57

CHAPTER 3: Kinematics in Two Dimensions; Vectors

Solution Guide to WebAssign Problems

3.1 [2] The truck has a displacement of 18 + !16( ) = 2 blocks north and 10 blocks east.

The resultant has a magnitude of 22+10

2= 10 blocks and a direction of

tan!1

2 10 = 11º north of east .

3.2 [9] (a) ( )( ) hkm 55041.5º coshkm 735north

==v ( )( ) hkm 48741.5º sinhkm 735west

==v

(b) ( )( ) km 1650h 00.3hkm 550northnorth

===! tvd

( )( ) km 1460h 00.3hkm 487westwest

===! tvd

3.3 [10] 20.6628.0ºsin 0.44 38.8528.0º cos 0.44 ==== yx AA

0.31270ºsin 31.0 0.0270º cos 0.31

21.9756.0ºsin 5.26 14.8256.0º cos 5.26

!====

==!=!=

yx

yx

CC

BB

(a)

!A +!B +!C( )

x

= 38.85 + !14.82( ) + 0.0 = 24.03 = 24.0

!A +!B +!C( )

y= 20.66 + 21.97 + !31.0( ) = 11.63 = 11.6

(b)

!A +!B +!C = 24.03( )

2

+ 11.63( )2

= 26.7 º8.2503.24

63.11tan

1==

!"

3.4 [13] 20.6628.0ºsin 0.44 38.8528.0º cos 0.44 ==== yx AA

0.31270ºsin 0.31 0.0270º cos 0.31

21.9756.0ºsin 26.5 14.8256.0º cos 5.26

!====

==!=!=

yx

yx

CC

BB

(a)

!A !!B +!C( )

x

= 38.85 ! !14.82( ) + 0.0 = 53.67

!A !!B +!C( )

y= 20.66 ! 21.97 + !31.0( ) = !32.31

Note that since the x component is positive and the y component is negative, the vector is in the 4th

quadrant.

PIM WebAssign Chapter 3 Answer Guide Physics: Principles with Applications, 6th Edition

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!A !!B +!C = 53.67( )

2

+ !32.31( )2

= 62.6 " = tan!1!32.31

53.67= 31.0º below + x axis

(b)

!A +!B !!C( )

x

= 38.85 + !14.82( ) ! 0.0 = 24.03

!A +!B !!C( )

y= 20.66 + 21.97 ! !31.0( ) = 73.63

( ) ( ) º9.7103.24

63.73tan 5.7763.7303.24

122===+=!+

!"CBA

!!!

(c)

!C !!A !!B( )

x

= 0.0 ! 38.85 ! !14.82( ) = !24.03

!C !!A !!B( )

y= !31.0 ! 20.66 ! 21.97 = !73.63

Note that since both components are negative, the vector is in the 3rd quadrant.

( ) ( ) axis below º9.7103.24

63.73tan 5.7763.7303.24

122x!=

!

!==!+!=!!

!"BAC

!!!

Note that the answer to (c) is the exact opposite of the answer to (b).

3.5 [14] 20.6628.0ºsin 0.44 38.8528.0º cos 0.44 ==== yx AA

31.0270ºsin 0.31 0.0270º cos 0.31

21.9756.0ºsin 5.26 82.1456.0º cos 5.26

!====

==!=!=

yx

yx

CC

BB

(a)

!B ! 2

!A( )

x= !14.82 ! 2 38.85( ) = !92.52

!B ! 2

!A( )

y= 21.97 ! 2 20.66( ) = !19.35

Note that since both components are negative, the vector is in the 3rd quadrant.

( ) ( ) axis below º8.1152.92

35.19tan 5.9435.1952.92

122x!=

!

!==!+!=!

!"A2B

!!

(b) 2!A ! 3

!B + 2

!C( )

x

= 2 38.85( ) ! 3 !14.82( ) + 2 0.0( ) = 122.16

2!A ! 3

!B+ 2

!C( )

y= 2 20.66( ) ! 3 21.97( ) + 2 !31.0( ) = !86.59

Note that since the x component is positive and the y component is negative, the vector is in the 4th

quadrant.

