Solucionario Fundamentos de Física 9na edición Capitulo 3

3Vectors and Two-Dimensional Motion

CLICKER QUESTIONS

Question A3.01

Description: Understanding acceleration: identifying nonzero acceleration.

Question

Consider the following situations:

• A car slowing down at a stop sign

• A ball being swung in a circle at constant speed

• A vibrating string

• The Moon orbiting the Earth

• A skydiver falling at terminal speed

• An astronaut in an orbiting space station

• A ball rolling down a hill

• A person driving down a straight section of highway at constant speed with her foot on the accelerator

• A molecule in the fl oor of this room

In how many of the situations is the object accelerating?

1. One of the situations 2. Two of the situations 3. Three of the situations 4. Four of the situations 5. Five of the situations 6. Six of the situations 7. Seven of the situations 8. Eight of the situations 9. Nine of the situations 10. None of the situations

Commentary

Purpose: To introduce the concept of acceleration, explore your preexisting defi nitions of the term, and contrast “everyday” usage of the word with the way physicists use it.

Discussion: In physics, “acceleration” means the rate of change of velocity with respect to time. The more the velocity changes during a certain interval (say, 1 second), the larger the acceleration is.

75

56157_03_ch03_p075-118.indd 7556157_03_ch03_p075-118.indd 75 1/4/08 8:46:56 PM1/4/08 8:46:56 PM

There are three ways in which the velocity can change: (1) the speed increases; (2) the speed decreases; or (3) the direction of motion changes. If any one of these occurs, then the object is said to be “accelerating.” In other words, unlike everyday usage, “accelerating” does not necessarily mean “speeding up.”

When a car is slowing down, we say it is accelerating. We generally avoid the phrases “deceleration” and “negative acceleration” as unnecessary and unhelpful. We avoid “negative acceleration” because an object can have a negative value for acceleration and still be speeding up—for example, if it is moving in the negative direction.

When a ball is moving in a circle, its direction of motion is constantly changing, so it is accelerating even if its speed is constant.

A vibrating string (except for its endpoints) is accelerating at most instants in time, though it happens so quickly that we cannot perceive the changes in velocity. The only time it is not accelerating is when its speed is maximum. When the string stops and reverses direction at the extremes of its motion, its accelera-tion is largest.

The Moon orbiting the Earth is moving in a circle, so it is accelerating.

A skydiver moving at terminal speed has a constant velocity and thus zero acceleration.

An astronaut in an orbiting space station may not be moving relative to the space station, but the station is not a proper frame of reference since it is accelerating as it orbits the Earth. (In this course, a proper frame of reference has no acceleration.) The Earth is a better frame of reference, and the astronaut is moving in a circle around it, so she is accelerating.

A ball rolling down a hill generally speeds up.

A person (or car, or any object) moving at constant speed along a straight line has constant velocity, and thus zero acceleration. This is true even though we must press the “accelerator” pedal to maintain constant velocity.

A molecule in a solid that is not at absolute zero temperature is vibrating, so like the vibrating string it is accelerating most of the time. The solid as a whole may still be at rest.

Key Points:

• An object’s acceleration is the rate at which its velocity is changing.

• If an object’s speed or direction of motion changes, we say it is “accelerating.”

• We avoid using the phrases “deceleration” and “negative acceleration” to mean “slowing down.”

For Instructors Only

A nice feature of this style of question (“how many of the following . . .”) is that even if a student answers the “correct” number of objects (7), there are 36 different combinations to arrive at this answer, so there is really no need to focus any attention at all on the correctness of students’ multiple-choice answers. Rather, class discussion should proceed immediately to considering each situation independently and applying the defi nition of acceleration.

We encourage students to wrestle with their own defi nitions at this stage. This question may be used prior to any formal presentation of acceleration, since everyone already “knows” the term; the disagreements that typically arise prepare students to appreciate and better understand the formal defi nition.

76 Chapter 3


Vectors and Two-Dimensional Motion 77

It is not too early to mention forces, at least informally. Many students already have some ideas about force that can help them sort out what you are trying to explain regarding acceleration. For example, in the cases that have nonzero acceleration, identify the force(s) causing it; in cases with zero acceleration, discuss how any forces present balance out.

Question A3.02

Description: Develop understanding of vector acceleration in curvilinear motion.

Question

A pendulum is released from rest at position A and swings toward the vertical under the infl uence of gravity as depicted below.

A

B

87

6 2

1

5 34

C

When at position B, which direction most nearly corresponds to the direction of the acceleration?

Enter (9) if the direction cannot be determined.

Commentary

Purpose: Develop your understanding of the vector nature of acceleration in curvilinear motion.

Description: Acceleration is the rate of change of velocity. If an object’s speed is changing, that means the length of its velocity vector is changing, so the acceleration must have a nonzero component parallel to the velocity vector. If an object’s direction of motion is changing, that means the direction of the velocity vector is changing, so the acceleration must have a nonzero component perpendicular to the velocity vector.

At point B, the pendulum bob is both speeding up and changing direction. At that moment its velocity points in a direction tangential to the circular path it follows: direction (3). Since it is speeding up, the acceleration must have a component that points in direction (3). Since it is traveling in a curved path, it must have an acceleration component that points towards the center of the curve (the so-called “centrip-etal acceleration”), direction (1). Thus, the acceleration vector must point somewhere between directions (1) and (3), so (2) is probably the best choice.

(To fi nd the exact direction, we would have to calculate the relative sizes of these two components of the acceleration.)


78 Chapter 3

Key Points:

• Acceleration is a vector that describes the rate of change of the velocity vector’s magnitude and its direction.

• It is often useful to divide the acceleration into components that are parallel (tangential) and perpendicular to the object’s direction of motion.

• A nonzero tangential component of acceleration indicates that the object is changing speed.

• A nonzero perpendicular component of acceleration indicates that the object is changing direction.

For Instructors Only

Students often neglect one component or the other in this question. Common answers include (1) (fi xated on the centripetal acceleration or the string), (3) (fi xated on tangential acceleration and the increase in speed), and (4) (fi xated on gravity).

Students choosing answer (5) might be revealing a misunderstanding of the fi ctitious “centrifugal force.”

Answer (9), impossible to determine, is defensible if chosen because students don’t know enough to determine the relative sizes of the two acceleration components and therefore can’t choose among directions (1), (2), and (3).

QUICK QUIZZES

1. (c). The largest possible magnitude of the resultant occurs when the two vectors are in the same direction. In this case, the magnitude of the resultant is the sum of the magnitudes of A

�� and B

��:

R = A + B = 20 units. The smallest possible magnitude of the resultant occurs when the two vec-tors are in opposite directions, and the magnitude is the difference of the magnitudes of A

�� and B

��:

R = |A − B| = 4 units.

2. Vector x-component y-component

A�� − +

B�� + −

A��

+ B�� − −

3. (b). If velocity is constant, the acceleration (rate of change in velocity) is zero. An object may have constant speed (magnitude of velocity) but still be accelerating due to a change in direction of the velocity. If an object is following a curved path, it is accelerating because the velocity is changing in direction.

4. (a). Any change in the magnitude and/or direction of the velocity is an acceleration. The gas pedal and the brake produce accelerations by altering the magnitude of the velocity. The steering wheel produces accelerations by altering the direction of the velocity.

5. (c). A projectile has constant horizontal velocity. Thus, if the runner throws the ball straight up into the air, the ball maintains the horizontal velocity it had before it was thrown (that is, the velocity of the runner). In the runner’s frame of reference, the ball appears to go straight upward and come straight downward. To a stationary observer, the ball follows a parabolic trajectory, moving with the same horizontal velocity as the runner and staying above the runner’s hand.



6. (b). The velocity is always tangential to the path while the acceleration is always directed vertically downward. Thus, the velocity and acceleration are perpendicular only where the path is horizontal. This only occurs at the peak of the path.

ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. At maximum height ( )max∆y h= , the vertical velocity of the stone will be zero. Thus, v vy y ya y2

02 2= + ( )∆ gives

ha g

y y

ymax

sin s=

−=

−−( ) =

−( )v v v202

02 2 2

2

0

2

45θ m s iin .

..

2 55 0

2 9 8069 3

°( )−( ) =

m sm

2

and we see that choice (c) is the correct answer.

2. The skier has zero initial velocity in the vertical direction ( )v0 0y = and undergoes a vertical displacement of ∆y = −3 20. m. The constant acceleration in the vertical direction is a gy = − , so we use ∆y t a ty y= +v0

12

2 to fi nd the time of fl ight as

− = + −( )3 20 01

29 80 2. . m m s2 t or t =

−( )−

=2 3 20

9 800 808

.

..

m

m s s2

During this time, the object moves with constant horizontal velocity v vx x= =0 22 0. m s. The horizontal distance traveled during the fl ight is

∆x tx= = ( )( ) =v 22 0 0 808 17 8. . .m s s m

which is choice (d).

3. Choose coordinate system with north as the positive y-direction and east as the positive x-direction. The velocity of the cruise ship relative to Earth is

�vCE = 4 50. m s due north, with

components of ( ) ( ) .� �v vCE CEx y= =0 4 50and m s. The velocity of the patrol boat relative to

Earth is �v EP = 5 20. m s at 45.0° north of west, with components of

� �v vE EP P( ) = − ° = −( )( ) = −

xcos . . . .45 0 5 20 0 707 3m s 668 m s

and

� �v vE EP P( ) = + ° = ( )( ) =

ysin . . . .45 0 5 20 0 707 3 68m s m s

Thus, the velocity of the cruise ship relative to the patrol boat is � � �v v vCE PECP = − , which has

components of

� � �v v vCP CE PE( ) = ( ) − ( ) = − −( ) = +

x x x0 3 68 3 68. .m s mm s

and

� � �v v vCP CE PE( ) = ( ) − ( ) = − = +

y y y4 50 3 68 0 82. . .m s m s 33 m s

Choice (a) is then the correct answer.


80 Chapter 3

4. For vectors in the x-y plane, their components their components have the signs indicated in the following table:

Quadrant of VectorFirst Second Third Fourth

x-component Positive Negative Negative Positive

y-component Positive Positive Negative Negative

Thus, a vector having components of opposite sign must lie in either the second or fourth quadrants and choice (e) is the correct answer.

5. The path followed (and distance traveled) by the athlete is shown in the sketch, along with the vectors for the initial position, fi nal position, and change in position.

The average speed for the elapsed time interval ∆t is

vav = d

t∆

and the magnitude of the average velocity for this time interval is

�

�v

rav =

∆∆t

The sketch clearly shows that d > ∆r�

in this case, meaning that vav av> �v and that (a) is the

correct choice.

6. Consider any two very closely spaced points on a circular path and draw vectors of the same length (to represent a constant velocity magnitude or speed) tangent to the path at each of these points as shown in the leftmost diagram below. Now carefully move the velocity vector

�vf at

the second point down so its tail is at the fi rst point as shown in the rightmost diagram. Then, draw the vector difference ∆� � �

v v v= −f i and observe that if the start of this vector were located on the circular path midway between the two points, its direction would be inward toward the center of the circle.

