64

Chapter 3

Vectors

Summary
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DEFINITIONS
Scalar quantities are those that have only a numerical value and no associated direction. Vector quantities have
both magnitude and direction and obey the laws of vector addition. The magnitude of a vector is always a positive
number.
CO N C E P T S A N D P R I N C I P L E S
When two or more vectors are added together, they
must all have the same units and all of them must be
S
the same type of quantity. We can add two vectors A
S
and B graphically. In this method (Active Fig. 3.6), the
S
S
S
S
resultant vector R ϭ A ϩ B runs from the tail of A to
S
the tip of B.
S

If a vector A has an x component Ax and a y component Ay, Sthe vector can be expressed in unit–vector
form as A ϭ Axˆi ϩ Ayˆj . In this notation, ˆi is a unit
vector pointing in the positive x direction and ˆj is a

S
3. O Figure Q3.3 shows two vectors, D1 and D2. Which of the
S
S
possibilities (a) through (d) is the vector D2 Ϫ 2D1, or (e) is
it none of them?

4. O The cutting tool on a lathe is given two displacements,
one of magnitude 4 cm and one of magnitude 3 cm, in
each one of five situations (a) through (e) diagrammed
in Figure Q3.4. Rank these situations according to the
magnitude of the total displacement of the tool, putting
the situation with the greatest resultant magnitude first. If
the total displacement is the same size in two situations,
give those letters equal ranks.

(a)

(b)

D1

(c)

(d)

(e)

Figure Q3.4
D2

(d) 10 (e) Ϫ10 (f) 14.1 m/s (g) undefined (ii) What
is the y component of this vector? (Choose from among
the same answers.)
7. O A submarine dives from the water surface at an angle
of 30° below the horizontal, following a straight path 50
m long. How far is the submarine then below the water
surface? (a) 50 m (b) sin 30° (c) cos 30° (d) tan 30°
(e) (50 m)/sin 30° (f) (50 m)/cos 30° (g) (50 m)/
tan 30° (h) (50 m)sin 30° (i) (50 m)cos 30°
(j) (50 m)tan 30° (k) (sin 30°)/50 m (l) (cos 30°)/50 m
(m) (tan 30°)/50 m (n) 30 m (o) 0 (p) none of these
answers
8. O (i) What is the x component of the vector shown in
Figure Q3.8? (a) 1 cm (b) 2 cm (c) 3 cm (d) 4 cm
(e) 6 cm (f) Ϫ1 cm (g) Ϫ2 cm (h) Ϫ3 cm (i) Ϫ4 cm

65

(j) Ϫ6 cm (k) none of these answers (ii) What is the y
component of this vector? (Choose from among the same
answers.)
y, cm
2
Ϫ4 Ϫ2

0

2

x, cm

30.0°) and (3.80 m, 120.0°). Determine (a) the Cartesian
coordinates of these points and (b) the distance between
them.
3. A fly lands on one wall of a room. The lower left-hand corner of the wall is selected as the origin of a two-dimensional
Cartesian coordinate system. If the fly is located at the
point having coordinates (2.00, 1.00) m, (a) how far is it
from the corner of the room? (b) What is its location in
polar coordinates?
4. The rectangular coordinates of a point are given by (2, y),
and its polar coordinates are (r, 30°). Determine y and r.
5. Let the polar coordinates of the point (x, y) be (r, u).
Determine the polar coordinates for the points (a) (Ϫx, y),
(b) (Ϫ2x, Ϫ2y), and (c) (3x, Ϫ3y).
2 = intermediate;

3 = challenging;

Ⅺ = SSM/SG;

ᮡ

Section 3.2 Vector and Scalar Quantities
Section 3.3 Some Properties of Vectors
6. A plane flies from base camp to lake A, 280 km away in
the direction 20.0° north of east. After dropping off supplies it flies to lake B, which is 190 km at 30.0° west of
north from lake A. Graphically determine the distance
and direction from lake B to the base camp.
7. A surveyor measures the distance across a straight river by
the following method: starting directly across from a tree
on the opposite bank, she walks 100 m along the riverbank to establish a baseline. Then she sights across to the

he skates all the way around the circle?
10. Arbitrarily define the “instantaneous vector height” of a
person as the displacement vector from the point halfway
between his or her feet to the top of the head. Make an
order-of-magnitude estimate of the total vector height of
all the people in a city of population 100 000 (a) at
10 o’clock on a Tuesday morning and (b) at 5 o’clock on
a Saturday morning. Explain your reasoning.
S
S
11. ᮡ Each of the displacement vectors A and B shown in
Figure P3.11 has a magnitude of 3.00 m. Graphically find
S
S
S
S
S
S
S
S
(a) A ϩ B, (b) A Ϫ B, (c) B Ϫ A, and (d) A Ϫ 2B. Report
all angles counterclockwise from the positive x axis.
9.

18.

19.

20.


