Sm PDF Chapter3

8/3/2019 Sm PDF Chapter3

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Chapter 3

Vectors and

Two-Dimensional Motion

Quick Quizzes

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

and B

: R =A + B = 20 units. The smallest possible magnitude of the resultant occurs

when the two vectors are in opposite directions, and the magnitude is the difference of themagnitudes of A and B : R = |A B|= 4 units.

a

A

2. (b). The resultant has magnitudeA + B when A

is oriented in the same direction as B

.

3. Vector x component y component

A

+

B

+

A

+ B

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

5. (a). Any change in the magnitude and/or direction of the velocity is an acceleration. Thegas 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.

6. (c). A projectile has constant horizontal velocity. Thus, if the runner throws the ballstraight up into the air, the ball maintains the horizontal velocity it had before it wasthrown (that is, the velocity of the runner). In the runners frame of reference, the ballappears 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 runnerand staying above the runners hand.

7. (b). The velocity is always tangential to the path while the acceleration is always directedvertically downward. Thus, the velocity and acceleration are perpendicular only wherethe path is horizontal. This only occurs at the peak of the path.

55


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56 CHAPTER 3

Answers to Even Numbered Conceptual Questions

2. The magnitudes add when andA

B

are in the same direction. The resultant will be zerowhen the two vectors are equal in magnitude and opposite in direction.

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

6. The component of a vector can only be equal to or less than the vector itself. It can neverbe greater than the vector.

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

10. They both start from rest in the downward direction and accelerate alike in the verticaldirection. Thus, they reach the ground with the same vertical speed. However, the ballthrown horizontally had an initial horizontal component of velocity which is maintainedthroughout the motion. Thus, the ball thrown horizontally moves with the greater speed.

12. The car can round a turn at a constant speed of 90 miles per hour. Its velocity will bechanging, however, because it is changing in direction.

14. The balls will be closest at the instant the second ball is projected. The first ball will alwaysbe going faster than the second ball. There will be a one second time interval between theircollisions with the ground. The two move with the same acceleration in the verticaldirection. Thus, changing their horizontal velocity can never make them hit at the sametime.

16. Let v and0x 0yv represent its original velocity components. We know that the vertical

component of velocity is zero at the top of the trajectory. Thus, 00 yv gt= and the time at

the top of the trajectory is0y g=t v .

(a)0

0

y

x

vx v

g

=

and

20

2

yvy

g=

(b) Its velocity is horizontal and equal to .0xv

(c) Its acceleration is vertically downward, g.

With air resistance, the answers to (a) and (b) would be smaller. As for (c) the magnitudewould be somewhat larger because the total acceleration would have a componenthorizontally backward in addition to the vertical component of g.

18. The equations of projectile motion are only valid for objects moving freely under the

influence 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.

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


3/44

Vectors and Two-Dimensional Motion 57

Answers to Even Numbered Problems

2. (a) Approximately 484 km (b) Approximately 18.1 N of W

4. Approximately 83 m at 33 N of W

6. (a) Approximately 6.1 units at 113(b) Approximately 15 units at 23

8. (a) Approximately 5.2 m at + 60 (b) Approximately 3.0 m at -30(c) Approximately 3.0 m at + 150 (d) Approximately 5.20 m at -60

10. 1.31 km north, 2.81 km east

12. 358 m at 2.00 S of E

14. 42.7 yards

16. 788 mi at 48.1 N of E

18. (a) 185 N at 77.8 (b) 185 N at 258

20. (a) 74.6 N of E (b) 470 km

22. 2.68 ft (0.817 m)

24. 3.19 s, 36 at 60.1 below the horizontal.1 m s

26. 2.8 m from base of table;5.0 m s, 5.4 m sx y

v v= =

28. 3 37.23 10 m, 1.68 10 mx y= =

30. (a) clears the bar by 0.85 m (b) falling, 13.4 m syv =

32. 18.7 m

34. 9.91 m s

36. 61 s

38. (a) 10.1 m s at 8.53 E of N (b) 45.0 m

40. (1. ), Yes81 m s, 5.43 m s

42. 15.3 m

44. 7.87 N at 97.8 counterclockwise from a horizontal line to the right


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58 CHAPTER 3

46. (a) 57.4m, 1.22 s (b) 57.4 m, 1.22 s

48. 0.344 m, 2.34 m

50. (a) 0.85 m s (b) 2 .1 m s

52. 14 m s

54. See Solution Section

56. 29.4 m s

58. 10.7 m s

60. (a) ,

(b) ,

132 cm at 69.6m = d

146 cm at 69.6m = d

111 cm at 70.0f = d

132 cm at 70.0f = d

cm at 65.8

m 14.0f = =d d d

64. (a) 1.20 m s , 0 (b) 0.960 m (c) 0.500 m s

66. 3.96 m s

68. 26 knots at 50 south of east; 20 knots due south

70. (a) 26.6 (b) 0.950

72. 4.12 m


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Problem Solutions

3.1 Your sketch should be drawn to scale, and be similar to that pictured below. The length

of R and the angle

can be measured to find, with use of your scale factor, the

magnitude and direction of the resultant displacement. The result should be:approximately 421 f at 3 below the horizontalt

q

Vertical

Horizontal

R

200 ft13

5ft 40

30

135ft

ur

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

2 2d x y= +

where ( )200x

(

km 300 km cos 30.0 460 km= + =

and )0 300y = + km sin 30.0 150 km =

2 2(460 km) (150 km) 484 kmd = + =

(b) 1 1150 km

tan tan 18.1 N of W460 km

y

x =

= =

3.3 The displacement vectors and

can be drawn to scale as at the right. The

vector represents the displacement that the man in themaze must undergo to return to his starting point. The

scale used to draw the sketch can be used to find

8.00 m westward=A

13.0 m north=B

C

C

to be

15 m at 58 S of E

d300km

B 200 km

C

A

f

q

8.00 m

13.0 m

N

ES

W

Cur


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60 CHAPTER 3

Rplane

A

B

BaseE

N

20.0

30.0

q 280k

m

190km

ur

q 15.0

R=175m

100 m

N

W

S

E

Bur

3.4 Your vector diagram should look like theone shown at the right. The initial

displacement and the

resultant areboth known. In order to reach the end point

of the run following the initialdisplacement, the jogger must follow the

path shown as B

100 m due west=A

175 m at 15.0 N of W= R

. The length of B

and the

angle can be measured. The results

should be 83 m at 33 N of W

3.5 Using a vector diagram, drawn to scale, likethat shown at the right, the finaldisplacement of the plane can be found to

be R . The

requested displacement of the base from

point B is

310 km at = 57 N of Eplane =

planeR

plane =R

, which has the same

magnitude but the opposite direction. Thus,the answer is

310 km at = 57 S of W

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

at the head of vector A . The

new vector has a magnitude of

+A B

6.1 units 113at from the positive x-axis.

