Post on 13-Apr-2015
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 2.
( )23 2 mx t t= − −
( )23 2 2 m/sdxv t tdt
= = − −
26 2 m/sdva t
dt= = −
(a) Time at a = 0.
00 6 2 0t= − =
013
t = 0 0.333 st =
(b) Corresponding position and velocity.
3 21 1 2 2.741 m
3 3x ⎛ ⎞ ⎛ ⎞= − − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ 2.74 mx = −
21 13 2 2 3.666 m/s
3 3v ⎛ ⎞ ⎛ ⎞= − − =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ 3.67 m/sv =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 3.
Position: 4 35 4 3 2 ftx t t t= − + −
Velocity: 3 220 12 3 ft/sdxv t t
dt= = − +
Acceleration: 2 260 24 ft/sdva t t
dt= = −
When 2 s,t =
( )( ) ( )( ) ( )( )4 35 2 4 2 3 2 2x = − − − 52 ft x =
( )( ) ( )( )3 220 2 12 2 3v = − + 115 ft/sv =
( )( ) ( )( )260 2 24 2a = − 2192 ft/sa =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 4.
Position: 4 3 26 8 14 10 16 in.x t t t t= + − − +
Velocity: 3 224 24 28 10 in./sdxv t t t
dt= = + − −
Acceleration: 2 272 48 28 in./sdva t t
dt= = + −
When 3 s,t =
( )( ) ( )( ) ( )( ) ( )( )4 3 26 3 8 3 14 3 10 3 16x = + − − + 562 in.x = !
( )( ) ( )( ) ( )( )3 224 3 24 3 28 3 10v = + − − 770 in./s v = !
( )( ) ( )( )272 3 48 3 28a = + − 2764 in./sa = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 5.
Position: 500sin mmx kt=
Velocity: 500 cos mm/sdxv k ktdt
= =
Acceleration: 2 2500 sin mm /sdva k kt
dt= = −
When 0.05 s, and 10 rad/st k= =
( )( )10 0.05 0.5 radkt = =
( )500sin 0.5x = 240 mmx = !
( )( ) ( )500 10 cos 0.5v = 4390 mm/sv = !
( )( ) ( )2500 10 sin 0.5a = − 3 224.0 10 mm/sa = − × !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 6.
Position: ( )21 250sin mmx k t k t= −
Where 2
1 21 rad/s and 0.5 rad/sk k= = Let
2 21 2 0.5 radk t k t t tθ = − = −
( )2
221 rad/s and 1 rad/sd dt
dt dtθ θ= − = −
Position: 50sin mmx θ=
Velocity: 50cos mm/sdx dvdt dt
θθ= =
Acceleration: dvadt
=
22
2250cos 50sin mm/sd da
dtdtθ θθ θ ⎛ ⎞= − ⎜ ⎟
⎝ ⎠
When 0,v = either cos 0θ =
or 1 0 1 sd t tdtθ = − = =
Over 0 2 s, values of cos are:t θ≤ ≤
( )st 0 0.5 1.0 1.5 2.0
( )radθ 0 0.375 0.5 0.375 0
cosθ 1.0 0.931 0.878 0.981 1.0
No solutions cos 0 in this range.θ =
For 1 s,t = ( )( )21 0.5 1 0.5 radθ = − =
( )50sin 0.5x = 24.0 mmx =
( )( ) ( )( )50cos 0.5 1 50sin 0.5 0a = − − 243.9 mm/sa = −
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 7.
Given: 3 26 9 5x t t t= − + +
Differentiate twice. 23 12 9dxv t t
dt= = − +
6 12dva tdt
= = −
(a) When velocity is zero. 0v =
( )( )23 12 9 3 1 3 0t t t t− + = − − =
1 s and 3 st t= = (b) Position at t = 5 s.
( ) ( )( ) ( )( )3 25 5 6 5 + 9 5 + 5x = − 5 25 ftx =
Acceleration at t = 5 s.
( )( )5 6 5 12a = − 2
5 18 ft/sa = Position at t = 0.
0 5 ftx = Over 0 ≤ t < 1 s x is increasing. Over 1 s < t < 3 s x is decreasing. Over 3 s < t ≤ 5 s x is increasing.
Position at t = 1 s.
( ) ( )( ) ( )( )3 21 1 6 1 9 1 5 9 ftx = − + + =
Position at t = 3 s.
( ) ( )( ) ( )( )3 23 3 6 3 9 3 5 5 ftx = − + + =
Distance traveled. At t = 1 s 1 1 0 9 5 4 ftd x x= − = − =
At t = 3 s 3 1 3 1 4 5 9 8 ftd d x x= + − = + − =
At t = 5 s 5 3 5 3 8 25 5 28 ftd d x x= + − = + − =
5 28 ftd =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 8.
( )32 2 ftx t t= − −
( )22 3 2 ft/sdxv t tdt
= = − −
(a) Positions at v = 0.
( )2 22 3 2 3 14 12 0t t t t− − = − + − =
214 (14) (4)( 3)( 12)
(2)( 3)t
− ± − − −=
−
1 21.1315 s and 3.535 st t= =
1At 1.1315 s, t = 1 1.935 ftx = 1 1.935 ftx =
2At 3.535 s,t = 2 8.879 ftx = 2 8.879 ftx =
(b) Total distance traveled.
0At 0,t t= = 0 8 ftx =
4At 4 s,t t= = 4 8 ftx =
Distances traveled.
10 to :t 1 1.935 8 6.065 ftd = − =
1 2to :t t 2 8.879 1.935 6.944 ftd = − =
2 4to :t t 3 8 8.879 0.879 ftd = − =
Adding, 1 2 3d d d d= + + 13.89 ftd =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 9.
0.23 ta e−=
0 0v tdv a dt=∫ ∫
0.2 0.2
00
30 30.2
tt t tv e dt e− −− = =
−∫
( ) ( )0.2 0.215 1 15 1t tv e e− −= − − = −
At t = 0.5 s, ( )0.115 1v e−= − 1.427 ft/sv =
0 0x tdx v dt=∫ ∫
( )0.2 0.20
0
10 15 1 150.2
tt t tx e dt t e− −⎛ ⎞− = − = +⎜ ⎟
⎝ ⎠∫
( )0.215 5 5tx t e−= + −
At 0.5 s,t = ( )0.115 0.5 5 5x e−= + − 0.363 ftx =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 10.
Given: 2
0 05.4sin ft/s , 1.8 ft/s, 0, 3 rad/sa kt v x k= − = = =
0 0 0 0
5.45.4 sin costt tv v a dt kt dt kt
k− = = − =∫ ∫
( )5.41.8 cos 1 1.8cos 1.83
v kt kt− = − = −
Velocity: 1.8cos ft/sv kt=
0 0 0 0
1.81.8 cos sintt tx x v dt kt dt kt
k− = = =∫ ∫
( )1.80 sin 0 0.6sin3
x kt kt− = − =
Position: 0.6sin ftx kt=
When 0.5 s,t = ( )( )3 0.5 1.5 radkt = =
1.8cos1.5 0.1273 ft/sv = = 0.1273 ft/sv =
0.6sin1.5 0.5985 ft x = = 0.598 ftx =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 11.
Given: 23.24sin 4.32cos ft/s , 3 rad/sa kt kt k= − − =
0 00.48 ft, 1.08 ft/sx v= =
( ) ( )
0 0 0 0
0 0
3.24 sin 4.32 cos
3.24 4.321.08 cos sin
3.24 4.32cos 1 sin 0
3 3
1.08cos 1.08 1.44sin
t t t
t t
v v a dt kt dt kt dt
v kt ktk k
kt kt
kt kt
− = = − −
− = −
= − − −
= − −
∫ ∫ ∫
Velocity: 1.08cos 1.44sin ft/sv kt kt= −
( ) ( )
0 0 0 0
0 0
1.08 cos 1.44 sin
1.08 1.440.48 sin cos
1.08 1.44sin 0 cos 1
3 30.36sin 0.48cos 0.48
t t t
t t
x x v dt kt dt kt dt
x kt ktk k
kt kt
kt kt
− = = −
− = +
= − + −
= + −
∫ ∫ ∫
Position: 0.36sin 0.48cos ftx kt kt= +
When 0.5 s,t = ( )( )3 0.5 1.5 radkt = =
1.08cos1.5 1.44sin1.5 1.360 ft/sv = − = − 1.360 ft/sv = − !
0.36sin1.5 0.48cos1.5 0.393 ftx = + = 0.393 ftx = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 12.
Given: 2mm/s where is a constant.a kt k=
At 0,t = 400 mm/s; at 1 s, 370 mm/s, 500 mmv t v x= = = =
2
400 0 012
v t tdv a dt kt dt kt= = =∫ ∫ ∫
2 21 1400 or 4002 2
v kt v kt− = = +
At 1 s,t = ( )2 31400 1 370, 60 mm/s2
v k k= + = = −
Thus 2400 30 mm/sv t= −
At 7 s,t = ( )( )27 400 30 7v = − 7 1070 mm/sv = −
2 2 2When 0, 400 30 0. Then 13.333 s , 3.651 sv t t t= − = = = For 0 3.651 s,t≤ ≤ 0 and is increasing.v x> For 3.651 s,t > 0 and is decreasing.v x<
( )2500 1 1 400 30x t tdx v dt t dt= = −∫ ∫ ∫
( )3 31
500 400 10 400 10 390t
x t t t t− = − = − −
Position: 3400 10 110 mmx t t= − +
At 0,t = 0 110 mmx x= =
At 3.651 s,t = ( )( ) ( )( )3
max 400 3.651 10 3.651 110 1083.7 mmx x= = − + =
At 7 s,t = ( )( ) ( )( )37 400 7 10 7 110x x= = − + 7 520 mmx = −
Distances traveled:
Over 0 3.651 s,t≤ ≤ 1 max 0 973.7 mmd x x= − =
Over 3.651 7 s,t≤ ≤ 2 max 7 1603.7 mmd x x= − =
Total distance traveled: 1 2 2577.4 mmd d d= + = 2580 mmd =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 13.
Determine velocity. 0.15 2 2 0.15v t tdv a dt dt− = =∫ ∫ ∫
( ) ( )( )0.15 0.15 0.15 2v t− − = −
0.15 0.45 m/sv t= −
At 5 s,t = ( )( )5 0.15 5 0.45v = − 5 0.300 m/s v =
When 0,v = 0.15 0.45 0 3.00 st t− = = For 0 3.00 s,t≤ ≤ 0, is decreasing.v x≤ For 3.00 5 s,t≤ ≤ 0, is increasing.v x≥
Determine position. ( )10 0 0 0.15 0.45x t tdx v dt t dt− = = −∫ ∫ ∫
( ) ( )2 2
010 0.075 0.45 0.075 0.45
tx t t t t− − = − = −
20.075 0.45 10 mx t t= − −
At 5 s,t = ( )( ) ( )( )25 0.075 5 0.45 5 10 10.375 mx = − − = −
5 10.38 mx = −
At 0,t = 0 10 m (given)x = −
At 3.00 s,t = ( )( ) ( )( )2
3 min 0.075 3.00 0.45 3.00 10 10.675 mmx x= = − − = −
Distances traveled: Over 0 3.00 s,t≤ ≤ 1 0 min 0.675 md x x= − =
Over 3.00 s 5 s,t< < 2 5 min 0.300 md x x= − =
Total distance traveled: 1 2 0.975 md d d= + = 0.975 md =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 14.