S
possible
ways
of
adding
these
vectors:
R
ϭ
A
ϩ
B ϩ C;
1
S
S
S
S
S
S
S
S
R2 ϭ B ϩ C ϩ A; R3 ϭ C ϩ B ϩ A. Explain what you can
conclude from comparing the diagrams.
13. A roller-coaster car moves 200 ft horizontally and then
rises 135 ft at an angle of 30.0° above the horizontal. It
next travels 135 ft at an angle of 40.0° downward. What is
its displacement from its starting point? Use graphical
techniques.
14. ⅷ A shopper pushing a cart through a store moves 40.0 m
down one aisle, then makes a 90.0° turn and moves 15.0 m.


rear bumper and the minivan’s front bumper should not
decrease. Can the camper be driven to satisfy this requirement? Explain your answer.
A girl delivering newspapers covers her route by traveling
3.00 blocks west, 4.00 blocks north, and then 6.00 blocks
east. (a) What is her resultant displacement? (b) What is
the total distance she travels?
Obtain expressions in component form for the position
vectors having the following polar coordinates: (a) 12.8 m,
150° (b) 3.30 cm, 60.0° (c) 22.0 in., 215°
A displacement vector lying in the xy plane has a magnitude of 50.0 m and is directed at an angle of 120° to the
positive x axis. What are the rectangular components of
this vector?
While exploring a cave, a spelunker starts at the entrance
and moves the following distances. She goes 75.0 m
north, 250 m east, 125 m at an angle 30.0° north of east,
and 150 m south. Find her resultant displacement from
the cave entrance.
A map suggests that Atlanta is 730 miles in a direction of
5.00° north of east from Dallas. The same map shows that
Chicago is 560 miles in a direction of 21.0° west of north
from Atlanta. Modeling the Earth as flat, use this information to find the displacement from Dallas to Chicago.
A man pushing a mop across a floor causes it to undergo
two displacements. The first has a magnitude of 150 cm
and makes an angle of 120° with the positive x axis. The
resultant displacement has a magnitude of 140 cm and is
directed at an angle of 35.0° to the positive x axis. Find
the magnitude and direction of the second displacement.
S
S

horizontal. When a ski jumper plummets onto the hill, a
parcel of splashed snow projects to a maximum position
of 5.00 m at 20.0° from the vertical in the uphill direction
as shown in Figure P3.26. Find the components of its
maximum position (a) parallel to the surface and (b) perpendicular to the surface.

20.0°

35.0°

Figure P3.26

27. A particle undergoes the following consecutive displacements: 3.50 m south, 8.20 m northeast, and 15.0 m west.
What is the resultant displacement?

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Problems

28. In a game of American football, a quarterback takes the
ball from the line of scrimmage, runs backward a distance
of 10.0 yards, and then runs sideways parallel to the line
of scrimmage for 15.0 yards. At this point, he throws a forward pass 50.0 yards straight downfield perpendicular to
the line of scrimmage. What is the magnitude of the football’s resultant displacement?
29. A novice golfer on the green takes three strokes to sink


F2 ϭ
80.0 N
75.0Њ

60.0Њ

x

39.

Figure P3.31
S

S

32. Use the component method to add the vectors A and B
S
S
shown in Figure P3.11. Express the resultant A ϩ B in
unit–vector notation.
S
33. Vector B has x, y, and z components of 4.00, 6.00, and
S
3.00 units, respectively. Calculate the magnitude of B and
S
the angles B makes with the coordinate axes.
S
34. Consider
the three displacement

north, through one-quarter of a circle of radius 3.70 cm
that lies in a north–south vertical plane. Find (a) the mag2 = intermediate;

3 = challenging;

Ⅺ = SSM/SG;

ᮡ

67

40.

41.

42.

43.

nitude of the total displacement of the object and (b) the
angle the total displacement makes with the vertical.
S
The vector A has x, y, and z components of 8.00, 12.0, and
–4.00 units, respectively. (a) Write a vector expression for
S
A in unit–vector notation. (b) Obtain a unit–vector expresS
S
sion for a vector B one-fourth the length of A pointing in
S
the same direction as A. (c) Obtain a unit–vector expresS

S
Vector A has a negative x component 3.00 units in
length and a positive y component 2.00 units in length.
S
(a) Determine an expression for A in unit–vector notaS
tion. (b) Determine the magnitude and direction of A.
S
S
(c) What vector B when added to A gives a resultant vector with no x component and a negative y component
4.00 units in length?
As it passes over Grand Bahama Island, the eye of a hurricane is moving in a direction 60.0° north of west with a
speed of 41.0 km/h. Three hours later the course of the
hurricane suddenly shifts due north, and its speed slows
to 25.0 km/h. How far from Grand Bahama is the eye
4.50 h after it passes over the island?
ᮡ Three displacement vectors of a croquet ball are shown
S
S
in Figure P3.43, where 0 A 0 ϭ 20.0 units, 0 B 0 ϭ 40.0 units,
S
and 0 C 0 ϭ 30.0 units. Find (a) the resultant in unit–vector

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68

S
S
S
and b such that a A ϩ b B ϩ C ϭ 0. (b) A student has
learned that a single equation cannot be solved to determine values for more than one unknown in it. How would
you explain to him that both a and b can be determined
from the single equation used in part (a)?
45. ⅷ Are we there yet? In Figure P3.45, the line segment represents a path from the point with position vector
15ˆi ϩ 3ˆj 2 m to the point with location 116ˆi ϩ 12ˆj 2 m.
Point A is along this path, a fraction f of the way to the
destination. (a) Find the position vector of point A in
terms of f. (b) Evaluate the expression from part (a) in
the case f ϭ 0. Explain whether the result is reasonable.
(c) Evaluate the expression for f ϭ 1. Explain whether the
result is reasonable.
S

S

y

49.

50.

map of the successive displacements. (b) What total distance did she travel? (c) Compute the magnitude and
direction of her total displacement. The logical structure
of this problem and of several problems in later chapters
was suggested by Alan Van Heuvelen and David Maloney,
American Journal of Physics 67(3) 252–256, March 1999.

directly from the tail of the snake to its head, and Olaf
starts from the same place at the same moment but runs
along the snake. If both children run steadily at 12.0 km/h,
Inge reaches the head of the snake how much earlier
than Olaf?