(b) The vector difference A B

is found by placing the negative of vector at the head

of vector . The resultant vector

B

A

A B

has magnitude 15 units at 23 from the

positive x-axis.

x

y

A A B

BB

A + B

ur

ur ur ur ur ur

ur


7/44


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

at the

tip of vector

. The vector sum,A +A B

, is then found asshown in the vector diagram and should be

5.0 units at 53+ = A B

(b) To find the vector difference, form the vector B

(same

magnitude, opposite direction) and add it to vector A

asshown in the diagram. You should find that

5.0 units at 53 = + A B

3.8 (a) Drawing these 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

by 180. The resultant is

3.0 m at 30 .

(c) Maintain the direction of B

, but reverse the direction of A

. The resultant is

3.0 m at 150+ .

(d) Maintain the direction of A

, reverse the direction of B

, and multiply its magnitude

by two. The resultant is 5.2 m at 60 .

+xA

B

B

A B

A + B

ur ur ur

ur

ur ur ur

E

5.4

0m

R

q

S

6.00 m

ur

3.9 Using the vector diagram given at the right, we find

( ) ( )2 2

6.00 m 5.40 m 8.07 mR = + =

and ( )1 15.40 m

tan tan 0.900 42.06.00 m

= =

=

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


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62 CHAPTER 3

3.10 The total displacement is D . The north and east components ofthis displacement are:

3.10 km at 25.0 N of E=

( )3.10 km sinyD 25.0= = 1.31 km north

and ( )3.10 km cosxD = 25.0 = 2.81 km east

3.11 (a) Her net x (east-west) displacement is 3.00 0 6.00 3.00 blocks + + =+

0 4.00 blocks

, while her nety

(north-south) displacement is 0 4.00+ + =+ . The magnitude of the

resultant displacement is

( ) ( ) ( ) ( )2

3.00R x y= + =2 2

4.0+2

0 5.00 blocks=

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

( ) 53.11 14.00

tan tan 1.333.00

= =

=

The resultant displacement is then 5.00 blocks at 53.1 N of E

(b) The total distance traveled is 3.00 4.00 6.00+ + = 13.0 blocks

3.12 ,eastwardx+ = northwardy+ =

( )250x = m 125 m cos3+(

0.0 358 m = )75.0y = m 125 m sin+

( ) ( )

30.0 150 m 12.5 m=

( ) ( )2 22 2

m 12.5 m 358+ =358y+ = md x=

1 -1 12=tan 35

y

x

.5= 2.00

8 =

tan 358 m at 2d .00 S of E=

3.13 25.0xA = 40.0yA =

( ) ( )2 22 2 40.0x yA +25.0= A A= + = 47.2 units

58.0

From the triangle, we find that = ,

so that 122= 25.0

x

y

A

40.0

qf

ur


9/44


3.14 Let be the vector corresponding to the 10.0 yd run, BA

to the 15.0 yd run, and C

to the50.0 yd pass. Also, we choose a coordinate system with the +y direction downfield, andthe + x direction toward the sideline to which the player runs.

The components of the vectors are then

0xA=

10.0 ydsyA=

0yB

15.0xB = yds =

0xC = 50.0 ydsyC = +

From these, 15.0 ydsxR x= = , and 40.0 ydyR y s= = ,

and ( ) ( )2 2

40.0 yds+ =2 2x yR= + 15.0= ydsR R 42.7 yards

3.15 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 twodisplacements are:

Displacement x-component (eastward) y-component (northward)123 km -61.5 km +107 km37.5 km 0 +37.5 km

Resultant -61.5 km 144 km

Therefore, the eye of the hurricane is now

( ) ( )2 2

61.5 km 144 km 157 km from the islandR = + =

3.16 Choose the positive x direction to be eastward and positivey as northward. Then, thecomponents of the resultant displacement from Dallas to Chicago are:

( ) ( )730 mi cos 5.00 560 mi sin 21.0 527 mixR x= = =

(

and ) ( )730 mi sin 5.00 560 mi cos 21.0 586 miyR y= = + =

( ) ( )

2 22 2

527 mi 586 mi 788 mix yR R R= + = + =

( )1 1tan tan 1.11 48.1y

x

= = =

Thus, the displacement from Dallas to Chicago is 788 mi at 48.1 N of E= R


10/44

64 CHAPTER 3

3.17 The components of the displacements a

, b

, and c

are:

and

Thus, , and

cos30.0 152 kmxa a= = +

cos110 51.3 kmxb b= =

cos180 190 kmxc c= =

sin 30.0 87.5 kmya a= = +

sin110 141 kmyb b= = +

sin180 0yc c= =

89. kmx x x xR a b c= + + =

7 228 km y y y yR a cb= + + = +

so 2 2 245 kmx yR R R= + = , and ( )1 1tan 1.11x

y

R

R tan 21.4

= = =

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

x east

y north

a

b

c

R

30.0

20.0

110

q

ur

ur

ur

ur

3.18 (a) ( ) ( )1 1 1120 N 60.0 N 104 Nx yF = =F F

(

=

) ( )2 2 280.0 N 20.7 N 77.3 Nx yF = =F F

( ) ( )

=

( ) ( )22 2 2

39.3 N 181 N 185 NR x yF F F= + = + =

and ( )1 1181 N

tan tan 4.61 77.8

39.3 N

= = =

The resultant force is 185 N at 77.8 from the x-axiR

=

sF

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

185 N at 258 from the x-axis


11/44


3.19 The resultant displacement is = +R A B

, where A

is the 150 cm displacement at 120

and B is the required second displacement. Solving for B

: ( )= = + B R A R A

The components of are

and B R

B

= 190 cmx x xB R A= + 49.6y y cmyA= =

Hence, 2 2 196 cmx yB B= + =B and ( ).262 14.71 1tan 0

y

x

B

B tan

= = =

196 cm a 14.7 be= B

t low the x-axis

3.20 Let +x = East, +y = North,

Displacement x (km) y (km)300 km, due E 300 0350 km, 30 W of N 175 303150 km, due N 0 150Resultant x = 125 y = 453

(a) 1tan 74.6 N of Ey

x

= = (b) ( ) ( )

22470 kmy= + =R x

3.21 (a) Your first displacement takes you to point A, so 1 A=r r

. In the second displacement,

you go one-half the distance from A toward B, so ( B A1

2 r r r )=

and your current

position vector is ( ) A B2 1 A B A1

=2 2

+= + + =r r r r r r

r r

. On the next leg of the hunt,

your displacement is ( )C 21

3 = r r r

and your new position vector becomes

( ) A B3 2 2 C1

3C

3C 2 2

1 2

3 3

+ += + = + =

r rr r r r r r = +r r

r

. The next displacement is

( )D 31

=4

r r r

and your position vector changes to

( ) A B D3 D1

4C

44 3 D 3 3

1 3

4 4

+ + +r r

= + =r r r + =

r rr r = +r r r

. On the final leg of the hunt,

the displacement is ( )E 41

=5

r r r

. Therefore, the position vector of the treasure is

( )5 4

4 1

5 5

= +

= +

r r r

r r

4

E4 E

= +

+

r

r

E 4

A B C D

1

5

5

+ + +=

r r

r r r r


12/44

66 CHAPTER 3

To determine the coordinates of this location, we consider

A B C D= + + + +R r r r r r

E

Vector x component (m) y component (m)