Given: 29 3a t= −
Separate variables and integrate.
( )20 0 9 3 9v tdv a dt t dt= = − =∫ ∫ ∫
30 9v t t− = − ( )29v t t= −
(a) When v is zero. 2(9 ) 0t t− =
0 and 3 s (2 roots)t t= = 3 st = (b) Position and velocity at 4 s.t =
( )35 0 0 9x t tdx v dt t t dt= = −∫ ∫ ∫
2 49 15
2 4x t t− = −
2 49 15
2 4x t t= + −
At 4 s,t = ( ) ( )2 44
9 15 4 42 4
x ⎛ ⎞ ⎛ ⎞= + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 4 13 mx =
( )( )24 4 9 4v = − 4 28 m/sv = −
(c) Distance traveled.
Over 0 3 s,t< < v is positive, so x is increasing.
Over 3 s 4 s,t< ≤ v is negative, so x is decreasing.
At 3 s,t = ( ) ( )2 43
9 15 3 3 25.25 m2 4
x ⎛ ⎞ ⎛ ⎞= + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
At 3 st = 3 3 0 25.25 5 20.25 md x x= − = − =
At 4 st = 4 3 4 3 20.25 13 25.25 32.5 md d x x= + − = + − = 4 32.5 md =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 15.
Given: 2dva kt
dt= =
Separate variables dv = kt2 dt
Integrate using v = –10 m/s when t = 0 and v = 10 m/s when t = 2 s.
10 2 210 0dv kt dt− =∫ ∫
10 3
100
13
t
v kt− =
[ ] ( )31(10) ( 10) 2 03
k ⎡ ⎤− − = −⎢ ⎥⎣ ⎦
(a) Solving for k, ( )( )3 20
8k =
47.5 m/sk =
(b) Equations of motion.
Using upper limit of v at t,
( )3 310
0
1 110 7.53 3
tvv kt v t−
⎛ ⎞= + = ⎜ ⎟⎝ ⎠
310 2.5 m/sv t= − +
Then, 310 2.5dx v t
dt= = − +
Separate variables and integrate using x = 0 when t = 2 s.
( )310 2.5dx t dt= − +
( )30 2 10 2.5x tdx t dt= − +∫ ∫
4
20 10 0.625
tx t t⎡ ⎤− = − +⎣ ⎦
( )( ) ( )( )4410 0.0625 10 2 0.625 2t t ⎡ ⎤⎡ ⎤= − + − − +⎢ ⎥⎣ ⎦ ⎣ ⎦
[ ]410 0.625 10t t= − + − −
410 10 0.625 mx t t= − +
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 16.
Note that is a given function of .a x ( )40 160 160 0.25a x x= − = −
( ) Note that is maximum when 0, or 0.25 ma v a x= =
( )Use 160 0.25 with the limitsv dv a dx x dx= = −
max0.3 m/s when 0.4 m and when 0.25 mv x v v x= = = =
( )max 0.250.3 0.4 160 0.25v v dv x dx= −∫ ∫
( ) ( )
0.252 22 2max
0.4
0.25 0.150.3 160 160 0 1.82 2 2 2
xv ⎡ ⎤− −⎢ ⎥− = − = − − =⎢ ⎥⎣ ⎦
2 2 2max 3.69 m /sv = max 1.921 m/s v =
( ) Note that is maximum or minimum when 0.b x v =
( )Use 160 0.25 with the limitsv dv a dx x= = −
0.3 m/s when 0.4 m, and 0 when mv x v x x= = = =
( )00.3 0.4 160 0.25mxv dv x dx= −∫ ∫
( ) ( ) ( ) ( )( )
2 22 2
0.4
0.3 0.250 160 80 0.25 80 0.15
2 2
mx
mx
x−
− = − = − − + −
( )20.25 0.02306 0.25 0.1519 mm mx x− = − = ±
0.0981 m and 0.402 mmx =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 17.
is a function of :a x ( ) 2100 0.25 m/sa x= −
( )Use 100 0.25 with limitsv dv a dx x dx= = − 0 when 0.2 mv x= =
( )0 0.2100 0.25
v xv dv x dx= −∫ ∫
( )( )22
0.2
1 10 100 0.25
2 2
x
v x− = − −
( )250 0.25 0.125x= − − +
So
( ) ( )2 22 0.25 100 0.25 or 0.5 1 400 0.25v x v x= − − = ± − −
Use ( )2
or 0.5 1 400 0.25
dx dxdx v dt dt
v x= = =
± − −
Integrate: ( )
0 0.2 20.5 1 400 0.25
t x dxdt
x= ±
− −∫ ∫
Let ( )20 0.25 ; when 0.2 = 1 and 20u x x u du dx= − = = −
So 1 1
1 21
1 1sin sin
10 10 210 1
uu du
t u uu
π− − = = = − −
∫m m m
Solve for .u 1sin 10
2u t
π− = m
( )sin 10 cos 10 cos102
u t t tπ = = ± =
m
( )cos 10 20 0.25u t x= = −
continued
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Solve for and .x v
10.25 cos10
20x t= −
1
sin102
v t=
Evaluate at 0.2 s.t =
( )( )( )10.25 cos 10 0.2
20x = − 0.271 m x =
( )( )( )1sin 10 0.2
2v = 0.455 m/s v =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 18.
Note that is a given function of a x
Use ( ) ( )2 3600 1 600 600v dv a dx x kx dx x kx dx= = + = +
Using the limits 7.5 ft/s when 0,v x= =
and 15 ft/s when 0.45 ft,v x= =
( )15 0.45 37.5 0 600 600v dv x kx dx= +∫ ∫
15 0.4522 4
07.5
600 6002 2 4v x kx⎡ ⎤ ⎡ ⎤= +⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦
( ) ( ) ( )( ) ( ) ( )
2 22 415 7.5
300 0.45 150 0.452 2
k− = +
84.375 60.75 6.1509k= +
Solving for ,k 23.84 ft k −=
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 19.
Note that is a given function of .a x
Use ( )3800 3200v dv a dx x x dx= = +
Using the limit 10 ft/s when 0,v x= =
( )310 0
800 3200v x
v dv x x dx= +∫ ∫
( )22
2 410400 800
2 2
vx x− = +
2 4 2 21600 800 100 Let v x x u x= + + =
Then ( )( )2 21 21600 800 100 1600 ,v u u u u u u= + + = − −
1 2where and are the roots ofu u 21600 800 100 0u u+ + =
Solving the quadratic equation,
( ) ( )( )( )( )( )
2
1,2
800 800 4 1600 100 800 00.25 0
2 1600 3200u
− ± − − ±= = = − ±
2
1 2 0.25 ftu u= = −
So ( ) ( )222 2 2 2 21600 0.25 1600 0.5 ft /sv u x= + = +
Taking square roots, ( )2 240 0.5 ft/sv x= ± +
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Use ( )2 2 or
40 0.5
dx dxdx v dt dt
v x= = = ±
+
2 240 Use limit 0 when 0
0.5
dxdt x t
x= ± = =
+
1
2 20 0
140 tan
0.5 0.50.5
t x dx xdt
x−= ± = ±
+∫ ∫
( ) ( )1 140 2.0 tan 2 or tan 2 20t x x t− −= ± = ±
( ) ( )2 tan 20 or 0.5 tan 20x t x t= ± = ±
( ) ( ) ( )2 20.5 sec 20 20 10 sec 20dx
v t tdt
= = ± = ±
At 0, 10 ft/s, which agrees with the given data if the minus sign is rejected.t v= = ±
Thus, ( ) ( )210 sec 20 ft/s, and 0.5 tan 20 ftv t x t= =
At 0.05 s,t = 20 1.0 radt =
( )22
1010sec 1.0
cos 1.0v = = 34.3 ft/s v =
( )0.5 tan 1.0x = 0.779 ft x =
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 20.
Note that is a given function of .a x 27
12 28 12 m/s3
a x x = − = −
7Use 12 with the limits
3v dv a dx x dx
= = −
8 m/s when 0.v x= =
22
8 08 0
7 12 712
3 2 2 3
xvv x vv dv x dx x
= − = − ∫ ∫
2 22 28 12 7 7
2 2 2 3 3
vx
− = − −
2 2 22 2 7 7 7 4
8 12 123 3 3 3
v x x = + − − = − −
27 4
123 3
v x = ± − −
Reject minus sign to get 8 m/s at 0.v x= =
(a) Maximum value of .x max0 when v x x= =
2 27 4 7 1
12 0 or 3 3 3 9
x x − − = − =
max max7 1 8 2
2 m and m 2 m3 3 3 3
x x x− = ± = = =
Now observe that the particle starts at 0 with 0 and reaches 2 m. At 2 m, 0 andx v x x v= > = = =
20, so that becomes negative and decreases. Thus, 2 m is never reached.
3a v x x< =
max 2 mx = !
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(b) Velocity when total distance traveled is 3 m.
The particle will have traveled total distance 3 md = when max maxd x x x− = − or 3 2 2 x− = − or 1 m.x =
Using
27 4
123 3
v x = − − −
, which applies when x is decreasing, we get
27 4
12 1 203 3
v = − − − = −
4.47 m/s v = − !
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Chapter 11, Solution 21.
Note that is a function of .a x ( )1 xa k e−= −
( )Use 1 with the limits 9 m/s when 3 m, and 0 when 0.xv dv a dx k e dx v x v x−= = − = = − = =
( )0 09 3 1 xv dv k e dx−
−= −∫ ∫
( )02 0
39
2xv k x e−
−
⎛ ⎞= +⎜ ⎟⎜ ⎟
⎝ ⎠
( )2
390 0 1 3 16.08552
k e k⎡ ⎤− = + − − − = −⎣ ⎦
(a) 2.5178k = 22.52 m/sk =
( ) ( )Use 1 2.5178 1 with the limit 0 when 0.x xv dv a dx k e dx e dx v x− −= = − = − = =
( )0 0 2.5178 1v x xv dv e dx−= −∫ ∫
( ) ( )2
02.5178 2.5178 1
2
xx xv x e x e− −= + = + −
( ) ( )1/22 5.0356 1 2.2440 1x xv x e v x e− −= + − = ± + −
(b) Letting 2 m,x = −
( )1/ 222.2440 2 1 4.70 m/sv e= ± − + − = ±
Since begins at 2 m and ends at 0, 0.x x x v= − = >
Reject the minus sign.
