Vectors and Two-Dimensional Motion

3.1 VECTORS AND THEIR PROPERTIES

3.1 Vectors and Their Properties 55

Subtracting Vectors. Vector subtraction makes use of the definition of the nega-tive of a vector. We define the operation ! as the vector ! added to the vec-tor :

! " # (! ) [3.1]

Vector subtraction is really a special case of vector addition. The geometric con-struction for subtracting two vectors is shown in Figure 3.5.

Multiplying or Dividing a Vector by a Scalar. Multiplying or dividing a vector by ascalar gives a vector. For example, if vector is multiplied by the scalar number 3,the result, written 3 , is a vector with a magnitude three times that of and point-ing in the same direction. If we multiply vector by the scalar ! 3, the resultis ! 3 , a vector with a magnitude three times that of and pointing in the op-posite direction (because of the negative sign).

A:

A:

A:

A:

A:

A:

B:

A:

B:

A:

A:

B:

B:

A:

R =

A +

B +

C +

D

A

C

B

D

Figure 3.4 A geometric construc-tion for summing four vectors.The resultant vector is the vectorthat completes the polygon.

R:

The magnitudes of two vectors and are 12 units and 8 units, respectively. Whatare the largest and smallest possible values for the magnitude of the resultantvector " # ? (a) 14.4 and 4; (b) 12 and 8; (c) 20 and 4; (d) none of these.B

:A:

R:

B:

A:

Quick Quiz 3.1

If vector is added to vector , the resultant vector # has magnitude A # Bwhen and are (a) perpendicular to each other; (b) oriented in the same direc-tion; (c) oriented in opposite directions; (d) none of these answers.

B:

A:

B:

A:

A:

B:

Quick Quiz 3.2

A – B– B

A

B

Figure 3.5 This constructionshows how to subtract vector fromvector . The vector ! has thesame magnitude as the vector , butpoints in the opposite direction.

B:

B:

A:

B:

EXAMPLE 3.1 Taking a TripGoal Find the sum of two vectors by using a graph.

Problem A car travels 20.0 km due north and then 35.0 km in a direction 60$ westof north, as in Figure 3.6. Using a graph, find the magnitude and direction of a sin-gle vector that gives the net effect of the car’s trip. This vector is called the car’s re-sultant displacement.

Strategy Draw a graph, and represent the displacement vectors as arrows. Graphi-cally locate the vector resulting from the sum of the two displacement vectors. Mea-sure its length and angle with respect to the vertical.

SolutionLet represent the first displacement vector, 20.0 km north, the second displace-ment vector, extending west of north. Carefully graph the two vectors, drawing aresultant vector with its base touching the base of and extending to the tip of .Measure the length of this vector, which turns out to be about 48 km. The angle %,measured with a protractor, is about 39$ west of north.

Remarks Notice that ordinary arithmetic doesn’t work here: the correct answer of48 km is not equal to 20.0 km # 35.0 km " 55.0 km!

Exercise 3.1Graphically determine the magnitude and direction of the displacement if a man walks 30.0 km 45$ north of east andthen walks due east 20.0 km.

Answer 46 km, 27$ north of east

B:

A:

R:

B:

A:

y(km)

40

20

60.0°

x(km)0

!

N

S

W E

B

–20

R

A

Figure 3.6 (Example 3.1) A graph-ical method for finding the resultantdisplacement vector " # .B

:A:

R:

44337_03_p53-80 10/13/04 2:24 PM Page 55

3.4 MOTION IN TWO DIMENSIONS

3.4 Motion in Two Dimensions 61

path of a projectile in Earth’s gravity field is curved in the shape of a parabola, asshown in Active Figure 3.14.

The positive x -direction is horizontal and to the right, and the y-direction is ver-tical and positive upward. The most important experimental fact about projectilemotion in two dimensions is that the horizontal and vertical motions are com-pletely independent of each other. This means that motion in one direction hasno effect on motion in the other direction. If a baseball is tossed in a parabolicpath, as in Active Figure 3.14, the motion in the y -direction will look just like a balltossed straight up under the influence of gravity. Active Figure 3.15 shows theeffect of various initial angles; note that complementary angles give the same hori-zontal range.