Thus, for an object following the circular path at constant speed, its instantaneous acceleration,

� �a v= ( )

→lim∆

∆ ∆t

t0

, at the point midway between your initial and end points is directed toward the

center of the circle, and the only correct choice for this question is (d).



7. Whether on Earth or the Moon, a golf ball is in free fall with a constant downward acceleration of magnitude determined by local gravity from the time it leaves the tee until it strikes the ground or other object. Thus, the vertical component of velocity is constantly changing while the horizontal component of velocity is constant. Note that the speed (or magnitude of the velocity)

v v v= = +�v x y

2 2

will change in time since vy changes in time. Thus, the only correct choices for this question are (b) and (d).

8. At maximum altitude, the projectile’s vertical component of velocity is zero. The time for the projectile to reach its maximum height is found from v vy y ya t= +0 as

ta g g

y y h y

ymax

maxsin sin

=−

=−

−==

v v v v∆ 00 0 0 00 θ θ

Since the acceleration of gravity on the Moon is one-sixth that of Earth, we see that (with v0 0and θ kept constant)

tg gmax

sin sin sinMoon

Moon Earth

= = =v v v0 0 0 0 0

66

θ θ θθ0 6g

tEarth

Eart

⎛⎝⎜

⎞⎠⎟

= max

and the correct answer for this question is (e).

9. The boat moves with a constant horizontal velocity (or its velocity relative to Earth has components of ( ) ( )

� �v vBE BEconstant, andx y= = =v0 0), where the y-axis is vertical and the x-axis

is parallel to the keel of the boat. Once the wrench is released, it is a projectile whose velocity relative to Earth has components of

�vWE( ) = + = + =

x x xa tv v v0 0 00 and �vWE( ) = + = − = −

y y ya t gt gtv0 0

The velocity of the wrench relative to the boat ( )� � �v v vWB WE BE= − has components of

� � �v v vWB WE BE( ) = ( ) − ( ) = − =

x x xv v0 0 0 and

� � �v v vWB WE BE( ) = ( ) − ( ) = − − = −

y y ygt gt0

Thus, the wrench has zero horizontal velocity relative to the boat and will land on the deck at a point directly below where it was released (i.e., at the base of the mast). The correct choice is (b).

10. While in the air, the baseball is a projectile whose velocity always has a constant horizontal component ( )v vx x= 0 and a vertical component that changes at a constant rate ( )∆ ∆vy yt a g= = − . At the highest point on the path, the vertical velocity of the ball is momentarily zero. Thus,

at this point, the resultant velocity of the ball is horizontal and its acceleration continues to be directed downward ( , )a a gx y= = −0 . The only correct choice given for this question is (c).


82 Chapter 3

11. Note that for each ball, v0 0y = , Thus, the vertical velocity of each ball when it reaches the ground ( )∆y h= − is given by v vy y ya y2

02 2= + ( )∆ as

vy g h gh= − + −( ) −( ) = −0 2 2

and the time required for each ball to reach the ground is given by v vy y ya t= +0 as

ta

gh

g

h

gy y

y

=−

=− −

−=

v v0 2 0 2

The speeds ( )i.e., magnitudes of total velocities of the balls at ground level are

Red Ball: v v v v vR x y x gh gh= + = + −( ) = +2 202 2

022 2

Blue Ball: v v v vB x y x gh gh gh= + = + −( ) = + =2 202 2

2 0 2 2

Therefore, we see that the two balls reach the ground at the same time but with different speeds ( )v vR B> , so only choice (b) is correct.

12. When the apple fi rst comes off the tree, it is moving forward with the same horizontal velocity as the truck. Since, while in free fall, the apple has zero horizontal acceleration, it will maintain this constant horizontal velocity as it falls. Also, while in free fall, the apple has constant downward

acceleration ( )a gy = − , so its downward speed increases uniformly in time.

(i) As the truck moves left to right past an observer stationary on the ground, this observer will see both the constant velocity horizontal motion and the uniformly accelerated downward motion of the apple. The curve that best describes the path of the apple as seen by this observer is (a).

(ii) An observer on the truck moves with the same horizontal motion as does the apple. This observer does not detect any horizontal motion of the apple relative to him. However, this observer does detect the uniformly accelerated vertical motion of the apple. The curve best describing the path of the apple as seen by the observer on the truck is (b).

13. Of the choices listed, the quantities that have magnitude or size, but no direction, associated with them (i.e., scalar quantities) are (b) temperature, (c) volume, and (e) height. The other quantities, (a) velocity of a sports car and (d) displacement of a tennis player who moves from the court’s backline to the net, have both magnitude and direction associated with them, and are both vector quantities.

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS

2. The components of a vector will be equal in magnitude if the vector lies at a 45° angle with the two axes along which the components lie.

4. The minimum sum for two vectors occurs when the two vectors are opposite in direction. If they are unequal, their sum cannot be zero.

6. The balls will be closest at the instant the second ball is projected. The fi rst ball will always be going faster than the second ball. There will be a one second time interval between their collisions with the ground. The two move with the same acceleration in the vertical direction. Thus, changing their horizontal velocity can never make them hit at the same time.



8. The equations of projectile motion are only valid for objects moving freely under the infl uence of gravity. The only acceleration such an object has is the free-fall acceleration, g, directed vertically downward. Of the objects listed, only a and d meet this requirement.

10. The passenger sees the ball go into the air and come back in the same way he would if he were at rest on Earth. An observer by the tracks would see the ball follow the path of a projectile. If the train were accelerating, the ball would fall behind the position it would reach in the absence of the acceleration.

PROBLEM SOLUTIONS

3.1 We are given that R A B��

= + . When two vectors are added graphically, the second vector is positioned with its tail at the tip of the fi rst vector. The resultant then runs from the tail of the fi rst vector to the tip of the second vector. In this case, vector A

�� will be positioned with its tail at the origin and its tip at the

point 0 29,( ). The resultant is then drawn, starting at the origin (tail of fi rst vector) and going 14 units in the negative y-direction to the point 0 14, −( ). The second vector, B

��, must then start from the tip of A

�� at point 0 29,( ) and

end on the tip of R��

at point 0 14, −( ) as shown in the sketch at the right. From this, it is seen that

B��

is 43 units in the negative -directiony

3.2 (a) Using graphical methods, place the tail of vector B

�� at the head of

vector A��

. The new vector A B��

+ has a magnitude of 6.1 units at 113° from the positive x-axis.

(b) The vector difference A B��

− is found by placing the negative of vector B

��

(a vector of the same magnitude as B��

, but opposite direction) at the head of vector A��

. The resultant vector A B��

− has magnitude 15 units at 23° from the positive x-axis.

3.3 (a) In your vector diagram, place the tail of vector B��

at the tip of vector A

��. The vector sum, A B

�� + , is then found as

shown in the vector diagram and should be

A B��

+ = − °5.0 units at 53

(b) To fi nd the vector difference, form the vector −B��

(same magnitude as B

��, opposite direction) and add it to vector

A��

as shown in the diagram. You should fi nd that

A B��

− = + °5.0 units at 53


84 Chapter 3

3.4 Sketches of the scale drawings needed for parts (a) through (d) are given below. Following the sketches is a brief comment on each part with its answer.

(a) (b) (c) (d)

(a) Drawing the vectors to scale and maintaining their respective directions yields a resultant

of 5.2 m at 60+ ° .

(b) Maintain the direction of A��

, but reverse the direction of B��

to produce −B��

. The resultant

is 3.0 m at 30− ° .

(c) Maintain the direction of B��

, but reverse the direction of A��

to produce −A��

. The resultant

is 3.0 m at 150+ ° .

(d) Maintain the direction of A��

, reverse the direction of B��

and double its magnitude, to produce −2B

��. The resultant is 5.2 m at 60− ° .

3.5 Your sketch should be drawn to scale, similar to that pictured below. The length of R��

and the angle θ can be measured to fi nd, with use of your scale factor, the magnitude and direction of the resultant displacement. The result should be

approximately 421 ft at 3° below the horizontal

→

3.6 (a) The distance d from A to C is

d x y= +2 2

where x = + ( ) ° =200 300 30 0 460km km kmcos .

and y = + ( ) ° =0 300 30 0 150km kmsin .

∴ = + = km km kmd ( ) ( )460 150 4842 2

(b) φ = ⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ =− −tan tan .1 1 150

18y

x

km

460 km11° N of W

(c) Because of the curvature of the Earth, the pplane doesn’t travel along straight lines . Thus,

the answer computed above is only approximately correct.



3.7 Using a vector diagram, drawn to scale, like that shown at the right, the fi nal displacement of the plane can be found to be R

��plane 310 km at 57 N of E= °θ = . The

requested displacement of the base from point B is − R

��plane, which has the same magnitude but the

opposite direction. Thus, the answer is

− =R��

plane 310 km at 57 S of W=θ °

3.8 Your vector diagram should look like the one shown at the right. The initial

displacement A��

= 100 m due west and the resultant R

��= °175 m at 15.0 N of W are

both known. In order to reach the end point of the run following the initial displacement, the jogger must follow the path shown as B

��.

The length of B��

and the angle θ can be measured. The results should be 83 m at 33 N of W° .

3.9 The displacement vectors A��

= 8.00 m westward and B

��= 1 0 m north3. can be drawn to scale as at the right.

The vector C��

represents the displacement that the man in the maze must undergo to return to his starting point. The scale used to draw the sketch can be used to

fi nd C��

to be 15 m at 58 S of E° .

3.10 The x- and y-components of vector A��

are its projections on lines parallel to the x- and y-axis, respectively, as shown in the sketch. The magnitude of these components can be computed using the sine and cosine functions as shown below:

Ax = ° = + ° = ( ) ° =A A

�� cos cos . cos .325 35 35 0 35 28 7 uunits

and

Ay = ° = − ° = − ( ) ° = −A A��

sin sin . sin .325 35 35 0 35 20 11 units

3.11 Using the vector diagram given at the right, we fi nd

R = ( ) + ( ) =6 00 5 40 8 072 2

. . . m m m

and

θ = ⎛⎝⎜

⎞⎠⎟ = ( ) =− −tan

.