(16, 12)

A
(5, 3)
O

x

Figure P3.45 Point A is a fraction f of the distance from the initial point (5, 3) to the final point (16, 12).

Additional Problems
46. On December 1, 1955, Rosa Parks (1913–2005), an icon of
the early civil rights movement, stayed seated in her bus
seat when a white man demanded it. Police in Montgomery, Alabama, arrested her. On December 5, blacks
began refusing to use all city buses. Under the leadership
of the Montgomery Improvement Association, an efficient
system of alternative transportation sprang up immediately, providing blacks with approximately 35 000 essential
trips per day through volunteers, private taxis, carpooling,
and ride sharing. The buses remained empty until they
were integrated under court order on December 21, 1956.
In picking up her riders, suppose a driver in downtown
Montgomery traverses four successive displacements represented by the expression

1Ϫ6.30b2 ˆi Ϫ 14.00b cos 40° 2 ˆi Ϫ 14.00b sin 40°2 ˆj

are the new speed and direction of the aircraft relative to
the ground?
S
54. ⅷ Let A ϭ 60.0 cm at 270° measured from the horizontal.
S
Let B ϭ 80.0 cm at some angle u. (a) Find the magnitude
S
S
of A ϩ B as a function of u. (b) From the answer to part
S
S
(a), for what value of u does 0 A ϩ B 0 take on its maximum

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Ⅵ = symbolic reasoning;

ⅷ = qualitative reasoning


Problems

value? What is this maximum value? (c) From the answer
S
S
to part (a), for what value of u does 0 A ϩ B 0 take on its
minimum value? What is this minimum value? (d) Without reference to the answer to part (a), argue that the
answers to each of parts (b) and (c) do or do not make
sense.
55. After a ball rolls off the edge of a horizontal table at time

calculations by making a particular choice for the directions of the x and y axes. What is your choice? Then add
the vectors by the component method.
57. A person going for a walk follows the path shown in Figure P3.57. The total trip consists of four straight-line
paths. At the end of the walk, what is the person’s resultant displacement measured from the starting point?
y
Start 100 m

x

300 m

End
200 m
60.0Њ

30.0Њ
150 m

Figure P3.57

58. ⅷ The instantaneous position of an object is specified by
its position vector Sr leading from a fixed origin to the
location of the object, modeled as a particle. Suppose for
a certain object the position vector is a function of time,
S
given by r ϭ 4ˆi ϩ 3ˆj Ϫ 2t ˆ
k , where r is in meters and t is
in seconds. Evaluate drS>dt. What does it represent about
the object?
59. ⅷ Long John Silver, a pirate, has buried his treasure on

does not depend on the order in which the trees are
labeled.

B
E

y

x
C
A
D
Figure P3.59

60. ⅷ Consider a game in which N children position themselves at equal distances around the circumference of a
circle. At the center of the circle is a rubber tire. Each
child holds a rope attached to the tire and, at a signal,
pulls on his or her rope. All children exert forces of the
same magnitude F. In the case N ϭ 2, it is easy to see that
the net force on the tire will be zero because the two
oppositely directed force vectors add to zero. Similarly, if
N ϭ 4, 6, or any even integer, the resultant force on the
tire must be zero because the forces exerted by each pair
of oppositely positioned children will cancel. When an
odd number of children are around the circle, it is not as
obvious whether the total force on the central tire will be
zero. (a) Calculate the net force on the tire in the case
N ϭ 3 by adding the components of the three force vectors. Choose the x axis to lie along one of the ropes.
(b) What If? State reasoning that will determine the net
force for the general case where N is any integer, odd or

S
of A and B is the vector 6.00ˆj . Determine the angle
S
S
between A and B.
62. A rectangular parallelepiped has dimensions a, b, and c as
shown in Figure P3.62. (a)
Obtain a vector expression for
S
the face diagonal vector R1. What is the magnitude of this
vector? (b)SObtain a vectorSexpression Sfor the body diagonal vector R2. Notice that R1, c ˆ
k , and R2 make a right triS
angle. Prove that the magnitude of R2 is 1 a 2 ϩ b 2 ϩ c 2.

z
a

b

O
x

R2

c

R1
y
Figure P3.62


S
component of B is negative.

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4.1

The Position, Velocity, and Acceleration Vectors

4.2

Two-Dimensional Motion with Constant Acceleration

4.3

Projectile Motion

4.4

The Particle in Uniform Circular Motion

4.5

Tangential and Radial Acceleration


coordinate system to the location of the particle in the xy plane, as in Figure 4.1
S
(page 72). At time ti, the particle is at point Ꭽ, described by position vector r i. At
S
some later time tf , it is at point Ꭾ, described by position vector r f . The path from
71


72

Chapter 4

Motion in Two Dimensions

Ꭽ to Ꭾ is not necessarily a straight line. As the particle moves from Ꭽ to Ꭾ in the
S
S
time interval ⌬t ϭ tf Ϫ ti, its position vector changes from r i to r f . As we learned in
Chapter 2, displacement is a vector, and the displacement of the particle is the difference between its final position and its initial position. We now define the disS
placement vector ¢r for a particle such as the one in Figure 4.1 as being the difference between its final position vector and its initial position vector:
Displacement vector

¢r ϵ r f Ϫ r i
S

ᮣ

S

S


Ꭾ tf
Path of
particle
x

Figure 4.1 A particle moving in the
xy plane is located with the position
S
vector r drawn from the origin
to the particle. The displacement of
the particle as it moves from Ꭽ to Ꭾ
in the time interval ⌬t ϭ tf Ϫ ti is
S
S
S
equal to the vector ¢ r ϭ r f Ϫ r i.