In general, the equations of constant acceleration developed in Chapter 2follow separately for both the x -direction and the y -direction. An important differ-ence is that the initial velocity now has two components, not just one as in thatchapter. We assume that at t ! 0, the projectile leaves the origin with an initialvelocity . If the velocity vector makes an angle "0 with the horizontal, where "0 iscalled the projection angle, then from the definitions of the cosine and sinefunctions and Active Figure 3.14, we have

v0x ! v0 cos "0 and v0y ! v0 sin "0

where v0x is the initial velocity (at t ! 0) in the x -direction and v0y is the initialvelocity in the y -direction.

Now, Equations 2.6, 2.9, and 2.10 developed in Chapter 2 for motion with con-stant acceleration in one dimension carry over to the two-dimensional case; thereis one set of three equations for each direction, with the initial velocities modifiedas just discussed. In the x -direction, with ax constant, we have

v:0

xv0x

!0

v0y

v0x

!vy

v0xvy = 0

v0x

!vy

0

v0y

v0x

y

!

!0!

g

v

v

v

v

x(m)

50

100

150

y(m)

75°

60°

45°

30°

15°

vi = 50 m/s

50 100 150 200 250

ACTIVE FIGURE 3.14The parabolic trajectory of a particlethat leaves the origin with a velocityof . Note that changes with time.However, the x-component of the ve-locity, vx , remains constant in time.Also, vy ! 0 at the peak of the trajec-tory, but the acceleration is alwaysequal to the free-fall acceleration andacts vertically downward.

Log into PhysicsNow atwww.cp7e.com, and go to ActiveFigure 3.14, where you can changethe particle’s launch angle and initialspeed. You can also observe thechanging components of velocityalong the trajectory of the projectile.

v:v:0

ACTIVE FIGURE 3.15A projectile launched from the originwith an initial speed of 50 m/s at vari-ous angles of projection. Note thatcomplementary values of the initialangle " result in the same value of R(the range of the projectile).

Log into PhysicsNow atwww.cp7e.com, and go to Active Figure 3.15, where you canvary the projection angle to observethe effect on the trajectory andmeasure the flight time.

TIP 3.4 Acceleration at theHighest PointThe acceleration in the y -direction isnot zero at the top of a projectile’s tra-jectory. Only the y -component of thevelocity is zero there. If the accelera-tion were zero, too, the projectilewould never come down!

44337_03_p53-80 10/13/04 2:25 PM Page 61

The most important experimental fact about projectile motion in two dimensions is that the horizontal and vertical motions are completely independent of each other.

3.4 Motion in Two Dimensions 61

path of a projectile in Earth’s gravity field is curved in the shape of a parabola, asshown in Active Figure 3.14.

The positive x -direction is horizontal and to the right, and the y-direction is ver-tical and positive upward. The most important experimental fact about projectilemotion in two dimensions is that the horizontal and vertical motions are com-pletely independent of each other. This means that motion in one direction hasno effect on motion in the other direction. If a baseball is tossed in a parabolicpath, as in Active Figure 3.14, the motion in the y -direction will look just like a balltossed straight up under the influence of gravity. Active Figure 3.15 shows theeffect of various initial angles; note that complementary angles give the same hori-zontal range.

In general, the equations of constant acceleration developed in Chapter 2follow separately for both the x -direction and the y -direction. An important differ-ence is that the initial velocity now has two components, not just one as in thatchapter. We assume that at t ! 0, the projectile leaves the origin with an initialvelocity . If the velocity vector makes an angle "0 with the horizontal, where "0 iscalled the projection angle, then from the definitions of the cosine and sinefunctions and Active Figure 3.14, we have

v0x ! v0 cos "0 and v0y ! v0 sin "0

where v0x is the initial velocity (at t ! 0) in the x -direction and v0y is the initialvelocity in the y -direction.

Now, Equations 2.6, 2.9, and 2.10 developed in Chapter 2 for motion with con-stant acceleration in one dimension carry over to the two-dimensional case; thereis one set of three equations for each direction, with the initial velocities modifiedas just discussed. In the x -direction, with ax constant, we have

v:0

xv0x

!0

v0y

v0x

!vy

v0xvy = 0

v0x

!vy

0

v0y

v0x

y

!

!0!

g

v

v

v

v

x(m)

50

100

150

y(m)

75°

60°

45°

30°

15°

vi = 50 m/s

50 100 150 200 250

ACTIVE FIGURE 3.14The parabolic trajectory of a particlethat leaves the origin with a velocityof . Note that changes with time.However, the x-component of the ve-locity, vx , remains constant in time.Also, vy ! 0 at the peak of the trajec-tory, but the acceleration is alwaysequal to the free-fall acceleration andacts vertically downward.