4.70 m/s v =
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Chapter 11, Solution 22.
0.000576.8 xdva v e
dx−= =
0.00057
0 0 6.8v x xv dv e dx−=∫ ∫
20.00057
0
6.802 0.00057
xxv e−− =−
( )0.0005711930 1 xe−= −
When 30 m/s.v =
( ) ( )
20.0005730
11930 12
xe−= −
0.000571 0.03772xe−− =
0.00057 0.96228xe− =
0.00057 ln (0.96228) 0.03845x− = = −
67.5 mx =
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Chapter 11, Solution 23.
Given: 0.4dva v vdx
= = −
or 0.4dvdx
= −
Separate variables and integrate using 75 mm/s when 0.v x= =
75 00.4 75 0.4v xdv v x= − − = −∫ ∫
(a) Distance traveled when 0v =
0 75 0.4x− = − 187.5 mmx =
(b) Time to reduce velocity to 1% of initial value.
(0.01)(75) 0.75v = =
0.752.5ln75
t = − 11.51 st =
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Chapter 11, Solution 24.
Given: dva v kvdx
= = − 2
Separate variables and integrate using 9 m/s when 0.v x= =
9 0v xdv k dx
v= −∫ ∫
ln9v kx= −
Calculate using 7 m/s when 13 m.k v x= =
( )( ) 3 17ln 13 19.332 10 m9
k k − −= − = ×
Solve for .x 1 ln 51.728 ln
9 9v vx
k= − = −
(a) Distance when 3 m/s.v =
351.728 ln9
x ⎛ ⎞= − ⎜ ⎟⎝ ⎠
56.8 m x =
(b) Distance when 0.v =
( )51.728 ln 0x = − x = ∞
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Chapter 11, Solution 25.
0 0, 0, 25 ft/sv dv a dx k vdx x v= = − = =
1/21dx v dv
k= −
0 00
3/21 23
vx vx v
vdx vdv v
k k= − = −∫ ∫
( ) ( )3/23/2 3/2 3/2 3/20 0
2 2 2 or 25 1253 3 3
x x v v x v vk k k
⎡ ⎤ ⎡ ⎤− = − = − = −⎢ ⎥ ⎣ ⎦⎣ ⎦
Noting that 6 ft when 12 ft/s,x v= =
3/2 32 55.626 125 12 or 9.27 ft/s3
kk k⎡ ⎤= − = =⎣ ⎦
Then, ( )( ) ( )3/2 3/22 125 0.071916 1253 9.27
x v v⎡ ⎤= − = −⎣ ⎦
3/2 125 13.905v x= −
( ) Whena 8 ft,x = ( )( ) ( )3/23/2 125 13.905 8 13.759 ft/sv = − =
5.74 ft/s v =
( )b dv a dt k vdt= = −
1/ 21 dvdtk v
= −
( )0
1/21/2 1/20
1 22v
vt v v v
k k⎡ ⎤= − ⋅ = −⎣ ⎦
At rest, 0v = ( )( )1/21/2
0 2 2529.27
vtk
= = 1.079 s t =
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Chapter 11, Solution 29.
x as a function of v.
0.000571
154xv e−= −
20.00057 1
154x ve− ⎛ ⎞= − ⎜ ⎟
⎝ ⎠
2
0.00057 ln 1154
vx⎡ ⎤⎛ ⎞− = −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
2
1754.4 ln 1154
vx⎡ ⎤⎛ ⎞= − −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦ (1)
a as a function of x.
( )2 0.0005723716 1v e−= −
( )( )2
0.000511858 0.000572
xdv d va v edx dx
−⎛ ⎞= = =⎜ ⎟⎜ ⎟
⎝ ⎠
2
0.000576.75906 6.75906 1154
x va e−⎡ ⎤⎛ ⎞= = −⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦ (2)
(a) v = 20 m/s.
From (1), x = 29.843 x = 29.8 m From (2), a = 6.64506 a = 6.65 m/s2
(b) v = 40 m/s.
From (1), x = 122.54 x = 122.5 m From (2), a = 6.30306 a = 6.30 m/s2
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Chapter 11, Solution 30.
( )0.3Given: 7.5 1 0.04 with units km and km/hv x= −
(a) Distance at 1 hr.t =
0.3Using , we get 7.5(1 0.04 )
dx dxdx v dt dtv x
= = =−
Integrating, using 0t = when 0,x =
( ) ( ) ( )( ) { }0.700.30 0 0
1 1 1 or [ ] 1 0.047.5 7.5 0.7 0.041 0.04
xt x tdxdt t x−= = ⋅ −−
∫ ∫
( ){ }0.74.7619 1 1 0.04t x= − − (1)
Solving for ,x ( ){ }1/0.725 1 1 0.210x t= − −
When 1 h,t = ( )( ){ }1/0.725 1 1 0.210 1x ⎡ ⎤= − −⎣ ⎦ 7.15 km x =
(b) Acceleration when 0.t =
0.7 0.7(7.5)(0.3)( 0.04)(1 0.04 ) 0.0900(1 0.04 )dv x x
dx− −= − − = − −
When 0t = and 0,x = 17.5 km/h, 0.0900 hdvvdx
−= −
2(7.5)( 0.0900) 0.675 km/hdva v
dx= = − = −
2
2(0.675)(1000) m/s
(3600)= −
6 252.1 10 m/s a −= − ×
(c) Time to run 6 km.
Using 6 kmx = in equation (1),
( )( ){ }0.74.7619 1 1 0.04 6 0.8323 ht ⎡ ⎤= − − =⎣ ⎦
49.9 mint =
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Chapter 11, Solution 31.
The acceleration is given by 2
2dv gRv adr r
= = −
Then, 2
2gR drv dv
r= −
Integrating, using the conditions esc0 at , and v r v v= = ∞ = at r R=
esc
0 22v R
drv dv gRr
∞= −∫ ∫
esc
02 21 1
2 v R
v gRr
∞⎛ ⎞= ⎜ ⎟⎝ ⎠
2 2esc
1 10 02
v gRR
⎛ ⎞− = −⎜ ⎟⎝ ⎠
esc 2v gR=
6 2Now, 3960 mi 20.909 10 ft and 32.2 ft/s .R g= = × =
Then, ( )( )( )6esc 2 32.2 20.909 10v = × 3
esc 36.7 10 ft/s v = ×
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Chapter 11, Solution 32.
The acceleration is given by 6
2
20.9 10
32.2
1 ya
×
−=⎡ ⎤⎛ ⎞+ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
6
2
20.9 10
32.2
1 y
dyvdv ady
×
−= =⎡ ⎤⎛ ⎞+ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦
20 maxIntegrate, using the conditions at 0 and 0 at . Also, use 32.2 ft/s andv v y v y y g= = = = =
620.9 10 ft.R = ×
( ) ( )0
0 22 20 0
1v y
R
dy dyv dv g gRR y
∞ ∞= − = −++
∫ ∫ ∫ max
0
02 2
0
1 12
y
vv gR
R y⎛ ⎞
= ⎜ ⎟+⎝ ⎠
( )2 2 2max0 0 max max
max max
1 1 10 22
gRyv gR v R y gRyR y R R y
⎡ ⎤− = − = − + =⎢ ⎥+ +⎣ ⎦
maxSolving for ,y 20
max 202
RvygR v
=−
Using the given numerical data, ( )( )( )6 2 6 2
0 0max 9 26 2
00
20.9 10 20.9 101.34596 102 32.2 20.9 10
v vyvv
× ×= =× −× −
0( ) 2400 ft/s,a v = ( )( )
( ) ( )
26
max 29
20.9 10 2400
1.34596 10 2400y
×=
× − 3
max 89.8 10 ft y = ×
0( ) 4000 ft/s,b v = ( )( )
( ) ( )
26
max 29
20.9 10 4000
1.34596 10 4000y
×=
× − 3
max 251 10 ft y = ×
0( ) 40000 ft/s,c v = ( )( )
( ) ( )
26
max 29
20.9 10 40000negative
1.34596 10 40000y
×= =
× −
Negative value indicates that 0v is greater than the escape velocity.
maxy = ∞
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Chapter 11, Solution 33.
( )( ) Given: sin na v v tω ϕ′= +
At 0,t = 0
0 sin or sinv
v v vv
ϕ ϕ′= = =′ (1)
Let x be maximum at 1t t= when 0.v =
Then, ( ) ( )1 1sin 0 and cos 1n nt tω ϕ ω ϕ+ = + = ± (2)
Using or dx
v dx v dtdt
= =
Integrating, ( )cos nn
vx C tω ϕ
ω′
= − +
At 0,t = 0 0cos or cosn n
v vx x C C xϕ ϕ
ω ω′ ′
= = − = +
Then, ( )0 cos cos nn n
v vx x tϕ ω ϕ
ω ω′ ′
= + − + (3)
max 0 1cos using cos 1nn
v vx x tϕ ω ϕ
ω ω′ ′
= + + + = −
Solving for cos ,ϕ ( )max 0cos 1nx x
v
ωϕ
−= −
′
max 0With 2 ,x x= 0cos 1nx
v
ωϕ = −′ (4)
Using
2 22 2 0 0sin cos 1, or 1 1nv x
v v
ωϕ ϕ + = + − = ′ ′
Solving for givesv′ ( )2 2 2
0 0
0
(5) 2
n
n
v xv
x
ω
ω
+′ =
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( ) Acceleration:b ( )cosn ndv
a v tdt
ω ω ϕ′= = +
2Let be maximum at when 0.v t t a= =
Then, ( )2cos 0ntω ϕ+ =
From equation (3), the corresponding value of x is
( )
00 0 0
2 2 2 20 0 0
0 0 20 0
cos 1 2
3 12
2 2 2
n
n n n
n
n n n
v v x vx x x x
v
v x vx x
x x
ωϕω ω ω
ωω ω ω
′ ′ ′ = + = + − = − ′
+= − = −
( )0
0
2
0
3
2
n
vx
xω
−
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Chapter 11, Solution 34.
0( ) 1 sindx ta v vdt T
π⎡ ⎤= = −⎢ ⎥⎣ ⎦
0Integrating, using 0 when 0,x x t= = =
00 0 0 1 sinx t t tdx v dt v dtTπ⎡ ⎤= = −⎢ ⎥⎣ ⎦
∫ ∫ ∫
000
0
cost
x v T tx v tTπ
π⎡ ⎤= +⎢ ⎥⎣ ⎦
0 00 cosv T t v Tx v t
Tπ
π π= + − (1)
When 3 ,t T= ( )0 00 0
23 cos 3 3v T v Tx v T v TT
ππ π
⎛ ⎞= + − = −⎜ ⎟⎝ ⎠
02.36 x v T=
0 cosdv v tadt T T
π π= = −
When 3 ,t T= 0 cos3va
Tπ π= − 0va
Tπ=
( ) Using equation (1) with ,b t T=
0 01 0 0
2cos 1v T v Tx v T v Tππ π π
⎛ ⎞= + − = −⎜ ⎟⎝ ⎠
Average velocity is
1 0
ave 021x x xv v
t T πΔ − ⎛ ⎞= = = −⎜ ⎟Δ ⎝ ⎠
ave 00.363 v v=
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Chapter 11, Solution 35.