vavg ϵ

S

¢r
¢t

(4.2)

Multiplying or dividing a vector quantity by a positive scalar quantity such as ⌬t
changes only the magnitude of the vector, not its direction. Because displacement
is a vector quantity and the time interval is a positive scalar quantity, we conclude


ᎮЉ
ᎮЈ
Ꭾ
O

x


Section 4.1

The Position, Velocity, and Acceleration Vectors

73

becomes smaller and smaller—that is, as Ꭾ is moved to Ꭾ¿ and then to Ꭾ– , and
so on—the direction of the displacement approaches that of the line tangent to
S
the path at Ꭽ. The instantaneous velocity v is defined as the limit of the average
S
velocity ¢ r >¢t as ⌬t approaches zero:
S

v ϵ lim

S

¢tS0

S

neous velocity vector changes from vi at time ti to vf at time tf. Knowing the velocity
at these points allows us to determine the average acceleration of the particle. The
S
average acceleration aavg of a particle is defined as the change in its instantaneous
S
velocity vector ¢v divided by the time interval ⌬t during which that change occurs:
vf Ϫ vi

S

aavg ϵ

S

S

tf Ϫ ti

S

ϭ

¢v
¢t

(4.4)

Because aavg is the ratio of a vector quantity ¢v and a positive scalar quantity ⌬t, we
S
conclude that average acceleration is a vector quantity directed along ¢v. As indiS


(4.5)

In other words, the instantaneous acceleration equals the derivative of the velocity
vector with respect to time.
Various changes can occur when a particle accelerates. First, the magnitude of the
velocity vector (the speed) may change with time as in straight-line (one-dimensional)
motion. Second, the direction of the velocity vector may change with time even if
its magnitude (speed) remains constant as in two-dimensional motion along a
curved path. Finally, both the magnitude and the direction of the velocity vector
may change simultaneously.

y

⌬v

Ꭽ

vf

vi
–vi

Ꭾ
vf

ri
rf

O


Quick Quiz 4.1 Consider the following controls in an automobile: gas pedal,
brake, steering wheel. What are the controls in this list that cause an acceleration
of the car? (a) all three controls (b) the gas pedal and the brake (c) only the
brake (d) only the gas pedal

4.2

Two-Dimensional Motion with
Constant Acceleration

In Section 2.5, we investigated one-dimensional motion of a particle under constant acceleration. Let us now consider two-dimensional motion during which the
acceleration of a particle remains constant in both magnitude and direction. As we
shall see, this approach is useful for analyzing some common types of motion.
Before embarking on this investigation, we need to emphasize an important
point regarding two-dimensional motion. Imagine an air hockey puck moving in a
straight line along a perfectly level, friction-free surface of an air hockey table. Figure 4.4a shows a motion diagram from an overhead point of view of this puck.
Recall that in Section 2.4 we related the acceleration of an object to a force on the
object. Because there are no forces on the puck in the horizontal plane, it moves
with constant velocity in the x direction. Now suppose you blow a puff of air on
the puck as it passes your position, with the force from your puff of air exactly in
the y direction. Because the force from this puff of air has no component in the x
direction, it causes no acceleration in the x direction. It only causes a momentary
acceleration in the y direction, causing the puck to have a constant y component
of velocity once the force from the puff of air is removed. After your puff of air on
the puck, its velocity component in the x direction is unchanged, as shown in Figure 4.4b. The generalization of this simple experiment is that motion in two
dimensions can be modeled as two independent motions in each of the two perpendicular directions associated with the x and y axes. That is, any influence in the y
direction does not affect the motion in the x direction and vice versa.
The position vector for a particle moving in the xy plane can be written
r ϭ xˆi ϩ yˆj

x
(b)
Figure 4.4 (a) A puck moves across a horizontal air hockey table at constant velocity in the x direction.
(b) After a puff of air in the y direction is applied to the puck, the puck has gained a y component of
velocity, but the x component is unaffected by the force in the perpendicular direction. Notice that the
horizontal red vectors, representing the x component of the velocity, are the same length in both parts
of the figure, which demonstrates that motion in two dimensions can be modeled as two independent
motions in perpendicular directions.


Section 4.2

Two-Dimensional Motion with Constant Acceleration

S

Because the acceleration a of the particle is assumed constant in this discussion,
its components ax and ay also are constants. Therefore, we can model the particle
as a particle under constant acceleration independently in each of the two directions and apply the equations of kinematics separately to the x and y components of
the velocity vector. Substituting, from Equation 2.13, vxf ϭ vxi ϩ axt and vyf ϭ vyi ϩ
ayt into Equation 4.7 to determine the final velocity at any time t, we obtain
vf ϭ 1vxi ϩ axt2 ˆi ϩ 1vyi ϩ ayt2 ˆj ϭ 1vxiˆi ϩ vyiˆj 2 ϩ 1axˆi ϩ ayˆj 2t

S

v f ϭ v i ϩ at

S

S

r f ϭ 1x i ϩ v xit ϩ 12a xt 2 2 ˆi ϩ 1y i ϩ v yit ϩ 12a yt 2 2 ˆj

S

ϭ 1x iˆi ϩ y iˆj 2 ϩ 1v xiˆi ϩ v yiˆj 2t ϩ 12 1a xˆi ϩ a yˆj 2t 2
(4.9)

r f ϭ r i ϩ vit ϩ 12 at 2

S

S

S

S

which is the vector version of Equation 2.16. Equation 4.9 tells us that the position
S
S
vector r f of a particle is the vector sum of the original position r i, a displacement
S
S
vit arising from the initial velocity of the particle and a displacement 21 at 2 resulting
from the constant acceleration of the particle.
Graphical representations of Equations 4.8 and 4.9 are shown in Active Figure
4.5. The components of the position and velocity vectors are also illustrated in the
S
figure. Notice from Active Figure 4.5a that vf is generally not along the direction
S

x

yi

ri
x
vxit

xi
vxf

at 2

vit

axt

vxi

(a)