Log into PhysicsNow atwww.cp7e.com, and go to ActiveFigure 3.14, where you can changethe particle’s launch angle and initialspeed. You can also observe thechanging components of velocityalong the trajectory of the projectile.

v:v:0

ACTIVE FIGURE 3.15A projectile launched from the originwith an initial speed of 50 m/s at vari-ous angles of projection. Note thatcomplementary values of the initialangle " result in the same value of R(the range of the projectile).

Log into PhysicsNow atwww.cp7e.com, and go to Active Figure 3.15, where you canvary the projection angle to observethe effect on the trajectory andmeasure the flight time.

TIP 3.4 Acceleration at theHighest PointThe acceleration in the y -direction isnot zero at the top of a projectile’s tra-jectory. Only the y -component of thevelocity is zero there. If the accelera-tion were zero, too, the projectilewould never come down!

44337_03_p53-80 10/13/04 2:25 PM Page 61

62 Chapter 3 Vectors and Two-Dimensional Motion

vx ! v0x " axt [3.11a]

#x ! v0x t " axt2 [3.11b]

vx2 ! v0x

2 " 2ax #x [3.11c]

where v0x ! v0 cos . In the y -direction, we have

vy ! v0y " ayt [3.12a]

#y ! v0yt " ayt2 [3.12b]

vy2 ! v0y

2 " 2ay#y [3.12c]

where v0y ! v0 sin $0 and ay is constant. The object’s speed v can be calculatedfrom the components of the velocity using the Pythagorean theorem:

The angle that the velocity vector makes with the x -axis is given by

This formula for $, as previously stated, must be used with care, because the in-verse tangent function returns values only between % 90& and " 90&. Adding 180& isnecessary for vectors lying in the second or third quadrant.

The kinematic equations are easily adapted and simplified for projectiles closeto the surface of the Earth. In that case, assuming air friction is negligible, the ac-celeration in the x-direction is 0 (because air resistance is neglected). This meansthat ax ! 0, and the projectile’s velocity component along the x -direction remainsconstant. If the initial value of the velocity component in the x -direction isv0x ! v0 cos $0, then this is also the value of vx at any later time, so

vx ! v0x ! v0 cos $0 ! constant [3.13a]

while the horizontal displacement is simply

x ! v0x t ! (v0 cos $0)t [3.13b]

For the motion in the y -direction, we make the substitution ay ! %g and v0y ! v0 sin $0 in Equations 3.12, giving

vy ! v0 sin $0 % gt [3.14a]

[3.14b]

vy2 ! (v0 sin $0)2 % 2g #y [3.14c]

The important facts of projectile motion can be summarized as follows:

1. Provided air resistance is negligible, the horizontal component of the velocityvx remains constant because there is no horizontal component of acceleration.

2. The vertical component of the acceleration is equal to the free fall acceleration % g.3. The vertical component of the velocity vy and the displacement in the

y -direction are identical to those of a freely falling body.4. Projectile motion can be described as a superposition of two independent mo-

tions in the x - and y -directions.

#y ! (v0 sin $0)t % 12 gt 2

#

$ ! tan%1 ! vy

vx"

v ! !v 2x " v 2

y

12

$0

12

EXAMPLE 3.4 Projectile Motion with DiagramsGoal Approximate answers in projectile motion using a motion diagram.

Problem A ball is thrown so that its initial vertical and horizontal components of velocity are 40 m/s and 20 m/s,respectively. Use a motion diagram to estimate the ball’s total time of flight and the distance it traverses beforehitting the ground.

A water fountain. The individualwater streams follow parabolictrajectories. The horizontal rangeand maximum height of a givenstream of water depend on theelevation angle of that stream’s initialvelocity as well as its initial speed.

HIRB

/Inde

x Sto

ck

44337_03_p53-80 10/14/04 2:12 PM Page 62 62 Chapter 3 Vectors and Two-Dimensional Motion

vx ! v0x " axt [3.11a]

#x ! v0x t " axt2 [3.11b]

vx2 ! v0x

2 " 2ax #x [3.11c]

where v0x ! v0 cos . In the y -direction, we have

vy ! v0y " ayt [3.12a]

#y ! v0yt " ayt2 [3.12b]

vy2 ! v0y

2 " 2ay#y [3.12c]

where v0y ! v0 sin $0 and ay is constant. The object’s speed v can be calculatedfrom the components of the velocity using the Pythagorean theorem:

The angle that the velocity vector makes with the x -axis is given by

This formula for $, as previously stated, must be used with care, because the in-verse tangent function returns values only between % 90& and " 90&. Adding 180& isnecessary for vectors lying in the second or third quadrant.