.tan . .1 15 40

6 000 900 42 0

m

m°°

Thus, the required displacement is 8.07 m at 42.0 S of E° .

planeplane

→→


86 Chapter 3

3.12 (a) The skater’s displacement vector, d��

, extends in a straight line from her starting point A to the end point B. When she has coasted half way around a circular path as shown in the sketch at the right, the displacement vector coincides with the diameter of the circle and has magnitude

�d = = ( ) =2 2 5 00 10 0r . . m m

(b) The actual distance skated, s, is one half the circumference of the circular path of radius r. Thus,

s r= ( ) = ( ) =1

22 5 00 15 7π π . . m m

(c) When the skater skates all the way around the circular path, her end point, B, coincides with the start point, A. Thus, the displacement vector has zero length, or

�d = 0

3.13 (a) Her net x (east–west) displacement is − + + = +3.00 0 6.00 3.00 blocks, while her net y (north–south) displacement is 0 0 0 4.00 blocks+ + = +4 0. . The magnitude of the resultant displacement is

R x y= ( ) + ( ) = ( ) + ( ) =Σ Σ2 2 2 23 00 4 00 5 00. . . blocks

and the angle the resultant makes with the x-axis (eastward direction) is

θ =⎛⎝⎜

⎞⎠⎟

= ⎛⎝

⎞⎠ =− − −tan tan

.

.tan .1 1 14 00

3 001

ΣΣ

y

x333 53 1( ) = . °

The resultant displacement is then 5 00 53 1. .blocks at N of E° .

(b) The total distance traveled is 3.00 0 6.00+ + =4 0. 13.0 blocks .

3.14 (a) The resultant displacement is R A B C D��

= + + + , where A��

= 75 0. m due north, B��

= 250 mdue east, C

��= 125 m at 30.0° north of east, and D

��= 150 m due south. Choosing east as the

positive x-direction and north as the positive y-direction, we fi nd the components of the resultant to be

R A B C Dx x x x x= + + + = + + ( ) ° + =0 250 125 30 0 0 35m m cos . 88 m

and

R A B C Dy y y y y= + + + = + + ( ) ° + −75 0 0 125 30 0 15. sin .m m 00 12 5m m( ) = − .

The magnitude and direction of the resultant are then

R R Rx y= + = ( ) + −( ) =2 2 2 2

358 12 5 358 m m m.

and

θ =⎛⎝⎜

⎞⎠⎟

= −⎛⎝⎜

⎞⎠⎟ = −− −tan tan

.1 1 12 5

358

R

Ry

x

m

m22 00. °

Thus, R��

= 358 m at 2.00° south of east .

(b) Because of the commutative property of vector addition, the net displacement is the same regardless of the order in which the individual displacements are executed.

C

B Ad

r�5.00 m

C

B Ad

r�5.00 m



3.15 Ax = −25 0. Ay = 40 0.

A A Ax y= + = −( ) + ( ) =2 2 2 2

25 0 40 0. . 47.2 units

From the triangle, we fi nd that

φ =⎛

⎝⎜⎞

⎠⎟= ⎛

⎝⎞⎠ = °− −tan tan

.

..1 1 40 0

25 058 0

A

Ay

x

, so θ φ= ° − = °180 122

Thus, A��

= 47 2. units at 122 counterclockwise fro° mm + -axisx .

3.16 Let A��

be the vector corresponding to the 10.0-yd run, B��

to the 15.0-yd run, and �C to the

50.0-yd pass. Also, we choose a coordinate system with the + y direction downfi eld, and the + x direction toward the sideline to which the player runs.

The components of the vectors are then

Ax = 0 Ay = −10 0. yds

Bx = 15 0. yds By = 0

Cx = 0 Cy = + 50 0. yds

From these, R xx = =Σ 15 0. yds, and R yy = =Σ 40 0. yds, and

R R Rx y= + = ( ) + ( ) =2 2 2 215 0 40 0. . yds yds 42.7 yardss

3.17 After 3.00 h moving at 41.0 km/h, the hurricane is 123 km at 60.0° N of W from the island. In the next 1.50 h, it travels 37.5 km due north. The components of these two displacements are as follows:

Displacement x-component (eastward) y-component (northward)

123 km - 61.5 km +107 km

37.5 km 0 +37.5 km

Resultant - 61.5 km 144 km

Therefore, the eye of the hurricane is now

R = − + ( ) =( )61 5 144

2 2. km km 157 km from the islaand


88 Chapter 3

3.18 Choose the positive x-direction to be eastward and positive y as northward. Then, the components of the resultant displacement from Dallas to Chicago are

R xx = = ( ) ° − ( ) ° =Σ 730 5 00 560 21 0 527 mi micos . sin . mi

and

R yy = = ( ) ° + ( ) ° =Σ 730 5 00 560 21 0 586mi misin . cos . mi

R R Rx y= + = ( ) + ( ) =2 2 2 2

527 586 788 mi mi mi

θ =⎛⎝⎜

⎞⎠⎟

= ( ) = °− −tan tan . .1 1 1 11 48 1ΣΣ

y

x

Thus, the displacement from Dallas to Chicago is

R��

= 788 mi at 48.1 N of E° .

3.19 The components of the displacements �a,

�b, and

�c are

a ax = ⋅ ° = +cos .30 0 152 km

b bx = ⋅ ° = −cos .110 51 3 km

c cx = ⋅ ° = −cos180 190 km

and

a ay = ⋅ ° = +sin . .30 0 87 5 km

b by = ⋅ ° = +sin110 141 km

c cy = ⋅ ° =sin180 0

Thus,

R a b cx x x x= + + = − 89.7 km, and R a b cy y y y= + + = + 2 km28

so

R R Rx y= + =2 2 245 km, and θ =⎛

⎝⎜⎞

⎠⎟= ( ) = °− −tan tan . .1 1 1 11 21 4

R

Rx

y

City C is 245 km at 21.4 W of N° from the starting point.

(north)

(east)

(north)

(east)



3.20 (a) F F x1 1120 120 60 0 60 0= = ( ) ° =N N Ncos . . N NF y1 120 60 0 104= ( ) ° =sin .

F F x2 280 0 80 0 75 0 20 7= = − ( ) ° = −. . cos . .N N NN N NF y2 80 0 75 0 77 3= ( ) ° =. sin . .

F F FR x y= ( ) + ( ) = ( ) + ( ) =Σ Σ2 2 2 2

39 3 181 185. N N N

and θ = ⎛⎝⎜

⎞⎠⎟ = ( ) = °− −tan

.tan . .1 1181

39 34 61 77 8

N

N

The resultant force is FR

� ��= °185 N at 77.8 from the -axisx

(b) To have zero net force on the mule, the resultant above must be cancelled by a force equal in magnitude and oppositely directed. Thus, the required force is

185 N at 258 from the -axis° x

3.21 The single displacement required to sink the putt in one stroke is equal to the resultant of the three actual putts used by the novice. Taking east as the positive x-direction and north as the positive y-direction, the components of the three individual putts and their resultant are

Ax = 0 Ay = +4 00. m

Bx = ( ) ° = +2 00 45 0 1 41. cos . .m m

By = ( ) ° = +2 00 45 0 1 41. sin . .m m

Cx = − ( ) ° = −1 00 30 0 0 500. sin . .m m Cy = − ( ) ° = −1 00 30 0 0 866. cos . .m m

R A B Cx x x x= + + = + 0 914. m R A B Cy y y y= + + = +4 55. m

The magnitude and direction of the desired resultant is then

R R Rx y= + =2 2 4 64. m and θ =⎛⎝⎜

⎞⎠⎟

= + °−tan .1 78 6R

Ry

x

Thus, R = 4 64. m at 78.6° north of east .

3.22 v0 1010 477

145 1x = ( )⎛

⎝⎜⎞⎠⎟

=mi hm s

mi hm s

.. and ∆x = ( )⎛

⎝⎜⎞⎠⎟ =60 5

118 4. . ft

m

3.281 ft m

The time to reach home plate is

tx

x

= = =∆v0

18.4 m

45.1 m s0.408 s

In this time interval, the vertical displacement is

∆y t a ty y= + = + −( )( ) = −v0

2 2 21

20

1

20 408 09.80 m s s. ..817 m

Thus, the ball drops vertically 0 817. m (3.281 ft 1 m ) 2.68 ft= .


90 Chapter 3

3.23 (a) With the origin chosen at point O as shown in Figure P3.23, the coordinates of the original

position of the stone are x y0 00 50 0= = + and m. .

(b) The components of the initial velocity of the stone are v v0 018 0 0x y= + =. m s and .

(c) The components of the stone’s velocity during its fl ight are given as functions of time by

v v vx x x xa t t= + = + ( ) =0 18 0 0 18 0. .m s or m s

and

v v vy y y xa t g t t= + = + −( ) = −( )0 0 9 80or m s2.

(d) The coordinates of the stone during its fl ight are

x x t a t t tx x= + + = + ( ) + ( )0 02 21

20 18 0

1

20v . m s or m sx t= ( )18 0.

and

y y t a t t g ty y= + + = + ( ) + −( )0 02 21

250 0 0

1

2v . m or m m s2y t= − ( )50 0 4 90 2. .

(e) We fi nd the time of fall from ∆y t a ty y= +v021

2 with v0 0y = :

t

y

a=

( ) =−( )

−=

2 2 50 0

9 80

∆ .

.

m

m s3.19 s2

(f ) At impact, v vx x= =0 18 0. m s, and the vertical component is

v vy y ya t= + = + −( )( ) = −0 0 9 80 3 19 31 3. . .m s s m s2

Thus,

v v v= + = ( ) + −( ) =x y2 2 2 2

18 0 31 3 36 1. . .m s m s m s

and

θ =⎛⎝⎜

⎞⎠⎟

= −⎛⎝

⎞⎠ = − °− −tan tan

.

..1 1 31 3

18 060 1

v

vy

x

or �v = °36 1. m s at 60.1 below the horizontal .

3.24 The constant horizontal speed of the falcon is

vx =⎛⎝⎜

⎞⎠⎟

=2000 447

189 4

mi

h

m s

mi hm s

..

The time required to travel 100 m horizontally is

tx

x

= = =∆v

1001 12

m

89.4 m ss.

The vertical displacement during this time is

∆y t a ty y= + = + −( )( ) = −v02 21

20

1

29 80 1 12 6. . .m s s2 113 m

or the falcon has a vertical fall of 6.13 m .