1
2

rf

yf

1 a t2
2 x

x

Conceptualize The components of the initial velocity tell
us that the particle starts by moving toward the right and
downward. The x component of velocity starts at 20 m/s
and increases by 4.0 m/s every second. The y component of velocity never changes from its initial value of
Ϫ15 m/s. We sketch a motion diagram of the situation
in Figure 4.6. Because the particle is accelerating in the
ϩx direction, its velocity component in this direction
increases and the path curves as shown in the diagram.
Figure 4.6 (Example 4.1) Motion diagram for the particle.
Notice that the spacing between successive images
increases as time goes on because the speed is increasing. The placement of the acceleration and velocity vectors in
Figure 4.6 helps us further conceptualize the situation.
Categorize Because the initial velocity has components in both the x and y directions, we categorize this problem as one
involving a particle moving in two dimensions. Because the particle only has an x component of acceleration, we model
it as a particle under constant acceleration in the x direction and a particle under constant velocity in the y direction.
Analyze

To begin the mathematical analysis, we set vxi ϭ 20 m/s, vyi ϭ Ϫ15 m/s, ax ϭ 4.0 m/s2, and ay ϭ 0.
vf ϭ vi ϩ at ϭ 1vxi ϩ axt2 ˆi ϩ 1vyi ϩ ayt2 ˆj

Use Equation 4.8 for the velocity vector:

S

Substitute numerical values:

S


vxf

b ϭ tanϪ1 a

Ϫ15 m>s
40 m>s

b ϭ Ϫ21°

vf ϭ 0 vf 0 ϭ 2vxf2 ϩ vyf 2 ϭ 2 1402 2 ϩ 1Ϫ15 2 2 m>s ϭ 43 m>s
S

Finalize The negative sign for the angle u indicates that the velocity vector is directed at an angle of 21° below the
S
positive x axis. Notice that if we calculate vi from the x and y components of vi, we find that vf Ͼ vi. Is that consistent
with our prediction?
(C) Determine the x and y coordinates of the particle at any time t and its position vector at this time.
SOLUTION
Analyze
Use the components of Equation 4.9 with xi ϭ yi ϭ 0 at
t ϭ 0:

x f ϭ v xit ϩ 12 axt 2 ϭ 120t ϩ 2.0t 2 2 m
yf ϭ v yit ϭ 1Ϫ15t2 m


Section 4.3

r f ϭ xfˆi ϩ yfˆj ϭ 3 120t ϩ 2.0t 2 2 ˆi Ϫ 15tˆj 4 m


lows directly from Equation 4.9, with a ϭ g:
r f ϭ ri ϩ vit ϩ 12 gt2

S

S

S

S

(4.10)

y

vᎮ

vy

vy ϭ 0 vᎯ

g

Ꭿ

൳

u
vi


S

S

The parabolic path of a projectile that leaves the origin with a velocity vi. The velocity vector v changes
with time in both magnitude and direction. This change is the result of acceleration in the negative y
direction. The x component of velocity remains constant in time because there is no acceleration along
the horizontal direction. The y component of velocity is zero at the peak of the path.
Sign in at www.thomsonedu.com and go to ThomsonNOW to change launch angle and initial speed.
You can also observe the changing components of velocity along the trajectory of the projectile.

1

This assumption is reasonable as long as the range of motion is small compared with the radius of the
Earth (6.4 ϫ 106 m). In effect, this assumption is equivalent to assuming that the Earth is flat over the
range of motion considered.

2

77

S

Express the position vector of the particle at any time t:
Finalize

Projectile Motion

This assumption is generally not justified, especially at high velocities. In addition, any spin imparted
to a projectile, such as that applied when a pitcher throws a curve ball, can give rise to some very interesting effects associated with aerodynamic forces, which will be discussed in Chapter 14.

process follow parabolic paths.

y
1
2

vit

gt 2

(x, y)
rf

x

O
S

Figure 4.8 The position vector r f of
a projectile launched from the origin
whose initial velocity at the origin is
S
S
vi. The vector vit would be the displacement of the projectile if gravity
S
were absent, and the vector 12 gt 2 is its
vertical displacement from a straightline path due to its downward gravitational acceleration.

(4.11)


component as shown in Figure 4.9 and returns to the same horizontal level. Two
points are especially interesting to analyze: the peak point Ꭽ, which has Cartesian
coordinates (R/2, h), and the point Ꭾ, which has coordinates (R, 0). The distance
R is called the horizontal range of the projectile, and the distance h is its maximum
height. Let us find h and R mathematically in terms of vi , ui , and g.
We can determine h by noting that at the peak vy Ꭽ ϭ 0. Therefore, we can use
the y component of Equation 4.8 to determine the time t Ꭽ at which the projectile
reaches the peak:
v yf ϭ v yi ϩ a yt
0 ϭ v i sin u i Ϫ gt Ꭽ

y

vi

Ꭽ

tᎭ ϭ

vy Ꭽ ϭ 0

h
ui

Ꭾx

O
R

Figure 4.9 A projectile launched


Section 4.3

Projectile Motion

79

y (m)

150

vi ϭ 50 m/s
75Њ

100

60Њ
45Њ

50

30Њ
15Њ
50

100

150

200


The maximum value of R from Equation 4.13 is Rmax ϭ vi 2>g. This result makes
sense because the maximum value of sin 2ui is 1, which occurs when 2ui ϭ 90°.
Therefore, R is a maximum when ui ϭ 45°.
Active Figure 4.10 illustrates various trajectories for a projectile having a given
initial speed but launched at different angles. As you can see, the range is a maximum for ui ϭ 45°. In addition, for any ui other than 45°, a point having Cartesian
coordinates (R, 0) can be reached by using either one of two complementary values of ui , such as 75° and 15°. Of course, the maximum height and time of flight
for one of these values of ui are different from the maximum height and time of
flight for the complementary value.