Ar

30.0 20.0Br

60.0 80.0

Cr

10.0 10.0

Dr

40.0 30.0

Er 70.0 60.0

R

50.0x = 80.0y =

The position vector of the treasure is 55

=R

r

and its coordinates are then seen to be

110.0 m

5

xx R= = + and1

16.0 m

5

yy R= = +

(b) From the solution of part (a), the position vector of the treasures location is seen todepend on the sum of the position vectors of the individual trees:

A B C D E5

5 5

+ + + += =

r r r r rRr

. Interchanging the trees would only change the order of

the vectors in this sum. Since a vector sum is independent of the order in which thevectors are added, the answer found in part (a) does not depend on the order of thetrees.

3.22 0 101 mi h 45.1 m sxv = = and 60.5 ft 18.4 mx = =

The time to reach home plate is0

18.4 m

45.1 m sx

x

v0.408 st

= = =

In this time interval, the vertical displacement is

( )( )229.80 m s 0.408 s 0. = 201 1

0 817 m2 2

y yy v t a t = + = +

Thus, the ball drops vertically3.281 ft

17 m 2.68 ft

1 m

=

0.8


13/44


3.23 The constant horizontal speed of the falcon is

0.447 m smi200 89.4 m s

h 1 mi hxv

= =

The time required to travel 100 m horizontally is 100 m 1.12 s89.4 m sx

xv

t = = = . The vertical

displacement during this time is

( )( )22 201 1

0 9.80 m s 1.12 s 6.13 m2 2

y yy v t a t = + = + =

or the falcon has a vertical fall of 6.13 m

3.24 We find the time of fall from2

0

1

2yy v t a t = + y with 0 0yv = :

( ) ( )2

2 2 50.0 m3.19 s

9.80 m s

yt

a

= = =

At impact, 0 18.0 m sx xv v= = , and the vertical component is

( )( )20 m s 3.19 s 3= 0 0 9.8 y y yv v a t= + = + 1.3 m s

Thus, ( ) ( )2 22 2

18.0 m s 31.3 m s 36.1 m sx yv= + = + =v v

and 1 131.3

tan tan 60.118.0

y

x

v

v

= = =

or 36.1 m s at 60.1 below the horizontal= v


14/44

68 CHAPTER 3

3.25 At the maximum height , and the time to reach this height is found from0yv =

0 y y yv v a= + t as0 0 0 0 y y y y

y

v v v vt

a g=

( )

g

= =

The vertical displacement that has occurred during this time is

( )2

0 0 0

x av

0

2 2

0

2ma y y y y y

y

v v v v v y v t

g g

+ + = = =

t =

Thus, if ( )max

12 ft 3.7 m

= =

( )

1 m

3.281 fty , then

( )( )2max

2 9.80 m s 3.7 m 8.5 m sv g y= =

45

0y 2=

and if the angle of projection is =

, the launch speed is

0 8.5 m s12 m s

n sin 45

yv

= = =

0

siv

3.26 The time of flight for Tom is found from 201

2y yy v t a t = + with v :0 0y =

( ) ( )2

2 2 1.5 m0.55 s

9.80 m s

y

a

= = =

t

The horizontal displacement during this time is

( )( )5.0 m s 0.55 s 2.8 mxt = =0x v =

Thus, he lands 2.8 m from the base of the table

The horizontal component of velocity does not change during the flight, so

0 5.0 m sx xv v= = . The vertical component of velocity is found as

( )( )20 0 9.80 m s 0.55 s 5.4 m s y y yv v a t= + = =


15/44


3.27 When ( )max

, 0y y y v = =

0

.

Thus, y y yv v a t= + 0yields or0 sin3.00v g= t0 sin3.00vt

g

=

The vertical displacement is 2012

yy v t a t = +

( )

y . At the maximum height, this becomes

( )2

2 20 0 0sin 3.00 sin 3.00

2 2

v v vg

g g

=

0max

sin3.00y v =sin3.00 1

g

If ( )max

0.330 m =y , the initial speed is

( ) ( )( )2s 0.330 m48.6 m s

3.00

=

max2 9.80 m

0 sin

= =02

sin3.0

g yv

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

3.28 The horizontal displacement at 42.0 st = is

( ) ( )( )( ) 30 0 cos 300 m s cos55.0 42.0 s 7.23 10 mxx v t v t= = = =

42.0 st

The vertical displacement at = is

( )2 20 01 1sin2 2

y y y v t a t v t gt= + =

( )( )( ) ( )( )22 31

9.80 m s 42.0 s 1.68 10 m2

+ =300 m s sin 55.0 42.0 s=

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

To findH, we use 201

2y yy v t a t = + , which becomes

( )( )( ) ( )( 2215 m s sin 25 3.0 s m s 3.0 sH = )1

9.802

+ , or 25 mH=


16/44

70 CHAPTER 3

3.30 The components of the initial velocity are:

0 0 cos 53.0 12.0 m sxv v= = , and 0 0 sin 53.0 16.0 m syv v= =

(a) The time required for the ball to reach the position of the crossbar is

0

36.0 m3.00 s

12.0 m sx

xt

v

= = =

At this time, the height of the football above the ground is

( ) ( )22

0 2

1 m 1 m16.0 3.00 s 9.80 3.00 s 3.90 m

2 s 2 sy yy v t a t

= + = + =

Thus, the ball clears the crossbar by 3.90 m 3.05 m 0.85m =

(b) The vertical component of the velocity of the ball as it moves over the crossbar is

( ) ( )( )20 16.0 m s 9.80 m s 3.00 s 13.4 m s y y yv v a t= + = + = . The negative sign

indicates the ball is moving downward or falling .

