1

Motion in Two Dimensions

SOLUTIONS TO PROBLEMS Section 3.1 The Position, Velocity, and Acceleration Vectors P3.1

x m( )0

!3 000!1 270!4 270 m

y m( )!3 600

01 270!2 330 m

(a)

Net displacement = x2 + y2

= 4.87 km at 28.6° S of W

FIG. P3.1

(b) Average speed =

20.0 m s( ) 180 s( ) + 25.0 m s( ) 120 s( ) + 30.0 m s( ) 60.0 s( )180 s + 120 s + 60.0 s

= 23.3 m s

(c) Average velocity =

4.87 ! 103 m360 s

= 13.5 m s along rR

2 Motion in Two Dimensions

Section 3.2 Two-Dimensional Motion with Constant Acceleration P3.3

rv i = 4.00i + 1.00 j( ) m s and

rv 20.0( ) = 20.0i ! 5.00 j( ) m s

(a) ax =

! vx

! t=

20.0 " 4.0020.0

m s2 = 0.800 m s2

ay =

! vy

! t="5.00 " 1.00

20.0 m s2 = "0.300 m s2

(b) ! = tan"1 "0.300

0.800#$%

&'(= "20.6° = 339° from + x axis

(c) At t = 25.0 s

x f = xi + vxit +12

axt2 = 10.0 + 4.00 25.0( ) + 12

0.800( ) 25.0( )2 = 360 m

y f = yi + vyit +12

ayt2 = !4.00 + 1.00 25.0( ) + 12!0.300( ) 25.0( )2 = !72.7 m

vxf = vxi + axt = 4 + 0.8 25( ) = 24 m svyf = vyi + ayt = 1 ! 0.3 25( ) = !6.5 m s

" = tan!1 vy

vx

#

$%&

'(= tan!1 !6.50

24.0#$%

&'(= !15.2°

P3.5

ra = 3.00 j m s2 ; rv i = 5.00i m s ;

rri = 0i + 0 j

(a)

rrf =rri +

rv it +12

ra t2 = 5.00ti + 12

3.00t2 j!"#

$%&

m

rv f =

rv i +ra t = 5.00i + 3.00tj( ) m s

(b) t = 2.00 s , rrf = 5.00 2.00( ) i + 1

23.00( ) 2.00( )2 j = 10.0i + 6.00 j( ) m

so x f = 10.0 m , y f = 6.00 m

rv f = 5.00i + 3.00 2.00( ) j = 5.00i + 6.00 j( ) m s

v f =rv f = vxf

2 + vyf2 = 5.00( )2 + 6.00( )2 = 7.81 m s

Chapter 3 3

P3.6 (a) For the x-component of the motion we have x f = xi + vxit +

12

axt2 .

0.01 m = 0 + 1.80 ! 107 m s( )t + 12

8 ! 1014 m s2( )t2

4 ! 1014 m s2( )t2 + 1.80 ! 107 m s( )t " 10"2 m = 0

t ="1.80 ! 107 m s ± 1.8 ! 107 m s( )2 " 4 4 ! 1014 m s2( ) "10"2 m( )

2 4 ! 1014 m s2( )="1.8 ! 107 ± 1.84 ! 107 m s

8 ! 1014 m s2

We choose the + sign to represent the physical situation

t = 4.39 ! 105 m s

8 ! 1014 m s2 = 5.49 ! 10"10 s .

Here

y f = yi + vyit +

12

ayt2 = 0 + 0 + 12

1.6 ! 1015 m s2( ) 5.49 ! 10"10 s( )2 = 2.41 ! 10"4 m .

So,

rrf = 10.0 i + 0.241 j( ) mm .

(b)

rv f =rv i +

rat = 1.80 ! 107 m s i + 8 ! 1014 m s2 i + 1.6 ! 1015 m s2 j( ) 5.49 ! 10"10 s( )= 1.80 ! 107 m s( ) i + 4.39 ! 105 m s( ) i + 8.78 ! 105 m s( ) j

= 1.84 ! 107 m s( ) i + 8.78 ! 105 m s( ) j

(c) rv f = 1.84 ! 107 m s( )2 + 8.78 ! 105 m s( )2 = 1.85 ! 107 m s

(d) ! = tan"1 vy

vx

#

$%&

'(= tan"1 8.78 ) 105

1.84 ) 107#

$%&

'(= 2.73°

4 Motion in Two Dimensions

Section 3.3 Projectile Motion

*P3.9 At the maximum height vy = 0 , and the time to reach this height is found from

vyf = viy + ayt as t =

vyf ! viy

ay=

0 ! viy

!g=

viy

g.