3.25 At the maximum height vy = 0, and the time to reach this height is found from

v vy y ya t= +0 as ta g g

y y

y

y y=−

=−−

=v v v v0 0 00

The vertical displacement that has occurred during this time is

∆y t tyy y y( ) = ( ) =

+⎛⎝⎜

⎞⎠⎟

=+⎛

⎝⎜⎞⎠⎟max av

vv v v0 0

2

0

2

vv v0 02

2y y

g g

⎛⎝⎜

⎞⎠⎟

=

Thus, if ( ) ( .∆y max ft 1 m 3.281 ft ) m= =12 3 7 , then

v0 2 2 9 80 3 7 8 5y g y= ( ) = ( )( ) =∆max

2m s m m s. . .

and if the angle of projection is θ = °45 , the launch speed is

v

v0

0 8 5

4512= =

°=y

sin

.

sinθm s

m s

3.26 (a) When a projectile is launched with speed v0 at angle θ0 above the horizontal, the initial velocity components are v v v v0 0 0 0 0 0x y= =cos sinθ θand . Neglecting air resistance, the vertical velocity when the projectile returns to the level from which it was launched (in this case, the ground) will be v vy y= − 0 . From this information, the total time of fl ight is found from v vy y ya t= +0 to be

ta g g

tyf y

y

y y ytotal t= or=

−=

− −−

v v v v v0 0 0 02ootal =

2 0 0v sinθg

Since the horizontal velocity of a projectile with no air resistance is constant, the horizontal distance it will travel in this time (i.e., its range) is given by

R tg gx= = ( )⎛

⎝⎜⎞⎠⎟

=v vv v

0 0 00 0 0

222total cos

sinsθ θiin cos

sinθ θ

θ0 0

02

02( ) =( )v

g

Thus, if the projectile is to have a range of R = 81 1. m when launched at an angle of θ0 45 0= °. , the required initial speed is

v002

81 1 9 80

90 0= ( ) =

( )( )( )° =Rg

sin

. .

sin .θm m s2

228 2. m s

(b) With v0 28 2= . m s and θ0 45 0= °. the total time of fl ight (as found above) will be

tgtotal

m s

m= =

( ) ( )°2 2 28 2 45 0

9 800 0v sin . sin .

.

θss

s2 = 4 07.

(c) Note that at θ0 45 0= °. , sin ( )2 10θ = and that sin ( )2 0θ will decrease as θ0 is increased above this optimum launch angle. Thus, if the range is to be kept constant while the

launch angle is increased above 45.0°, we see from v0 02= ( )Rg sin θ that

the required initial velocity will increase ..

Observe that for θ0 90< °, the function sinθ0 increases as θ0 is increased. Thus, increas-ing the launch angle above 45.0° while keeping the range constant means that both v0 0and sinθ will increase. Considering the expression for t total given above, we see that

the total time of flight will increase .


92 Chapter 3

3.27 When ∆ ∆y y y= =( ) ,max v 0.

Thus, v vy y ya t= +0 yields 0 3 000= ° −v sin . gt, or

tg

=°v0 3 00sin .

The vertical displacement is ∆y t a ty y= +v021

2. At the maximum height, this becomes

∆yg

g( ) = ( )° ⎛⎝⎜

⎞⎠⎟

−max

°v

v v0

0 03 003 00 1

2sin .

sin . ssin . sin .3 00 3 00

2

2

02 2° °

g g

⎛⎝⎜

⎞⎠⎟

=v

If ( ) .∆y max m= 0 330 , the initial speed is

v0

2

3 00

2 9 80 0 330=

( )°

=( )( )g y∆

max2m s m

sin .

. .

sinn ..

3 0048 6

°= m s

Note that it was unnecessary to use the horizontal distance of 12.6 m in this solution.

3.28 (a) With the origin at ground level directly below the window, the original coordinates of the

ball are ( , ) ( , )x y y= 0 0 .

(b) v v0 0 0 8 00 20 0 7 52x = = ( ) −( )° = +cos . cos . .θ m s m s

v v0 0 0 8 00 20 0 2 74y = = ( ) −( )° = −sin . sin . .θ m s m s

(c) x x t a t t tx x= + + = + ( ) + ( )0 02 21

20 7 52

1

20v . m s orr m sx t= ( )7 52.

y y t a t y ty y= + + = + −( ) + −0 02

0

1

22 74

1

29 80v . .m s m s2(( )t 2

or y y t t= − ( ) − ( )022 74 4 90. . m s m s2

(d) Since the ball hits the ground at t = 3 00. s, the x-coordinate at the landing site is

m s s landing s

x xt

= = ( )( ) ==3 007 52 3 00 22 6

.. . . mm

(e) Since y = 0 when the ball reaches the ground at t = 3 00. s, the result of (c) above gives

y y t tt

022 74 4 90= + ⎛

⎝⎜⎞⎠⎟ + ⎛

⎝⎜⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

. . m

s

m

s2

==

= + ⎛⎝⎜

⎞⎠⎟ ( ) + ⎛

⎝3 00

0 2 74 3 00 4 90.

. . . s

2 m

s s

m

s⎜⎜⎞⎠⎟ ( )3 00

2. s

or y0 52 3= . m .

(f ) When the ball has a vertical displacement of ∆y = −10 0. m, it will be moving downward with a velocity given by v vy y ya y2

02 2= + ( )∆ as

v vy y ya y= − + ( ) = − −( ) + −( ) −02 2

2 2 74 2 9 80∆ . .m s m s2 110 0 14 3. .m m s( ) = −

The elapsed time at this point is then

ta

y y

y

=−

=− − −( )

−=

v v0 14 3 2 74

9 801 1

. .

..

m s m s

m s2 88 s



3.29 We choose our origin at the initial position of the projectile. After 3.00 s, it is at ground level, so the vertical displacement is ∆y H= − .

To fi nd H, we use ∆y t a ty y= +v021

2, which becomes

− = ( )⎡⎣ ⎤⎦° ( ) + −( )H 15 25 3 01

29 80 3m s s m s2sin . . .00 2s( ) , or H = 25 m

3.30 The initial velocity components of the projectile are

v0 300 55 0 172x = ( ) =m s ° m scos . and v0 300 55 0 246y = ( ) =m s ° m ssin .

while the constant acceleration components are

ax = 0 and a gy = − = −9 80. m s2

The coordinates of where the shell strikes the mountain at t = 42 0. s are

x t a tx x= + = ( )( ) + = ×v02 31

2172 42 0 0 7 22 10m s s m. . == 7 22. km

and

y t a ty y= +

= ( )( ) + −(

v021

2

246 42 01

29 80m s s m s2. . ))( ) = × =42 0 1 69 10 1 692 3. . .s m km

3.31 The speed of the car when it reaches the edge of the cliff is

v v= + ( ) = + ( )( ) =0

2 2 0 2 4 00 50 0 20 0a x∆ . . .m s m m s2

Now, consider the projectile phase of the car’s motion. The vertical velocity of the car as it reaches the water is

v vy y ya y= − + ( ) = − −( ) °⎡⎣ ⎤⎦ +02 2

2 20 0 24 0 2∆ . sin .m s −−( ) −( )9 80 30 0. .m s m2

or

vy = − 25 6. m s

(b) The time of fl ight is

t

ay y

y

=−

=− − − ( )⎡⎣ ⎤⎦°

−v v0 25 6 2 24 0. sin .m s 0.0 m s

99 801 78

..

m ss2 =

(a) The horizontal displacement of the car during this time is

∆x tx= = ( )⎡⎣ ⎤⎦° ( ) =v0 20 0 24 0 1 78 32 5. cos . . .m s s mm


94 Chapter 3

3.32 The components of the initial velocity are

v0 40 0 30 0 34 6x = ( ) ° =. cos . .m s m s

and

v0 40 0 30 0 20 0y = ( ) ° =. sin . .m s m s

The time for the water to reach the building is

tx

x

= = =∆v0

50 01 44

..

m

34.6 ms

The height of the water at this time is

∆y t a ty y= + = ( )( ) + −v021

220 0 1 44

1

29 80. . .m s s m s22 s m( )( ) =1 44 18 72. .

3.33 (a) At the highest point of the trajectory, the projectile is moving horizontally with velocity components of vy = 0 and

v v vx x= = = ( ) ° =0 0 60 0 30 0 52 0cos . cos . .θ m/s m s

(b) The horizontal displacement is ∆x tx= = ( )( ) =v0 52 0 4 00 208. .m s s m and, from ∆y t a ty= +( sin )v0

12

2θ , the vertical displacement is

∆y = ( )( )° ( ) + −(60 0 30 0 4 001

29 80. sin . . .m s s m s2 ))( ) =4 00 41 62. .s m

The straight line distance is

d x y= ( ) + ( ) = ( ) + ( ) =∆ ∆2 2 2 2208 4 212m 1.6 m m

3.34 The horizontal kick gives zero initial vertical velocity to the rock. Then, from ∆y t a ty y= +v012

2, the time of fl ight is

ty

ay

=( ) =

−( )−

=2 2 40 0

9 808 16

∆ .

..

m

m s s2

The extra time ∆t = − =3 00 8 16 0 143. . .s s s is the time required for the sound to travel in a straight line back to the player. The distance the sound travels is

d x y t= ( ) + ( ) = = ( )( ) =∆ ∆ ∆2 2343 0 143 49vsound m s s. ..0 m

where ∆x represents the horizontal displacement of the rock when it hits the water. Thus,

∆ ∆x d y= − ( ) = ( ) − −( ) =2 2 2 2

49 0 40 0 28 3. . .m m m

The initial velocity given the ball must have been

v v0 0

28 39 91= = = =x

x

t

∆ ..

m

8.16 sm s



3.35 (a) The jet moves at 3 00 102. × mi h due east relative to the air. Choosing a coordinate system with the positive x-direction eastward and the positive y-direction northward, the compo-nents of this velocity are

� �v vJA JAmi h and( ) = × ( ) =

x y3 00 10 02.

(b) The velocity of the air relative to Earth is 1 00 102. × mi h at 30.0° north of east. Using the coordinate system adopted in (a) above, the components of this velocity are

� �v vAE AE mi h cos30.0 = 86.( ) = = ×( ) °

xcos .θ 1 00 102 66 mi h

and

� �v vAE AE mi h sin30.0 = 50.( ) = = ×( ) °

ysin .θ 1 00 102 00 mi h

(c) Carefully observe the pattern of the subscripts in Equation 3.16 of the textbook. There, two objects (cars A and B) both move relative to a third object (Earth, E). The velocity of object A relative to object B is given in terms of the velocities of these objects relative to E as

� � �v v vAB AE BE= − . In the present case, we have two objects, a jet (J) and the air (A), both

moving relative to a third object, Earth (E). Using the same pattern of subscripts as that in Equation 3.16, the velocity of the jet relative to the air is given by

� � �v v vJA JE AE

= −

(d) From the expression for �vJA found in (c) above, the velocity of the jet relative to the ground

is � � �v v vJE JA AE= + . Its components are then

� � �v v vJE JA AE mi h 86.6 m( ) = ( ) + ( ) = × +

x x x3 00 102. ii h mi h= ×3 87 102.

and

� � �v v vJE JA AE 50.0 mi h mi h( ) = ( ) + ( ) = + =

y y y0 50 0.

This gives the magnitude and direction of the jet’s motion relative to Earth as

� � �v v vJE JE JE mi h mi h= + = ×( ) +

x y

2 2 2 23 87 10 50 0. .(( ) = ×2 23 90 10. mi h

and

θ =( )( )

⎛

⎝⎜

⎞

⎠⎟ =− −tan tan

.

.1 1 50 0

3

�

�v

vJE

JE

mi hy

x887 10

7 372×⎛⎝⎜

⎞⎠⎟

= °mi h

.

Therefore, �vJE mi h at 7.37° north of east= ×3 90 102. .