2!A ! 3

!B + 2

!C = 122.16( )

2

+ !86.59( )2

= 149.7 " = tan!1!86.59

122.16= 35.3º below + x axis

3.6 [17] Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps

from the rock. In the horizontal direction, sm 5.30=

xv and .0=

xa In the vertical direction,

PIM WebAssign Chapter 3 Answer Guide Physics: Principles with Applications, 6th Edition

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,0 ,sm 80.9 ,0 0

2

0 === yav yy and the final location m. 5.6=y The time for the tiger to reach the ground

is found from applying Eq. 2-11b to the vertical motion.

y = yo+ v

oyt + 1

2ayt

2 ! 6.5m = 0 + 0 + 1

29.8 m s

2( )t 2 ! t =

2 6.5 m( )

9.8 m s2= 1.15 sec

The horizontal displacement is calculated from the constant horizontal velocity.

( )( ) m 0.4sec 1.15sm 5.3 ===! tvxx

3.7 (18) Choose downward to be the positive y direction. The origin will be at the point where the diver dives

from the cliff. In the horizontal direction, sm 8.10=

xv and .0=

xa In the vertical direction,

,0 ,sm 80.9 ,0 0

2

0 === yav yy and the time of flight is s. 0.3=t The height of the cliff is found from

applying Eq. 2-11b to the vertical motion.

y = yo+ v

oyt + 1

2ayt

2 ! y = 0 + 0 + 1

29.80 m s

2( ) 3.0 s( )2

= 44 m

The distance from the base of the cliff to where the diver hits the water is found from the horizontal motion at

constant velocity:

( )( ) m 4.5s 3sm 8.1 ===! tvxx

3.8 [21] Choose downward to be the positive y direction. The origin will be at the point where the ball is thrown

from the roof of the building. In the vertical direction, ,0 ,sm 80.9 ,0 0

2

0 === yav yy and the displacement

is 45.0 m. The time of flight is found from applying Eq. 2-11b to the vertical motion.

y = yo+ v

oyt + 1

2ayt

2 ! 45.0 m = 1

29.80 m s

2( )t 2 ! t =

2 45.0 m( )

9.80 m s2= 3.03 sec

The horizontal speed (which is the initial speed) is found from the horizontal motion at constant velocity:

.sm 92.7s 3.03m 0.24 ==!="=! txvtvxxx

3.9 [22] Choose the point at which the football is kicked the origin, and choose upward to be the positive y

direction. When the football reaches the ground again, the y displacement is 0. For the football,

( ) 2

0 sm 80.9 ,sm35.0ºsin 0.18 !== yy av and the final y velocity will be the opposite of the starting y

PIM WebAssign Chapter 3 Answer Guide Physics: Principles with Applications, 6th Edition

60

velocity (reference problem 3-28). Use Eq. 2-11a to find the time of flight.

( ) ( )s 11.2

sm 80.9

sm35.0ºsin 0.18sm35.0ºsin 0.18

2=

!

!!=

!="+=

a

vvtatvv

oyy

oyy

3.10 [23] Choose downward to be the positive y direction. The origin is the point where the ball is thrown from the

roof of the building. In the vertical direction, ,0 ,0 00 == yvy and .sm 80.92

=ya The initial horizontal

velocity is 22.2 m/s and the horizontal range is 36.0 m. The time of flight is found from the horizontal motion

at constant velocity.

s 62.1sm 22.2m 0.36 ==!="=!xxvxttvx

The vertical displacement, which is the height of the building, is found by applying Eq. 2-11b to the vertical

motion.

y = yo+ v

oyt + 1

2ayt

2 ! y = 0 + 0 + 1

29.80 m s

2( ) 1.62 s( )2

= 12.9 m

3.11 [31] Choose the origin to be at ground level, under the place where the projectile is launched, and upwards to

be the positive y direction. For the projectile, ,125 , ,º0.37 ,sm 0.65 000 =!=== ygav y" and

000sin !vvy =

(a) The time taken to reach the ground is found from Eq. 2-11b, with a final height of 0.

y = yo+ v

oyt + 1

2ayt

2 ! 0 = 125 + v

0 sin "

0t # 1

2gt

2 !

t =#v

0 sin "

0± v

0

2 sin

2 "

0# 4 #

1

2g( ) 125( )

2 #1

2g( )

=#39.1± 63.1

#9.8= 10.4 s, # 2.45 s = 10.4 s

Choose the positive sign since the projectile was launched at time .0t =

(b) The horizontal range is found from the horizontal motion at constant velocity.

