaluate your answer.If the diagram is drawn to scale, compare the algebraic results with the
drawing. The calculated magnitude seems reasonable because the distance
from the base camp to the ranger’s tower is longer than the distance hiked
during the first day and slightly longer than the distance hiked during the
second day. The calculated direction of the resultant seems reasonable
because the angle in Figure 3-13 looks to be about 30°.
q = (3.0 × 101)° north of east
d = 44 km
5.
3.
4.
6.
96 Chapter 3
Copyright © by Holt, Rinehart and Winston. All rights reserved.97Two-Dimensional Motion and Vectors
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 direc-
tion of the runner’s total displacement?
2. A plane travels 2.5 km at an angle of 35° to the ground, 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 35° east of north, 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 dis-
placements. 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?
PRACTICE 3C
Adding vectors algebraically
Section Review
1. Identify a convenient coordinate system for analyzing each of the follow-
ing 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 component vectors along the directions noted in parentheses.
a. a car displaced northeast 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)
4. Find the resultant displacement of a fox that heads 55° north of west for
10.0 m, then turns and heads west for 5.0 m.
Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 398
3-3Projectile motion
TWO-DIMENSIONAL MOTION
In the last section, quantities such as displacement and velocity were shown to
be vectors that can be resolved into components. In this section, these compo-
nents will be used to understand and predict the motion of objects thrown
into the air.
Use of components avoids vector multiplication
How can you know the displacement, velocity, and acceleration of a ball at any
point in time during its flight? All of the kinematic equations from Chapter 2
could be rewritten in terms of vector quantities. However, when an object is
propelled into the air in a direction other than straight up or down, the veloc-
ity, acceleration, and displacement of the object do not all point in the same
direction. This makes the vector forms of the equations difficult to solve.
One way to deal with these situations is to avoid using the complicated vec-
tor forms of the equations altogether. Instead, apply the technique of resolving
vectors into components. Then you can apply the simpler one-dimensional
forms of the equations for each component. Finally, you can recombine the
components to determine the resultant.
Components simplify projectile motion
When a long jumper approaches his jump, he runs along a straight line, which
can be called the x-axis. When he jumps, as shown in Figure 3-16, his velocity
has both horizontal and vertical components. Movement in this plane can be
depicted using both the x- and y-axes.
Note that in Figure 3-17(b) the jumper’s velocity vector is resolved into its
two component vectors. This way, the jumper’s motion can be analyzed using
the kinematic equations applied to one direction at a time.
3-3 SECTION OBJECTIVES
• Recognize examples of pro-jectile motion.
• Describe the path of a pro-jectile as a parabola.
• Resolve vectors into theircomponents and apply thekinematic equations to solveproblems involving projectilemotion.
v
(a)
Figure 3-17(a) A long jumper’s velocity whilesprinting along the runway can berepresented by a single vector.(b) Once the jumper is airborne,the jumper’s velocity at any instantcan be described by the compo-nents of the velocity.
Figure 3-16When the long jumper is in the air,his velocity has both a horizontaland a vertical component.
vx
vy
(b)
Copyright © by Holt, Rinehart and Winston. All rights reserved.99Two-Dimensional Motion and Vectors
In this section, we will focus on the form of two-dimensional motion called
projectile motion. Objects that are thrown or launched into the air and are
subject to gravity are called projectiles. Some examples of projectiles are soft-
balls, footballs, and arrows when they are projected through the air. Even a long
jumper can be considered a projectile.
Projectiles follow parabolic trajectories
The path of a projectile is a curve called a parabola, as shown in Figure 3-18(a). Many people mistakenly believe that projectiles eventually fall straight
down, much like a cartoon character does after running off a cliff. How-
ever, if an object has an initial horizontal velocity in any given time interval,
there will be horizontal motion throughout the flight of the projectile. Note that
for the purposes of samples and exercises in this book, the horizontal velocity of
the projectile will be considered constant. This velocity would not be constant if
we accounted for air resistance. With air resistance, a projectile slows down as it
collides with air particles. Hence, as shown in Figure 3-18(b), the true path of a
projectile traveling through Earth’s atmosphere is not a parabola.
Projectile motion is free fall with an initial horizontal velocity
To understand the motion a projectile undergoes, first examine Figure 3-19.The red ball was dropped at the same instant the yellow ball was launched
horizontally. If air resistance is disregarded, both balls hit the ground at the
same time.
By examining each ball’s position in relation to the horizontal lines and to
one another, we see that the two balls fall at the same rate. This may seem
impossible because one is given an initial velocity and the other begins from
rest. But if the motion is analyzed one component at a time, it makes sense.
First, consider the red ball that falls straight down. It has no motion in the
horizontal direction. In the vertical direction, it starts from rest (vy,i = 0 m/s)
and proceeds in free fall. Thus, the kinematic equations from Chapter 2 can be
applied to analyze the vertical motion of the falling ball. Note that the acceler-
ation, a, can be rewritten as −g because the only vertical component of accel-
eration is free-fall acceleration. Note also that ∆y is negative.
projectile motion
free-fall with an initial horizontalvelocity
Path with air resistance
Path without air resistance
(a)(b)
Figure 3-18(a) Without air resistance, a soccerball being headed into the air wouldbe represented as traveling along aparabola. (b) With air resistance,the soccer ball would travel along ashorter path, which would not be aparabola.
Figure 3-19This is a strobe photograph of twotable-tennis balls released at thesame time. Even though the yellowball is given an initial horizontal veloc-ity and the red ball is simply dropped,both balls fall at the same rate.
TOPIC: Projectile motionGO TO: www.scilinks.orgsciLINKS CODE: HF2032
NSTA
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The greatest distance a regulation-size baseball has ever been thrownis 135.9 m, by Glen Gorbous in 1957.
Chapter 3100
Now consider the components of motion of the yellow ball that is launched
in Figure 3-19. This ball undergoes the same horizontal displacement during
each time interval. This means that the ball’s horizontal velocity remains con-
stant (if air resistance is assumed to be negligible). Thus, when using the kine-
matic equations from Chapter 2 to analyze the horizontal motion of a
projectile, the initial horizontal velocity is equal to the horizontal velocity
throughout the projectile’s flight. A projectile’s horizontal motion is described
by the following equation.
Next consider the initial motion of the launched yellow ball in Figure 3-19.Despite having an initial horizontal velocity, the launched ball has no initial
velocity in the vertical direction. Just like the red ball that falls straight down,
the launched yellow ball is in free fall. Its vertical motion is described by the
same free-fall equations. In any time interval, the launched ball undergoes the
same vertical displacement as the ball that falls straight down. This is why
both balls reach the ground at the same time.
To find the velocity of a projectile at any point during its flight, find the
vector sum of the components of the velocity at that point. Use the Pythago-
rean theorem to find the magnitude of the velocity, and use the tangent func-
tion to find the direction of the velocity.
HORIZONTAL MOTION OF A PROJECTILE
vx = vx,i = constant
∆x = vx ∆t
VERTICAL MOTION OF A PROJECTILE THAT FALLS FROM REST
vy,f = −g∆t
vy,f2 = −2g∆y
∆y = − 12
g(∆t)2
Projectile Motion
M A T E R I A L S L I S T
✔ 2 identical balls
✔ slope or ramp
SAFETY CAUTION
Perform this experiment away from wallsand furniture that can be damaged.