96 Chapter 3

3.36 We use the following notation:

�vBS = velocity of boat relative to the shore

�vBW = velocity of boat relative to the water,

and �vWS = velocity of water relative to the shore.

If we take downstream as the positive direction, then �vWS m s= +1 5. for both parts of the trip.

Also, �vBW m s= +10 while going downstream and

�vBW m s= −10 for the upstream part of

the trip.

The velocity of the boat relative to the boat relative to the water is � � �v v vBW BS WS= − , so the

velocity of the boat relative to the shore is � � �v v vBS BW WS= + .

While going downstream, �vBS 1 m s m s= +0 1 5. and the time to go 300 m downstream is

t

ddown

BS

m

10 1.5 m ss= =

+( ) =�v

30026

When going upstream, �vBS 1 m s m s m s= − + = −0 1 5 8 5. . and the time required to move 300 m

upstream is

td

upBS

m

8.5 m ss= = =�

v300

35

The time for the round trip is t t t= + = +( ) =down up s s26 35 61 .

3.37 Prior to the leap, the salmon swims upstream through water fl owing at speed

�vWE m s= 1 50. relative to Earth. The fi sh swims at

�vFW m s= 6 26. relative

to the water in such a direction to make its velocity relative to Earth, �vFE,

vertical. Since � � �v v vFE FW WE= + , as shown in the diagram at the right, we fi nd

that

θ =⎛

⎝⎜⎞

⎠⎟=

⎛⎝⎜

⎞− −cos cos.

.1 1 1 50

6 26

��v

vWE

FW

m s

m s⎠⎠⎟= °76 1.

and the vertical velocity of the fi sh as it leaves the water is

v0 6 26 76 1 6 08y = = = ( ) ° =� �v vFE FW m s m ssin . sin . .θ

The height of the salmon above the water at the top of its leap (that is, when vy = 0) is given by

∆ya

y y

y

=−

=− ( )−( ) =

v v202 2

2

0 6 08

2 9 801 8

.

..

m s

m s288 m

→→



3.38 (a) The velocity of the boat relative to the water is

� � �v v vBW BS WS= −

where �vBS is the velocity of the boat relative to shore, and

�vWS is

the velocity of the water relative to shore. Thus, we may write

� � �v v vBS BW WS= + . Note, in the vector diagram at the right, that

these vectors form a right triangle. The Pythagorean theorem then gives the magnitude of the resultant

�vBS as

� � �v v vBS BW WS m s m s= + = ( ) + ( ) =2 2 2 2

10 0 1 50 10 1. . . m s

also,

θ =⎛

⎝⎜⎞

⎠⎟=

⎛⎝

− −tan tan.

.1 1 1 50

10 0

��v

vWS

BW

m s

m s⎜⎜⎞⎠⎟

= °8 53.

so �vBS m s at 8.53° east of north= 10 1. .

(b) The boat has a constant northward velocity of �vBW m s= 10 0. , and must travel a distance

d = 300 m northward to reach the opposite shore. The time required will be

td= = =�

vBW

m

m ss

300

10 030 0

..

During this time, the boat drifts downstream at a constant rate of �vWS m s= 1 50. .

The distance it drifts downstream during the crossing is

drift t= ⋅ = ( )( ) =v

�WS m s s m1 50 30 0 45 0. . .

3.39 �vBW = velocity of boat relative to the water

�vWS = velocity of water relative to the shore

�vBS = velocity of boat relative to the shore

� � �v v vBS BW WS= + (as shown in the diagram)

The northward (that is, cross-stream) component of �vBS is

�vBS north BW mi h( ) = ( ) ° + = ( )v sin . . sin62 5 0 3 30 62.. .5 0 2 93° + = mi h

The time required to cross the stream is then

t = =0 5050 173

..

mi

2.93 mi h h

The eastward (that is, downstream) component of �vBS is

�vBS east BW WS( ) = −( ) ° +v vcos .62 5

= − ( ) ° + = −3 30 62 5 1 25 0 274. cos . . .mi h mi h mi h

Since the last result is negative, it is seen that the boat moves upstream as it crosses the river. The distance it moves upstream is

d t= ( ) = ( )( ) = × −�vBS east

mi h h0 274 0 173 4 72 10. . . 22 5 280249mi

ft

1 mift( )⎛

⎝⎞⎠ =

north

east

VBW VBS

VWS

north

east

VBW VBS

VWS


98 Chapter 3

3.40 If the salmon (a projectile) is to have vy = 0 when ∆y = +1 50. m, the required initial velocity in the vertical direction is given by v vy y ya y2

02 2= + ∆ as

v v02 2 0 2 9 80 1 50y y ya y= + − = − −( ) +( ) =∆ . .m s m2 5 42. m s

The elapsed time for the upward fl ight will be

∆ta

y y

y

=−

= −−

=v v0 0 5 42

9 800 553

.

..

m s

m ss2

If the horizontal displacement at this time is to be ∆x = +1.00 m, the required constant horizontal component of the salmon’s velocity must be

v0

1 00x

x

t= = =∆

∆. m

0.553 s 1 81. m s

The speed with which the salmon must leave the water is then

v v v0 0

202 2 2

1 81 5 42 5 72= + = ( ) + ( ) =x y . . .m s m s m s

Yes , since v0 6 26< . m s the salmon is capable of making this jump.

3.41 (a) Both the student (S) and the water (W) move relative to Earth (E). The velocity of the student relative to the water is given by

� � �v v vSW SE WE= − , where

� �v vSE WEand are the

velocities of the student relative to Earth and the water relative to Earth, respectively. If we choose downstream as the positive direction, then

�vWE m s= + 0 500. ,

�vSW m s= −1 20.

when the student is going up stream, and �vSW m s= +1 20. when the student moves

downstream.

The velocity of the student relative to Earth for each leg of the trip is

� � �v v vSE upstream WE SW upstream

m s( ) = + ( ) = 0 500. ++ −( ) = −1 20 0 700. .m s m s

and

� � �v v vSE downstream WE SW downstream

( ) = + ( ) = 0 50. 00 1 20 1 70m s m s m s+ +( ) = +. .

The distance (measured relative to Earth) for each leg of the trip is d = =1 00. km 1.00 × 103 m. The times required for each of the two legs are

td

upstreamSE upstream

m

m s= = ×

�v

1 00 10

0 700

3.

.== ×1 43 103. s

and

td

downstreamSE downstream

m= = ×�v

1 00 10

1 70

3.

. m ss= ×5 88 10. s

so the time for the total trip is

t t ttotal upstream downstream s= + = × +1 43 10 5 83. . 88 10 2 02 103× = ×s s s.

continued on next page



(b) If the water is still �vWE =( )0 , the speed of the student relative to Earth would have been the

same for each leg of the trip, � � �v v vSE SE upstream SE downstream

m s= = = 1 20. . In this case, the time for each leg, and the total time would have been

td

legSE

m

m ss= = × = ×�

v1 00 10

1 208 33 10

33.

.. aand stotal legt t= = ×2 1 67 103.

(c)

The time savings going downstream with the ccurrent is always less than the extra timerequired to go the same distance against tthe current.

3.42 (a) The speed of the student relative to shore is v v vup = − s while swimming upstream and v v vdown = + s while swimming downstream. The time required to travel distance d upstream is then

td d

sup

up

= =−v v v

(b) The time required to swim the same distance d downstream is

td d

sdown

down

= =+v v v

(c) The total time for the trip is therefore

t t td d d d

as s

s s= + =−

++

=+( ) + −( )−up down v v v v

v v v vv vss s s s

d d

( ) +( ) =−

=−v v

vv v

vv v

2 2

12 2 2 2

(d) In still water, vs = 0 and the time for the complete trip is seen to be

t td d

b as

= =−

==v

vv v0 2

2

1 0

2

(e) Note that td d

b = =2 22vv

v and that t

da

s

=−

22 2

vv v

.

Thus, when there is a current ( )vs > 0 , it is always true that ta > tb .


100 Chapter 3

3.43 (a) The bomb starts its fall with v v v0 00 275y = = =and m splanex . Choosing the origin at the location of the plane when the bomb is released and upward as positive, the y coordinate of the bomb at ground level is y h= − = − ×3 00 103. m. The time required for the bomb to fall is given by y y t a ty y= + +0 0

12

2v as

− × = + + −( )3 00 10 0 01

29 803 2. .m m s2

fallt or tfall 2

m

m ss=

×( )=

2 3 00 10

9 8024 7

3.

..

With ax = 0, the horizontal distance the bomb travels during this time is

d t= = ( )( ) = × =v0

3275 24 7 6 80 10 6 80x fall m s s m. . . km

(b) While the bomb is falling, the plane travels in the same horizontal direction with the same constant horizontal speed, v v vx x= =0 plane, as the bomb. Thus, the plane remains directly above the bomb as the bomb falls to the ground. When impact occurs, the plane is

directly over the impact point, at an altituude of 3.00 km.

(c) The angle, measured in the forward direction from the vertical, at which the bombsight must have been set is

θ = ⎛⎝⎜

⎞⎠⎟ = ⎛

⎝⎜⎞⎠⎟ =− −tan tan

.

.1 1 6 80

3 006

d

h

km

km66 2. °

3.44 (a) The time required for the woman, traveling at constant speed v1 relative to the ground, to

travel distance L relative to the ground is t Lwoman = v1 .

(b) With both the walkway (W) and the man (M) moving relative to Earth (E), we know that the velocity of the man relative to the moving walkway is

� � �v v vMW ME WE= − . His velocity

relative to Earth is then � � �v v vME MW WE= + . Since all of these velocities are in the same

direction, his speed relative to Earth is � � �v v vME MW WE= + = +v v2 1. The time required for

the man to travel distance L relative to the ground is then

tL L

manME

= =+�

v v v1 2

3.45 Choose the positive direction to be the direction of each car’s motion relative to Earth. The velocity of the faster car relative to the slower car is given by

� � �v v vFS FE SE= − , where

�vFE km h= + 60 0. is the velocity of the faster car relative to Earth, and

�vSE km h= 40 0. is the

velocity of the slower car relative to Earth.

Thus, �vFS km h km h km h= + − = +60 0 40 0 20 0. . . and the time required for the faster car to

move 100 m (0.100 km) closer to the slower car is

td= = = × −

vFS

km

20.0 km hh

3600 s

1

0 1005 00 10 3.

.hh

s⎛⎝

⎞⎠ = 18 0.

3.46 The vertical displacement from the launch point (top of the building) to the top of the arc may be found from v vy y ya y2

02 2= + ∆

with vy = 0 at the top of the arc. This yields

∆y

ay y

y

=−

=− ( )−( ) = +

v v202 2

2

0 12 0

2 9 807

.

..

m s

m s2335 m

and ∆y y y= −max 0 gives

y y y ymax .= + = +0 0 7 35∆ m




(a) If the origin is chosen at the top of the building, then y y0 0 7 35= =and mmax . .