The vertical displacement that has occurred during this time is

!y( )max = vy ,avgt =

vyf + viy

2"

#$%

&'t =

0 + viy

2"

#$%

&'viy

g"

#$%

&'=

viy2

2g.

Thus, if !y( )max = 12 ft

1 m3.281 ft

"#$

%&'= 3.66 m , then

viy = 2g !y( )max = 2 9.80 m s2( ) 3.66 m( ) = 8.47 m s , and if the angle of projection is ! = 45° , the launch speed is

vi =

viy

sin!=

8.47 m ssin 45°

= 12.0 m s .

P3.11 Take the origin at the mouth of the cannon.

x f = vxi t 2 000 m = 1 000 m s( )cos!it

Therefore, t = 2.00 s

cos!i

y f = vyi t + 1

2ay t2 :

800 m = 1 000 m s( )sin!i t + 1

2"9.80 m s2( )t2

800 m = 1 000 m s( )sin!i

2.00 scos!i

"

#$%

&'(

12

9.80 m s2( ) 2.00 scos !i

"

#$%

&'

2

800 m cos2 !i( ) = 2 000 m sin!i cos!i( ) " 19.6 m

19.6 m + 800 m cos2 !i( ) = 2 000 m 1 " cos2 !i cos!i( )

384 + (31360)cos2 !i + (640000)cos4 !i = (4 000000)cos2 !i " (4 000000)cos4 !i

4 640000cos4 !i " 3 968 640cos2 !i + 384 = 0

cos2 !i =

3 968 640 ± (3 968 640)2 " 4(4 640000)(384)9 280000

cos!i = 0.925 or 0.00984

!i = 22.4° or 89.4° (Both solutions are valid.)

Chapter 3 5 P3.12 (a)

x f = vxi t = 8.00cos 20.0° 3.00( ) = 22.6 m

(b) Taking y positive downwards, and the final point just before ground impact,

y f = vyit +12

g t2

y f = 8.00sin 20.0° 3.00( ) + 12

9.80( ) 3.00( )2 = 52.3 m .

(c) Now take the final point 10 m below the window.

10.0 = 8.00 sin 20.0°( )t + 1

29.80( )t2

4.90t2 + 2.74t ! 10.0 = 0

t = !2.74 ± 2.74( )2 + 1969.80

= 1.18 s

P3.13 Consider the motion from original zero height to maximum height h:

vyf2 = vyi

2 + 2ay y f ! yi( ) gives 0 = vyi2 ! 2g h ! 0( ) or

vyi = 2gh Now consider the motion from the original point to half the maximum height:

vyf2 = vyi

2 + 2ay y f ! yi( ) gives vyh

2 = 2gh + 2 !g( ) 12

h ! 0"#$

%&'

so

vyh = gh

At maximum height, the speed is vx =

12

vx2 + vyh

2 =12

vx2 + gh

Solving, vx =

gh3

Now the projection angle is !i = tan"1 vyi

vx= tan"1 2gh

gh/3= tan"1 6 = 67.8° .

6 Motion in Two Dimensions

P3.15 (a) We use the trajectory equation: y f = x f tan!i "

gx f2

2vi2 cos2 !i

. With x f = 36.0 m ,

vi = 20.0 m s , and ! = 53.0° we find

y f = 36.0 m( ) tan 53.0° !

9.80 m s2( ) 36.0 m( )2

2 20.0 m s( )2 cos2 53.0°( )= 3.94 m . The ball clears the bar by

3.94 ! 3.05( ) m = 0.889 m . (b) The time the ball takes to reach the maximum height is

t1 =

vi sin!ig

=20.0 m s( ) sin53.0°( )

9.80 m s2 = 1.63 s .

The time to travel 36.0 m horizontally is t2 =

x f

vix

t2 =

36.0 m(20.0 m s ) cos 53.0°( )

= 2.99 s .

Since t2 > t1

the ball clears the goal on its way down .

P3.16 The horizontal component of displacement is

x f = vxit = vi cos!i( )t . Therefore, the time required to

reach the building a distance d away is t = d

vi cos!i. At this time, the altitude of the water is

y f = vyit +

12

ayt2 = vi sin!id

vi cos!i

"

#$%

&'(

g2

dvi cos!i

"

#$%

&'

2

.

Therefore the water strikes the building at a height h above ground level of

h = y f = d tan!i "

gd2

2vi2 cos2 !i

.

Section 3.4 The Particle in Uniform Circular Motion

P3.23 ac =

v2

r=

20.0 m s( )2

1.06 m= 377 m s2

The mass is unnecessary information.