The kinematic equations are easily adapted and simplified for projectiles closeto the surface of the Earth. In that case, assuming air friction is negligible, the ac-celeration in the x-direction is 0 (because air resistance is neglected). This meansthat ax ! 0, and the projectile’s velocity component along the x -direction remainsconstant. If the initial value of the velocity component in the x -direction isv0x ! v0 cos $0, then this is also the value of vx at any later time, so

vx ! v0x ! v0 cos $0 ! constant [3.13a]

while the horizontal displacement is simply

x ! v0x t ! (v0 cos $0)t [3.13b]

For the motion in the y -direction, we make the substitution ay ! %g and v0y ! v0 sin $0 in Equations 3.12, giving

vy ! v0 sin $0 % gt [3.14a]

[3.14b]

vy2 ! (v0 sin $0)2 % 2g #y [3.14c]

The important facts of projectile motion can be summarized as follows:

1. Provided air resistance is negligible, the horizontal component of the velocityvx remains constant because there is no horizontal component of acceleration.

2. The vertical component of the acceleration is equal to the free fall acceleration % g.3. The vertical component of the velocity vy and the displacement in the

y -direction are identical to those of a freely falling body.4. Projectile motion can be described as a superposition of two independent mo-

tions in the x - and y -directions.

#y ! (v0 sin $0)t % 12 gt 2

#

$ ! tan%1 ! vy

vx"

v ! !v 2x " v 2

y

12

$0

12

EXAMPLE 3.4 Projectile Motion with DiagramsGoal Approximate answers in projectile motion using a motion diagram.

Problem A ball is thrown so that its initial vertical and horizontal components of velocity are 40 m/s and 20 m/s,respectively. Use a motion diagram to estimate the ball’s total time of flight and the distance it traverses beforehitting the ground.

A water fountain. The individualwater streams follow parabolictrajectories. The horizontal rangeand maximum height of a givenstream of water depend on theelevation angle of that stream’s initialvelocity as well as its initial speed.

HIRB

/Inde

x Sto

ck

44337_03_p53-80 10/14/04 2:12 PM Page 62

ax = 0 ay = g

62 Chapter 3 Vectors and Two-Dimensional Motion

vx ! v0x " axt [3.11a]

#x ! v0x t " axt2 [3.11b]

vx2 ! v0x

2 " 2ax #x [3.11c]

where v0x ! v0 cos . In the y -direction, we have

vy ! v0y " ayt [3.12a]

#y ! v0yt " ayt2 [3.12b]

vy2 ! v0y

2 " 2ay#y [3.12c]

where v0y ! v0 sin $0 and ay is constant. The object’s speed v can be calculatedfrom the components of the velocity using the Pythagorean theorem:

The angle that the velocity vector makes with the x -axis is given by

This formula for $, as previously stated, must be used with care, because the in-verse tangent function returns values only between % 90& and " 90&. Adding 180& isnecessary for vectors lying in the second or third quadrant.

The kinematic equations are easily adapted and simplified for projectiles closeto the surface of the Earth. In that case, assuming air friction is negligible, the ac-celeration in the x-direction is 0 (because air resistance is neglected). This meansthat ax ! 0, and the projectile’s velocity component along the x -direction remainsconstant. If the initial value of the velocity component in the x -direction isv0x ! v0 cos $0, then this is also the value of vx at any later time, so

vx ! v0x ! v0 cos $0 ! constant [3.13a]

while the horizontal displacement is simply

x ! v0x t ! (v0 cos $0)t [3.13b]

For the motion in the y -direction, we make the substitution ay ! %g and v0y ! v0 sin $0 in Equations 3.12, giving

vy ! v0 sin $0 % gt [3.14a]

[3.14b]

vy2 ! (v0 sin $0)2 % 2g #y [3.14c]

The important facts of projectile motion can be summarized as follows:

1. Provided air resistance is negligible, the horizontal component of the velocityvx remains constant because there is no horizontal component of acceleration.