!x " vxt = v

0 cos #

0( )t = 65.0 m s( )cos 37.0º 10.4 s( ) = 541 m

(c) At the instant just before the particle reaches the ground, the horizontal component of its velocity is the

constant ( ) .sm 9.51º0.37cossm 0.65 cos 00

=== !vvx

The vertical component is found from Eq.

2-11a.

vy= v

oy+ at = v

0 sin !

0" gt = 65.0 m s( ) sin 37.0º " 9.80 m s

2( ) 10.4 s( )

= "63.1 m s

PIM WebAssign Chapter 3 Answer Guide Physics: Principles with Applications, 6th Edition

61

(d) The magnitude of the velocity is found from the x and y components calculated in part c) above.

( ) ( ) sm 7.81sm 1.63sm 9.512222=!+=+= yx vvv

(e) The direction of the velocity is ,º6.509.51

1.63tantan

11!=

!==

!!

x

y

v

v" and so the object is moving 50.6º

below the horizon .

(f) The maximum height above the cliff top reached by the projectile will occur when the y-velocity is 0,

and is found from Eq. 2-11c.

( )

( )( )

m 1.78sm 80.92

37.0º sin sm 0.65

2

sin

2 sin 0 2

22

0

22

0

max

max0

22

00

22

===

!="!+=

g

vy

gyvyyavv yoyy

#

#

3.12 [33] Choose the origin to be where the projectile is launched, and upwards to be the positive y direction. The

initial velocity of the projectile is ,0v the launching angle is , ,0 gay !=" and .sin 00

!vvyo = The range of

the projectile is given by the range formula from Example 3-8, .2sin

0

2

0

g

vR

!= The maximum height of the

projectile will occur when its vertical speed is 0. Apply Eq. 2-11c.

( )g

vygyvyyavv yyoy

2

sin 2 sin 0 2

0

22

0

maxmax0

22

00

22 !! ="#="#+=

Now find the angle for which .maxyR =

76º4 tan 4 tan sin cos 4 2

sin cos sin 2

2

sin2sin

2

sin 2sin v

1

0000

0

2

00

0

2

0

0

22

00

2

0

max

==!=!=!=

!=!=!=

"####

###

##

##

g

v

gyR

3.13 [37] Call the direction of the flow of the river the x direction, and the

direction of Huck walking relative to the raft the y direction.

PIM WebAssign Chapter 3 Answer Guide Physics: Principles with Applications, 6th Edition

62

!v

Huck

rel. bank

=!v

Huck

rel. raft

+!v

raft rel.

bank

= 0, 0.60( )m s + 1.70, 0( )m s

= 1.70, 0.60( )m s

Magnitude: vHuck

rel. bank

= 1.702+ 0.60

2= 1.80 m s

Direction: ! = tan"1 0.60

1.70= 19º relative to river

3.14 [39] Call the direction of the flow of the river the x direction, and the direction the

boat is headed the y direction.

(a) sm 59.230.220.1222

water

rel.boat

2

shore

rel.water

shore

rel.boat =+=+= vvv

shore torelative º4.6290º ,º6.2730.2

20.1tan

1=!===

! "#"

(b) The position of the boat after 3.00 seconds is given by

!d = vboat rel.shore

t = 1.20, 2.30( )m s"# $% 3.00 sec( )

= 3.60 m downstream, 6.90 m across the river( )

As a magnitude and direction, it would be 7.8 m away from the starting point, at an angle of 62.4o

relative to the shore.

3.15 [44] Call the direction of the boat relative to the water the x

direction, and upward the y direction. Also see the diagram.

( )

( ) ( )

sm 89.1354.0854.1

sm354.0,854.1sm0,50.1

sm45ºsin ,0.5045º cos 50.0

22

waterrel.passenger

waterrel.boat

boat rel.passenger

waterrel.passenger

=+=

=+

=

+=

v

vvv!!!