10 km/h 2.7778 m/s= 100 km/h 27.7778 m/s=
(a) Acceleration during start test.
dvadt
= 8.2 27.77780 2.7778a dt v dt=∫ ∫
8.2 27.7778 2.7778a = − 23.05 m/sa =
(b) Deceleration during braking.
dva vdx
= =
44 00 27.7778a dx v dv= =∫ ∫
( ) ( )0
44 20
27.7778
12
a x v=
( )2144 27.77782
a = −
28.77 m/sa = − deceleration 28.77 m/sa= − =
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Chapter 11, Solution 36.
10 km/h 2.7778 m/s= 100 km/h 27.7778 m/s=
(a) Distance traveled during start test.
dvadt
= 00t v
va dt dv=∫ ∫
0at v v= − 0v va
t−=
227.7778 2.7778 3.04878 m/s
8.2a −= =
0 2.7778 3.04878v v at t= + = +
)8.20 0 2.7778 3.04878tx v dv t dt= = +∫ ∫
( )( ) ( )( )22.7778 8.2 1.52439 8.2= + 125.3 mx =
(b) Elapsed time for braking test.
dva vdx
= 00x v
va dx v dv=∫ ∫
2 2
0
2 2v vax = −
( ) ( )( ) ( )2 2 20
1 1 0 27.77782 2 44
a v vx
= − = −
28.7682 m/s= −
dvadt
= 00t v
va dt dv=∫ ∫
0at v v= −
0 0 27.7778
8.7682v vt
a− −= =
− 3.17 st =
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Chapter 11, Solution 37. Constant acceleration. 0 00, 0A Av v x x= = = =
0v v at at= + = (1)
2 2
0 01 12 2
x x v t at at= + + = (2)
At point ,B 2700 ft and 30 sBx x t= = =
(a) Solving (2) for a, ( )( )( )2 2
2 2700230
xat
= = 26 ft/sa =
(b) Then, ( )( )6 30Bv at= = 180 ft/s Bv =
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Chapter 11, Solution 38. Constant acceleration. 0 0x =
0v v at= + (1)
2
0 012
x x v t at= + + (2)
Solving (1) for a, 0v va
t−= (3)
Then, ( ) ( )20
0 0 0 0 01 1 12 2 2
v vx x v t t x v v t v v tt
−= + + = + + = +
At 6 s,t = 0 6
1 and 540 ft2
v v x= =
( )0 0 0 0
0
1 1 540540 6 4.5 or 120 ft/s2 2 4.51 60 ft/s2
v v v v
v v
⎛ ⎞= + = = =⎜ ⎟⎝ ⎠
= =
Then, from (3), 2 260 120 60 ft/s 10 ft/s6 6
a −= = − = −
Substituting into (1) and (2), 120 10v t= −
( ) 210 120 102
x t t= + −
At stopping, 0 or 120 10 0 12 ss sv t t= − = =
( )( ) ( )( )210 120 12 10 12 720 ft2
x = + − =
( ) Additional time for stopping 12 s 6 sa = − 6 s tΔ =
( ) Additional distance for stopping 720 ft 540 ftb = − 180 ftdΔ =
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Chapter 11, Solution 39.
2
0 01( ) During the acceleration phase 2
a x x v t at= + +
0 0Using 0, and 0, and solving for givesx v a= =
22xat
=
Noting that 130 m when 25 s,x t= =
( )( )
( )22 130
25a = 0.416 m/s a =
(b) Final velocity is reached at 25 s.t =
( )( )0 0 0.416 25fv v at= + = + 10.40 m/s fv =
(c) The remaining distance for the constant speed phase is
400 130 270 mxΔ = − =
For constant velocity, 270 25.96 s
10.40xt
vΔΔ = = =
Total time for run: 25 25.96t = + 51.0 s t =
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Chapter 11, Solution 40. Constant acceleration. Choose 0t = at end of powered flight.
Then, 21 27.5 m 9.81 m/sy a g= = − = −
(a) When y reaches the ground, 0 and 16 s.fy t= =
2 2
1 1 1 11 12 2fy y v t at y v t gt= + + = + −
( )( )221 1
1 2 21
0 27.5 9.81 1676.76 m/s
16fy y gt
vt
− + − += = =
1 76.8 m/s v =
(b) When the rocket reaches its maximum altitude max,y
0v =
( ) ( )2 2 21 1 1 12 2v v a y y v g y y= + − = − −
2 2
11 2
v vy yg
−= −
( )
( )( )2
max0 76.76
27.52 9.81
y−
= − max 328 m y =
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Chapter 11, Solution 41. Place origin at 0.
Motion of auto. ( ) ( ) 20 00, 0, 0.75 m/sA A Ax v a= = =
( ) ( ) ( )2 20 0
1 10 0 0.752 2A A A Ax x v t a t t⎛ ⎞= + + = + + ⎜ ⎟
⎝ ⎠
20.375 mAx t=
Motion of bus. ( ) ( )0 0?, 6 m/s, 0B B Bx v a= = − =
( ) ( ) ( )0 0 0 6 mB B B Bx x v t x t= − = −
At 20 , 0.Bt s x= =
( ) ( )( )00 6 20Bx= − ( )0 120 mBx =
Hence, 120 6Bx t= −
When the vehicles pass each other, .B Ax x=
2120 6 0.375t t− =
20.375 6 120 0t t+ − =
( )( )( )
( )( )26 (6) 4 0.375 120
2 0.375t
− ± − −=
6 14.697 11.596 s and 27.6 s0.75
t − ±= = −
Reject the negative root. 11.60 st =
Corresponding values of xA and xB.
( )( )20.375 11.596 50.4 mAx = =
( )( )120 6 11.596 50.4 mBx = − = 50.4 mx =
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Chapter 11, Solution 42.
Place the origin at A when t = 0.
Motion of A: ( ) ( ) 20 00, 15 km/h = 4.1667 m/s, 0.6 m/sA A Ax v a= = =
( )0 4.1667 0.6A A Av v a t t= + = +
( ) ( ) 2 20 0
1 4.1667 0.32A A A Ax x v t a t t t= + + = +
Motion of B: ( ) ( ) 20 025 m, 23 km/h = 6.3889 m/s, 0.4 m/sB B Bx v a= = = −
( )0 6.3889 0.4B B Bv v a t t= + = −
( ) ( ) 2 2
0 01 25 6.3889 0.22B B B Bx x v t a t t t= + + = + −
(a) When and where A overtakes B. A Bx x=
2 24.1667 0.3 25 6.3889 0.2t t t t+ = + −
20.5 2.2222 25 0t t− − =
( )( )( )
( )( )22.2222 2.2222 4 0.5 25
2 0.5t
± − −=
2.2222 7.4120 9.6343 s and 5.19 st = ± = −
Reject the negative root. . 9.63 st =
( )( ) ( )( )24.1667 9.6343 0.3 9.6343 68.0 mAx = + =
( )( ) ( )( )225 6.3889 9.6343 0.2 9.6343 68.0 mBx = + − =
moves 68.0 mA
moves 43.0 mB (b) Corresponding speeds.
( )( )4.1667 0.6 9.6343 9.947 m/sAv = + = 35.8 km/hAv =
( )( )6.3889 0.4 9.6343 2.535 m/sBv = − = 9.13 km/hBv =
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Chapter 11, Solution 43. Constant acceleration ( )1 2 and a a for horses 1 and 2.
Let 0x =
and 0t = when the horses are at point A.
Then, 2
012
x v t at= +
Solving for , a ( )02
2 x v ta
t−
=
Using 1200 ftx = and the initial velocities and elapsed times for each horse,
( )( )( )
21 11 2 2
1
2 1200 20.4 61.50.028872 ft/s
61.5x v ta
t
⎡ ⎤−− ⎣ ⎦= = = −
( )( )
( )22 2
2 2 22
2 1200 21 62.00.053070 ft/s
62.0x v ta
t
⎡ ⎤−− ⎣ ⎦= = = −
1 2Calculating ,x x− ( ) ( ) 21 2 1 2 1 2
12
x x v v t a a t− = − + −
( ) ( ) ( ) 2
1 2
2
120.4 21 0.028872 0.0530702
0.6 0.012099
x x t t
t t
⎡ ⎤− = − + − − −⎣ ⎦
= − +
At point B, 21 2 0 0.6 0.012099 0B Bx x t t− = − + =
(a) 0.6 49.59 s0.012099Bt = =
Calculating Bx using data for either horse,
Horse 1: ( )( ) ( )( )2120.4 49.59 0.028872 49.592Bx = + − 976 ftBx =
Horse 2: ( )( ) ( )( )2121 49.59 0.05307 49.59 976 ft2Bx = + − =
When horse 1 crosses the finish line at 61.5 s,t =
(b) ( )( ) ( )( )21 2 0.6 61.5 0.012099 61.5x x− = − + 8.86 ftxΔ =
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Chapter 11, Solution 44. Choose x positive upward. Constant acceleration a g= −
Rocket launch data: Rocket :A 00, , 0x v v t= = =
Rocket : B
00, , 4 sBx v v t t= = = =
Velocities:
Rocket :A 0Av v gt= −
Rocket : B ( )0B Bv v g t t= − −
Positions: 20
1Rocket :
2AA x v t gt= −
( ) ( )20
1Rocket : ,
2B B B BB x v t t g t t t t= − − − ≥
For simultaneous explosions at 240 ft when ,A B Ex x t t= = =
( ) ( )22 2 20 0 0 0
1 1 1 1
2 2 2 2E E E B E B E B E E B Bv t gt v t t g t t v t v t gt gt t gt− = − − − = − − + −
0Solving for , v 0 2B
Egt
v gt= − (1)
Then, when , Et t= 21,
2 2B
A E E Egt
x gt t gt = − −
or 2 20A
E B Ex
t t tg
− − =
Solving for , Et ( )( )( ) ( ) ( )( )( )( )22 4 1 2 2402
32.24 1 4 4
6.35 s2 2
AxB B g
E
t tt
± + ± += = =
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(a) From equation (1), ( )( ) ( )( )0
32.2 432.2 6.348
2v = − 0 140.0 ft/sv =
At time ,Et 0A Ev v gt= − ( )0B E Bv v g t t= − −
(b) ( )( )32.2 4B A Bv v gt− = = / 128.8 ft/sB Av =
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Chapter 11, Solution 45. (a) Acceleration of A.