Roll a ball off a table. At the instant therolling ball leaves the table, drop a secondball from the same height above the floor.Do the two balls hit the floor at the same
time? Try varying the speed at which youroll the first ball off the table. Does vary-ing the speed affect whether the two ballsstrike the ground at the same time? Nextroll one of the balls down a slope. Dropthe other ball from the base of the slopeat the instant the first ball leaves the slope.Which of the balls hits the ground first inthis situation?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Copyright © by Holt, Rinehart and Winston. All rights reserved.101Two-Dimensional Motion and Vectors
SAMPLE PROBLEM 3D
Projectiles launched horizontally
P R O B L E MThe Royal Gorge Bridge in Colorado rises 321 m above the Arkansas River.Suppose you kick a little rock horizontally off the bridge. The rock hits thewater such that the magnitude of its horizontal displacement is 45.0 m.Find the speed at which the rock was kicked.
S O L U T I O NGiven: ∆y = −321 m ∆x = 45.0 m ay = g = 9.81 m/s2
Unknown: vi = ?
Diagram: The initial velocity vector of the
rock has only a horizontal compo-
nent. Choose the coordinate system
oriented so that the positive y direc-
tion points upward and the positive
x direction points to the right.
Choose the equation(s) or situation:Because air resistance can be neglected, the rock’s
horizontal velocity remains constant.
∆x = vx∆t
Because there is no initial vertical velocity, the following equation applies.
∆y = − 12
g(∆t)2
Rearrange the equation(s) to isolate the unknown(s):Note that the time interval is the same for the vertical and horizontal dis-
placements, so the second equation can be rearranged to solve for ∆t.
∆t =�2
−∆�g
y� where ∆y is negative
vx = ∆∆
x
t = ��
2
−∆�g
y��∆x
Substitute the values into the equation(s) and solve:The value for vx can be either positive or negative because of the square root.
Because the direction was not asked for, use the positive root for v.
vx = �� (45.0 m) =
To check your work, estimate the value of the time interval for ∆x and solve
for ∆y. If vx is about 5.5 m/s and ∆x = 45 m, ∆t ≈ 8 s. If you use an approxi-
mate value of 10 m/s2 for g, ∆y ≈ −320 m, almost identical to the given value.
5.56 m/s−9.81 m/s2
(2)(−321 m)
1. DEFINE
2. PLAN
3. CALCULATE
4. EVALUATE
vx = vxi = vi = ?
ay = g =9.81 m/s2
45.0 m
–321 m
yx
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Use components to analyze objects launched at an angle
Let us examine a case in which a projectile is launched at an angle to the hori-
zontal, as shown in Figure 3-20. The projectile has an initial vertical compo-
nent of velocity as well as a horizontal component of velocity.
Suppose the initial velocity vector makes an angle q with the horizontal.
Again, to analyze the motion of such a projectile, the object’s motion must be
resolved into its components. The sine and cosine functions can be used to
find the horizontal and vertical components of the initial velocity.
vx,i = vi (cos q) and vy,i = vi (sin q)
We can substitute these values for vx,i and vy,i into the kinematic equations
from Chapter 2 to obtain a set of equations that can be used to analyze the
motion of a projectile launched at an angle.
As we have seen, the velocity of a projectile launched at an angle to the
ground has both horizontal and vertical components. The vertical motion is
similar to that of an object that is thrown straight up with an initial velocity.
PROJECTILES LAUNCHED AT AN ANGLE
vx = vi (cos q) = constant
∆x = vi(cos q)∆t
vy,f = vi(sin q) − g∆t
vy,f2 = vi
2(sin q)2 − 2g∆y
∆y = vi(sin q)∆t − 12
g(∆t)2
1. An autographed baseball rolls off of a 0.70 m high desk and strikes the
floor 0.25 m away from the base of the desk. How fast was it rolling?
2. A cat chases a mouse across a 1.0 m high table. The mouse steps out of
the way, and the cat slides off the table and strikes the floor 2.2 m from
the edge of the table. What was the cat’s speed when it slid off the table?
3. A pelican flying along a horizontal path drops a fish from a height of
5.4 m. The fish travels 8.0 m horizontally before it hits the water below.
What is the pelican’s initial speed?
4. If the pelican in item 3 was traveling at the same speed but was only
2.7 m above the water, how far would the fish travel horizontally before
hitting the water below?
PRACTICE 3D
Projectiles launched horizontally
vivy,i
vx,i
θ
Figure 3-20An object is projected with an initialvelocity, vi, at an angle of q . Byresolving the initial velocity into itsx and y components, the kinematicequations can be applied todescribe the motion of the projec-tile throughout its flight.
Chapter 3102
PHYSICSPHYSICSModule 3“Two-Dimensional Motion”provides an interactive lessonwith guided problem-solvingpractice to teach you aboutanalyzing motion at angles,including projectile motion.
Copyright © by Holt, Rinehart and Winston. All rights reserved.103Two-Dimensional Motion and Vectors
SAMPLE PROBLEM 3E
Projectiles launched at an angle
P R O B L E MA zookeeper finds an escaped monkey hanging from a light pole. Aimingher tranquilizer gun at the monkey, the zookeeper kneels 10.0 m from thelight pole, which is 5.00 m high. The tip of her gun is 1.00 m above theground. The monkey tries to trick the zookeeper by dropping a banana,then continues to hold onto the light pole. At the moment the monkeyreleases the banana, the zookeeper shoots. If the tranquilizer dart travelsat 50.0 m/s, will the dart hit the monkey, the banana, or neither one?
S O L U T I O NSelect a coordinate system.
As shown in Figure 3-21, the positive
y-axis points up along the tip of the
gun, and the positive x-axis points
along the ground toward the light
pole. Because the dart leaves the gun at
a height of 1.00 m, the vertical distance
to the monkey (and to the banana) is
4.00 m rather than 5.00 m.
Use the tangent function to find the initial angle that the initial velocitymakes with the x-axis.
q = tan−1�∆∆
x
y� = tan−1�41.
0
0
.
0
0
m
m�
q = 21.8°
Choose a kinematic equation to solve for time.First rearrange the equation for motion along the x-axis to isolate the
unknown, ∆t, which is the time it takes the bullet to travel the horizontal
distance from the tip of the gun to the pole.
∆x = vi(cos q)∆t
∆t = vic
∆o
x
sq =
∆t = 0.215 s
Using your knowledge of free fall, find out how far each object will fallduring this time.For the banana:
This is a free-fall problem, where vi = 0.
∆y = − 12
g(∆t)2 = − 12
(9.81 m/s2)(0.215 s)2
∆y = −0.227 m
Thus, the banana will be 0.227 m below its starting point.
10.0 m(50.0 m/s)(cos 21.8°)
1.
2.
3.
4.
continued onnext page
θ
y
x
vi
10.0 m
4.00 m
1.00 m
Figure 3-21
Copyright © by Holt, Rinehart and Winston. All rights reserved.