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

( ) ( )( )2 20 2 0 2 4.00 m s 50.0 m 20.0 m sv v a x= + = + =

( )

Now, consider the projectile phase of the cars motion. The vertical velocity of the car asit reaches the water is

( ) ( )( )22 2

0 2 20.0 m s sin 24.0 2 9.80 m s 30.0 m y y yv v a y = + = +

or 25.6 m syv =

(b) The time of flight is

( )02

25.6 m s 20.0 m s sin 24.01.78 s

9.80 m s

y y

y

v vt

a

= = =

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

( ) ( )0 20.0 m s cos 24.0 1.78 s 32.5 mxx v t = = =


17/44


3.32 The components of the initial velocity are

( )0 40.0 m s cos 30.0 34.6 m sxv = = , and

( )0 40.0 m s sin 30.0 20.0 m syv = =

The time for the water to reach the building is

0

50.0 m1.44 s

34.6 mx

xt

v

= = =

The height of the water at this time is

( )( ) ( )( )22 20 .80 m s 1.44 s 18.7 my v = =1 1

20.0 m s 1.44 s 92 2

y yt a t+ = +

40.0m

/s

30.0

y

50.0 m

3.33 (a) At the highest point of the trajectory, the projectile is moving horizontally withvelocity components of and0yv =

( )= =0 0 cos 60.0 cos30.0 52.0 m sx xv v v = =m/s

(b) The horizontal displacement is ( )( )0 52.0 m s 4.00 s 208 mxx v t = = = and, from

( ) 201

sin2

y y v t a t = + , the vertical displacement is

( )( )( ) ( )( )221

60.0 m s sin 30.0 4.00 s 9.80 m s 4.00 s 41.6 m2

y = + =

( ) ( )

The straight line distance is

( ) ( )22 2 2

208 m 41.6 m 212 md x y= + = + =


18/44

72 CHAPTER 3

3.34 The horizontal kick gives zero initial vertical velocity to the ball. Then, from

20

1

2y yy v t a t = + , the time of flight is

( ) ( )2

2 2 40.0 m8.16 s

9.80 m sy

yt

a

= = =

The extra time 3.00 s 8.16 s 0.143 st = =

( ) ( )

is the time required for the sound to travel

in a straight line back to the player. The distance the sound travels is2

sound y v t = x2

d x= + where represents the horizontal displacement of the ball

when it hits the water. Thus,

( ) ( )( ) ( )22 22 343 m 0.143 s 40.0 m 28.3 mx d y = = = s

The initial velocity given the ball must have been

0 0

28.3 m9.9 s

8.16 sx

xv v

t

= = = = 1 m

3.35 The velocity of the plane relative to the ground is the vector sum of the velocity of thep the air and the velocity of the air relative to the ground, or

.lane relative to

PG PA AG= +v v v

30.0q

east

north

vPA = 300 mph

vAG = 100 mphvPG

ur

ur

ur

The components of this velocity are

( ) ( )PG 300 mi h 100 mi h cos30.0 387 mi heast = + =v

(

and ) ( )PG 0 100 mi h sin30.0 50.0 mi hnorth = + =v


19/44


Thus, ( ) ( ) ( ) ( )2 2

PG PG PG

2 2387 50.0 mi h 390 mi heast north

= + =v v

(

v = +

and)

( )

PG1 1

PG

50.0tan tan 7.37

387north

east

= = =

v

v

The plane moves at 390 mi h at 7.37 N of E re lative to the ground

3.36 We use the following notation:

velocity of boat relative to the shore

velocity of boat relative to the water,

and velocity of water relative to the shore.

If we take downstream as the positive direction, then

BS=v

BW=v

WS=v

WS1.5 m s= +v

for both parts of

the trip. Also,BW

10 m s= +v

while going downstream and BW 10 m s= v

for the

upstream part of the trip.

The velocity of the boat relative to the shore is given byBW WS

= +v v vBS

While going downstream, BS 10 m s 1.5 m sv = + and the time to go 300 m downstream

is( )

300 m26 s

.5 m sdownt = =

10+1

When going upstream, BS 10 m s 1.5 m s 8.5 m= + = sv and the time required to

move 300 m upstream is300 m

35 s8.5 m s

upt

= =

( )

The time for the round trip is 26down upt= + = 35 s+ 61 s=t t


20/44

74 CHAPTER 3

3.37 Prior to the leap, the salmon swims upstream through water flowing at

speedWE

1.50 m s=v

relative to Earth. The fish swims atFW

6.26 m s=v

relative to the water in suc ake its velocity relative to

Earth, v , vertical. Since as shown in the diagram at the

right, we find that

h a direction to m

FE

WEFE FW= +v v v

WE

FW

cos 76.1 = = =

v

v

1 1 1.50 m scos6.26 m s

and the vertical velocity of the fish as it leaves the water is

( )0yv FE FW sin 6.26 m s sin76.1 6.08 m s= = = =v v

The height of the salmon above the water at the top of its leap (that is, when v 0y = ) is

given by

( )

( )

22 20

2

0 6.08 m s1.88 m

2 2 9.80 m s

y y

y

v v

a

= =

y =

vFWvFE

vWE

ur

ur

ur

q

q

3.38BW

10 m s=v

, directed northward, is the velocity of the boat

relative to the water.

WS1.5 m s=v

BSv

BSv v

, directed eastward, is the velocity of the water

relative to shore.

is the velocity of the boat relative to shore, and directed at an

angle of , relative to the northward direction as shown.

BW WS= + v

vBW

q

east

north

vBS

vWS

ur

ur

ur

The northward component ofBS

v

is BS BWcos 10 m sv v = = (1)

The eastward component is BS WSsin 1.5 m sv v = = (2)


21/44


(a) Dividing equation (2) by equation (1) gives

1 1WS

BW

1.50tan tan 8.53

10.0

v

v

= = =

From equation (1), BS10 m s 10.1 m s

cos8.53v = =

Therefore, BS 10.1 m s at 8.53 E of N=

v

(b) The time to cross the river isBS

300 m 300 m30.0 s

cos 10.0 m sv = = =t and the downstream

drift of the boat during this crossing is

( ) ( )( )BS sin 1.50drift v t= = m s 30.0 s 45.0 m=

3.39 velocity of boat relative to the water,

velocity of water relative to the shore

and velocity of boat relative to the shore.

as shown in the diagram.

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

BW=v

WS=v

v

BS=

BS BW WS= +v v v

BSv

is

( ) ( ) ( )BS BWnorth v sin62.5 0 3.30 mi h sin 0 2.93 mi h= + = + =62.5v

The time required to cross the stream is then0.5 i

0.173 h2.9 h

= =

v

05 m

3 mi

BS

t

The eastward (that is, downstream) component of

is

( ) ( )BWBSeast

= WScos62.5v v +v

( )3.30= mi h cos62.5 1.25 mi h 74 mi h + 0.2=

vWS

62.5

north

east

vBW

vBSur

ur

ur


22/44

76 CHAPTER 3

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

( ) ( )( ) ( )2BS5280 ft

0.274 mi h 0.173 h 4.72 10 mi 249 ft1 mieast

d t

= = = =

v

3.40 If the salmon (a projectile) is to have 0yv = when 1.50 my = +

2 20 2 y y ya y

, the required initial

velocity in the vertical direction is given by v v= + as

( )( )2 2s +0 2 0 2 y y yv v a y= + = 9.80 m 1.50 m = 5.42 m s

The elapsed time for the upward flight will be

0

2

0 5.42 m s

9.80 m s

y y

y

v vt

a

= = =

0.553 s

If the horizontal displacement at this time is to be .00 mx = +1 , the required constanthorizontal component of the salmons velocity must be

0

1.00 m

0.553 sx

xv

t

= = =

1.81 m s

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

( ) ( )2 2

m s2 20 0 0 1.x yv v v= + = 81 m s 5.42+ 5.72 m s=

Yes , since 0 6.26 m sv < the salmon is capable of making this jump.