2. The vertical component of the acceleration is equal to the free fall acceleration % g.3. The vertical component of the velocity vy and the displacement in the

y -direction are identical to those of a freely falling body.4. Projectile motion can be described as a superposition of two independent mo-

tions in the x - and y -directions.

#y ! (v0 sin $0)t % 12 gt 2

#

$ ! tan%1 ! vy

vx"

v ! !v 2x " v 2

y

12

$0

12

EXAMPLE 3.4 Projectile Motion with DiagramsGoal Approximate answers in projectile motion using a motion diagram.

Problem A ball is thrown so that its initial vertical and horizontal components of velocity are 40 m/s and 20 m/s,respectively. Use a motion diagram to estimate the ball’s total time of flight and the distance it traverses beforehitting the ground.

A water fountain. The individualwater streams follow parabolictrajectories. The horizontal rangeand maximum height of a givenstream of water depend on theelevation angle of that stream’s initialvelocity as well as its initial speed.

HIRB

/Inde

x Sto

ck

44337_03_p53-80 10/14/04 2:12 PM Page 62

62 Chapter 3 Vectors and Two-Dimensional Motion

vx ! v0x " axt [3.11a]

#x ! v0x t " axt2 [3.11b]

vx2 ! v0x

2 " 2ax #x [3.11c]

where v0x ! v0 cos . In the y -direction, we have

vy ! v0y " ayt [3.12a]

#y ! v0yt " ayt2 [3.12b]

vy2 ! v0y

2 " 2ay#y [3.12c]

where v0y ! v0 sin $0 and ay is constant. The object’s speed v can be calculatedfrom the components of the velocity using the Pythagorean theorem:

The angle that the velocity vector makes with the x -axis is given by

This formula for $, as previously stated, must be used with care, because the in-verse tangent function returns values only between % 90& and " 90&. Adding 180& isnecessary for vectors lying in the second or third quadrant.

The kinematic equations are easily adapted and simplified for projectiles closeto the surface of the Earth. In that case, assuming air friction is negligible, the ac-celeration in the x-direction is 0 (because air resistance is neglected). This meansthat ax ! 0, and the projectile’s velocity component along the x -direction remainsconstant. If the initial value of the velocity component in the x -direction isv0x ! v0 cos $0, then this is also the value of vx at any later time, so

vx ! v0x ! v0 cos $0 ! constant [3.13a]

while the horizontal displacement is simply

x ! v0x t ! (v0 cos $0)t [3.13b]

For the motion in the y -direction, we make the substitution ay ! %g and v0y ! v0 sin $0 in Equations 3.12, giving

vy ! v0 sin $0 % gt [3.14a]

[3.14b]

vy2 ! (v0 sin $0)2 % 2g #y [3.14c]

The important facts of projectile motion can be summarized as follows:

1. Provided air resistance is negligible, the horizontal component of the velocityvx remains constant because there is no horizontal component of acceleration.

2. The vertical component of the acceleration is equal to the free fall acceleration % g.3. The vertical component of the velocity vy and the displacement in the

y -direction are identical to those of a freely falling body.4. Projectile motion can be described as a superposition of two independent mo-

tions in the x - and y -directions.

#y ! (v0 sin $0)t % 12 gt 2

#

$ ! tan%1 ! vy

vx"

v ! !v 2x " v 2

y

12

$0

12

EXAMPLE 3.4 Projectile Motion with DiagramsGoal Approximate answers in projectile motion using a motion diagram.

Problem A ball is thrown so that its initial vertical and horizontal components of velocity are 40 m/s and 20 m/s,respectively. Use a motion diagram to estimate the ball’s total time of flight and the distance it traverses beforehitting the ground.

A water fountain. The individualwater streams follow parabolictrajectories. The horizontal rangeand maximum height of a givenstream of water depend on theelevation angle of that stream’s initialvelocity as well as its initial speed.

HIRB

/Inde

x Sto

ck

44337_03_p53-80 10/14/04 2:12 PM Page 62

62 Chapter 3 Vectors and Two-Dimensional Motion

vx ! v0x " axt [3.11a]

#x ! v0x t " axt2 [3.11b]

vx2 ! v0x

2 " 2ax #x [3.11c]

where v0x ! v0 cos . In the y -direction, we have

vy ! v0y " ayt [3.12a]

#y ! v0yt " ayt2 [3.12b]

vy2 ! v0y

2 " 2ay#y [3.12c]

where v0y ! v0 sin $0 and ay is constant. The object’s speed v can be calculatedfrom the components of the velocity using the Pythagorean theorem:

The angle that the velocity vector makes with the x -axis is given by

This formula for $, as previously stated, must be used with care, because the in-verse tangent function returns values only between % 90& and " 90&. Adding 180& isnecessary for vectors lying in the second or third quadrant.