The dart has an initial vertical component of velocity equal to vi(sinq), so:
∆y = vi(sin q)∆t − 12
g(∆t)2
∆y = (50.0 m/s)(sin 21.8°)(0.215 s) − 12
(9.81 m/s2)(0.215 s)2
∆y = 3.99 m − 0.227 m = 3.76 m
The dart will be 3.76 m above its starting point.
Analyze the results.By rearranging our equation for displacement, we can find the final height
of both the banana and the dart.
yf = yi + ∆y
ybanana, f = 5.00 m + (−0.227 m) =
ydart, f = 1.00 m + 3.76 m =
The dart hits the banana. The slight difference is due to rounding.
4.76 m above the ground
4.77 m above the ground
5.
PRACTICE 3E
Projectiles launched at an angle
Chapter 3104
1. In a scene in an action movie, a stuntman jumps from the top of one
building to the top of another building 4.0 m away. After a running start,
he leaps at an angle of 15° with respect to the flat roof while traveling at a
speed of 5.0 m/s. Will he make it to the other roof, which is 2.5 m shorter
than the building he jumps from?
2. A golfer can hit a golf ball a horizontal distance of over 300 m on a good
drive. What maximum height will a 301.5 m drive reach if it is launched
at an angle of 25.0° to the ground? (Hint: At the top of its flight, the ball’s
vertical velocity component will be zero.)
3. A baseball is thrown at an angle of 25° relative to the ground at a speed
of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long
was it in the air? How high was the tallest spot in the ball’s path?
4. Salmon often jump waterfalls to reach their breeding grounds. Starting
2.00 m from a waterfall 0.55 m in height, at what minimum speed must a
salmon jumping at an angle of 32.0° leave the water to continue upstream?
5. A quarterback throws the football to a stationary receiver who is 31.5 m
down the field. If the football is thrown at an initial angle of 40.0° to the
ground, at what initial speed must the quarterback throw the ball for it
to reach the receiver? What is the ball’s highest point during its flight?
Copyright © by Holt, Rinehart and Winston. All rights reserved.105Two-Dimensional Motion and Vectors
Section Review
1. Which of the following are examples of projectile motion?
a. an airplane taking off
b. a tennis ball lobbed over a net
c. a plastic disk sailing across a lawn
d. a hawk diving to catch a mouse
e. a parachutist drifting to Earth
f. a frog jumping from land into the water
2. Which of the following exhibit parabolic motion?
a. a flat rock skipping across the surface of a lake
b. a three-point shot in basketball
c. the space shuttle while orbiting Earth
d. a ball bouncing across a room
e. a cliff diver
f. a life preserver dropped from a stationary helicopter
g. a person skipping
3. An Alaskan rescue plane drops a package
of emergency rations to a stranded party
of explorers, as illustrated in Figure 3-22.The plane is traveling horizontally at
100.0 m/s at a height of 50.0 m above the
ground. What horizontal distance
does the package travel before striking
the ground?
4. Find the velocity (magnitude and direction) of the package in item 3 just
before it hits the ground.
5. During a thunderstorm, a tornado lifts a car to a height of 125 m above
the ground. Increasing in strength, the tornado flings the car horizon-
tally with an initial speed of 90.0 m/s. How long does the car take to
reach the ground? How far horizontally does the car travel before hitting
the ground?
6. Physics in Action Streams of water in a fountain shoot from one
level to the next. A particle of water in a stream takes 0.50 s to travel between
the first and second level. The receptacle on the second level is a horizontal
distance of 1.5 m away from the spout on the first level. If the water is pro-
jected at an angle of 33°, what is the initial speed of the particle?
7. Physics in Action If a water particle in a stream of water in a foun-
tain takes 0.35 s to travel from spout to receptacle when shot at an angle
of 67° and an initial speed of 5.0 m/s, what is the vertical distance
between the levels of the fountain?
50.0 m
vplane = 100.0 m/s
Figure 3-22
Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 3106
3-4Relative motion
3-4 SECTION OBJECTIVES
• Describe situations in termsof frame of reference.
• Solve problems involving rel-ative velocity.
(a)
(b)
Figure 3-23When viewed from the plane (a), the stunt dummy (representedby the maroon dot) falls straight down. When viewed from a sta-tionary position on the ground (b), the stunt dummy follows aparabolic projectile path.
FRAMES OF REFERENCE
If you are moving at 80 km/h north and a car passes you going 90 km/h, to
you the faster car seems to be moving north at 10 km/h. Someone standing on
the side of the road would measure the velocity of the faster car as 90 km/h
toward the north. This simple example demonstrates that velocity measure-
ments depend on the frame of reference of the observer.
Velocity measurements differ in different frames of reference
Observers using different frames of reference may measure different displace-
ments or velocities for an object in motion. That is, two observers moving with
respect to each other would generally not agree on some features of the motion.
Let us return to the example of the stunt dummy that is dropped from an air-
plane flying horizontally over Earth with a constant velocity. As shown in
Figure 3-23(a), a passenger on the airplane would describe the motion of the
dummy as a straight line toward Earth, whereas an observer on the ground
would view the trajectory of the dummy as that of a projectile, as shown in
Figure 3-23(b). Relative to the ground, the dummy would have a vertical com-
ponent of velocity (resulting from free-fall acceleration and equal to the velocity
measured by the observer in the airplane) and a horizontal component of veloc-
ity given to it by the airplane’s motion. If the airplane continued to move hori-
zontally with the same velocity, the dummy would enter the swimming pool
directly beneath the airplane (assuming negligible air resistance).
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Like velocity, displacement andacceleration depend on the frame inwhich they are measured. In somecases, it is instructive to visualizegravity as the ground acceleratingtoward a projectile rather than the projectile accelerating towardthe ground.
107Two-Dimensional Motion and Vectors
1. Elevator acceleration A boy bounces asmall rubber ball in an elevator that is going down. Ifthe boy drops the ball as the elevator is slowingdown, is the ball’s acceleration relative to the eleva-tor less than or greater than its acceleration relativeto the ground?
RELATIVE VELOCITY
The case of the faster car overtaking your car was easy to solve with a mini-
mum of thought and effort, but you will encounter many situations in which
a more systematic method of solving such problems is beneficial. To develop
this method, write down all the information that is given and that you want to
know in the form of velocities with subscripts appended.
vse = +80 km/h north (Here the subscript se means the velocity
of the slower car with respect to Earth.)
vfe = +90 km/h north (The subscript fe means the velocity
of the fast car with respect to Earth.)
We want to know vfs, which is the velocity of the fast car with respect to the
slower car. To find this, we write an equation for vfs in terms of the other
velocities, so on the right side of the equation the subscripts start with f and
eventually end with s. Also, each velocity subscript starts with the letter that
ended the preceding velocity subscript.
vfs = vfe + ves
The boldface notation indicates that velocity is a vector quantity. This
approach to adding and monitoring subscripts is similar to vector addition, in
which vector arrows are placed head to tail to find a resultant.