3.41 Choose the positive direction to be the direction of each cars motion relative to Earth.

The velocity of the faster car relative to the slower car is given by FS FE ES= +v v v

, where

FE60.0 km h= +v

is the velocity of the faster car relative to Earth and

ES SE 40.0 km h= = v v

is the velocity of Earth relative to the slower car.

Thus,FS

60.0 km h 40.0 km h 20.0 km h= + =+v

and the time required for the faster

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

3

FS

0.100 km 3600 s5.00 10 h 18.0 s

20.0 km h 1 h

dt

v = = = =


23/44


3.42 the velocity of the ball relative to the car

velocity of the car relative to Earth

BC =v

CE =v

10 m s=

the velocity of the ball relative to Earth

These velocities are related by the equation

BE=v

BEv v

CEBC= + v

as illustrated in the diagram.

Considering the horizontal components, we see that

orBE CEcos60.0v v =CE

BC .0 m scos60.0

vv = = =

10.0 m s20

cos60.0

From the vertical components, the initial velocity of the ball relative to Earth is

BEv vBC sin 60.0 17.3 m= =

(

s

Using )2 20 2 y y yv v a y= + , with 0yv = when the ball is at maximum height, we find

( )( )

( )

( )

2

2

.3 m s015.3 m

2 2 .80 m s

v

a g

= =

20

max

0 y

y

vy

=

2BE

17

2 9=

as the maximum height the ball rises.

3.43 Since R A , then B R and the components of thesecond displacement are:

= + B

xB R=

(

= A

x xA

) ( )140 cm cos35.0

( )

150 cm cos120 190 cm =+=

and

( )n 35.0 150 cm sin 120 y yA 140 cm si= 49.6 cm=yB R=

Thus, 2 2x yB 196 cm=B B= + , and ( )62 14.71 1tan tan 0.2

y

x

B

B

= =

=

The second displacement is 196 cm at 14.7 below the positive -x= axisB

60.0

vBE

vBC

vCE

ur

ur

ur

A

B

R

35.0

120

x

yq

ur

urur


24/44

78 CHAPTER 3

3.44 Observe that when one chooses the x andy axesas shown in the drawing, each of the four forces

lie along one of the axes. The resultant, R

, is easilycomputed as

12.0 N 8.40 N 3.60 Nx xR F= = + = +

31.0 N 24.0 N 7.00 Ny yR F= = + = +

( ) ( )2 22 2 3.60 N 7.00 Nx yR R R= + = + 7.87= N

1tan 62.8y

x

R

R

= =

62.8 35.0 + , or = fr97.8 om the horizontal

7.87 N at 97.8 counterclockwise f= R r eom th horizontal line to the right

3.45 the velocity of the car relative to Earth

the velocity of the water relative to the car

the velocity of the water relative to Earth

These velocities are related as shown in the diagram at the right

CE=v

WC=v

WE=v

horizontal

x

y

31.0 N

8.40 N

24.0 N

12.0 N

R

q

35.0

ur

60.0

vCE

vWE = vCE + vWC

vWC

vWE

ur

ur

ur

ur ur ur

(a) Since v is vertical,WE

WC CEsin60.0 50.0 km h = =v v

orWC

57.7 km h at 60.0 west of vertical= v

(b) Since v has zero vertical component,CE

( )WEv vWC cos 60.0 57.7 km h cos 60.0 28.9 km h downward= = =


25/44


3.46 The vertical displacement from the launchpoint (top of the building) to the top of

the arc may be found from 2 20 2 y y yv v a y= +

with v at the top of the arc. This yields0y =

( )( )

2

2

s7.35 m

s= +

7.35 m+

2 20

2

y y

y

v v

a

=

max

0 12.0 m

2 9.80 my

= =

max 0 y y y

0 0 y y y y= + =

and gives

50.0 m

+x

v0y

v0xhmax

v0y = 12.0 m/sv0x = 9.0 m/s

(a) If the origin is chosen at the top of the building, then 0 max0 and 7.35 my y= =

Thus, the maximum height above the ground is

max max50.0 m 50.0 m 7.35 m 57.4 mh y= + = + =

0

The elapsed time from the point of release to the top of the arc is found from

y y yv v a t= + as

0

2

0 12.0 m s1.22 s

9.80 m s

y y

y

v vt

a

= = =

(b) If the origin is chosen at the base of the building (ground level), then

and , giving

0 50.0 my = +

max maxh y=

max 0h y= + 50.0 m 7.35 m 57.4 my = + =

The calculation for the time required to reach maximum height is exactly the same

as that given above. Thus, 1.22 st =


26/44

80 CHAPTER 3

3.47 ( )( )1 90.0 km h 2.50 h 225 kmAC v t= = =

cos40.0 80.0 km=92.4BD AD AB AC= = km

From the triangle BCD,

( ) ( )2 2

BC BD DC= +

( )

( )22

92.4 km sin 40.0 =172 kAC m= +

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

2

172 km68.6 km h

2.50 hv = =

A B80.0 km

C

D

40.0

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

20 1.50 m sx xv v= =

Thus, when the speed of the water is 3.00 m sv = , the vertical component of its velocity

will be

( ) ( )2 22 2 3.00 m s 1.50 m s 2.60 m sy xv v v= = =

The vertical displacement of the water at this point is

( )

( )

22 20

2

2.60 m s 00.344 m

2 2 9.80 m s

y y

y

v vy

a

= = =

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

( )

( )

2200 0 6.26 m s

2.00 m2 2 -9.80 m s

y

leapy

v

y a

= = =

Therefore, the maximum height waterfall the salmon can clear is

max 0.344 m 2.34 mleaph y= + =


27/44


3.49 The distance, s, moved in the first 3.00 seconds is given by

( )( ) ( )( )22 20 100 m s 3.00 s 30.0 m s 3.00 s 435 m2 2

s v t at= + = + =

1 1cos53.0 262 m, and sin53.0 347 mx s y s= = = =

1 1

At the end of powered flight, the coordinates of the rocket are:

The speed of the rocket at the end of powered flight is

( )( )21 0 100 m s 30.0 m s 3.00 s 190 m sv v at= + = + =

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

0 1x cos 53.0 114 m s= =v v and 0 1 sin 53.0 152 m syv v= =

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

00 0 15

9.80 m

y

rise

y

vt

a

= =

22 m

15.5 ss

=

The vertical displacement occurring during this time is

( )s 1=0 30 152 m s 15.5 .17 10 m

2 2

y y

rise

v vy t

+ + = =

The maximum altitude reached is then

3 3.52 101 347 m 1.17 10 m 1 m H y y= + = + =

(b) After reaching the top of the arc, the rocket falls to the ground, starting

with zero vertical velocity

31.52 10 m

( )0 0yv = . The time for this fall is found from2

0

1

2y yy v t a t = +

( )

as

( )32

2 1.52 10 m217.6 s

-9.80 m sfall

y

yt

a

= = =

( )

The total time of flight is

3.00 15.5 17.6 s powered rise fallt t t t= + + = + + 36.1 s=


28/44

82 CHAPTER 3

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

( )2 15.5 17.6 s 33.1 srise fallt t t= + = + =

The horizontal displacement occurring during this time is

( )( )0 2 114 m s 33.1 s 3.78xx v t = = =310 m

and the full horizontal range is

3 31 262 m 3.78 10 m 4.05 10R x x= + = + = m

3.50 The velocity of a canoe relative to the shore is given by CS CW WS= +v v v

, where CWv

is

the velocity of the canoe relative to the water and WSv

is the velocity of the water

relative to shore.

Applied to the canoe moving upstream, this gives

CW WS1.2 m s v v = + (1)

and for the canoe going downstream

CW WS2 .9 m s v v+ = + + (2)

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

WS2 1.7 mv = s , so WS 0.85 m sv =

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

CW2 4.1 mv = s , or CW 2.1 m sv =


29/44


3.51 The time of flight is found from 201

2yy v t a t = + y

with 0y = , as02 yv

g=t . This gives the

range as0 0

0

2 x yx

v vR v t

g= =

On Earth this becomes0 02 x y

Earth

Earth

v vR

g= , and on the moon,

0 02 x yMoon

Moon

v v

g=R

Dividing MoonR by , we findEarthREarth

Moon Earth

Moon

gR R

g

=

. With

1

6 Moon Earthg

=

g , this gives

( )6 6 m 18MoonR R =3.0 mEarth= =

Similarly,3.0 m

7.9 m0.38

EarthR R= = =Earth

Mars

g

g

Mars

3.52 The time to reach the opposite side is0 0

10 m

cos15x

x

v vt

= =

When the motorcycle returns to the original level, the vertical displacement is .

Using this in the relation

0y =

20

1

2yy v t a t+ y = gives a second relation between the takeoff

speed and the time of flight as:

( ) ( ) 20 10 sin15 2v t= + g t or 0 2sin15gv t

=

Substituting the time found earlier into this result yields the required takeoff speed as

( )( )( )( )

2

0

9.80 m s 10 m14 m s

2 sin15 cos15v = =

3.53 (a) If A is in the first quadrant, then

0 and 0x yA A> >

(b) When B is in the second quadrant, then

0 and 0x yB B< >


30/44

84 CHAPTER 3

Let C A , then= + B

andx x x y y yC A . SinceB C A B= + = + 0 and 0x xA B>

orx x

< xC, may be either

positive or negative (depending on whether A B has the larger absolute value.

Because 0 and 0y yA B> > yC >, then . Therefore the vector C 0

must be in either the first or second quadrant

3.54 (a) Velocity vector at several points (b) Acceleration vector at several points

3.55 (a) The time to reach the fence is0 0 0

130 m 159 mcos35x

x

v v vt = = =

At this time, the ball must be 20 m above its launch position.

20

1

2yy v t a t = + y gives

( ) ( )2

20

0 0

159 m 159 m20 m sin 35 4.90 m sv

v v

=

From which, 0 42 m sv =

(b) From above,0

159 m 159 m3.8 s

42 m sv=t = =

(c) ( )0 42 m s cos35 34 m sx x= = =v v

( ) ( )( )20 42 m s sin 35 9.80 m s 3.8 s 13 m s y y yv v a t= + = =

( ) ( )2 22 2 34.1 m s 13.4 m s 37 m sx yv v v= + = + =


31/44


3.56 We shall first find the initial velocity of the ball thrown vertically upward. At itsmaximum height, and0yv = 1.50 st = . Hence, 0 y y yv v a t= + gives

( )( )1.50 s200 9 syv= .80 m , or 0 14yv .7 m s=

In order for the second ball to reach the same vertical height as the first, the second musthave the same initial vertical velocity. Thus, we find as0v

0

0

14.7 m s29.4

sin 30.0 0.500

yvv = = =

m s

3.57 The time of flight of the ball is given by 201

2yy v t a t = + y , with 0y = , as

( ) (2 21

0 20 m s sin 30 9.80 m s2t t = + )

or t = 2.0 s

The horizontal distance the football moves in this time is

( ) ( )0 20 m s cos30 2.0 s 35 mxx v t = = =

Therefore, the receiver must run a distance of (35 m 20 m) = 15 m 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

15 m 7.5 m s2.0 s

xvt

= = =

3.58 The horizontal component of the initial velocity is v v0 0 cos 40 0.766x v0= = and the time

required for the ball to move 10.0 m horizontally is

0 0

10.0 m 13.1 m

0.766x

xt

v v

= = =

(

0v

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

)3.05 2.00 m 1.05y = = m


32/44

86 CHAPTER 3

Thus, 201

2yy v t a t+ y = becomes

( ) ( )( )

2

20 2

0 0

13.1 m13.1 m 11.05 m sin 40.0 9.80 m s

2v

v v= +

or3 2

20

835 m s1.05 m 8.39 m

v=

which yields 0 10.7 m sv =

3.59 Choose an origin where the projectile leaves the gunand let they-coordinates of the projectile and the targetat time t be labeled andp Ty y , respectively.

Then, ( ) ( ) 20 0n2

pp

g y y t =

( )

0 siv = t , and

202

gh = TTy y = t or

2

2T

g y h t=

The time when the projectile will have the same x-

coordinate as the target is 0

0 0 cosx

xx

v v 0t

= =

q0

v0

x0

h

Target

ur

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

( ) 200 00 0

sincos 2 2

gxv

v

=

2gt h t which reduces to 00

tanh

x =

This requirement is satisfied provided that the gun is aimed at the initial location of thetarget. Thus, a collision is guaranteed if the shooter aims the gun in this manner.