Thus, the maximum height above the ground is

h ymax max. . . .= + = + =50 0 50 0 7 35 57 4 m m m m

The elapsed time from the point of release to the top of the arc is found from v vy y ya t= +0 as

ta

y y

y

=−

= −−

=v v0 0 12 0

9 801 22

.

..

m s

m ss2

(b) If the origin is chosen at the base of the building (ground level), then y0 50 0= + . m and h ymax max= , giving

h y ymax . . .= + = + =0 50 0 7 35 57 4∆ m m m

The calculation for the time required to reach maximum height is exactly the same as that given above. Thus, t = 1 22. s .

3.47 (a) The known parameters for this jump are: θ0 10 0= − °. , ∆ x = 108 m, ∆y = −55 0. m, ax = 0, and a gy = − = −9 80. m s.

Since ax = 0, the horizontal displacement is ∆x t tx= = ( )v v0 0 0cosθ where t is the total time of the fl ight. Thus, t x= ( )∆ v0 0cosθ .

The vertical displacement during the fl ight is given by

∆y t a t tgt

y y= + = ( ) −v v02

0 0

21

2 2sinθ

or

∆ ∆ ∆y

x g x= ( )⎛

⎝⎜

⎞

⎠⎟ −

⎛⎝⎜

⎞v

v v0 00 0 0 02

sincos cos

θθ θ ⎠⎠⎟

= ( ) −( )⎡

⎣⎢

⎤

⎦⎥

2

0

2

20 0

22

1∆∆

xg x

tancos

θθ v

Thus,

∆ ∆∆

y xg x

− ( )⎡⎣ ⎤⎦ = −( )⎡

⎣⎢

⎤

⎦⎥tan

cosθ

θ0

2

20 0

22

1

v

or

v0

2

02

02

9 80=

− ( )− ( )⎡⎣ ⎤⎦

=−g x

y x

∆∆ ∆ tan cos

.

θ θm s2(( )( )

− − ( ) −( )°⎡⎣ ⎤⎦

108

2 55 0 108 10 0

2m

m m. tan . coos .2 10 0−( )°

yielding

v0

51 143 10

69 7540 5= − ×

−=.

..

m s

mm s

3 2

(b) Rather than falling like a rock, the skier glides through the air much like a bird, prolonging the jump.


102 Chapter 3

3.48 The cup leaves the counter with initial velocity components of ( , )v v v0 0 0x i y= = and has acceleration components of ( , )a a gx y= = −0 while in fl ight.

(a) Applying ∆y t a ty y= +v02 2 from when the cup leaves the counter until it reaches the fl oor

gives

− = +−( )

hg

t02

2

so the time of the fall is

th

g= 2

(b) If the cup travels a horizontal distance d while falling to the fl oor, ∆x tx= v0 gives

v vi x

x

t

d

h g= = =0 2

∆ or vi d

g

h=

2

(c) The components of the cup’s velocity just before hitting the fl oor are

v v vx x i dg

h= = =0 2

and v vy y ya t gh

ggh= + = − = −0 0

22

Thus, the total speed at this point is

v v v= + = +x y

d g

hgh2 2

2

22

(d) The direction of the cup’s motion just before it hits the fl oor is

θ =⎛⎝⎜

⎞⎠⎟

=−⎛

⎝⎜⎞⎠⎟

=− − −tan tan tan1 1 12

2

v

vy

x

gh

d g h

−− ⎛

⎝⎜⎞

⎠⎟⎛

⎝⎜⎜

⎞

⎠⎟⎟

= −⎛⎝

⎞⎠

−12

2 21

dg h

h

g

h

dtan

3.49 AL t= = ( )( ) =v1 90 0 2 50 225. .km h h km

BD AD AB AL= − = ° − =cos . .40 0 80 0 km 92.4 km

From the triangle BCD,

BL BD DL

AL

= ( ) + ( )= ( ) + °( )

2 2

2 240 092.4 km =172sin . km

Since car 2 travels this distance in 2.50 h, its constant speed is

v2

17268 6= =km

2.50 hkm h.



3.50 After leaving the ledge, the water has a constant horizontal component of velocity.

v vx x= =0 1 50. m s2

Thus, when the speed of the water is v = 3 00. m s, the vertical component of its velocity will be

v v vy x= − − = − ( ) − ( ) = −2 2 2 2

3 00 1 50 2 60. . .m s m s m s

The vertical displacement of the water at this point is

∆ya

y y

y

=−

=−( ) −

−( ) = −v v2

02 2

2

2 60 0

2 9 800

.

.

m s

m s2..344 m

or the water is 0 344. m below the ledge.

If its speed leaving the water is 6 26. m s, the maximum vertical leap of the salmon is

∆ya

y

yleap

m s

2 9.80 m s=

−=

− ( )−( ) =

0

2

0 6 2620

2 2v ..000 m

Therefore, the maximum height waterfall the salmon can clear is

h ymax . .= + =∆ leap m m0 344 2 34

3.51 The distance, s, moved in the fi rst 3.00 seconds is given by

s t at= + = ( )( ) + ( )v021

2100 3 00

1

230 0 3m s s m s2. . .000 4352s m( ) =

Choosing the origin at the point where the rocket was launched, the coordinates of the rocket at the end of powered fl ight are:

x s1 53 0 262= ( )° =cos . m and my s1 53 0 347= ( )° =sin .

The speed of the rocket at the end of powered fl ight is

v v1 0 100 30 0 3 00 190= + = + ( )( ) =at m s m s s m s2. .

so the initial velocity components for the free-fall phase of the fl ight are

v v0 1 53 0 114x = ° =cos . m s and v v0 1 53 0 152y = ° =sin . m s

(a) When the rocket is at maximum altitude, vy = 0. The rise time during the free-fall phase can be found from v vy y ya t= +0 as

ta

y

yrise 2

m

m ss=

−= −

−=

0 0 152

9 8015 50v

..

The vertical displacement occurring during this time is

∆y ty y=+⎛

⎝⎜⎞⎠⎟

= +⎛⎝

⎞⎠ ( )

v v0

2

0 152

21rise

m s5.5 s == ×1 17 103. m

The maximum altitude reached is then

H y y= + = + × = ×1

3 3347 1 17 10 1 52 10∆ m m m. .



104 Chapter 3

(b) After reaching the top of the arc, the rocket falls 1 52 103. × m to the ground, starting with zero vertical velocity ( )v0 0y = . The time for this fall is found from ∆y t a ty y= +v0

12

2 as

ty

ayfall 2

m

9.80 m s=

( ) =− ×( )−

=2 2 1 52 10

17 63∆ .

. ss

The total time of fl ight is

t t t t= + + = + +( ) =powered rise fall s3 00 15 5 17 6 3. . . 66 1. s

(c) The free-fall phase of the fl ight lasts for

t t t2 15 5 17 6 33 1= + = +( ) =rise fall s s. . .

The horizontal displacement occurring during this time is

∆x tx= = ( )( ) = ×v0 23114 33 1 3 78 10m s s m. .

and the full horizontal range is

R x x= + = + × = ×1

3 3262 3 78 10 4 05 10∆ m m m. .

3.52 Taking downstream as the positive direction, the velocity of the water relative to shore is �vWS WS= +v , where vWS is the speed of the fl owing water. Also, if vCW is the common speed of the two canoes relative to the water, their velocities relative to the water are

�vCW downstream CW( ) = +v and

�vCW upstream CW( ) = −v

The velocity of a canoe relative to the water can also be expressed as v v v� � �

CW CS WS= − .

Applying this to the canoe moving downstream gives

+ = + −v vCW WSm s2 9. [1]

and for the canoe going upstream

− = − −v vCW WSm s1 2. [2]

(a) Adding equations [1] and [2] gives

2 1 7vWS m s= . , so vWS m s= 0 85.

(b) Subtracting [2] from [1] yields

2 4 1vCW m s= . , or vCW m s= 2 1.



3.53 The time of fl ight is found from ∆y t a ty y= +v012

2 with ∆y = 0, as

tg

y=2 0v

This gives the range as

R tgxx y= =v

v v0

0 02

On Earth this becomes

Rg

x yEarth

Earth

=2 0 0v v

and on the moon,

Rg

x yMoon

Moon

=2 0 0v v

Dividing RMoon by REarth, we fi nd R g g RMoon Earth Moon Earth= ( ) . With g gMoon Earth= ( )1 6 , this gives

R RMoon Earth m 18 m= = ( ) =6 6 3 0. .

Similarly,

Rg

gRMars

Earth

MarsEarth

m

0.387=

⎛⎝⎜

⎞⎠⎟

= =3 09

.. m .

3.54 The time to reach the opposite side is

tx

x

= =∆v v0 0

10

15

m

°cos

When the motorcycle returns to the original level, the vertical displacement is ∆y = 0. Using this in the relation ∆y t a ty y= +v0

12

2 gives a second relation between the takeoff speed and the time of fl ight as

0 151

202= ( )° + −( )v sin t g t or v0 2 15

=⎛⎝⎜

⎞⎠⎟

gt

sin °

Substituting the time found earlier into this result yields

vv0

02 15

10

15=

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

g

sin cos°

m

° or v0

9 80 10

2 15 1514=

( )( )( )° ( )° =.

sin cos

m s mm s

2


106 Chapter 3

3.55 (a) The time for the ball to reach the fence is

tx

x

= = =∆v v v0 0 0

130

35

159m

°

m

cos

At this time, the ball must be ∆y = − =21 1 0 20m m m. above its launch position, so ∆y t a ty y= +v0

12

2 gives

20 35159

4 90159

00

mm

m s2= ( )° ⎛

⎝⎜

⎞

⎠⎟ − ( )v

vsin .

mm

v0

2⎛⎝⎜

⎞⎠⎟

or

159 35 204 90 159 2

02m m

m s m2

( ) ° − =( )( )

sin.

v

From which, v0

24 90 159

159 35 2042=

( )( )( ) ° −

=.

sin

m s m

m m

2

mm s

(b) From above, t = = =159 1593 8

0

m m

42 m ss

v. .

(c) When the ball reaches the wall (at t = 3 8. s),

v vx x= = ( ) ° =0 42 35 34m s m scos

v vy y ya t= + = ( ) ° − ( )( ) =0 42 35 9 80 3 8m s m s s2sin . . −−13 m s

v v v= + = ( ) + −( ) =x y2 2 2 2

34 13 37m s m s m s

3.56 We shall fi rst fi nd the initial velocity of the ball thrown vertically upward, recognizing that it takes descend from its maximum height as was required to reach this height. Thus, the elapsed time when it reaches maximum height is t = 1 50. s. Also, at this time, vy = 0, and, v vy y ya t= +0 gives

0 9 80 1 500= − ( )( )v y . .m s s2 or v0 14 7y = . m s

In order for the second ball to reach the same vertical height as the fi rst, the second must have the same initial vertical velocity as the fi rst. Thus, we fi nd v0 as

vv

00

30 0

14 7

0 50029 4=

°= =y

sin .