If we take north to be the positive direction, we know that ves = −vse
because an observer in the slow car perceives Earth as moving south at a
velocity of 80 km/h while a stationary observer on the ground (Earth) views
the car as moving north at a velocity of 80 km/h. Thus, this problem can be
solved as follows:
vfs = +90 km/h − 80 km/h = +10 km/h
The positive sign means that the fast car appears (to the occupants of the
slower car) to be moving north at 10 km/h.
There is no general equation to work relative velocity problems; instead,
you should develop the necessary equations on your own by following the
above technique for writing subscripts.
2. Aircraft carrier Why doesa plane landing on an aircraft carrierapproach the carrier from the rear instead offrom the front?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
SAMPLE PROBLEM 3F
Relative velocity
P R O B L E MA boat heading north crosses a wide river with a velocity of 10.00 km/hrelative to the water. The river has a uniform velocity of 5.00 km/h dueeast. Determine the boat’s velocity with respect to an observer on shore.
S O L U T I O NGiven: vbr = 10.00 km/h due north (velocity of the
boat, b, with respect to the river, r)
vre = 5.00 km/h due east (velocity of the
river, r, with respect to Earth, e)
Unknown: vbe = ?
Diagram:
Choose an equation(s) or situation: To find vbe, write the
equation so that the subscripts on the right start with b and end with e.
vbe = vbr + vre
As in Section 3-2, we use the Pythagorean theorem to calculate the magni-
tude of the resultant velocity and the tangent function to find the direction.
(vbe)2 = (vbr)
2 + (vre)2
tan q = v
v
b
re
r
Rearrange the equation(s) to isolate the unknown(s):
vbe = �(v�br�)2� +� (�vr�e)�2�
q = tan−1�v
v
b
re
r�
Substitute the known values into the equation(s) and solve:
vbe = �(1�0.�00� k�m�/h�)2� +� (�5.�00� k�m�/h�)2�
q = tan−1�150.0.000�
The boat travels at a speed of 11.18 km/h in the direction 26.6° east of north
with respect to Earth.
q = 26.6°
vbe = 11.18 km/h
1. DEFINE
2. PLAN
3. CALCULATE
108 Chapter 3
4. EVALUATE
vre
vbr vbe
x
y
θN
S
W E
Copyright © by Holt, Rinehart and Winston. All rights reserved.109Two-Dimensional Motion and Vectors
PRACTICE 3F
Relative velocity
1. A passenger at the rear of a train traveling at 15 m/s relative to Earth
throws a baseball with a speed of 15 m/s in the direction opposite the
motion of the train. What is the velocity of the baseball relative to Earth
as it leaves the thrower’s hand? Show your work.
2. A spy runs from the front to the back of an aircraft carrier at a velocity of
3.5 m/s. If the aircraft carrier is moving forward at 18.0 m/s, how fast
does the spy appear to be running when viewed by an observer on a
nearby stationary submarine? Show your work.
3. A ferry is crossing a river. If the ferry is headed due north with a speed of
2.5 m/s relative to the water and the river’s velocity is 3.0 m/s to the east,
what will the boat’s velocity be relative to Earth? (Hint: Remember to
include the direction in describing the velocity.)
4. A pet-store supply truck moves at 25.0 m/s north along a highway.
Inside, a dog moves at 1.75 m/s at an angle of 35.0° east of north. What is
the velocity of the dog relative to the road?
Section Review
1. Describe the motion of the following objects if they are observed from
the stated frames of reference:
a. a person standing on a platform viewed from a train traveling north
b. a train traveling north viewed by a person standing on a platform
c. a ball dropped by a boy walking at a speed of 1 m/s viewed by the boy
d. a ball dropped by a boy walking 1 m/s as seen by a nearby viewer who
is stationary
2. A woman on a 10-speed bicycle travels at 9 m/s relative to the ground as
she passes a little boy on a tricycle going in the opposite direction. If the
boy is traveling at 1 m/s relative to the ground, how fast does the boy
appear to be moving relative to the woman? Show your work.
3. A girl at an airport rolls a ball north on a moving walkway that moves
east. If the ball’s speed with respect to the walkway is 0.15 m/s and the
walkway moves at a speed of 1.50 m/s, what is the velocity of the ball
relative to the ground?
Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 3110
In Section 3-4, you learned that velocity measurements are not absolute; every
velocity measurement depends on the frame of reference of the observer with
respect to the moving object. For example, imagine that someone riding a bike
toward you at 25 m/s (v) throws a softball toward you. If the bicyclist measures
the softball’s speed (u�) to be 15 m/s, you would perceive the ball to be moving
toward you at 40 m/s (u) because you have a different frame of reference than
the bicyclist does. This is expressed mathematically by the equation u = v + u�,
which is also known as the classical addition of velocities.
The speed of light
As stated in the “Time dilation” feature in Chapter 2, according to Einstein’s
special theory of relativity, the speed of light is absolute, or independent of all
frames of reference. If, instead of a softball, the bicyclist were to shine a beam
of light toward you, both you and the bicyclist would measure the light’s
speed as 3.0 × 108 m/s. This would remain true even if the bicyclist were
moving toward you at 99 percent of the speed of light. Thus, Einstein’s theory
requires a different approach to the addition of velocities. Einstein’s modifi-
cation of the classical formula, which he derived in his 1905 paper on special
relativity, covers both the case of the softball and the case of the light beam.
u = 1 +
v
(
+
vu
u
�/
�
c2)
In the equation, u is the velocity of an object in a reference frame, u� is the
velocity of the same object in another reference frame, v is the velocity of one
reference frame relative to another, and c is the speed of light.
The universality of Einstein’s equation
How does Einstein’s equation cover both cases? First we shall consider the
bicyclist throwing a softball. Because c2 is such a large number, the vu�/c2
term in the denominator is very small for velocities typical of our everyday
experience. As a result, the denominator of the equation is nearly equal to 1.
Hence, for speeds that are small compared with c, the two theories give nearly
the same result, u = v + u�, and the classical addition of velocities can be used.
111
Table 3-1 Classical and relativistic addition of velocities
c � 2.997 925 84 � 108 m/s Classical Relativisticaddition addition
Speed between Speed in Speed in Speed inframes (v) A (u�) B (u) B (u)
000 025 m/s 000 0 15 m/s 000 000 040 m/s 000 000 040 m/s
100 000 m/s 100 000 m/s 000 200 000 m/s 000 200 000 m/s
50% of c 50% of c 299 792 584 m/s 239 834 067 m/s
90% of c 90% of c 539 626 651 m/s 298 136 271 m/s
99.99% of c 99.99% of c 599 525 210 m/s 299 792 582 m/s
Figure 3-24According to Einstein’s relativisticequation for the addition of velo-cities, material particles can neverreach the speed of light.
However, when speeds approach the speed of light, vu�/c2 increases, and
the denominator becomes greater than 1 and less than 2. When this occurs,
the difference between the two theories becomes significant. For example, if a
bicyclist moving toward you at 80 percent of the speed of light were to throw
a ball to you at 70 percent of the speed of light, you would observe the ball
moving toward you at about 96 percent of the speed of light rather than the
150 percent of the speed of light predicted by classical theory. In this case, the
difference between the velocities predicted by each theory cannot be ignored,
and the relativistic addition of velocities must be used.