( ) ( )0 0, 168 km/h 46.67 m/sA A A Av v a t v= + = =
At 8 s,t = 228 km/h 63.33 m/sAv = =
( )0 63.33 46.67
8A A
Av v
at
− −= = 22.08 m/s Aa =
(b) ( ) ( ) 20 0
12A A A Ax x v t a t= + + ( ) ( ) 2
0 012B B B Bx x v t a t= + +
( ) ( ) ( ) ( ) ( ) 20 0 0 0
12A B A B A B A Bx x x x v v t a a t⎡ ⎤− = − + − + −⎣ ⎦
When 0,t = ( ) ( )0 0 38 mA Bx x− = and ( ) ( )0 0 0B Av v− =
When 8 s,t = 0A Bx x− =
Hence, ( )( )210 38 8 , or 1.18752 A B A Ba a a a= + − − = −
1.1875 2.08 1.1875B Aa a= + = + 23.27 m/s Ba =
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Chapter 11, Solution 46. (a) Acceleration of A.
( ) ( ) ( ) 20 0 0
1 and
2A A A A A A Av v a t x x v t a t= + = + =
Using ( ) ( )0 00 and 0 givesA Av x= =
21
and 2A A A Av a t x a t= =
When cars pass at 1, 90 mAt t x= =
( )( )2
1 12 902 180
and AA A
A A A
xt v a t
a a a= = = =
For 0 5 s,t≤ ≤ ( )0 96 km/h 26.667 m/sB Bv v= = − = −
For 5 s,t > ( ) ( ) ( )0
15 26.667 5
6B B B Av v a t a t= + − = − + −
When vehicles pass, A Bv v= −
( )1 11
26.667 56A Aa t a t= − −
1 17 5 160
26.667 or 7 56 6A A
A
a t a ta
− = − =
Using 1180 7 180 160
gives 5A AA
ta aa
= − =
Let 1
,A
ua
= 27 180 5 160u u− =
or 2160 7 180 5 0u u− + =
continued
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Solving the quadratic equation,
( )( ) ( )( )( )( )( )
7 180 49 180 4 160 5 93.915 74.967
2 160 320
0.0592125 and 0.52776
u± − ±
= =
=
2
1285.2 m/s and 3.590 m/sAa
u= =
The corresponding values for 1t are
1 1180 180
0.794 s, and 7.08 s285.2 3.590
t t= = = =
Reject 0.794 s since it is less than 5 s.
Thus, 23.59 m/sAa =
(b) Time of passing. 1 7.08 s t t= =
(c) Distance d.
( ) ( )0 00 5 s, 26.667B B Bt x x v t d t≤ ≤ = − = −
At 5 s,t = ( )( )22.667 5 133.33Bx d d= − = −
For 5 s,t > ( ) ( ) ( )2
0
1133.33 5 5
2B B Bx d v t a t= − + − + −
( ) ( )21 3.59133.33 26.667 5 5
2 6Bx d t t = − − − + −
1When 7.08 s,t t= = 90B Ax x= =
( )( ) ( )( )( )( )
23.59 2.08
90 133.33 26.667 2.082 6
d= − − +
90 133.33 55.47 1.29d = + + − 278 m d =
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Chapter 11, Solution 47.
For 0,t > ( ) ( ) ( )2 2 20 0
1 10 0 6.5 or 3.252 2A A A A Ax x v t a t t x t= + + = + + =
For 2 s,t > ( ) ( ) ( ) ( ) ( )( )2 20 0
1 12 2 0 0 11.7 22 2B B B Bx x v t a t t= + − + − = + + −
or ( )2 25.85 2 5.85 23.4 23.4Bx t t t= − = − +
For ,A Bx x= 2 23.25 5.85 23.4 23.4,t t t= − +
or 22.60 23.4 23.4 0t t− + =
Solving the quadratic equation, 1.1459 and 7.8541 st t= =
Reject the smaller value since it is less than 5 s.
( )a 7.85 st =
( )( )23.25 7.8541A Bx x= = 200 ftx =
( )b ( ) ( )( )0 0 6.5 7.8541A A Av v a t= + = + 51.1 ft/sAv =
( ) ( ) ( )( )0 2 0 11.7 7.8541 2B B Bv v a t= + − = + − 68.5 ft/sBv =
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Chapter 11, Solution 48. Let x be the position relative to point P.
Then, ( ) ( )0 00 and 0.62 mi 3273.6 ftA Bx x= = =
Also, ( ) ( )0 068 mi/h 99.73 ft/s and 39 mi/h 57.2 ft/sA Bv v= = = − = −
(a) Uniform accelerations.
( ) ( )( ) ( )0 02
20 0
21 or 2
A A AA A A A A
x x v tx x v t a t a
t
⎡ ⎤− −⎣ ⎦= + + =
( )( )
( )2
2
2 3273.6 0 99.73 400.895 ft/s
40Aa
⎡ ⎤− −⎣ ⎦= = − 20.895 ft/sAa =
( ) ( )( ) ( )0 02
20 0
21 or 2
B B BB B B B B
x x v tx x v a t a
t
⎡ ⎤− −⎣ ⎦= + + =
( )( )
( )2
2
2 0 3273.6 57.2 420.988 ft/s
42Ba
⎡ ⎤− − −⎣ ⎦= = − 20.988 ft/sBa =
(b) When vehicles pass each other .A Bx x=
( ) ( ) ( ) ( )2 20 0 0 0
1 12 2A A A B B Bx v t a t x v t a t+ + = + +
( ) ( )2 21 10 99.73 0.895 3273.6 57.2 0.9882 2
t t t t+ + − = − + −
20.0465 156.93 3273.6 0t t− − + =
Solving the quadratic equation, 20.7 st = and 3390 s−
Reject the negative value. Then, 20.7 st =
(c) Speed of B.
( ) ( )( )0 57.2 0.988 20.7 77.7 ft/sB B Bv v a t= + = − + − = −
77.7 ft/sBv =
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Chapter 11, Solution 49. Let x be positive downward for all blocks and for point D.
1 m/sAv =
Constraint of cable supporting A: ( ) constantA A Bx x x+ − =
( )( )2 0 or 2 2 1 2 m/sA B B Av v v v− = = = =
Constraint of cable supporting B: 2 constantB Cx x+ =
( )( )2 0 or 2 2 2 4 m/sC B C Bv v v v+ = = − = − = −
(a) 4 m/sC =v
(b) / 2 1B A B Av v v= − = − / 1 m/sB A =v
(c) constant, 0D C D Cx x v v+ = + =
4 m/sD Cv v= − =
/ 4 1D A D Av v v= − = − / 3 m/sD A =v
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Chapter 11, Solution 50. Let x be positive downward for all blocks.
Constraint of cable supporting A: ( ) constantA A Bx x x+ − =
2 0 or 2 and 2A B B A B Av v v v a a− = = =
Constraint of cable supporting B: 2 constantB Cx x+ =
2 0, or 2 , and 2 4B C C B C B Av v v v a a a+ = = − = − = −
Since Cv and Ca are down, Av and Aa are up, i.e. negative.
( ) ( )220 02A A A A Av v a x x⎡ ⎤− = −⎣ ⎦
( )( )
( )( )( )
2 2220
0
0.2 0( ) 0.04 m/s
2 0.52A A
AA A
v va a
x x
− −= = = −
⎡ ⎤ −−⎣ ⎦
20.04 m/sAa =
4C Aa a= − 20.16 m/sCa =
( )( ) 2( ) 2 2 0.04 0.08 m/sB Ab a a= = − = −
( )( )0.08 2 0.16 m/sB Bv a tΔ = = − = − 0.16 m/sBvΔ =
( )( )221 1 0.08 2 0.16 m2 2B Bx a tΔ = = − = − 0.16 mBxΔ =
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Chapter 11, Solution 51.
Let xA, xB, xC, and xD be the displacements of blocks A, B, C, and D relative to the upper supports, increasing downward.
Constraint of cable AB: constantA Bx x+ =
0A Bv v+ = B Av v= −
Constraint of cable BED: 2 constantB Dx x+ =
1 12 0 or 2 2B D D B Av v v v v+ = = − =
Constraint of cable BCD: ( ) ( ) constantC B C Dx x x x− + − =
12 0 or 2 02C B D C A Av v v v v v− − = + − =
(a) Velocity of block A.
1 2 (2)(4)2 A Cv v= − = − 8 ft/sAv = − 8 ft/sAv =
(b) Velocity of block D.
1 4 ft/s2D Av v= = − 4 ft/sDv =
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Chapter 11, Solution 52.
Let xA, xB, xC, xD, and xE be the displacements of blocks A, B, C, and D and cable point E relative to the upper supports, increasing downward.
Constraint of cable AB: constantA Bx x+ =
0A Bv v+ = B Av v= −
0A Ba a+ = B Aa a= −
Constraint of cable BED: 2 constantB Dx x+ =
1 12 0
2 2B D D B Av v v v v+ = = − =
1 12 0
2 2B D D A Aa a a a a+ = = − =
Constraint of cable BCD: ( ) ( ) constantC B C Dx x x x− + − =
2 0 2 0C B D C Av v v v v− − = + =
12 0 2 0
2C B D C Aa a a a a− − = + =
1
4C Aa a= −
Since block C moves downward, vC and aC are positive.
Then, vA and aA are negative, i.e. upward.
Also, vD and aD are negative.
Relative motion: /1
2A D A D Av v v v= − =
/1
2A D A D Aa a a a= − =
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(a) Acceleration of block C.
/ 2/
2 (2)(8)2 3.2 ft/s
5A D
A A D
va a
t= = = =
23.2 ft/sAa = −
210.8 ft/s
4C Aa a= − = 20.8 ft/sCa =
Constraint of cable portion BE: constantB Ex x+ =
0B Ev v+ = 0B Ea a+ =
(b) Acceleration of point E.
23.2 ft/sE B Aa a a= − = = − 23.2 ft/sEa =
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Chapter 11, Solution 53.
Let x be position relative to the right supports, increasing to the left.
Constraint of entire cable: ( )2 constantA B B Ax x x x+ + − =
2 0 2B A A Bv v v v+ = = −
Constraint of point C of cable: 2 constantA Cx x+ =
2 0 2A C C Av v v v+ = = −
(a) Velocity of collar A.
( )( )2 2 300 600 mm/sA Bv v= − = − = − 600 mm/sAv =
(b) Velocity of point C of cable.
( )( )2 2 600 1200 mm/sC Av v= − = − − = 1200 mm/sCv =
(c) Velocity of point C relative to collar B.
/ 1200 300 900 mm/sC B C Bv v v= − = − = / 900 mm/sC Bv =
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Chapter 11, Solution 54.