Quick Quiz 4.3 Rank the launch angles for the five paths in Active Figure 4.10
with respect to time of flight, from the shortest time of flight to the longest.

P R O B L E M - S O LV I N G S T R AT E G Y

Projectile Motion

We suggest you use the following approach when solving projectile motion problems:
1. Conceptualize. Think about what is going on physically in the problem. Establish
the mental representation by imagining the projectile moving along its trajectory.
2. Categorize. Confirm that the problem involves a particle in free fall and that air
resistance is neglected. Select a coordinate system with x in the horizontal direction and y in the vertical direction.
3. Analyze. If the initial velocity vector is given, resolve it into x and y components.
Treat the horizontal motion and the vertical motion independently. Analyze the

PITFALL PREVENTION 4.3
The Height and Range Equations
Equation 4.13 is useful for calculating R only for a symmetric path as
shown in Active Figure 4.10. If the
path is not symmetric, do not use this

the final height is the same as the initial height, we further categorize
this problem as satisfying the conditions for which Equations 4.12 and
4.13 can be used. This approach is the most direct way to analyze this
problem, although the general methods that have been described will
always give the correct answer.
Analyze
Use Equation 4.13 to find the range of the jumper:

Mike Powell/Allsport/Getty Images

Conceptualize The arms and legs of a long jumper move in a complicated way, but we will ignore this motion. We conceptualize the motion
of the long jumper as equivalent to that of a simple projectile.

Figure 4.11 (Example 4.2) Mike Powell, current
holder of the world long-jump record of 8.95 m.

Rϭ

111.0 m>s2 2 sin 2 120.0°2
vi 2 sin 2u i
ϭ
ϭ 7.94 m
g
9.80 m>s2

hϭ

111.0 m>s2 2 1sin 20.0°2 2
v i 2 sin2 ui
ϭ

81

y
Target
1
2

© Thomson Learning/Charles D. Winters

ht

ne

ig
fs

o

Li

0

gt 2
x T tan ui

Point of
collision

yT



(2)

yP ϭ yiP ϩ v yi Pt Ϫ 12gt 2 ϭ 0 ϩ 1v i P sinui 2 t Ϫ 12gt 2 ϭ 1v i P sinui 2t Ϫ 12gt 2
x P ϭ x iP ϩ v xi Pt ϭ 0 ϩ 1v i P cos ui 2t ϭ 1v iP cos ui 2t

Write an expression for the x coordinate
of the projectile at any moment:

tϭ

Solve this expression for time as a function
of the horizontal position of the projectile:
Substitute this expression into Equation
(2):

(3)

yP ϭ 1v iP sin ui 2 a

xP
v i P cos ui

xP
b Ϫ 12gt 2 ϭ x P tan ui Ϫ 12gt 2
v iP cos ui

Compare Equations (1) and (3). We see that when the x coordinates of the projectile and target are the same—that
is, when xT ϭ xP—their y coordinates given by Equations (1) and (3) are the same and a collision results.
Finalize Note that a collision can result only when v i P sin ui Ն 1gd>2, where d is the initial elevation of the target


v i ϭ 20.0 m/s

y
(0, 0)

x
Ꭽ

ui ϭ 30.0Њ

45.0 m

Figure 4.13 (Example
4.4) A stone is thrown
from the top of a building.

v xi ϭ v i cos ui ϭ 120.0 m>s2cos 30.0° ϭ 17.3 m>s
v yi ϭ v i sin ui ϭ 120.0 m>s2sin 30.0° ϭ 10.0 m>s
yf ϭ yi ϩ v yi t ϩ 12a y t 2

Express the vertical position of the stone from the vertical component of Equation 4.9:

Ϫ45.0 m ϭ 0 ϩ 110.0 m>s2t ϩ 12 1Ϫ9.80 m>s2 2t 2

Substitute numerical values:

t ϭ 4.22 s

Solve the quadratic equation for t :

Conceptualize We can conceptualize this problem based on memories of observing winter Olympic ski competitions. We estimate the skier to be airborne for perhaps 4 s and to travel a distance of about 100 m horizontally. We


Section 4.3

should expect the value of d, the distance traveled along the
incline, to be of the same order of magnitude.