In this last example, it is significant that classical addition predicts a speed
greater than the speed of light (1.5c), while the relativistic addition predicts a
speed less than the speed of light (0.96c). In fact, no matter how close the
speeds involved are to the speed of light, the relativistic equation yields a
result less than the speed of light, as seen in Table 3-1.How does Einstein’s equation cover the second case, in which the bicyclist
shines a beam of light toward you? Einstein’s equation predicts that any
object traveling at the speed of light (u� = c) will appear to travel at the speed
of light (u = c) for an observer in any reference frame:
u = 1 +
v
(
+
vu
u
�/
�
c2) =
1 +
v
(
+vc
c
/c2) =
1 +v +
(v
c
/c) =
(c
v
++v
c
)/c = c
This corresponds with our earlier statement that the bicyclist measures the
beam of light traveling at the same speed that you do, 3.0 × 108 m/s, even
though you have a different reference frame than the bicyclist does. This
occurs regardless of how fast the bicycle is moving because v (the bicycle’s
speed) cancels from the equation. Thus, Einstein’s relativistic equation suc-
cessfully covers both cases. So, Einstein’s equation is a more general case of
the classical equation, which is simply the limiting case.
TOPIC: Speed of lightGO TO: www.scilinks.orgsciLINKS CODE: HF2033
NSTA
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Chapter 3112
KEY IDEAS
Section 3-1 Introduction to vectors• A scalar is a quantity completely specified by only a number with appropri-
ate units, whereas a vector is a quantity that has magnitude and direction.
• Vectors can be added graphically using the triangle method of addition, in
which the tail of one vector is placed at the head of the other. The resultant is
the vector drawn from the tail of the first vector to the head of the last vector.
Section 3-2 Vector operations• The Pythagorean theorem and the tangent function can be used to find
the magnitude and direction of a resultant vector.
• Any vector can be resolved into its component vectors using the sine and
cosine functions.
Section 3-3 Projectile motion• Neglecting air resistance, a projectile has a constant horizontal velocity
and a constant downward free-fall acceleration.
• In the absence of air resistance, projectiles follow a parabolic path.
Section 3-4 Relative motion• If the frame of reference is denoted with subscripts (vab is the velocity of a
with respect to b), then the velocity of an object with respect to a different
frame of reference can be found by adding the known velocities so that the
subscript starts with the letter that ends the preceding velocity subscript,
vab = vac + vcb.• If the order of the subscripts is reversed, there is a change in sign, for exam-
ple, vcd = −vdc.
CHAPTER 3Summary
KEY TERMS
components of a vector (p. 92)
projectile motion (p. 99)
resultant (p. 85)
scalar (p. 84)
vector (p. 84)
Variable symbols
Quantities Units
d (vector) displacement m meters
v (vector) velocity m/s meters/second
a (vector) acceleration m/s2 meters/second2
∆x (scalar) horizontal component m meters
∆y (scalar) vertical component m meters
Diagram symbols
displacement vector
velocity vector
acceleration vector
resultant vector
component
Copyright © by Holt, Rinehart and Winston. All rights reserved.
113Two-Dimensional Motion and Vectors
VECTORS AND THE GRAPHICALMETHOD
Review questions
1. The magnitude of a vector is a scalar. Explain thisstatement.
2. If two vectors have unequal magnitudes, can theirsum be zero? Explain.
3. What is the relationship between instantaneousspeed and instantaneous velocity?
4. What is another way of saying −30 m/s west?
5. Is it possible to add a vector quantity to a scalarquantity? Explain.
6. Vector A is 3.00 units in length and points along thepositive x-axis. Vector B is 4.00 units in length andpoints along the negative y-axis. Use graphicalmethods to find the magnitude and direction of thefollowing vectors:
a. A + Bb. A − Bc. A + 2Bd. B − A
7. Each of the displacement vectors A and B shown inFigure 3-25 has a magnitude of 3.00 m. Graphicallyfind the following:
a. A + Bb. A − Bc. B − Ad. A − 2B
Figure 3-25x
y
3.00 m
3.00 m
30.0°
A
B
8. A dog searching for a bone walks 3.50 m south, then8.20 m at an angle of 30.0° north of east, and finally15.0 m west. Use graphical techniques to find thedog’s resultant displacement vector.
9. A man lost in a maze makes three consecutive dis-placements so that at the end of the walk he is backwhere he started, as shown in Figure 3-26. The firstdisplacement is 8.00 m westward, and the second is13.0 m northward. Use the graphical method tofind the third displacement.
Conceptual questions
10. If B is added to A, under what conditions does theresultant have the magnitude equal to A + B?
11. Give an example of a moving object that has a veloc-ity vector and an acceleration vector in the samedirection and an example of one that has velocityand acceleration vectors in opposite directions.
12. A student accurately uses the method for combin-ing vectors. The two vectors she combines havemagnitudes of 55 and 25 units. The answer that shegets is either 85, 20, or 55. Pick the correct answer,and explain why it is the only one of the three thatcan be correct.
13. If a set of vectors laid head to tail forms a closedpolygon, the resultant is zero. Is this statement true?Explain your reasoning.
Figure 3-26
X
CHAPTER 3Review and Assess
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 3114
25. A shopper pushing a cart through a store moves40.0 m south down one aisle, then makes a 90.0°turn and moves 15.0 m. He then makes another90.0° turn and moves 20.0 m. Find the shopper’stotal displacement. Note that you are not given thedirection moved after any of the 90.0° turns, sothere could be more than one answer.(See Sample Problem 3A.)
26. A submarine dives 110.0 m at an angle of 10.0° belowthe horizontal. What are the horizontal and verticalcomponents of the submarine’s displacement? (See Sample Problem 3B.)
27. A person walks 25.0° north of east for 3.10 km.How far would another person walk due north anddue east to arrive at the same location?(See Sample Problem 3B.)
28. A roller coaster travels 41.1 m at an angle of 40.0°above the horizontal. How far does it move hori-zontally and vertically?(See Sample Problem 3B.)
29. A person walks the path shown in Figure 3-27. Thetotal trip consists of four straight-line paths. At theend of the walk, what is the person’s resultant dis-placement measured from the starting point?(See Sample Problem 3C.)
Figure 3-27
PROJECTILE MOTION
Review questions
30. A bullet is fired horizontally from a pistol, andanother bullet is dropped simultaneously from thesame height. If air resistance is neglected, whichbullet hits the ground first?
31. If a rock is dropped from the top of a sailboat’smast, will it hit the deck at the same point whetherthe boat is at rest or in motion at constant velocity?
N
S
W E300.0 m
100.0 m
200.0 m
150.0 m30.0°60.0°
?
VECTOR OPERATIONS
Review questions
14. Can a vector have a component equal to zero andstill have a nonzero magnitude?
15. Can a vector have a component greater than itsmagnitude?
16. Explain the difference between vector addition andvector resolution.
17. If the component of one vector along the directionof another is zero, what can you conclude aboutthese two vectors?
18. How would you add two vectors that are not per-pendicular or parallel?
Conceptual questions
19. If A + B = 0, what can you say about the compo-nents of the two vectors?
20. Under what circumstances would a vector havecomponents that are equal in magnitude?
21. The vector sum of three vectors gives a resultantequal to zero. What can you say about the vectors?
Practice problems
22. A girl delivering newspapers travels three blockswest, four blocks north, then six blocks east.
a. What is her resultant displacement? b. What is the total distance she travels?