.

..

m sm s

3.57 The time of fl ight of the ball is given by ∆y t a ty y= +v012

2, with ∆y = 0, as

0 20 301

29 80 2= ( )⎡⎣ ⎤⎦° + −( )m s m s2sin .t t

or t = 2.0 s.

The horizontal distance the football moves in this time is

∆x tx= = ( )⎡⎣ ⎤⎦° ( ) =v0 20 30 2 0 35m s s mcos .

Therefore, the receiver must run a distance of ∆x = −35 m m = 15 m20 away from the quarterback, in the direction the ball was thrown to catch the ball. He has a time of 2.0 s to do this, so the required speed is

v = = =∆x

t

157 5

m

2.0 sm s.



3.58 The horizontal component of the initial velocity is v v v0 0 040 0 766x = ° =cos . , and the time required for the ball to move 10.0 m horizontally is

tx

x

= = =∆v v v0 0

10 0

0 766

13 1.

.

.m m

0

At this time, the vertical displacement of the ball must be

∆y y y= − = − =0 3 05 2 00 1 05. . . m m m

Thus, ∆y t a ty y= +v012

2 becomes

1 05 40 013 1 1

29 80

130

0

. sin ..

.mm

m s2= ( )° + −( )vv

..1 2

02

m( )v

or

1 05 8 39835

02. .m m

m s3 2

= −v

which yields

v0

835

8 39 1 0510 7=

−=m s

m mm s

3 2

. ..

3.59 Choose an origin where the projectile leaves the gun and let the y-coordinates of the projectile and the target at time t be labeled y yp T and , respectively.

Then, ( ) ( sin )∆y y tg

tp p= − = −020 0

2v θ , and

( )∆y y hg

tT T= − = −02

2 or y hg

tT = −2

2

The time when the projectile will have the same x-coordinate as the target is

tx x

x

= =∆v v0

0

0 0cosθ

For a collision to occur, it is necessary that y yp T= at this time, or

vv0 0

0

0 0

2 2

2 2sin

cosθ

θ( )⎛

⎝⎜

⎞

⎠⎟ − = −

x gt h

gt

which reduces to

tanθ00

= h

x

This requirement is satisfi ed provided that the gun is aimed at the initial location of the target.

Thus, a collision is guaranteed if the shooter aimms the gun in this manner.

Target

h

x0

q0

Target

h

x0

q0


108 Chapter 3

3.60 (a) The components of the vectors are as follows:

Vector x-component (cm) y-component (cm)

�d1m 0 104

�d2m 46.0 19.5

�d1f 0 84.0

�d2f 38.0 20.2

The sums � � �d d dm 1m 2m= + and

� � �d d df 1f 2f= + are computed as

dm cm and = +( ) + +( ) = = −0 46 0 104 19 5 1322 2

. . tanθ 11 104 19 5

0 46 069 6

++

⎛⎝⎜

⎞⎠⎟ = °.

..

df cm and = +( ) + +( ) = = −0 38 0 84 0 20 2 111

2 2. . . tanθ 11 84 0 20 2

0 38 070 0

. .

..

++

⎛⎝⎜

⎞⎠⎟ = °

or � �d dm fcm at 69.6° and cm at= =132 111 770.0° .

(b) To normalize, multiply each component in the above calculation by the appropriate scale factor. The scale factor required for the components of

� �d d1m 2mand is

sm

cm

180 cm= =200

1 11.

and the scale factor needed for components of � �d d1f 2fand is

sf

cm

168 cm= =200

1 19.

After using these scale factors and recomputing the vector sums, the results are

′ = ′ =� �d dm f cm at 69.6 and cm a146 132° tt 70.0°

The difference in the normalized vector sums is ∆ ′ = ′ − ′� � �d d dm f.

Vector x-component (cm) y-component (cm)

′

�dm 50.9 137

− ′�df –45.1 –124

∆ ′�d Σx =5.74 Σy =12.8

Therefore, ∆ Σ Σ′ = ( ) + ( ) = ( ) + ( ) =d x y2 2 2 25 74 12 8 14 0. . .cm cm, and

θ =⎛⎝⎜

⎞⎠⎟

= ⎛⎝

⎞⎠ = °− −tan tan

.

..1 1 12 8

5 7465 8

ΣΣ

y

x or ∆ ′ = °

�d 14 0. cm at 65.8



3.61 To achieve maximum range, the projectile should be launched at 45° above the horizontal. In this case, the initial velocity components are

v vv

0 00

2x y= =

The time of fl ight may be found from v vy y gt= −0 by recognizing that when the projectile returns to the original level, v vy y= − 0 .

Thus, the time of fl ight is

tg g g g

y y y=− −

−= = ⎛

⎝⎜⎞⎠⎟ =

v v v v v0 0 0 0 02 2

2

2

The maximum horizontal range is then

R tg gx= = ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=vv v v

00 0 0

2

2

2 [1]

Now, consider throwing the projectile straight upward at speed v0. At maximum height, vy = 0,and the time required to reach this height is found from v vy y gt= −0 as 0 0= −v gt, which

yields t g= v0 .

Therefore, the maximum height the projectile will reach is

∆y tg gy( ) = ( ) =

+⎛⎝

⎞⎠

⎛⎝⎜

⎞⎠⎟

=max av

vv v v0

2 20 0 0

2

Comparing this result with the maximum range found in equation [1] above reveals that ( )∆y Rmax = 2 provided the projectile is given the same initial speed in the two tosses.

If the girl takes a step when she makes the horizontal throw, she can likely give a higher initial speed for that throw than for the vertical throw.

3.62 (a) x tx= v0 , so the time may be written as t x x= v0 .

Thus, y t gty= −v012

2 becomes

yx

gx

yx x

=⎛⎝⎜

⎞⎠⎟

−⎛⎝⎜

⎞⎠⎟

vv v0

0 0

21

2

or

yg

x xx

y

x

= −⎛⎝⎜

⎞⎠⎟

+⎛⎝⎜

⎞⎠⎟

+2

002

2 0

0v

v

v

(b) Note that this result is of the general form y ax bx c= + +2 with

ag

bx

y

x

= −⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟2 0

20

0v

v

v, , andd c = 0

45°

R

v0

45°

R

v0

45°

R

v0

45°

R

v0


110 Chapter 3

3.63 In order to cross the river in minimum time, the velocity of the boat relative to the water (

�vBW) must be perpendicular to the banks

(and hence perpendicular to the velocity �vWS of the water relative

to shore).

The velocity of the boat relative to the water is � � �v v vBW BS WS= − ,

where �vBS is the velocity of the boat relative to shore. Note that this vector

equation can be rewritten as � � �v v vBS BW WS= + . Since

�vBW and

�vWS are to be perpendicular in this

case, the vector diagram for this equation is a right triangle with �vBS as the hypotenuse.

Hence, velocity of the boat relative to shore must have magnitude

v v vBS BW WS km h km h km h= + = ( ) + ( ) =2 2 2 2

12 5 0 13.

and be directed at

θ =⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠

− −tan tan.

1 1 12

5 0

vv

BW

WS

km h

km h⎟⎟ = °67

to the direction of the current in the river (which is the same as the line of the riverbank).

The minimum time to cross the river is

t = = ⎛⎝

width of river km

12 km h

min

1 hBWv1 5 60. ⎞⎞

⎠ = 7 5. min

During this time, the boat drifts downstream a distance of

d t= = ( )( )⎛⎝

⎞⎠vWS

3

km h minh

60 min

10 m5 0 7 5

1. .

11 kmm

⎛⎝⎜

⎞⎠⎟

= ×6 3 102.



3.64 For the ball thrown at 45.0°, the time of fl ight is found from

∆y t a ty y= +v021

2 as 0

2 20

1 12= ⎛

⎝⎜⎞⎠⎟ −

vt

gt

which has the single nonzero solution of

tg1

0 2=

v

The horizontal range of this ball is

R tg gx1 0 1

0 0 02

2

2= = ⎛

⎝⎜⎞⎠⎟

⎛⎝⎜

⎞⎠⎟

=vv v v

Now consider the fi rst arc in the motion of the second ball, started at angle q with initial speed v0. Applied to this arc, ∆y t a ty y= +v0

12

2 becomes

020 21 21

2= ( ) −v sinθ tg

t

with nonzero solution

tg21

02=

v sinθ

Similarly, the time of fl ight for the second arc (started at angle q with initial speed v0 2) of this ball’s motion is found to be

tg g22

0 02 2=

( )=

v vsin sinθ θ

The horizontal displacement of the second ball during the fi rst arc of its motion is

R tgx21 0 21 0

0 022 2

= = ( )⎛⎝⎜

⎞⎠⎟

=v vv v

cossin sin c

θ θ θ oos sinθ θ( ) =( )

g g

v02 2

Similarly, the horizontal displacement during the second arc of this motion is

Rg g22

0

2

022 2 1

4

2=

( ) ( )=

( )v vsin sinθ θ

The total horizontal distance traveled in the two arcs is then

R R Rg2 21 22

025

4

2= + =

( )v sin θ

(a) Requiring that the two balls cover the same horizontal distance (that is, requiring that R R2 1= ) gives

5

4

202

02v vsin ( )θ

g g=

This reduces to sin ( )2 45θ = , which yields 2 53 1θ = °. , so θ = °26 6. is the required

projection angle for the second ball.



112 Chapter 3

(b) The total time of fl ight for the second ball is

t t tg g g2 21 22

0 0 02 3= + = + =

v v vsin sin sinθ θ θ

Therefore, the ratio of the times of fl ight for the two balls is

t

t

g

g2

1

0

0

3

2

3

2=

( )( ) =v

v

sinsin

θθ

With q = 26.6° as found in (a), this becomes

t

t2

1

3

226 6 0 950= ( )° =sin . .

3.65 The initial velocity components for the daredevil are v v0 0 45x = °cos and v v0 0 45y = °sin , or

v vv

0 00

2

25 0x y= = = . m s

2

The time required to travel 50.0 m horizontally is

tx

x

= = ( )=∆

v0

50 0

25 02 2

.

.

m 2

m ss

The vertical displacement of the daredevil at this time, and the proper height above the level of the cannon to place the net, is

∆y t a ty y= + = ⎛⎝⎜

⎞⎠⎟ ( ) −v0

21

2

25 0

22 2

1

29 80

..

m ss mm s s m2( )( ) =2 2 10 8

2.

3.66 The vertical component of the salmon’s velocity as it leaves the water is

v v0 0 6 26 45 0 4 43y = + = +( ) ° = +sin . sin . .θ m s m s

When the salmon returns to water level at the end of the leap, the vertical component of velocity will be v vy y= − = −0 4 43. m s.