3.16 [47] Call the direction of the flow of the river the x direction, and the direction straight across the river the y

direction. Call the location of the swimmer’s starting point the origin.

( ) ( )

( ) sm0.45 ,40.0

0 s,m 40.0sm 0.45 ,0shore

rel.water waterrel.

swimmershore rel.

swimmer

=

+=+= vvv!!!

!

PIM WebAssign Chapter 3 Answer Guide Physics: Principles with Applications, 6th Edition

63

(a) Since the swimmer starts from the origin, the distances covered in the x and y directions will be exactly

proportional to the speeds in those directions.

m 67 sm 45.0

sm 40.0

m 75 =!"=

!"==

!

!x

x

v

v

tv

tv

y

x

y

x

y

x

(b) The time is found from the constant velocity relationship for either the x or y directions.

s 170sm 0.45

m 75 ==

!="=!

y

yv

yttvy

3.17 [55] Assume a constant upward slope, and so the deceleration is along a straight line. The starting velocity

along that line is .sm 3.33hkm 6.3

sm 1hkm 120 =!!

"

#$$%

& The ending velocity is 0 m/s. The acceleration is found

from Eq. 2-11a.

( ) 2

0sm 56.5

s 0.6

sm 3.33 s 0.6sm 33.30 !=!="+="+= aaatvv

The horizontal acceleration is ( ) . sm 7.432 cossm 56.5 cos 22

horiz!=°!== "#a

The vertical acceleration is ( ) sm 8.230º sinsm 56.5sin 2

vert!=!== "aa

The horizontal acceleration is to the left in the textbook diagram, and the vertical acceleration is down.

3.18 [69] The proper initial speeds will be those for which the ball has traveled a horizontal distance somewhere

between 10.78 m and 11.22 m while it changes height from 2.10 m to 2.60 m with a shooting angle of 38.0o.

Choose the origin to be at the shooting location of the basketball, with upward as the positive y direction. Then

the vertical displacement is ,sin ,sm 80.9 m, 5.0 0

2!ooyy vvay ="== and the (constant) x velocity is

. cos 0

!oxvv = See the diagram (not to scale).

For the horizontal motion at constant velocity,

tvtvxox 0

cos !==" and so . cos

00!v

xt

"=

For the vertical motion, applying Eq. 2-11b.

y = yo+ v

oyt + 1

2ayt

2= v

o sin !t " 1

2gt

2

Substitute the expression for the time of flight and solve for the initial velocity.

PIM WebAssign Chapter 3 Answer Guide Physics: Principles with Applications, 6th Edition

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y = v0 sin !t " 1

2gt

2= v

o sin !

#xvo cos !

0

" 1

2g

#xv

0 cos !

0

$

%&'

()

2

= #x tan ! "g #x( )

2

2v0

2 cos

2 !

0

v0=

g #x( )2

2 cos2 !

0"y + #x tan !( )

For m, 78.10=!x the shortest shot:

v0=

9.80 m s2( ) 10.78 m( )

2

2 cos2 38.0º !0.5 m + 10.78 m( ) tan 38.0º( )"# $%

= 10.8 m s .

For m, 22.11=!x the longest shot:

v0=

9.80 m s2( ) 11.22 m( )

2

2 cos2 38.0º !0.5 m + 11.22 m( ) tan 38.0º( )"# $%

= 11.0 m s .

3.19 [71] (a) Choose the origin to be the location where the car leaves the ramp, and choose upward to be the

positive y direction. At the end of its flight over the 8 cars, the car must be at m. 5.1!=y Also for the car,

, , ,0 00 vvgav xyy =!== and m. 20=!x The time of flight is found from the horizontal motion at constant

velocity: . 0vxttvx

x!="=! That expression for the time is used in Eq. 2-11b for the vertical motion.

y = yo+ v

oyt + 1

2ayt

2 ! y = 0 + 0 + 1

2"g( )

#xv

0

$

%&'

()

2

!

v0=

"g #x( )2

2 y( )=

" 9.80 m s2( ) 20 m( )

2

2 "1.5 m( )= 36 m s

(b) Again choose the origin to be the location where the car leaves the ramp, and choose upward to

be the positive y direction. The y displacement of the car at the end of its flight over the 8 cars must again

be m. 5.1!=y For the car, , cos , ,sin 000 !! vvgavv xyooy ="== and m. 20=!x The launch angle is

º.100=! The time of flight is found from the horizontal motion at constant velocity.