The kinematic equations are easily adapted and simplified for projectiles closeto the surface of the Earth. In that case, assuming air friction is negligible, the ac-celeration in the x-direction is 0 (because air resistance is neglected). This meansthat ax ! 0, and the projectile’s velocity component along the x -direction remainsconstant. If the initial value of the velocity component in the x -direction isv0x ! v0 cos $0, then this is also the value of vx at any later time, so

vx ! v0x ! v0 cos $0 ! constant [3.13a]

while the horizontal displacement is simply

x ! v0x t ! (v0 cos $0)t [3.13b]

For the motion in the y -direction, we make the substitution ay ! %g and v0y ! v0 sin $0 in Equations 3.12, giving

vy ! v0 sin $0 % gt [3.14a]

[3.14b]

vy2 ! (v0 sin $0)2 % 2g #y [3.14c]

The important facts of projectile motion can be summarized as follows:

1. Provided air resistance is negligible, the horizontal component of the velocityvx remains constant because there is no horizontal component of acceleration.

2. The vertical component of the acceleration is equal to the free fall acceleration % g.3. The vertical component of the velocity vy and the displacement in the

y -direction are identical to those of a freely falling body.4. Projectile motion can be described as a superposition of two independent mo-

tions in the x - and y -directions.

#y ! (v0 sin $0)t % 12 gt 2

#

$ ! tan%1 ! vy

vx"

v ! !v 2x " v 2

y

12

$0

12

EXAMPLE 3.4 Projectile Motion with DiagramsGoal Approximate answers in projectile motion using a motion diagram.

Problem A ball is thrown so that its initial vertical and horizontal components of velocity are 40 m/s and 20 m/s,respectively. Use a motion diagram to estimate the ball’s total time of flight and the distance it traverses beforehitting the ground.

A water fountain. The individualwater streams follow parabolictrajectories. The horizontal rangeand maximum height of a givenstream of water depend on theelevation angle of that stream’s initialvelocity as well as its initial speed.

HIRB

/Inde

x Sto

ck

44337_03_p53-80 10/14/04 2:12 PM Page 62

For higher level problems we need to

sub. from one eqn. into another. Usually for

time.

3.5 RELATIVE VELOCITY

Most of the time, we use a stationary frame of reference relative to Earth, but occasionally we use a moving frame of reference associated with a bus, car, or plane moving with constant velocity relative to Earth.

68 Chapter 3 Vectors and Two-Dimensional Motion

(c) Find the magnitude and direction of the velocity.

Find the magnitude using the Pythagorean theorem andthe results of parts (a) and (b): ! 411 m/s

v ! !v 2x " v 2

y ! !(# 1.40 $ 102 m/s)2 " (386 m/s)2

Use the inverse tangent function to find the angle: #19.9%& ! tan#1 ! vy

vx" tan#1 ! #1.40 $ 102 m/s

386 m/s " !

Remarks Notice the symmetry: The kinematic equations for the x- and y -directions are handled in exactly the sameway. Having a nonzero acceleration in the x -direction doesn’t greatly increase the difficulty of the problem.

Exercise 3.8Suppose a rocket-propelled motorcycle is fired from rest horizontally across a canyon 1.00 km wide. (a) What minimumconstant acceleration in the x -direction must be provided by the engines so the cycle crosses safely if the opposite side is0.750 km lower than the starting point? (b) At what speed does the motorcycle land if it maintains this constant hori-zontal component of acceleration? Neglect air drag, but remember that gravity is still acting in the negative y -direction.

Answers (a) 13.1 m/s2 (b) 202 m/s

In a stunt similar to that described in Exercise 3.8, motorcycle daredevil EvelKnievel tried to vault across Hells Canyon, part of the Snake River system in Idaho,on his rocket-powered Harley-Davidson X-2 “Skycycle.” (See the chapter-openingphoto on page 53). He lost consciousness at takeoff and released a lever, prema-turely deploying his parachute and falling short of the other side. He landed safelyin the canyon.