Let x be position relative to the right supports, increasing to the left.
Constraint of entire cable: ( )2 constant,A B B Ax x x x+ + − =
1 12 0, or , and 2 2B A B A B Av v v v a a+ = = − = −
(a) Accelerations of A and B.
/ /1 2 2 3B A B A A A A B Av v v v v v v= − = − − = −
( )2 610 406.67 mm/s3Av = − = −
( ) ( ) 20A0
406.67 0, or 50.8 mm/s8
A AA A A
v vv v a t a
t− − −− = = = = −
250.8 mm/sAa =
( )1 1 50.82 2B Aa a= − = − − 225.4 mm/sBa =
(b) Velocity and change in position of B after 6 s.
( ) ( )( )0 0 25.4 6B B Bv v a t= + = + 152.5 mm/sBv =
( ) ( ) ( )( )220 0
1 1 25.4 62 2B B B Bx x v t a t− = + = 458 mmBxΔ =
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Chapter 11, Solution 55.
Let x be position relative to left anchor. At the right anchor, .x d=
Constraint of cable: ( ) ( )2 constantB B A Ax x x d x+ − + − =
2 22 3 0 or and 3 3B A A B A Bv v v v a a− = = =
Constraint of point D of cable: ( ) constantA Dd x d x− + − =
0 or and A D D A D Av v v v a a+ = = − = −
(a) Accelerations of A and B.
( ) ( ) ( )0 026 in./s 6 4 in./s3B Av v= = =
( ) ( )220 02A A A A Av v a x x⎡ ⎤− = −⎣ ⎦
( )( )
( ) ( )( )( )
2 2 2220
0
2.4 40.512 in./s
2 102A A
AA A
v va
x x
− −= = = −
⎡ ⎤−⎣ ⎦ 2 0.512 in./sAa =
( ) 23 3 0.512 0.768 in./s2 2B Aa a= = = − 20.768 in./sBa =
(b) Acceleration of point D. ( )0.512D Aa a= − = − − 20.512 in./sDa =
(c) Velocity of block B after 4 s.
( ) ( )( )0 6 0.768 4B B Bv v a t= + = + − 2.93 in./sBv =
Change in position of block B.
( ) ( ) ( )( ) ( )( )220 0
1 16 4 0.768 42 2B B B Bx x v t a t− = + = + − 17.86 in.BxΔ =
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Chapter 11, Solution 56.
Let x be position relative to left anchor. At right anchor .x d=
Constraint of entire cable: ( ) ( )2 constantB B A Ax x x d x+ − + − = 2 3 0B Av v− =
(a) Velocity of A: ( )2 2 123 3A Bv v= = 8.00 in./sAv =
Constraint of point C of cable: constantB B Cx x x+ − = 2 0B Cv v− =
(b) Velocity of C: ( )2 2 12C Bv v= = 24 in./sCv =
Constraint of point D of cable: constantA Cd x d x− + − = 0,A Dv v+ =
(c) Velocity of D: 8.00 in./sD Av v= − = − 8.00 in./sDv =
(d) Relative velocity. / 24 8C A C Av v v= − = − / 16.00 in./sC Av =
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Chapter 11, Solution 57.
Let x be position relative to the anchor, positive to the right.
Constraint of cable: ( ) ( )3 constantB C B C Ax x x x x− + − + − =
4 2 3 0 4 2 3 0C B A C B Av v v a a a− − = − − = (1, 2)
When 0,t = ( )050 mm/s and 100 mm/sB av v= − =
(a) ( ) ( ) ( )( ) ( )( )001 12 3 2 50 3 1004 4C B Av v v⎡ ⎤ ⎡ ⎤= + = − +⎣ ⎦⎣ ⎦ ( )0 50 mm/sCv =
Constraint of point D: ( ) ( ) ( ) constantD A C A C B Bx x x x x x x− + − + − − =
2 2 2 0D C A Bv v v v+ − − =
(b) ( ) ( ) ( ) ( )( ) ( )( ) ( )( )0 0 02 2 2 2 100 + 2 50 2 50D A B Cv v v v= + − = − − ( )0 0Dv =
( ) ( ) 20 0
12C C C Cx x v t a t− = +
(c) ( ) ( ) ( )( )
( )0 0 2
2 2
2 2 40 50 230 mm/s
2C C C
Cx x v t
at
⎡ ⎤− − ⎡ ⎤−⎣ ⎦ ⎣ ⎦= = = −
230 mm/sCa =
Solving (2) for aA
( ) ( )( ) ( )( ) 21 14 2 4 30 2 0 40 mm/s3 3A C Ba a a ⎡ ⎤= − = − − = −⎣ ⎦
240 mm/sAa =
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Chapter 11, Solution 58.
Let x be position relative to the anchor, positive to the right.
Constraint of cable: ( ) ( )3 constantB C B C Ax x x x x− + − + − =
4 2 3 0 and 4 2 3 0C B A C B Av v v a a a− − = − − =
(a) Accelerations of B and C.
At 2 s,t = 420 mm/s and 30 mm/sA Bv v= = −
( ) ( )( ) ( )( )1 12 3 2 30 3 420 300 mm/s4 4C B Av v v ⎡ ⎤= + = − + =⎣ ⎦
( )0 0Cv =
( )0C C Cv v a t= + ( )0 300 0
2C C
Cv v
at
− −= = 2150 mm/sCa =
( ) ( )( ) ( )( ) 21 14 3 4 150 3 270 105 mm/s2 2B C Aa a a ⎡ ⎤= − = − = −⎣ ⎦
2105 mm/sBa =
(b) Initial velocities of A and B.
( )0A A Av v a t= − ( ) ( )( )0 420 270 2 120 mm/sA A Av v a t= − = − = −
( )0 120 mm/sAv =
( )0B B Bv v a t= − ( ) ( )( )0 30 105 2B B Bv v a t= − = − − − ( )0 180 mm/sBv =
Constraint of point E: ( ) ( )2 constantC A E Ax x x x− + − =
3 2 0E A Cv v v− + =
(c) ( ) ( ) ( ) ( )( ) ( )( )0 0 03 2 3 120 2 0 360 mm/sE A Cv v v= − = − − = −
( )0 360 mm/sEv =
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Chapter 11, Solution 59.
Define positions as positive downward from a fixed level.
Constraint of cable. ( ) ( ) ( )2 constantB A C A C Bx x x x x x− + − + − =
3 2 constantC B Ax x x− − =
3 2 0C B Av v v− − =
3 2 0C B Aa a a− − =
Motion of block C.
( ) ( )20 0
0, 3.6 in./s , 18 in./s, 0A A B B Bv a v v a= = − = = =
( ) ( ) ( )0 00
12 6 in./s
3C B Av v v = + =
( ) ( )( ) 21 12 0 2 3.6 2.4 in./s
3 3C B Aa a a = + = + − = −
( )06 1.2C C Cv v a t t= + = −
( ) ( ) 2 20 0
16 0.6
2C C C Cx x v t a t t t− = + = −
(a) Time at vC = 0.
0 6 2.4t= − 2.5 st =
(b) Corresponding position of block C.
( ) ( )( ) ( )( )2
0
16 2.5 2.4 2.5
2C Cx x − = + −
( )07.5 in.C Cx x− =
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Chapter 11, Solution 60.
Define positions as positive downward from a fixed level.
Constraint of cable: ( ) ( ) ( )2 constantB A C A C Bx x x x x x− + − + − =
3 2 constantC B Ax x x− − =
3 2 0C B Av v v− − =
3 2 0C B Aa a a− − =
Motion of block C.
( )0 0,Av = 22.5 in./s ,Aa t= − ( )0 0,Bv = 215 in./sBa =
( ) ( ) ( )0 001 2 03C B Av v v⎡ ⎤= + =⎣ ⎦
( ) 21 12 (15 5 ) in./s3 3C B Aa a a t= + = −
( ) 00t
C C Cv v a dt= + ∫
( )210 15 2.5 in./s3
t t= + −
( ) ( )2 30
1 7.5 0.83333 in.3C Cx x t t− = −
( ) Time at 0Ca v =
( )210 15 2.5 03
t t+ − = 0 and 6 st t= = 6 st =
(b) Corresponding position of block C.
( ) ( )( ) ( )( )2 30
10 7.5 6 0.83333 63C Cx x ⎡ ⎤− = + −⎢ ⎥⎣ ⎦
( )0 30 in.C Cx x− =
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Chapter 11, Solution 61.
Let x be position relative to the support taken positive if downward.
Constraint of cable connecting blocks A, B, and C:
2 2 constant, 2 2 0A B C A B Cx x x v v v+ + = + + =
2 2 0A B Ca a a+ + = (1)
Constraint of cable supporting block D:
( ) ( ) constant, 2 0D A D B D A Bx x x x v v v− + − = − − =
2 0D B Aa a a− − = (2)
Given: / 120 or 120C B C B C Ba a a a a= − = − = − (3)
Given: / 220 or 220D A D A D Aa a a a a= − = = + (4)
Substituting (3) and (4) into (1) and (2),
( )2 2 120 0 or 2 3 120A B B A Ba a a a a+ + − = + = (5)
( )2 220 0 or 440A A B A Ba a a a a+ − − = − = − (6)
Solving (5) and (6) simultaneously, 2 2240 mm/s and 200 mm/sA Ba a= − =
From (3) and (4), 2 280 mm/s and 20 mm/sC Da a= = −
(a) Velocity of C after 6 s.
( ) ( )( )0 0 80 6C C Cv v a t= + = + 480 mm/sCv =
(b) Change in position of D after 10 s.
( ) ( ) ( )( )220 0
1 10 20 10 1000 mm2 2D D D Dx x v t a t− = + = + − = −
1.000 mDxΔ =
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Chapter 11, Solution 62.