3.60 (a) The components of the vectors are

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

1md

0 104

2md

46.0 19.5

1fd

0 84.0

2fd 38.0 20.2


33/44


The sums d d and d dm 1m 2= + d

m 2ff 1f= + d

)

are computed as:

( ) (2 2

132 cm an=

)

1m

104 19.50 19.5 d tan 69.6

0 46.0d

+ = + + = =

+ 46.0 104+

( ) (

2 2111 cm an= 1f

84.0 20.20 20.2 d tan 70.00 38.0

d + = + + = = +

38.0 84.0+

or m f 111 cm=d d

132 cm at 69.6= and at 70.0

(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

m

200 cm1.11

180 cms = = , and the scale factor needed for components of d d is1f and 2f

f

200 cm

1.19168 cms= =

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

m fat 69.6 and 132 cm at 70.0 = =d d

146 cm

The difference in the normalized vector sums is m f- d d d =

vector x-component (cm) y-component (cm)

md

50.9 137

- d f

45.1 124

d

x = 5.74 y = 12.8

Therefore, ( ) ( ) ( ) ( )2 22 2

5.74 12.8 cm 14.0 cmd x y = + = + = , and

1 1 12.8tan 65.85.74

y

x = = =

tan , or 14.0 cm at 65.8 = d


34/44

88 CHAPTER 3

v045

R

ur

3.61 To achieve maximum range, the projectile shouldbe launched at 45 above the horizontal. In thiscase, the initial velocity components are:

00 0

2

x y

vv v= =

The time of flight may be found from 0y yv v gt=

0

by recognizing that when the

projectile returns to the original level, y yv v= . Thus, the time of flight is

0 0 0 002 22

2

y y yv v v vvt

g g g

= = = = g

. The maximum horizontal range is then

200 0

0

2

2x

vv vR v t

g g

= = =

0yv =

00 v gt=

(1)

Now, consider throwing the projectile straight upward at speed . At maximum

height, , and the time required to reach this height is found from as

which yields

0v

0y yv v g= t

0v

g=

( ) ( )

t . Therefore, the maximum height the projectile will reach

is20

2 2

v

g g

=

0 0v v max av

0y y v t = =

( )

+

Comparing this result with the maximum range found in equation (1) above reveals that

max 2

Ry = provided the projectile is given the same initial speed in the two tosses.

If the boy takes a step when he makes the horizontal throw, he can likely give a higherinitial speed for that throw than for the vertical throw.

3.62 (a) At the top of the arc v , and from0y = 0y yv v gt= , we find the time to reach the top

of the arc to be0y yv v 0 0 0in sin

v

g

0

0 svt

g g

= =

=

The vertical height, h, reached in this time is found from

( ) 02

y yv v+ av

yy v = t t= to be2 2

0 0 0sin sin

2 2

v v vh

g g0 00 sin 0 = =

+


35/44


(b) The total time of flight, ft , is double the time required to reach the top of the arc, or

00

2sinf

vt

g

=

. The horizontal range is given by

( ) ( ) ( )2 2

0 0 0 000 0 0 2sin cos sin 2cos sinx f v vvR v t v

g g0

g0 2

= = = =

3.63 The velocity of the boat relative to the shore isBS BW WS

= +v v v

WS

, where is the

velocity of the boat relative to the water andBW

v

v

is the velocity of the water relative to

shore.

In order to cross the river (flowing parallel to the banks) in minimum time, the velocityof the boat relative to the water must be perpendicular to the banks. That is, v must

be perpendicular to v . Hence, the velocity of the boat relative to the shore must be

BW

WS

( ) ( )2 2

km h 13 km=2 2WSvBS BW 12 km h 5.0 hv v= + = +

at 1 1tan BW

WS

12 km htan 67

5.0 km h

v

v

= =

= 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

BW

width of 1.5 km 60 min 7.5 min12 km h 1 h

tv= = =

river

During this time, the boat drifts downstream a distance of

( )( )3

WS

1 h 10 mh 7.5 min

60 1 kmd v t

= =

26.3 10 m= 5.0 kmmin

3.64 Taking upstream as positive, the velocity of the water relative to the ground is

WG0.500 m s= v

. The velocity of the skater relative to shore is

SG

0.560 m0.700 m s

0.800 s+= = +v

while moving upstream, andSG WG

0.500 m s= = v v

while drifting back downstream.

(a) At any time, , or the velocity of the skater relative to the water isSG SW WG

= +v v v

WGv

SW SG= v v


36/44

90 CHAPTER 3

(i) While going upstream, ( )SW 0.700 m s 0.500 m s 1.20 m s= + =v

(ii) While drifting down stream, ( )SW 0.500 m s 0.500 m s 0= =v

(b) ( )( )SW SW 1.20 m s 0.800 s 0.960 mt= = =

d v

(c) The time to go upstream t 0.800 sup = and the time to drift back downstream is

0.560 m1.12 s

0.500 m sdownt = =

( )

, giving the cycle time as 1.92 s.

Therefore, SWSW av 0.500 m scycle

dv

t=

0.960 m

1.92 s= =

3.65 The initial velocity components for the daredevil are

00 0

25.0 m s

2 2x y

vv v= = =

The time required to travel 50.0 m horizontally is

( )

0

50.0 m 22 2 s

25.0 m sx

xt

v

= = =

The vertical displacement of the daredevil at this time, and the proper height above the

level of the cannon to place the net, is

( ) ( )( )2

2 20

25.0 m s12 2 s 9.80 m s 2 2 s 10.8 m

2 2y yy v t a t

= + = =

3.66 The vertical component of the salmons velocity as it leaves the water is

( )0 0 sin 6.26 m s sin 45.0 4.43 m syv v = + = + = +

When the salmon returns to water level at the end of the leap, the vertical component of

velocity will be 0 4.43 m sy yv v= =

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

0

1 2

4.43 m s 4.43 m s0.903 s

9.80 m s

y y

y

v vt

a

= = =


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The horizontal distance traveled during the leap is

( ) ( ) ( )0 1 0 1cos 6.26 m s cos45.0 0.903 s 4.00 mxL v t v t= = = =

To travel this same distance underwater, at speed 3.58 m sv = , requires a time of

2

4.00 m1.12 s

3.58 m s

Lt

v= = =

The average horizontal speed for the full porpoising maneuver is then

( )av

1 2

2 4.00 m23.96 m s

0.903 s 1.12 stotal

total

x Lv

t t t

= = = =

+ +

3.67 (a) and (b)

Since the shot leaves the gun horizontally, the time it takes to reach the target is

0 0x

x xt

v v

= = . The vertical displacement occurring in this time is

2

20

0

1 10

2 2y y

xy y v t a t g

v

= = + =

, which gives the drop as

2

202

0 0

1with , where is the muzzle velocity

2 2

gx y g Ax A v

v v

= = =

(c) If x = 3.00 m, andy = 0.210 m, then( )

2 -

22

0.210 m2.33 10 m

3.00 m

yA

x= = = 1

and( )

2

0 2 -1

9.80 m s

2 2 2.33 10 m

gv

A = =

= 14.5 m s

3.68 We are given that:

(velocity of boat relative to Earth)

and

(velocity of wind relative to Earth)

BE 20 knots due north=v

WE17 knots due east=v

north

eastvWEur

vWB

ur vEB

ur

q


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92 CHAPTER 3

The velocity of the wind relative to the boat is

where is the velocity of Earth relative to the boat. The vector

diagram above shows this vector addition.