(See Sample Problem 3A.)
23. A golfer takes two putts to sink his ball in the holeonce he is on the green. The first putt displaces the ball6.00 m east, and the second putt displaces it 5.40 msouth. What displacement would put the ball in thehole in one putt? (See Sample Problem 3A.)
24. A quarterback takes the ball from the line of scrim-mage, runs backward for 10.0 yards, then runs side-ways parallel to the line of scrimmage for 15.0 yards.At this point, he throws a 50.0-yard forward passstraight down the field. What is the magnitude ofthe football’s resultant displacement?(See Sample Problem 3A.)
Copyright © by Holt, Rinehart and Winston. All rights reserved.115Two-Dimensional Motion and Vectors
32. Does a ball dropped out of the window of a movingcar take longer to reach the ground than onedropped at the same height from a car at rest?
33. A rock is dropped at the same instant that a ball atthe same elevation is thrown horizontally. Whichwill have the greater velocity when it reachesground level?
Practice problems
34. The fastest recorded pitch in Major League Baseballwas thrown by Nolan Ryan in 1974. If this pitchwere thrown horizontally, the ball would fall 0.809 m (2.65 ft) by the time it reached home plate,18.3 m (60 ft) away. How fast was Ryan’s pitch?(See Sample Problem 3D.)
35. A shell is fired from the ground with an initial speedof 1.70 × 103 m/s (approximately five times thespeed of sound) at an initial angle of 55.0° to thehorizontal. Neglecting air resistance, find
a. the shell’s horizontal rangeb. the amount of time the shell is in motion
(See Sample Problem 3E.)
36. A person standing at theedge of a seaside cliffkicks a stone over theedge with a speed of 18m/s. The cliff is 52 mabove the water’s surface,as shown in Figure 3-28.How long does it take forthe stone to fall to thewater? With what speeddoes it strike the water?(See Sample Problem 3D.)
37. A spy in a speed boat is being chased down a riverby government officials in a faster craft. Just as theofficials’ boat pulls up next to the spy’s boat, bothboats reach the edge of a 5.0 m waterfall. If the spy’sspeed is 15 m/s and the officials’ speed is 26 m/s,how far apart will the two vessels be when they landbelow the waterfall?
38. A place kicker must kick a football from a point36.0 m (about 40.0 yd) from the goal. As a result ofthe kick, the ball must clear the crossbar, which is3.05 m high. When kicked, the ball leaves theground with a speed of 20.0 m/s at an angle of 53°to the horizontal.
a. By how much does the ball clear or fall shortof clearing the crossbar?
b. Does the ball approach the crossbar while stillrising or while falling?
(See Sample Problem 3E.)
39. A daredevil is shot out of a cannon at 45.0° to thehorizontal with an initial speed of 25.0 m/s. A net ispositioned at a horizontal distance of 50.0 m fromthe cannon from which it is shot. At what heightabove the cannon should the net be placed in orderto catch the daredevil? (See Sample Problem 3E.)
40. When a water gun is fired while being held horizon-tally at a height of 1.00 m above ground level, thewater travels a horizontal distance of 5.00 m. Achild, who is holding the same gun in a horizontalposition, is also sliding down a 45.0° incline at aconstant speed of 2.00 m/s. If the child fires the gunwhen it is 1.00 m above the ground and the watertakes 0.329 s to reach the ground, how far will thewater travel horizontally?(See Sample Problem 3E.)
41. A ship maneuvers to within 2.50 × 103 m of anisland’s 1.80 × 103 m high mountain peak and fires aprojectile at an enemy ship 6.10 × 102 m on the otherside of the peak, as illustrated in Figure 3-29. If theship shoots the projectile with an initial velocity of2.50 × 102 m/s at an angle of 75.0°, how close to theenemy ship does the projectile land? How close (ver-tically) does the projectile come to the peak?(See Sample Problem 3E.)
Figure 3-29
x
y
vi = +18 m/s
h = 52 mg
Figure 3-28
vi =1.80 × 103 m
75.0°
2.50 × 103 m
2.50 × 102 m/s
6.10 × 102 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 3116
50. A hunter wishes to cross a river that is 1.5 km wideand that flows with a speed of 5.0 km/h. The hunteruses a small powerboat that moves at a maximumspeed of 12 km/h with respect to the water. What isthe minimum time necessary for crossing?(See Sample Problem 3F.)
51. A swimmer can swim in still water at a speed of9.50 m/s. He intends to swim directly across a riverthat has a downstream current of 3.75 m/s.
a. What must the swimmer’s direction be?b. What is his velocity relative to the bank?
(See Sample Problem 3F.)
MIXED REVIEW PROBLEMS
52. A motorboat heads due east at 12.0 m/s across a riverthat flows toward the south at a speed of 3.5 m/s.
a. What is the resultant velocity relative to anobserver on the shore?
b. If the river is 1360 m wide, how long does ittake the boat to cross?
53. A ball player hits a home run, and the baseball justclears a wall 21.0 m high located 130.0 m fromhome plate. The ball is hit at an angle of 35.0° to thehorizontal, and air resistance is negligible. Assumethe ball is hit at a height of 1.0 m above the ground.
a. What is the initial speed of the ball?b. How much time does it take for the ball to
reach the wall?c. Find the velocity components and the speed
of the ball when it reaches the wall.
54. A daredevil jumps a canyon 12 m wide. To do so, hedrives a car up a 15° incline.
a. What minimum speed must he achieve toclear the canyon?
b. If the daredevil jumps at this minimum speed,what will his speed be when he reaches theother side?
55. A 2.00 m tall basketball player attempts a goal 10.00 m from the basket (3.05 m high). If he shootsthe ball at a 45.0° angle, at what initial speed musthe throw the basketball so that it goes through thehoop without striking the backboard?
RELATIVE MOTION
Review questions
42. Explain the statement, “All motion is relative.”
43. What is a frame of reference?
44. When we describe motion, what is a common frameof reference?
45. A small airplane is flying at 50 m/s toward the east.A wind of 20 m/s toward the east suddenly begins toblow, giving the plane a velocity of 70 m/s east.
a. Which of these are component vectors?b. Which is the resultant? c. What is the magnitude of the wind velocity?
46. A ball is thrown upward in the air by a passenger ona train that is moving with constant velocity.
a. Describe the path of the ball as seen by thepassenger. Describe the path as seen by a sta-tionary observer outside the train.
b. How would these observations change if thetrain were accelerating along the track?
Practice problems
47. The pilot of a plane measures an air velocity of165 km/h south. An observer on the ground sees theplane pass overhead at a velocity of 145 km/h towardthe north. What is the velocity of the wind that isaffecting the plane?(See Sample Problem 3F.)
48. A river flows due east at 1.50 m/s. A boat crosses theriver from the south shore to the north shore bymaintaining a constant velocity of 10.0 m/s duenorth relative to the water.
a. What is the velocity of the boat as viewed byan observer on shore?
b. If the river is 325 m wide, how far downstreamis the boat when it reaches the north shore?