Let x be position relative to the support taken positive if downward. Constraint of cable connecting blocks A, B, and C:
2 2 constant,A B Cx x x+ + = 2 2 0,A B Cv v v+ + = 2 2 0A B Ca a a+ + =
( ) ( ) ( ) ( ) ( ) ( )0 0 0 00 00, ,A B C A B Cv v v x x x= = = = = ( ) ( )/ /0 00, 0B A B Ax v= =
( ) ( ) ( )2 2/ / / / /0 0
2B A B A P A B A B Av v a x x⎡ ⎤− = −⎣ ⎦
( )2/ /0 2 0B A B A B Av a x x− = − −
( ) ( )2 2
/ 2/
40 10 mm/s2 2 160 80
B AB A
B A
va
x x= = =
− −
( ) ( ) 2 2/ / / / /0 0
1 10 02 2B A B A B A B A B Ax x v t a t a t= + + = + +
( ) ( )/2
/ /
2 2 2 160 80, or 4 s
10B A B A
B A B A
x x xt t
a a− −
= = = =
( ) ( ) 20 0
12A A A Ax x v t a t− = +
(a) ( ) ( ) ( )
( )0 0
2 2
2 2 80 0
4A A A
Ax x v t
at
⎡ ⎤− − −⎣ ⎦= = 210 mm/sAa =
/ 10 10B A B Aa a a= + = + 220 mm/sBa =
( ) ( )( ) ( )( )2 2 2 20 2 10 60 mm/sC B Aa a a ⎡ ⎤= − + = − + = −⎣ ⎦
( ) ( )00
300 0 5 s60
C CC C C
C
v vv v a t t
a− − −= + = = =
−
Constraint of cable supporting block D: ( ) ( ) constant, 2 0D A D B D A Bx x x x v v v− + − = − − =
( ) ( )1 12 0, 10 20 15 mm/s2 2D A B D A Ba a a a a a− − = = + = + =
(b) ( ) ( ) ( )( )220 0
1 10 15 52 2D D D Dx x v t a t− = + = + 187.5 mmDxΔ =
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Chapter 11, Solution 63.
curvea t−
1 212 m/s, 8 m/sA A= − =
(a) curvev t−
6 4 m/sv = −
( )0 6 1 4 12v v A= − = − − − 8 m/s=
10 4 m/sv = −
(b) 14 10 2 4 8v v A= + = − + 14 4 m/sv =
3 416 m, 4 mA A= = −
5 616 m, 4 mA A= − = −
7 4 mA =
(a) curvex t−
0 0x =
4 0 3 16 mx x A= + =
6 4 4 12 mx x A= + =
10 6 5 4 mx x A= + = −
12 10 6 8 mx x A= + = −
(b) 14 12 7x x A= + 14 4 mx = −
Distance traveled:
0 4 s,t≤ ≤ 1 16 0 16 md = − =
4 s 12 s,t≤ ≤ 2 8 16 24 md = − − =
12 s 14 s,t≤ ≤ ( )3 4 8 4 md = − − − =
Total distance traveled: 16 24 4d = + + 44 md =
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Chapter 11, Solution 64.
(a) Construction of the curves.
curvea t−
1 212 m/s, 8 m/sA A= − =
curvev t−
0 8 m/sv =
( )6 0 1 8 12 4 m/sv v A= + = + − = −
10 6 4 m/sv v= = −
14 10 2 4 8 4 m/sv v A= + = − + =
3 416 m, 4 mA A= = −
5 616 m, 4 mA A= − = −
7 4 mA =
curvex t−
0 0x =
4 0 3 16 mx x A= + =
6 4 4 12 mx x A= + =
10 6 5 4 mx x A= + = −
12 10 6 8 mx x A= + = −
14 12 7 4 mx x A= + = −
(b) Time for 8 m.x >
From the x t− diagram, this is time interval 1 2to .t t
Over 0 6 s,t< < 8 2dx
v tdt
= = −
continued
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Integrating, using limits 0x = when 0t = and 8 mx = when
1t t=
8 2 21 10 0
8 or 8 8t
x t t t t = − = −
or 21 18 8 0t t− + =
Solving the quadratic equation,
( ) ( )( )( )( )( )
2
1
8 8 4 1 84 2.828 1.172 s and 6.828 s
2 1t
± −= = ± =
The larger root is out of range, thus 1 1.172 st =
Over 6 10,t< < ( )12 4 6 36 4x t t= − − = −
Setting 8,x = 2 28 36 4 or 7 st t= − =
Required time interval: ( )2 1 5.83 st t− =
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Chapter 11, Solution 65.
The a–t curve is just the slope of the v–t curve.
0 10 s,t< < 0a =
10 s < 18 s,t < 218 61.5 ft/s
18 10a
−= =
−
18 s < 30 s,t < 218 18
3 ft/s30 18
a− −
= = −−
30 s < 40 st < 0a =
Points on the x–t curve may be calculated using areas of the v–t curve.
1 (10)(6) 60 ftA = =
21
(6 18)(18 10) 96 ft2
A = + − =
31
(18)(24 18) 54 ft2
A = − =
41
( 18)(30 24) 54 ft2
A = − − = −
5 ( 18)(40 30) 180 ftA = − − = −
0 48 ftx = −
01 0 1 12 ftx x A= + =
81 10 2 108 ftx x A= + =
24 18 3 162 ftx x A= + =
30 24 4 108 ftx x A= + =
40 30 5 72 ftx x A= + = −
continued
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(a) Maximum value of x.
Maximum value of x occurs
When 0,v = i.e. 24 s.t =
max 162 ftx =
(b) Time s when 108 ft.x =
From the x–t curve,
18 s and 30 st t= =
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Chapter 11, Solution 66.
Data from problem 11.65: 0 48 ftx = −
The a–t curve is just the slope of the v–t curve.
0 10 s,t< < 0a = !
10 s < 18 s,t < 218 61.5 ft/s
18 10a
−= =−
!
18 s < 30 s,t < 218 18
3 ft/s30 18
a− −= = −
−!
30 s < 40 s,t < 0a = !
Points on the x–t curve may be calculated using areas of the v–t
curve. !
1 (10)(6) 60 ftA = =
21
(6 18)(18 10) 96 ft2
A = + − =
31
(18)(24 18) 54 ft2
A = − =
41
( 18)(30 24) 54 ft2
A = − − = −
5 ( 18)(40 30) 180 ftA = − − = −
0 48 ftx = − !
0 01 1 12 ftx x A= + = !
1018 2 108 ftx x A= + = !
24x = 18 3x A+ = 162 ft !
30x = 24 4x A+ = 108 ft !
40 30 5 72 ftx x A= + = − !
continued
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(a) Total distance traveled during 0 30 st≤ ≤ .
For 0 24 st≤ ≤ 1 24 0 210 ftd x x= − =
For 24 s 30 st≤ ≤ 2 30 24 54 ftd x x= − =
Total distance. 1 2d d d= + 264 ftd = !
(b) Values of t for which 0.x =
In the range 0 10 st≤ ≤
0 0 48 6x x v t t= + = − +
Set 0.x = 148 6 0t− + = 1 8 st = !
In the range 30 s 40 s,t< <
30 30 ( 30)x x v t= + −
108 ( 18)( 30)t= + − −
648 18t= −
Set 0.x = 2648 18 0t− = 2 36 st = !
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Chapter 11, Solution 67.
Sketch v t− curve as shown. Label areas 1 2, ,A A and 3A
( )( )1 3 20 60 in.A = =
1 12 in./sv at tΔ = =
( ) 22 1 1
1 in.2
A v t t= Δ =
( )( ) ( )3 1 1 120 2 20 in.A v t t t= Δ − = −
Distance traveled: 12 ft 144 in.xΔ = =
( )21 1 1total area, 144 60 2 20x t t tΔ = = + + −
or 21 140 84 0t t− + =
( )( )( )( )( )2
140 40 4 1 84
2.224 s and 37.8 s2 1
t± −
= =
Reject the larger root. 1 2.224 st =
12 4.45 in./sv tΔ = =
max 3 3 4.45v v= + Δ = + max 7.45 in./sv =
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Chapter 11, Solution 68.
Let x be the altitude. Then v is negative for decent and a is positive for deceleration.
Sketch the v t− and x t− curves using times 1 2, t t and 3t as shown. Use constant slopes in the v t− curve for the constant acceleration stages.
Areas of v t− curve:
( )1 1 11 180 44 112 ft2
A t t= − + = −
2 244A t= −
( )3 3 31 44 222
A t t= − = −
Changes in position: 1 1800 1900 100 ftxΔ = − = −
2 100 1800 1700 ftxΔ = − = −
3 0 100 100 ftxΔ = − = −
Using i ix AΔ = gives 1100 0.893 s112
t −= =−
21700 38.64 s
44t −= =
−
3100 4.55 s22
t −= =−
(a) Total time: 1 2 3 44.1 st t t+ + =
(b) Initial acceleration. ( ) ( )44 1800.893
vat
− − −Δ= =Δ
2152.3 ft/sa =
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Chapter 11, Solution 69.
Sketch the v t− curve
Data: 0 64 km/h 17.778 m/sv = =
32 4.8 km 4.8 10 mx = = ×
1 32 km/hr 8.889 m/sv = =
3 31 4.8 10 800 4.0 10 mx = × − = ×
2 450 st =
(a) Time 1t to travel first 4 km.
( ) ( )31 1 0 1 1 1
1 14.0 10 17.778 8.8892 2
x A v v t t= × = = + = + 1 300 st =
(b) Velocity 2.v
( )( ) ( )( )2 1 2 1 2 2 1 1 21 1800 450 3002 2
x x A v v t t v v− = = = + − = + −
2 1 10.667 mv v+ =
2 10.667 8.889v = − 2 1.778 m/sv =
(c) Final deceleration.
22 112
2 1
1.778 8.889 0.0474 m/s450 300
v vat t
− −= = = −− −
212 0.0474 m/sa =
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Chapter 11, Solution 70.
10 2010 min 20 s 0.1722 h60 3600
= + =
Sketch the v t− curve
60
25
35
a
b
c
ta
ta
ta
=
=
=
( )( ) ( )1 1 11 1 1 160 60 25 60 1800 312.52 2a bA t t t t
a a= − − = − −
But 1 5 miA =
1160 2112.5 5ta
− = (1)
( )2 1 1135 0.1722 35 6.0278 35 612.5cA t t ta
= − − = − −
But 2 8 5 3 miA = − =
1135 612.5 3.0278ta
+ = (2)
11Solving equations (1) and (2) for and ,ta
31 85.45 10 h 5.13 mint −= × =
6 21 60.23 10 h /mia
−= ×
( )( )
( )
33 2
2
16.616 10 528016.616 10 mi/h
3600a
×= × = 26.77 ft/sa =
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Chapter 11, Solution 71.
Sketch the curve as shown a t−
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
0 120 ft/s, 6 ft/sv v= = −
( )1 1
2 1
61 40 6 172
A t
1A t t
= −
= − − = −
1 0 1v v A A2= + +
1 16 20 6 17t t= − −
(a) 1 0.6087 st = 1 0.609 st =
2 1.4 st =
2 1 0.7913 st t− =
( )( )1 3 6 1.4 8.4 ft/sA A+ = − = −
( )( )2 17 0.6087 10.348 ft/sA = − = −
2 0 1 3 2 20 8.4 10.348v v A A A= + + + = − − 2 1.252 ft/sv =
(b) ( )2 0 0 2 1 3 13 2 2 by moment-area methodx x v t A A x A x= + + + +
( )0 2 1 3 2 2 2 11 102 3
v t A A t A t t⎛ ⎞ ⎛= + + + + −⎜ ⎟ ⎜⎝ ⎠ ⎝
⎞⎟⎠
( )( ) ( ) ( ) ( )1 0.60870 20 1.4 8.4 1.4 10.348 1.42 3
⎛ ⎞ ⎛ ⎞= + − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
2 9.73 ftx =
)
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
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Chapter 11, Solution 72.