0 cos

!

o

xv

xttvx

"=#="

That expression for the time is used in Eq. 2-11b for the vertical motion.

PIM WebAssign Chapter 3 Answer Guide Physics: Principles with Applications, 6th Edition

65

y = yo+ v

oyt + 1

2ayt

2 ! y = v

0 sin "

0

#xv

0 cos "

0

+ 1

2$g( )

#xv

0 cos "

0

%

&'(

)*

2

!

v0=

g #x( )2

2 #x tan "0$ y( ) cos

2 "

0

=9.80m s

2( ) 20 m( )2

2 20 m( ) tan 10º +1.5 m( )cos2

10º= 20 m s

3.20 [72] Choose the origin to be the point at ground level directly below where the ball was hit. Call upwards the

positive y direction. For the ball, we have m, 9.0 , ,º61 ,º61 ,sm 28 000 =!=!=== ygaav yy" and

m. 0.0=y

(a) To find the horizontal displacement of the ball, the horizontal velocity and the time of flight are needed.

The (constant) horizontal velocity is given by . cos 00

!vvx= The time of flight is found from Eq. 2-

11b.

y = yo+ v

oyt + 1

2ayt

2 ! 0 = y

0+ v

0sin "

0t # 1

2gt

2 !

t =#v

0 sin "

0± v

0

2 sin

2 "

0# 4 #

1

2g( )y0

2 #1

2g( )

=# 28 m s( ) sin61º ± 28 m s( )

2

sin2

61º # 4 #1

2( ) 9.80 m s2( ) 0.9 m( )

2 #1

2( ) 9.80 m s2( )

= 5.034 s, # 0.0365 s

Choose the positive time, since the ball was hit at .0=t The horizontal displacement of the ball will be

found by the constant velocity relationship for horizontal motion.

!x = vxt = v

0 cos "

0t = 28 m s( ) cos 61º( ) 5.034s( ) = 68.34 m # 68 m

(b) The center fielder catches the ball right at ground level. He ran m 36.66 m 68.34 m 105 =! to

catch the ball, so his average running speed would be

sm 3.7sm 282.75.034s

m 66.36!==

"=

t

dvavg

3.21 [74] Choose the origin to be the point at the top of the building from which the ball is shot, and call upwards

the positive y direction. The initial velocity is sm180=v at an angle of º.42

0=! The acceleration due to

gravity is .gay !=

(a) ( ) sm1313.3842º cossm 18 cos 00

!=== "vvx

PIM WebAssign Chapter 3 Answer Guide Physics: Principles with Applications, 6th Edition

66

( ) sm 1212.0442ºsin sm 18sin 000

!=== "vv y

(b) Since the horizontal velocity is known and the horizontal distance is known, the time of flight

can be found from the constant velocity equation for horizontal motion.

!x = vxt " t =

!x

vx

=55 m

13.38 m s= 4.111 s

With that time of flight, calculate the vertical position of the ball using Eq. 2-11b.

y = yo+ v

oyt + 1

2ayt

2= 12.04 m s( ) 4.111 s( ) + 1

2!9.80 m s

2( ) 4.111 s( )2

= !33.3 = !33 m

So the ball will strike 33 m below the top of the building.

3.22 [75] When shooting the gun vertically, half the time of flight is spent moving upwards. Thus the upwards

flight takes two seconds. Choose upward as the positive y direction. Since at the top of the flight, the vertical

velocity is zero, find the launching velocity from Eq. 2-11a.

vy= v

yo+ at ! v

yo= v

y" at = 0 = 9.8 m s

2( ) 2.0 s( ) = 19.6 m s

Using this initial velocity and an angle of 45o in the range formula will give the maximum range for the gun.

R =v

0

2 sin 2!

0

g=

19.6 m s( )2

sin 2 " 45º( )

9.80 m s2

= 39 m