(See Sample Problem 3F.)
49. The pilot of an aircraft wishes to fly due west in a50.0 km/h wind blowing toward the south. The speedof the aircraft in the absence of a wind is 205 km/h.
a. In what direction should the aircraft head?b. What should its speed be relative to the ground?
(See Sample Problem 3F.)
Copyright © by Holt, Rinehart and Winston. All rights reserved.117Two-Dimensional Motion and Vectors
56. A ball is thrown straight upward and returns to thethrower’s hand after 3.00 s in the air. A second ball isthrown at an angle of 30.0° with the horizontal. Atwhat speed must the second ball be thrown so that itreaches the same height as the one thrown vertically?
57. An escalator is 20.0 m long. If a person stands onthe escalator, it takes 50.0 s to ride from the bottomto the top.
a. If a person walks up the moving escalatorwith a speed of 0.500 m/s relative to the esca-lator, how long does it take the person to getto the top?
b. If a person walks down the “up” escalator withthe same relative speed as in item (a), howlong does it take to reach the bottom?
58. A ball is projected horizontally from the edge of atable that is 1.00 m high, and it strikes the floor at apoint 1.20 m from the base of the table.
a. What is the initial speed of the ball? b. How high is the ball above the floor when its
velocity vector makes a 45.0° angle with thehorizontal?
59. How long does it take an automobile traveling 60.0 km/h to become even with a car that is travel-ing in another lane at 40.0 km/h if the cars’ frontbumpers are initially 125 m apart?
60. The eye of a hurricane passes over Grand BahamaIsland. It is moving in a direction 60.0° north of westwith a speed of 41.0 km/h. Exactly three hours later,the course of the hurricane shifts due north, and itsspeed slows to 25.0 km/h, as shown in Figure 3-30.How far from Grand Bahama is the hurricane 4.50 hafter it passes over the island?
Figure 3-30
41.0 km/h
25.0 km/h
60.0°
N
S
EW
61. A car is parked on a cliff overlooking the ocean onan incline that makes an angle of 24.0° below thehorizontal. The negligent driver leaves the car inneutral, and the emergency brakes are defective. Thecar rolls from rest down the incline with a constantacceleration of 4.00 m/s2 and travels 50.0 m to theedge of the cliff. The cliff is 30.0 m above the ocean.
a. What is the car’s position relative to the baseof the cliff when the car lands in the ocean?
b. How long is the car in the air?
62. A boat moves through a river at 7.5 m/s relative tothe water, regardless of the boat’s direction. If thewater in the river is flowing at 1.5 m/s, how longdoes it take the boat to make a round trip consistingof a 250 m displacement downstream followed by a250 m displacement upstream?
63. A golf ball with an initial angle of 34° lands exactly240 m down the range on a level course.
a. Neglecting air friction, what initial speedwould achieve this result?
b. Using the speed determined in item (a), findthe maximum height reached by the ball.
64. A water spider maintains an average position on thesurface of a stream by darting upstream (against thecurrent), then drifting downstream (with the cur-rent) to its original position. The current in thestream is 0.500 m/s relative to the shore, and thewater spider darts upstream 0.560 m (relative to aspot on shore) in 0.800 s during the first part of itsmotion. Use upstream as the positive direction.
a. Determine both the velocity of the water spiderrelative to the water during its dash upstreamand its velocity during its drift downstream.
b. How far upstream relative to the water doesthe water spider move during one cycle of thisupstream and downstream motion?
c. What is the average velocity of the water spiderrelative to the water for one complete cycle?
65. A car travels due east with a speed of 50.0 km/h.Rain is falling vertically with respect to Earth. Thetraces of the rain on the side windows of the carmake an angle of 60.0° with the vertical. Find thevelocity of the rain with respect to the following:
a. the carb. Earth
Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 3118
68. A science student riding on a flatcar of a train mov-ing at a constant speed of 10.0 m/s throws a balltoward the caboose along a path that the studentjudges as making an initial angle of 60.0° with thehorizontal. The teacher, who is standing on theground nearby, observes the ball rising vertically.How high does the ball rise?
69. A football is thrown toward a receiver with an initialspeed of 18.0 m/s at an angle of 35.0° above the hor-izontal. At that instant, the receiver is 18.0 m fromthe quarterback. In what direction and with whatconstant speed should the receiver run to catch thefootball at the level at which it was thrown?
66. A shopper in a department store can walk up a sta-tionary (stalled) escalator in 30.0 s. If the normallyfunctioning escalator can carry the standing shop-per to the next floor in 20.0 s, how long would ittake the shopper to walk up the moving escalator?Assume the same walking effort for the shopperwhether the escalator is stalled or moving.
67. If a person can jump a horizontal distance of 3.0 mon Earth, how far could the person jump on themoon, where the free-fall acceleration is g/6 andg = 9.81 m/s2? How far could the person jump onMars, where the acceleration due to gravity is 0.38g?
Graphing calculatorsRefer to Appendix B for instructions on download-
ing programs for your calculator. The program
“Chap3” allows you to analyze a graph of height
versus time for a baseball thrown straight up.
Recall the following equation from your studies
of projectiles launched at an angle.
∆y = vi(sin q)∆t − 12
g(∆t)2
The program “Chap3” stored on your graphing
calculator makes use of the projectile motion equa-
tion. Given the initial velocity, your graphing calcu-
lator will use the following equation to graph the
height (Y1) of the baseball versus the time interval
(X) that the ball remains in the air. Note that the
relationships in this equation are the same as those
in the projectile motion equation shown above.
Y1 = VX − 4.9X2
a. The two equations above differ in that the lat-
ter does not include the factor sin q . Why has
this factor been disregarded in the second
equation?
Execute “Chap3” on the p menu, and press
e to begin the program. Enter the value for the
initial velocity (shown below), and press e to
begin graphing.
The calculator will provide the graph of the
displacement function versus time. The x value
corresponds to the time interval in seconds, and the
y value corresponds to the height in meters. If the
graph is not visible, press w and change the set-
tings for the graph window.
Press ◊, and use the arrow keys to trace
along the curve to the highest point of the graph.
The y value there is the greatest height that the ball
reaches. Trace the curve to the right, where the
y value is zero. The x value is the duration of the
ball’s flight.
For each of the following initial velocities,
identify the maximum height and flight time of a
baseball thrown vertically.
b. 25 m/s
c. 35 m/s
d. 75 m/s
e. Does the appearance of the graph represent
the actual trajectory of the ball? Explain.
Press @ q to stop graphing. Press e to
input a new value or ı to end the program.
Copyright © by Holt, Rinehart and Winston. All rights reserved.119Two-Dimensional Motion and Vectors
70. A rocket is launched at an angle of 53° above thehorizontal with an initial speed of 75 m/s, as shownin Figure 3-31. It moves for 25 s along its initial lineof motion with an overall acceleration of 25 m/s2.At this time its engines fail and the rocket proceedsto move as a free body.
a. What is the rocket’s maximum altitude?b. What is the rocket’s total time of flight?c. What is the rocket’s horizontal range? Figure 3-31
a = 25 m/s2
vi = 75 m/s
53°
Performance assessment1. Work in cooperative groups to analyze a game of
chess in terms of displacement vectors. Make a
model chessboard, and draw arrows showing all the
possible moves for each piece as vectors made of
horizontal and vertical components. Then have two
members of your group play the game while the
others keep track of each piece’s moves. Be prepared
to demonstrate how vector addition can be used to
explain where a piece would be after several moves.