3.5 RELATIVE VELOCITYRelative velocity is all about relating the measurements of two different observers,one moving with respect to the other. The measured velocity of an object dependson the velocity of the observer with respect to the object. On highways, for example,cars moving in the same direction are often moving at high speed relative to Earth,but relative each other they hardly move at all. To an observer at rest at the side ofthe road, a car might be traveling at 60 mi/h, but to an observer in a truck travelingin the same direction at 50 mi/h, the car would appear to be traveling only 10 mi/h.

So measurements of velocity depend on the reference frame of the observer. Ref-erence frames are just coordinate systems. Most of the time, we use a stationary frameof reference relative to Earth, but occasionally we use a moving frame of referenceassociated with a bus, car, or plane moving with constant velocity relative to Earth.

In two dimensions, relative velocity calculations can be confusing, so a system-atic approach is important and useful. Let E be an observer, assumed stationarywith respect to Earth. Let two cars be labeled A and B, and introduce the followingnotation (see Figure 3.21):

! the position of Car A as measured by E (in a coordinate system fixed withrespect to Earth).

! the position of Car B as measured by E.

! the position of Car A as measured by an observer in Car B.

According to the preceding notation, the first letter tells us what the vector ispointing at and the second letter tells us where the position vector starts. The posi-tion vectors of Car A and Car B relative to E, and , are given in the figure.How do we find , the position of Car A as measured by an observer in Car B?We simply draw an arrow pointing from Car B to Car A, which can be obtained bysubtracting from :

! # [3.15]

Now, the rate of change of these quantities with time gives us the relationship

r:BEr:AEr:AB

r:AEr:BE

r:AB

r:BEr:AE

r:AB

r:BE

r:AE

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3.5 Relative Velocity 69

between the associated velocities:

! " [3.16]

The coordinate system of observer E need not be fixed to Earth, although it oftenis. Take careful note of the pattern of subscripts; rather than memorize Equation3.15, it’s better to study the short derivation shown in Figure 3.21. Note also thatthe equation doesn’t work for observers traveling a sizeable fraction of the speedof light, when Einstein’s theory of special relativity comes into play.

v:BEv:AEv:AB

y

xE

B

A

r AE

rAB ! rAE " rBE

rBE

Figure 3.21 The position of Car A relative to Car Bcan be found by vector subtraction. The rate of changeof the resultant vector with respect to time is therelative velocity equation.

PROBLEM-SOLVING STRATEGY Relative Velocity1. Label each object involved (usually three) with a letter that reminds you of what it

is (for example, E for Earth).2. Look through the problem for phrases such as “The velocity of A relative to B,”

and write the velocities as ‘ ’. When a velocity is mentioned but it isn’t explicitlystated as relative to something, it’s almost always relative to Earth.

3. Take the three velocities you’ve found and assemble them into an equation justlike Equation 3.16, with subscripts in an analogous order.

4. There will be two unknown components. Solve for them with the x - and y -components of the equation developed in step 3.

v:AB

EXAMPLE 3.9 Pitching Practice on the TrainGoal Solve a one-dimensional relative velocity problem.

Problem A train is traveling with a speed of 15.0 m/s relative to Earth. A passenger standing at the rear of the trainpitches a baseball with a speed of 15.0 m/s relative to the train off the back end, in the direction opposite the motionof the train. What is the velocity of the baseball relative to Earth?

Strategy Solving these problems involves putting the proper subscripts on the velocities and arranging themas in Equation 3.16. In the first sentence of the problem statement, we are informed that the train travels at“15.0 m/s relative to Earth.” This quantity is , with T for train and E for Earth. The passenger throws the base-ball at “15 m/s relative to the train,” so this quantity is , where B stands for baseball. The second sentenceasks for the velocity of the baseball relative to Earth, . The rest of the problem can be solved by identifyingthe correct components of the known quantities and solving for the unknowns, using an analog of Equation 3.16.

SolutionWrite the x-components of the known quantities:

v:BE

v:BT

v:TE

! # 15 m/s(v:TE)x

! " 15 m/s(v:BT)x

Follow Equation 3.16: ! " (v:TE)x(v:BE)x(v:BT)x

Insert the given values, and solve: " 15 m/s ! " 15 m/s

! 0(v:BE)x

(v:BE)x

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Crossing a river is a common problem.