Since the vector triangle is a 90 triangle, we find the magnitude of to be

WB WE EB= +v v v

EB BE20 k= =v v

nots south

WBv

( ) ( )2 2

17 knots 20 knots 26 knotsv = + =2 2WB WE EBv v= +

and the direction is given by

1 1 20 knotstan 5017 knots

=

EB

WE

tanv

v

= =

Thus,WB

26 knots a=v

t 50 south of east

From the vector diagram above, the component of this velocity parallel to the motion of

the boat (that is, parallel to a north-south line) is seen to beEB

20 knots south= v vBE

=

3.69 The components of the three displacements are:

Displacement x-component (paces) y-component (paces)

75.0 paces @ 240 37.5 65.0125 paces @ 135 88.4 +88.4

100 paces @ 160 94.0 +34.2

Resultant 220 pacesx = 57.6 pacesy = +

The resultant displacement is then

( ) ( ) ( ) ( )2 2 22

220 paces 57.6 paces 227 pacesR x y= + = + + =

1 1 57.6

tan tan 165220

y

x

+ = = =

or 227 paces at 165 from the positive axisx= R


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3.70 For the ball thrown at 45.0, the time of flight is found from

20

1

2y yy v t a t = + as

201 10

22

gvt t

=

which has the single non-zero solution of 012v

g=t

The horizontal range of this ball is2

00 01 0 1

2

2x

vv vR v t

g g

= = =

Now consider the first arc in the motion of the second ball, started at angle with initial

speed . Applied to this arc,0v2

0

1

2y yy v t a t = + becomes

( ) 20 210 sin2gv t= 21t

with non-zero solution 0212 sinv

tg

=

Similarly, the time of flight for the second arc (started at angle with initial speed 0 2v )

of this balls motion is found to be

( )0 022

2 2 sin sinv vt

g g

= =

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

( )( ) ( )2 20 00

21 0 21 0

2sin cos sin 22 sincosx

v vvR v t v

g g g

= = = =

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

( ) ( ) ( )2 2

0 0

22

2 sin 2 sin 21

4

v vR

g g

= =

The total horizontal distance traveled in the two arcs is then

( )202 21 22

sin 25

4

vR R R

g

= + =


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94 CHAPTER 3

(a) Requiring that the two balls cover the same horizontal distance (that is, requiringthat ) gives2R R= 1

( )20 sin 2v 2054

v

g g

=

This reduces to ( )4

sin 25

= which yields 2 53.1 = , so 26.6 = is the required

projection angle for the second ball.

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

0 0 02 21 22

2 sin sin 3 sinv v vt t t

g g g

= + = + =

Therefore, the ratio of the times of flight for the two balls is

( )

( )02

1 0

3 sin 3sin

22

v gt

t v g

= =

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

( )2

1

3sin 26.6 0.950

2

t

t= =

3.71 (a) Applying 20 12y yy v t a t+ = to the vertical motion of the first snowball gives

( ) ( 2 211

0 25.0 m s sin70.0 9.80 m s2

t = + ) 1t which has the non-zero solution of

( )1 2

2 25.0 m s sin70.04.79 s

9.80 m st

= = as the time of flight for this snowball.

The horizontal displacement this snowball achieves is

( ) ( )0 1 25.0 m s cos70.0 4.79 s 4xx v t = = = 1.0 m


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Now consider the second snowball, also given an initial speed of 0 25.0 m sv = ,

thrown at angle , and is in the air for time . Applying2t21

2y yt a t+0y v = to its

vertical motion yields

( ) ( )10 25.0 m s sin 9.80 m s2t t = +

2 22 2

which has a non-zero solution of

( )( )2 2

2 25.0 m s sin5.10 s sin

9.80 m st

= =

We require the horizontal range of this snowball be the same as that of the first ball,

namely ( ) ( )s0 2 25.0 m s cos 5.10xx v t = = sin 41.0 = m . This yields the

equation

( )( )41.0 m

sin cos 0.32125.0 m s 5.10 s

= =

sin2 =2si

From the trigonometric identity n cos , this result becomes

( )sin 2 =2 0.321 0.642 = 2 40.0, so =

and the required angle of projection for the second snowball is

20.0 above the horizontal =

(b) From above, the time of flight for the first snowball is t1 4.79 s= and that for the

second snowball is ( ) ( )2 5.10 s sin 5.10 s sin 20.0t 1.74 s= = = .

Thus, if they are to arrive simultaneously, the time delay between the first andsecond snowballs should be

1 2 4.79t t t = = s 1.74 s= 3.05 s


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96 CHAPTER 3

3.72 First, we determine the velocity with which the dart leaves the gun by using the datafrom when the dart is fired horizontally ( 0 0yv = ) from a stationary gun. In this case,

210 2y yy v t a t = + gives the time of flight as

( )22 1.00 m s

2 0.459.80 m syyt a

= = = 2 s

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

DG

5.00 m11.1 m s

0.452 s

xv

t

= = =

When the dart is fired horizontally from a moving gun,the initial velocity of the dart relative to Earth is

where is the initial velocity of dart relative to gun

and is the velocity of the gun relative to Earth

0 DE DG G= = +v v v v

DGv

GEv

vDGur

45.0

vGE

ur

v0 = vDEur ur

E

From the vector diagram, observe that

( )0 GE sin 45.0 2.00 m s sin 45.0 1.41 m syv v= = =

and ( )0 DG GE cos 45.0 11.1 m s 2.00 m s cos 45.0 12.5 m sxv v v= + = + =

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

( ) ( )( )22 2

0 2 1.41 m s 2 9.80 m s 1.00 m 4.65 m s y y yv v a y= + = + =

and the time of flight is( )0

2

4.65 m s 1.41 m s0.330 s

9.80 m s

y y

y

v vt

a

= = =

The displacement during the flight is ( )( )0 12.5 m s 0.330 s 4.12 mxx v t = = =


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3.73 (a) First, use 210 2x xx v t a t+ = to find the time for the coyote to travel 70 m, starting

from rest with constant acceleration 215 m sxa =

( )1 2

2 70 m23.1 s

15 m sx

xt

a

= = =

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

1

70 m23 m s

3.1 s

xv

t

= = =

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

( )( )2

0 0 10 15 m s 3.1 s 46x xv v a t= = + = = m s

From 210 2y yy v t a t = + , the time for the coyote to drop 100 m with v is0 0y =

( )2 2

2 100 m24.52 s

9.80 m sy

yt

a

= = =

The horizontal displacement of the coyote during his fall is

( )( ) ( )( )22 21 12 2 15 m s 4.52 s 3.6+ = 20 2 2x xx v t a t = + = 46 m s 4.52 s 10 m


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98 CHAPTER 3

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