Projectile Motion

83

25.0 m/s
(0,0)

f ϭ 35.0Њ

Categorize We categorize the problem as one of a particle in projectile motion.
y

Analyze It is convenient to select the beginning of the jump as
the origin. The initial velocity components are vxi ϭ 25.0 m/s and
vyi ϭ 0. From the right triangle in Figure 4.14, we see that the
jumper’s x and y coordinates at the landing point are given by
xf ϭ d cos 35.0° and yf ϭϪd sin 35.0°.

d

x


d cos 35.0° 2
b
25.0 m>s

xf ϭ d cos 35.0° ϭ 1109 m2cos 35.0° ϭ 89.3 m

yf ϭ Ϫd sin 35.0° ϭ Ϫ 1109 m2sin 35.0° ϭ Ϫ62.5 m

Finalize Let us compare these results with our expectations. We expected the horizontal distance to be on the
order of 100 m, and our result of 89.3 m is indeed on this order of magnitude. It might be useful to calculate the
time interval that the jumper is in the air and compare it with our estimate of about 4 s.
What If? Suppose everything in this example is the same
except the ski jump is curved so that the jumper is projected upward at an angle from the end of the track. Is
this design better in terms of maximizing the length of
the jump?
Answer If the initial velocity has an upward component, the skier will be in the air longer and should
therefore travel further. Tilting the initial velocity vector
upward, however, will reduce the horizontal component
of the initial velocity. Therefore, angling the end of the
ski track upward at a large angle may actually reduce the
distance. Consider the extreme case: the skier is projected at 90° to the horizontal and simply goes up and
comes back down at the end of the ski track! This argument suggests that there must be an optimal angle
between 0° and 90° that represents a balance between
making the flight time longer and the horizontal velocity component smaller.
Let us find this optimal angle mathematically. We
modify equations (1) through (4) in the following way,

assuming that the skier is projected at an angle u with
respect to the horizontal over a landing incline sloped
with an arbitrary angle f:


Motion in Two Dimensions

PITFALL PREVENTION 4.4
Acceleration of a Particle in Uniform
Circular Motion
Remember that acceleration in
physics is defined as a change in
the velocity, not a change in the
speed (contrary to the everyday
interpretation). In circular motion,
the velocity vector is changing in
direction, so there is indeed an
acceleration.

4.4

The Particle in Uniform Circular Motion

Figure 4.15a shows a car moving in a circular path with constant speed v. Such
motion, called uniform circular motion, occurs in many situations. Because it
occurs so often, this type of motion is recognized as an analysis model called the
particle in uniform circular motion. We discuss this model in this section.
It is often surprising to students to find that even though an object moves at a
constant speed in a circular path, it still has an acceleration. To see why, consider
S
S
the defining equation for acceleration, a ϭ d v>dt (Eq. 4.5). Notice that the acceleration depends on the change in the velocity. Because velocity is a vector quantity,
an acceleration can occur in two ways, as mentioned in Section 4.1: by a change in
the magnitude of the velocity and by a change in the direction of the velocity. The

help us analyze the motion. The angle ⌬u between the two position vectors in Figure 4.15b is the same as the angle between the velocity vectors in Figure 4.15c
S
S
because the velocity vector v is always perpendicular to the position vector r .
Therefore, the two triangles are similar. (Two triangles are similar if the angle
between any two sides is the same for both triangles and if the ratio of the lengths
of these sides is the same.) We can now write a relationship between the lengths of
the sides for the two triangles in Figures 4.15b and 4.15c:

0 ¢vS 0
v

ϭ

Ꭽ
r
O

(a)

0 ¢rS 0
r

vi

Ꭾ
⌬r

vf
vi

The Particle in Uniform Circular Motion

where v ϭ vi ϭ vf and r ϭ ri ϭ rf . This equation can be solved for 0 ¢v 0 , and the
S
S
expression obtained can be substituted into Equation 4.4, aavg ϭ ¢v>¢t, to give the
magnitude of the average acceleration over the time interval for the particle to
move from Ꭽ to Ꭾ:
S
0 ¢vS 0
v 0 ¢r 0
0 Saavg 0 ϭ
ϭ
r ¢t
0 ¢t 0
S

Now imagine that points Ꭽ and Ꭾ in Figure 4.15b become extremely close
S
together. As Ꭽ and Ꭾ approach each other, ⌬t approaches zero, 0 ¢r 0 approaches
S
the distance traveled by the particle along the circular path, and the ratio 0 ¢r 0 >¢t
approaches the speed v. In addition, the average acceleration becomes the instantaneous acceleration at point Ꭽ. Hence, in the limit ⌬t S 0, the magnitude of the
acceleration is
v2
ac ϭ
(4.14)
r
An acceleration of this nature is called a centripetal acceleration (centripetal means
center-seeking). The subscript on the acceleration symbol reminds us that the acceleration is centripetal.

ᮤ

Period of circular motion

Quick Quiz 4.4 A particle moves in a circular path of radius r with speed v. It
then increases its speed to 2v while traveling along the same circular path. (i) The
centripetal acceleration of the particle has changed by what factor (choose one)?
(a) 0.25 (b) 0.5 (c) 2 (d) 4 (e) impossible to determine (ii) From the
same choices, by what factor has the period of the particle changed?
E XA M P L E 4 . 6

The Centripetal Acceleration of the Earth

What is the centripetal acceleration of the Earth as it moves in its orbit around the Sun?
SOLUTION
Conceptualize Think about a mental image of the Earth in a circular orbit around the Sun. We will model the Earth
as a particle and approximate the Earth’s orbit as circular (it’s actually slightly elliptical, as we discuss in Chapter 13).
Categorize The Conceptualize step allows us to categorize this problem as one of a particle in uniform circular motion.
Analyze We do not know the orbital speed of the Earth to substitute into Equation 4.14. With the help of Equation
4.15, however, we can recast Equation 4.14 in terms of the period of the Earth’s orbit, which we know is one year,
and the radius of the Earth’s orbit around the Sun, which is 1.496 ϫ 1011 m.