Chapter 3 7

P3.25 r = 0.500 m ;

vt =2! rT

=2! 0.500 m( )

60.0 s200 rev

= 10.47 m s = 10.5 m s

a = v2

R=

10.47( )2

0.5= 219 m s2 inward

P3.27 The satellite is in free fall. Its acceleration is due to gravity and is by effect a centripetal acceleration.

ac = g

so v2

r= g . Solving for the velocity, v = rg = 6, 400 + 600( ) 103 m( ) 8.21 m s2( ) = 7.58 ! 103 m s .

v = 2!r

T and

T =2! r

v=

2! 7,000 " 103 m( )7.58 " 103 m s

= 5.80 " 103 s

T = 5.80 " 103 s 1 min60 s

#$%

&'(= 96.7 min .

Section 3.5 Tangential and Radial Acceleration P3.28 (b) We do part (b) first. The tangential speed is described by v f = vi + att

0.7 m s = 0 + at 1.75 s( ) so at = 0.400 m s2 forward

(a) Now at t = 1.25 s , v f = vi + att = 0 + 0.4 m s2( )1.25 s

v f = 0.5 m s

so ac =

v2

r=

0.5 m s( )2

0.2 m= 1.25 m s2 toward the center

(c)

ra = rar +rat = 0.4 m s2 forward + 1.25 m s2 inward

ra = 0.42 + 1.252 forward and inward at ! = tan"1 1.250.4

#$%

&'(

ra = 1.31 m s2 forward and 72.3° inward

8 Motion in Two Dimensions

P3.30 (a) See figure to the right. (b) The components of the 20.2 and the 22.5 m s2 along the rope together

constitute the centripetal acceleration:

ac = 22.5 m s2( )cos 90.0° ! 36.9°( ) + 20.2 m s2( )cos 36.9° = 29.7 m s2

(c) ac =

v2

r so v = acr = 29.7 m s2 1.50 m( ) = 6.67 m s tangent to circle

rv = 6.67 m s at 36.9° above the horizontal

FIG. P3.30

Section 3.6 Relative Velocity

P3.33 Total time in still water t = d

v=

2 0001.20

= 1.67 ! 103 s . Total time = time upstream plus time downstream:

tup =1 000

(1.20 ! 0.500)= 1.43 " 103 s

tdown =1 000

1.20 + 0.500= 588 s.

Therefore, ttotal = 1.43 ! 103 + 588 = 2.02 ! 103 s, 21.0% more than if the water were still. .

*P3.34

rvce = the velocity of the car relative to the earth.

rvwc = the velocity of the water relative to the car.

rvwe = the velocity of the water relative to the earth.

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

(a) Since

rvwe is vertical, vwc sin 60.0° = vce = 50.0 km h or

rvwc = 57.7 km h at 60.0° west of vertical .

(b) Since

rvce has zero vertical component,

FIG. P3.34

vwe = vwc cos 60.0° = 57.7 km h( )cos 60.0° = 28.9 km h downward .

Chapter 3 9

Additional Problems P3.47

x f = vixt = vit cos 40.0°

Thus, when x f = 10.0 m , t = 10.0 m

vi cos 40.0°.

At this time, y f should be 3.05 m ! 2.00 m = 1.05 m .

Thus, 1.05 m =

vi sin 40.0°( )10.0 mvi cos 40.0°

+12!9.80 m s2( ) 10.0 m

vi cos 40.0°"

#$

%

&'

2

.

From this, vi = 10.7 m s .

10 Motion in Two Dimensions

P3.50 Measure heights above the level ground. The elevation yb of the ball follows

yb = R + 0 ! 1

2gt2

with x = vi t so yb = R !

gx2

2vi2 .

(a) The elevation yr of points on the rock is described by

yr2 + x2 = R2 .

We will have yb = yr at x = 0 , but for all other x we require the ball to be above the rock

surface as in yb > yr . Then yb2 + x2 > R2

R !gx2

2vi2

"

#$%

&'

2

+ x2 > R2

R2 !gx2R

vi2 +

g2x4

4vi4 + x2 > R2

g2x4

4vi4 + x2 >

gx2Rvi

2 .

If this inequality is satisfied for x approaching zero, it will be true for all x. If the ball’s

parabolic trajectory has large enough radius of curvature at the start, the ball will clear the

whole rock: 1 > gR

vi2

vi > gR .

(b) With vi = gR and yb = 0 , we have 0 = R !

gx2

2gR

or x = R 2 . The distance from the rock’s base is

x ! R = 2 ! 1( )R .