The time the salmon is out of the water is given by

ta

y y

y1

0 4 43 4 43

9 800 903=

−= − −

−=

v v . .

..

m s m s

m s2 s

The horizontal distance traveled during the leap is

L t tx= = ( ) = ( ) °v v0 1 0 1 6 26 45 0 0 903cos . cos . .θ m s s(( ) = 4 00. m

To travel this same distance underwater, at speed v = 3 58. m s, requires a time of

tL

2

4 001 12= = =

v.

.m

3.58 m ss

The average horizontal speed for the full porpoising maneuver is then

vavtotal

total

m

s= =

+= ( )∆

∆x

t

L

t t

2 2 4 00

0 9031 2

.

. ++=

1 123 96

..

sm s



3.67 (a) and (b)

Since the shot leaves the gun horizontally, v v0 0x = and the time required to reach the target is

tx x

x

= =∆v v0 0

.

The vertical displacement occurring in this time is

∆y y t a t gx

y y= − = + = −⎛⎝⎜

⎞⎠⎟

vv0

2

0

21

20

1

2

which gives the drop as

y gx

Ax Ag=

⎛⎝⎜

⎞⎠⎟

= =1

2 20

2

2

02 0v v

vwith , where iis the muzzle velocity

(c) If x = 3.00 m, and y = 0.210 m, then

m

3.00 mm 1A

y

x= =

( )= × − −

2 220 210

2 33 10.

.

and

v0 22

9 80

2 2 33 1014 5= =

×( ) =− −

g

A

.

..

m s

mm s

2

1

3.68 The velocity of the wind relative to the boat, �vWB, is given by

� � �v v vWB WE BE= − , where

�vWE and

�vBE are the velocities of the wind and the boat relative to Earth, respectively. Choosing the positive x-direction as east and positive y as north, these relative velocities have components of

�vWE knots( ) = +

x17

�vWE knots( ) =

y0

�vBE( ) =

x0

�vBE knots( ) = +

y20

so

� � �v v vWB WE BE knots( ) = ( ) − ( ) = +

x x x17

� � �v v vWB WE BE knots( ) = ( ) − ( ) = −

y y y20

Thus,

� � �v v vWB WB WB knots knot= ( ) + ( ) = ( ) + −

x y

2 2 217 20 ss knots( ) =2 26

and

θ =( )( )

⎛

⎝⎜

⎞

⎠⎟ = −− −tan

knots1 WB

WB

�

�v

vy

x

tan 1 20

17750

knots⎛⎝

⎞⎠ = − °

or

�vWB knots at south of east= °26 50

The component of this velocity parallel to the motion of the boat (that is, parallel to a north-south line) is ( ) .

�vWB knots, or knots southy = −20 20


114 Chapter 3

3.69 (a) Take the origin at the point where the ball is launched. Then x y0 0 0= = , and the coordinates of the ball at time t later are

x t tx= = ( )v v0 0 0cosθ and y t a t tg

ty y= + = ( ) − ⎛⎝

⎞⎠v v0

20 0

21

2 2sinθ

When the ball lands at x R= = 240 m, the y-coordinate is y = 0 and the elapsed time is found from

020 0

2= ( ) − ⎛⎝

⎞⎠v sinθ t

gt

for which the nonzero solution is

tg

=2 0 0v sinθ

Substituting this time into the equation for the x-coordinate gives

xg g

= = ( )⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

2402

0 00 0 0

2

m vv v

cossinθ θ

22 20 002

0sin cos sinθ θ θ( ) =⎛⎝⎜

⎞⎠⎟

( )vg

Thus, if v0 50 0= . m s, we must have

sin.

.2

240 240 9 80

50 00

02θ( ) = ( )

=( )( )m m m s2

g

v mm s( )= +2 0 941.

with solutions of 2 70 2 2 180 70 2 109 80 0θ θ= ° = ° − ° = °. . .and .

So, the two possible launch angles are θ θ0 035 1 54 9= ° = °. . and .

(b) At maximum height, vy = 0 and the elapsed time is given by

t

a g

y y

ypeak

peak or=( ) −

=−

−

v v v00 00 sinθ

peaktg

=v0 0sinθ

The y coordinate of the ball at this time will be

y tg

tmax sin sin= ( ) − ⎛⎝

⎞⎠ = ( )v v

v0 0

20 02

θ θpeak peak00 0 0

2 20

202 2

0

2 2

sin sin sinθ θ θg

g

g

⎛⎝⎜

⎞⎠⎟

− ⎛⎝

⎞⎠ =

v vgg

The maximum heights corresponding to the two possible launch angles are

( ). sin .

.maxy 1

2 250 0 35 1

2 9 8042=

( ) ( )°( ) =m s

m s2..2 m

and

( ). sin .

.maxy 2

2 250 0 54 9

2 9 8085=

( ) ( )°( ) =m s

m s2..4 m



3.70 (a) Consider the falling water to consist of droplets, each following a projectile trajectory. If the origin is chosen at the level of the pool and directly beneath the end of the channel, the parameters for these projectiles are

x0 0= y h0 2 35= = . m

v0 0 75x = . m s v0 0y =

ax = 0 a gy = −

The elapsed time when the droplet reaches the pool is found from y y t a ty y− = +0 012

2v as

0 02

2− = −hg

t p or th

gp = 2

The distance from the wall where the water lands is then

R x th

gx p x= = = = ( ) ( )max .

.

.v v0 0

20 750

2 2 35

9 80m s

m

m sm2 = 0 519.

This space is too narrow for a pedestrian walkway.

(b) It is desired to build a model in which linear dimensions, such as the height hmodel and horizontal range of the water Rmodel, are one-twelfth the corresponding dimensions in the actual waterfall. If vmodel is to be the speed of the water fl ow in the model, then we would have

R th

gpmodel model model modelmodel= ( ) =v v

2

or

vmodel modelmodel

actual

actual

= = (Rg

h

R g

h2 12 2 12)) =⎛

⎝⎜⎞

⎠⎟=1

12 2 12R

g

hactualactual

actualv

and the needed speed of fl ow in the model is

vv

modelactual m s

m s= = =12

0 750

120 217

..


116 Chapter 3

3.71 (a) Applying ∆y t a ty y= +v012

2 to the vertical motion of the fi rst snowball gives

0 25 0 70 01

29 801 1

2= ( )⎡⎣ ⎤⎦° + −( ). sin . .m s m s2t t

which has the nonzero solution of

t1

2 25 0 70 0

9 804 79=

( ) °=

. sin .

..

m s

m s s2

as the time of fl ight for this snowball.

The horizontal displacement this snowball achieves is

∆x tx= = ( )⎡⎣ ⎤⎦° ( ) =v0 1 25 0 70 0 4 79 41 0. cos . . .m s s m

Now consider the second snowball, also given an initial speed of v0 25 0= . m s, thrown at angle q , and is in the air for time t2. Applying ∆y t a ty y= +v0

12

2 to its vertical motion yields

0 25 01

29 802 2

2= ( )⎡⎣ ⎤⎦ + −( ). sin . m s m s2θ t t

which has a nonzero solution of

t2

2 25 0

9 805 10=

( )= ( ). sin

.. sin

m s

m s s2

θθ

We require the horizontal range of this snowball be the same as that of the fi rst ball, namely

∆x tx= = ( )⎡⎣ ⎤⎦ ( )[ ] =v0 2 25 0 5 10 41. cos .m s s sinθ θ ..0 m

This yields the equation

sin

m

m s sθ θcos

.

. ..= ( )( ) =41 0

25 0 5 100 321

From the trigonometric identity sin 2 2sinθ θ θ= cos , this result becomes

sin 2 2θ = 0 321 0 642. .( ) = so

2θ = °40 0.

and the required angle of projection for the second snowball is

θ = °20.0 above the horizontal

(b) From above, the time of fl ight for the fi rst snowball is t1 4 79= . s and that for the second snowball is

t2 5 10 5 10 20 0 1 74= ( ) = ( ) ° =. sin . sin . .s s sθ

Thus, if they are to arrive simultaneously, the time delay between the fi rst and second snowballs should be

∆t t t= − = − =1 2 4 79 1 74 3 05. . .s s s



3.72 First, we determine the velocity with which the dart leaves the gun by using the data collected when the dart was fi red horizontally (v0 0y = ) from a stationary gun. In this case, ∆y t a ty y= +v0

12

2 gives the time of fl ight as

ty

ay

= =−( )

−=2 2 1 00

0 452∆ .

. m s

9.80 m s s2

Thus, the initial speed of the dart relative to the gun is

v vDG

m

0.452 sm s= = = =0

5 0011 1x

x

t

∆ ..

At the instant when the dart is fi red horizontally from a moving gun, the velocity of the dart

relative to the gun may be expressed

as � � �v v vDG DE GE= − where

� �v vDE GEand are the velocities of the

dart and gun relative to Earth respectively. The initial velocity of the dart relative to Earth is therefore

� � � �v v v v0 DE DG GE= = +

From the vector diagram, observe that

v v0 45 0 2 00 45 0 1 41y = − °= −( ) ° = −GE m s msin . . sin . . ss

and

v v v0 45 0 11 1 2 00 45x = + ° = + ( )DG GE m s m scos . . . cos .00 12 5° = . m s

The vertical velocity of the dart after dropping 1.00 m to the ground is

v vy y ya y= − + = − −( ) + −( ) −02 2

2 1 41 2 9 80 1∆ . . .m s m s2 000 4 65m m s( ) = − .

and the time of fl ight is

ta

y y

y

=−

=− − −( )

−=

v v0 4 65 1 41

9 800 3

. .

..

m s m s

m s2 330 s

The displacement during the fl ight is ∆x tx= = ( )( ) =v0 12 5 0 330 4 12. . .m s s m .


118 Chapter 3

3.73 (a) First, use ∆ x t a tx x= +v012

2 to fi nd the time for the coyote to travel 70 m, starting from rest with constant acceleration ax = 15 m s2:

tx

ax1

2 2 70

153 1= = ( )

=∆ m

m ss2 .

The minimum constant speed the roadrunner must have to reach the edge in this time is

v = = =∆x

t1

7023

m

3.1 sm s

(b) The initial velocity of the coyote as it goes over the edge of the cliff is horizontal and equal to

v v0 0 10 15 3 1 46= = + = ( )( ) =x xa t m s s m s2 .

From ∆y t a ty y= +v012

2, the time for the coyote to drop 100 m with v0 0y = is

ty

ay2

2 2 100

9 804 52= = −( )

−=∆ m

m ss2.

.

The horizontal displacement of the coyote during his fall is

∆x t a tx x= + = ( )( ) + ( )v0 212 2

2 1246 4 52 15m s s m s2. 44 52 3 6 102 2. .s m( ) = ×


Solucionario Fundamentos de Física 9na edición Capitulo 3

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Transcript of Solucionario Fundamentos de Física 9na edición Capitulo 3