Combine Equations 4.14 and 4.15:

ac ϭ

Substitute numerical values:

ac ϭ




86

Chapter 4

Motion in Two Dimensions

4.5

Tangential and Radial Acceleration

Let us consider the motion of a particle along a smooth, curved path where the
velocity changes both in direction and in magnitude as described in Active Figure
4.16. In this situation, the velocity vector is always tangent to the path; the accelerS
ation vector a, however, is at some angle to the path. At each of three points Ꭽ, Ꭾ,
and Ꭿ in Active Figure 4.16, we draw dashed circles that represent the curvature
of the actual path at each point. The radius of the circles is equal to the path’s
radius of curvature at each point.
As the particle moves along the curved path in Active Figure 4.16, the direction
S
of the total acceleration vector a changes from point to point. At any instant, this
vector can be resolved into two components based on an origin at the center of
the dashed circle corresponding to that instant: a radial component ar along the
radius of the circle and a tangential component at perpendicular to this radius.
S
The total acceleration vector a can be written as the vector sum of the component
vectors:
Total acceleration



ᮣ

v2
r

(4.18)

where r is the radius of curvature of the path at the point in question. We recognize the radial component of the acceleration as the centripetal acceleration discussed in Section 4.4. The negative sign in Equation 4.18 indicates that the direction of the centripetal acceleration is toward the center of the circle representing
the radius of curvature. The direction is opposite that of the radial unit vector ˆ
r,
which always points away from the origin at the center of the circle.
S
S
S
Because ar and at are perpendicular component vectors of a, it follows that
S
the magnitude of a is a ϭ 1a r 2 ϩ a t 2 . At a given speed, ar is large when the radius
of curvature is small (as at points Ꭽ and Ꭾ in Fig. 4.16) and small when r is large
S
S
(as at point Ꭿ). The direction of at is either in the same direction as v (if v is
S
increasing) or opposite v (if v is decreasing).

Path of
particle

Ꭾ


Section 4.6

Relative Velocity and Relative Acceleration

87

In uniform circular motion, where v is constant, at ϭ 0 and the acceleration is
always completely radial as described in Section 4.4. In other words, uniform circular motion is a special case of motion along a general curved path. Furthermore, if
S
the direction of v does not change, there is no radial acceleration and the motion
is one dimensional (in this case, ar ϭ 0, but at may not be zero).

Quick Quiz 4.5 A particle moves along a path and its speed increases with time.
(i) In which of the following cases are its acceleration and velocity vectors parallel?
(a) when the path is circular (b) when the path is straight (c) when the path is
a parabola (d) never (ii) From the same choices, in which case are its acceleration and velocity vectors perpendicular everywhere along the path?

E XA M P L E 4 . 7

Over the Rise

A car exhibits a constant acceleration of 0.300 m/s2 parallel to
the roadway. The car passes over a rise in the roadway such that
the top of the rise is shaped like a circle of radius 500 m. At the
moment the car is at the top of the rise, its velocity vector is horizontal and has a magnitude of 6.00 m/s. What are the magnitude
and direction of the total acceleration vector for the car at this
instant?

at ϭ 0.300 m/s2
at

The radial acceleration is given by Equation 4.18, with v ϭ 6.00 m/s and r ϭ 500 m. The radial acceleration vector is
directed straight downward, and the tangential acceleration vector has magnitude 0.300 m/s2 and is horizontal.
16.00 m>s2 2
v2
ϭ Ϫ
ϭ Ϫ0.072 0 m>s2
ar ϭ Ϫ
r
500 m

Evaluate the radial acceleration:

2ar2 ϩ at2 ϭ 2 1Ϫ0.072 0 m>s2 2 2 ϩ 10.300 m>s2 2 2

S

Find the magnitude of a:

ϭ 0.309 m/s2
S

Find the angle f (see Fig. 4.17b) between a and the
horizontal:

4.6

Ϫ1

f ϭ tan


0

+5

xA

(a)

–5

A

P

0

+5

+5

+10

B

P

0

xA


x
vBA

Figure 4.20 A particle located at P is
described by two observers, one in
the fixed frame of reference SA and
the other in the frame SB, which
moves to the right with a constant
S
S
velocity vBA. The vector r PA is the particle’s position vector relative to SA,
S
and r PB is its position vector relative
to SB.

Galilean velocity
transformation

Let us conceptualize a sample situation in which there will be different observations for different observers. Consider the two observers A and B along the number line in Figure 4.18a. Observer A is located at the origin of a one-dimensional
xA axis, while observer B is at the position xA ϭ Ϫ5. We denote the position variable as xA because observer A is at the origin of this axis. Both observers measure
the position of point P, which is located at xA ϭ ϩ5. Suppose observer B decides
that he is located at the origin of an xB axis as in Figure 4.18b. Notice that the two
observers disagree on the value of the position of point P. Observer A claims point
P is located at a position with a value of ϩ5, whereas observer B claims it is located
at a position with a value of ϩ10. Both observers are correct, even though they
make different measurements. Their measurements differ because they are making the measurement from different frames of reference.
Imagine now that observer B in Figure 4.18b is moving to the right along the xB
axis. Now the two measurements are even more different. Observer A claims point
P remains at rest at a position with a value of ϩ5, whereas observer B claims the
position of P continuously changes with time, even passing him and moving

position vector r P B, both at time t. From Figure 4.20, we see that the vectors r P A
S
and r P B are related to each other through the expression
r P A ϭ r P B ϩ vBAt

S

S

(4.19)

S

S

By differentiating Equation 4.19 with respect to time, noting that vBA is constant, we obtain
S

S

d rP A
d rP B S
ϭ
ϩ vBA
dt
dt
uP A ϭ uP B ϩ vBA

ᮣ