2. Use a garden hose to investigate the laws of projectile
motion. Design experiments to investigate how the
angle of the hose affects the range of the water stream.
(Assume that the initial speed of water is constant and
is determined by the pressure indicated by the faucet’s
setting.) What quantities will you measure, and how
will you measure them? What variables do you need
to control? What is the shape of the water stream?
How can you reach the maximum range? How can
you reach the highest point? Present your results to
the rest of the class and discuss the conclusions.
3. How would a physics expert respond to the follow-
ing suggestions made by three airline executives?
Write a script of the expert’s response for perfor-
mance in front of the class.
Airline Executive A: Since the Earth rotates from
west to east, we could operate “static flights”—heli-
copters that begin by hovering above New York City
could begin their landing four hours later, when San
Francisco arrives below.
Airline Executive B: This could work for one-way
flights, but the return trip would take 20 hours.
Airline Executive C: That will never work. It’s like
when you throw a ball up in the air; it comes back to
the same point.
Airline Executive A: That’s only because the Earth’s
motion is not significant during that short a time.
Portfolio projects4. You are helping NASA engineers design a basketball
court for a colony on the moon. How do you antici-
pate the ball’s motion compared with its motion on
Earth? What changes will there be for the players—
how they move and how they throw the ball? What
changes would you recommend for the size of the
court, the basket height, and other regulations in
order to adapt the sport to the moon’s low gravity?
Create a presentation or a report presenting your
suggestions, and include the physics concepts
behind your recommendations.
5. There is conflicting testimony in a court case. A
police officer claims that his radar monitor indicated
that a car was traveling at 176 km/h (110 mi/h). The
driver argues that the radar must have recorded the
relative velocity because he was only going 88 km/h
(55 mi/h). Is it possible that both are telling the
truth? Could one be lying? Prepare scripts for expert
witnesses, for both the prosecution and the defense,
that use physics to justify their positions before the
jury. Create visual aids to be used as evidence to sup-
port the different arguments.
Alternative Assessment
Chapter 3120
VELOCITY OF A PROJECTILE
In this experiment, you will determine the horizontal velocity of a projectile
and compare the effects of different inclined planes on this projectile motion.
PREPARATION
1. Read the entire lab, and plan what measurements you will take.
2. Prepare a data table in your lab notebook with five columns and nine
rows. In the first row, label the columns Trial, Height of ramp (m), Length
of ramp (m), Displacement ∆x (m), Displacement ∆y (m). In the first col-
umn, label the second through ninth rows 1, 2, 3, 4, 5, 6, 7, and 8.
3. Set up the inclined plane at any angle, as shown in Figure 3-32(a). Tape the
aluminum to the end of the plane and to the table. Leave at least 5 cm
between the bottom end of the inclined plane and the edge of the table.
4. In front of the table, place the box to catch the ball after it bounces. Perform
a practice trial to find the correct placement of the box. Use masking tape to
secure the box to the floor. Cover the floor with white paper. Cover the white
paper with the carbon paper, carbon side down. Tape the paper down.
5. Use a piece of cord and tape to hang a washer from the edge of the table so
that the washer hangs a few centimeters above the floor. Mark the floor
directly beneath the washer with tape. Move the washer to another point
on the table edge, and repeat. Connect the two marks on the floor with
masking tape.
PROCEDURE
6. Use tape to mark a starting line near the top of the inclined plane. Mea-
sure the height of the ramp from the tabletop to the tape mark. Measure
the length along the ramp from the tape mark to the tabletop. Measure
CHAPTER 3Laboratory Exercise
OBJECTIVES
•Measure the velocity ofprojectiles in terms ofthe horizontal displace-ment during free fall.
•Compare the velocityand acceleration of pro-jectiles accelerateddown different inclinedplanes.
MATERIALS LIST✔ aluminum sheet, edges
covered with heavy tape✔ C-clamp✔ cardboard box✔ cord✔ inclined plane✔ several sheets carbon paper✔ several large sheets white
paper✔ masking tape✔ meterstick✔ packing tape✔ small metal ball✔ small metal washer✔ support stand and clamp✔ towel or cloth
SAFETY
• Tie back long hair, secure loose clothing, and remove loose jewelry toprevent their getting caught in moving or rotating parts.
• Perform this experiment in a clear area. Falling or dropped massescan cause serious injury.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
121Two-Dimensional Motion and Vectors
the distance from the top of the tabletop to the floor.
Enter these values in the data table for Trial 1 as Height
of ramp (m), Length of ramp (m), and Displacement ∆y.
7. Place the metal ball on the inclined plane at the tape
mark. Keep the area around the table clear of people and
obstructions. Release the ball from rest so that it rolls
down the inclined plane, off the table onto the carbon
paper, and bounces into the box.
8. Lift the carbon paper. There should be a heavy carbon
mark on the white paper where the ball landed. Label
the mark with the trial number.
9. Replace the carbon paper and repeat this procedure as Trial 2.
10. With the inclined plane in the same position, place another tape
mark about halfway down the inclined plane. Measure and record
the height and length of the inclined plane from this mark. Use this
mark as the starting point for Trial 3 and Trial 4. Record all data.
11. Raise or lower the inclined plane and repeat the procedure. Perform
two trials for each tape mark as Trials 5, 6, 7, and 8. For each trial,
measure the distance from the carbon mark to the tape line. Record
this distance as Displacement ∆x.
12. Clean up your work area. Put equipment away safely so it is ready to
be used again. Recycle or dispose of used lab materials as directed by your
teacher.
ANALYSIS AND INTERPRETATION
Calculations and data analysis
1. Organizing data Find the time interval for the ball’s motion from the
edge of the table to the floor using the equation for the vertical motion of
a projectile from page 100, where ∆y is the Displacement ∆y recorded in
your data table. The result is the time interval for each trial.
2. Organizing data Using the time interval from item 1 and the value for
Displacement ∆x, calculate the average horizontal velocity for each trial
during the ball’s motion from the edge of the table to the floor.
Conclusions
3. Inferring conclusions What is the relationship between the height of
the inclined plane and the horizontal velocity of the ball? Explain.
4. Inferring conclusions What is the relationship between the length of
the inclined plane and the horizontal velocity of the ball? Explain.
Figure 3-32Step 3: Use tape to cover the sharpedges of the aluminum sheet beforetaping it to the end of the plane. Thealuminum keeps the ball from bounc-ing as it rolls onto the table.Step 4: Place a soft cloth in the boxas shown. Throughout the lab, becareful not to have too much contactwith the carbon paper.Step 5: This tape line will serve as the baseline for measuring thehorizontal distance traveled by theprojectile.
(a)
(b)
Copyright © by Holt, Rinehart and Winston. All rights reserved.