Note that 1 5280
mile 660 ft8 8
= =
Sketch v t− curve for first 660 ft.
Runner A: 1 24 s, 25 4 21 st t= = − =
( )( ) ( )1 max max
14 2
2 A AA v v= =
( )2 max21 AA v=
1 25280 ft
= 660 ft8
A A x+ = ∆ =
( ) ( )max max23 660 or 28.696 ft/sA Av v= =
Runner B: 1 25 s, 25.2 5 20.2 st t= = − =
( )( ) ( )1 max max
15 2.5
2 B BA v v= =
( )2 max20.2 BA v=
1 2 660 ftA A x+ = ∆ =
( ) ( )max max22.7 660 or 29.075 ft/sB Bv v= =
Sketch v t− curve for second 660 ft. 3 30.3v a t t∆ = =
23 max 3 3 3 max 3
1660 or 0.15 660 0
2A v t vt t v t= − ∆ = − + =
( ) ( )( )( )( )( ) ( )
22max max
3 max max
4 0.15 6603.3333 396
2 0.15
v vt v v
± − = = ± −
Runner A: ( )max 28.696,A
v = ( )3 164.57 s and 26.736 sA
t =
Reject the larger root. Then total time (a) 25 26.736 51.736 sAt = + =
51.7 sAt =
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Runner B: ( )max 29.075,B
v = ( )3 167.58 s and 26.257 sB
t =
Reject the larger root. Then total time 25.2 26.257 51.457 sBt = + =
51.5 s Bt =
Velocity of A at 51.457 s:t =
( )( )1 28.696 0.3 51.457 25 20.759 ft/sv = − − =
Velocity of A at 51.736 s:t =
( )( )2 28.696 0.3 51.736 25 20.675 ft/sv = − − =
Over 51.457 s 51.736 s, runner covers a distance t A x≤ ≤ ∆
(b) ( ) ( )( )ave1
20.759 20.675 51.736 51.4572
x v t∆ = ∆ = + − 5.78 ftx∆ =
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Chapter 11, Solution 73.
Sketch the v t− curves.
At 12 min 720 s,t = =
( )( )truck
bus
bus
19.44 720 14000 m
14000 1200 15200 marea under curve
x
xx v t
= =
= + == −
( )( ) ( )( )1 11 120 27.78 720 27.78 152002
t t− + − =
1 225.8 st =
(a) When bus truck ,x x= areas under the v t− curves are equal.
( )( ) ( )1 2 1 21 27.78 120 27.78 19.442
t t t t− + − =
With 1 225.8 s,t = 2 576 st =
( )( )truck 19.44 576 11200 mx = = truck 11.20 kmx =
(b) 0bus
1
27.78 0120 225.8 120
v vat
− −= =− −
2bus 0.262 m/sa =
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Chapter 11, Solution 74.
( )0 32 km/h 8.889 m/s
24 km/h 6.667 m/sA
B
v
v
= =
= =
Sketch the v t− curves.
( )( )
( )( ) ( )
( )( )
1
2 /
/
1 20
10
6.667 45 300 m
1 12.222 45 452 250 22.5
A B
A B
A A
B B
A
A v
v
x x A A
x x A
= =
= +
= +
= + +
= +
( )/ / 20B A B Ax x A= −
( )b /0 60 50 22.5 A Bv= − − / 0.444 m/sA Bv =
/ 6.667 0.444 7.111 m/sA B A Bv v v= + = + =
(a) ( )0 7.111 8.889
45A A
Av v
at
− −= = 20.0395 m/sAa = −
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Chapter 11, Solution 75.
( )( )
0
0
22 mi/h 32.267 ft/s
13 mi/h 19.067 ft/sA
B
v
v
= =
= =
Sketch the v t− curves.
Slope of v t− curve for car A.
( )( )
2
1
1
2
13.2 0.14 ft/s
13.2 94.29 s0.141 13.2 94.29 622.3 m2
at
t
A
= − = −
= =
= =
( )( )
10
1 20
B B
A A
x x A
x x A A
= +
= + +
( ) ( )/ 2 20 0 , or 0B A B A B Ax x x x x A d A= − = − − = −
2d A= 622 md =
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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 76.
Construct the a t− curves for the elevator and the ball.
Limit on 1A is 24 ft/s. Using 1 4A t=
2 24 24 6 st t= =
Motion of elevator.
For 10 6 s,t≤ ≤ ( ) ( )0 00 0E Ex v= =
Moment of 1A about 1:t t= 211 14 2
2tt t=
( ) ( ) 2 21 1 10 0 2 2E E Ex x v t t t= + + =
Motion of ball. At 2,t = ( ) ( )0 040 ft 64 ft/sB Bx v= =
For 1 2 s,t > ( )2 132.2 2 ft/sA t= − −
Moment of 2A about 2 :t t= ( ) ( )211 1
232.2 2 16.1 22
tt t−⎛ ⎞− − = − −⎜ ⎟⎝ ⎠
( ) ( ) ( ) ( )
( ) ( )
21 10 0
21 1
2 16.1 2
40 64 2 16.1 2
B B Bx x v t t
t t
= + − − −
= + − − −
When ball hits elevator, B Ex x=
( ) ( )2 21 1 1
21 1
40 64 2 16.1 2 2 or
18.1 128.4 152.4 0
t t t
t t
+ − − − =
− + =
Solving the quadratic equation, 1 1.507 s and 5.59 st =
The smaller root is out of range, hence 1 5.59 st =
Since this is less than 6 s, the solution is within range.
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Chapter 11, Solution 77.
Let x be the position of the front end of the car relative to the front end of the truck.
Let dx
vdt
= and dv
adt
= .
The motion of the car relative to the truck occurs in 3 phases, lasting t1, t2, and t3 seconds, respectively.
Phase 1, acceleration. 21 2 m/sa =
Phase 2, constant speed. 2 90 km/h 54 km/hv = −
36 km/h = 10 m/s=
Phase 3, deceleration. 23 8 m/sa = −
Time of phase 1. 21
1
0 10 05 s
2
vt
a
− −= = =
Time of phase 3. 23
2
0 0 101.25 s
8
vt
a
− −= = =
Sketch the a t− curve.
Areas: 1 1 2 10 m/sA t v= =
3 3 10 m/sA t v= = −
Initial and final positions.
0 30 16 46 mx = − − = −
30 5 35 mfx = + =
Initial velocity. 0 0v =
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Final time. 1 2 3ft t t t= + +
0 0f f i ix x v t A t= + + ∑
1 11
2ft t t= −
25 1.25 2.5t= + + −
23.75 t= +
2 31
0.625 s2
t t= =
( )( ) ( )( )235 46 0 10 3.75 10 0.625t= − + + + + −
249.75
4.975 s10
t = =
1 2 3 11.225 sft t t t= + + =
Total time. 11.23 sft =
1 2 9.975 st t+ =
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Chapter 11, Solution 78.
Let x be the position of the front end of the car relative to the front end of the truck.
Let dxv
dt= and dv
adt
= .
The motion of the car relative to the truck occurs in two phases, lasting t1 and t2 seconds, respectively.
Phase 1, acceleration. 21 2 m/sa =
Phase 2, deceleration. 22 8 m/sa = −
Sketch the a–t curve.
Areas: 1 12A t=
2 28A t= −
Initial and final positions
0 30 16 46 mx = − − = −
30 5 35 mfx = + =
Initial and final velocities.
0 0fv v= =
0 1 2fv v A A= + +
1 20 0 2 8t t= + −
1 24t t=
0 0f f i ix x v t A t= + + ∑
1 2 1 21
32
t t t t= + =
2 21
2t t=
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( )( ) ( )2 2 2 2
135 46 0 2 4 3 8
2t t t t
= − + + + −
2281 20 t=
2 2.0125 st =
1 8.05 st =
1 2 10.0625 s.ft t t= + =
Maximum relative velocity.
( )( )1 1 2 8.05 16.10 m/smv a t= = =
60.0 km/hmv =
Maximum velocity relative to ground.
max 54 60.0Tv v v= + = +
max 112.0 km/hv = !
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Chapter 11, Solution 79.
Sketch acceleration curve.
Let jerk dajdt
= =
Then, ( )maxa j t= Δ
( ) ( )
( )
1 max max
2
1 22
A a t a t
j t
= Δ = Δ
= Δ
0 1 2
1 2
2 1
0 0fv v A A
A AA A
= + −
= + −=
( ) ( )( ) ( )( ) ( ) ( )
( )( )
0 1 2
3 3 3
3 3
4 3
0 3 2
0.36 0.49322 2 1.5
x v t A t A t
j t j t j t
xtj
Δ = Δ + Δ − Δ
= + Δ − Δ = Δ
ΔΔ = = =
(a) Shortest time: ( )( )4 4 0.4932 1.973 stΔ = =
(b) Maximum velocity: ( )2max 0 1 0v v A j t= + = + Δ
( )( )21.5 0.4932 0.365 m/s= =
Average velocity: ave0.36 0.1825 m/s
4 1.973xvt
Δ= = =Δ
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Chapter 11, Solution 80.
Sketch the a t− curve.
From the jerk limit, ( )1 maxj t aΔ = or ( ) max1
1.25 5 s.0.25
atj
Δ = = =
( )( )11 5 1.25 3.125 m/s2
A = =
( )( )max 1 2
2 max 1
22
max
32 km/hr 8.889 m/s 2
2 8.889 2 3.125 2.639 m/s
2.639 2.111 s1.25
v A A
A v A
Ata
= = = +
= − = − =
Δ = = =
Total distance is 5 km 5000 m.= Use moment-area formula.
( ) ( )
( )0 0 1 2 1 2 1 2 1 2
max 1 2
1 12 22 2
0 0 2
f f f
f
x x v t A A t t t A A t t
v t t t
⎛ ⎞ ⎛ ⎞= + + + − Δ − Δ − + Δ + Δ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= + + − Δ − Δ
(a) ( )( )1 2max
50002 2 5 2.111 10 2.111 562.5 575 s8.889
ff
xt t t
v= Δ + Δ + = + + = + + =
9.58 minft =
(b) ave5000 8.70 m/s575
f
f
xv
t= = = ave 31.3 km/hv =
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Chapter 11, Solution 81.
Indicate areas 1 2and A A on the a t− curve.
( )11 0.6 0.1 m/s2 3
TA T= =
( )21 20.6 0.2 m/s2 3
TA T= =
By moment-area formula,
( )
( )( )
0 1 2
2 2 2 2
2 2
7 49 9
7 8 15 140 090 90 90 6
40 6 240 s
x v t A T A T
T T T T
T
⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= + + = =
= =
(a) 15.49 sT =
max 0 1 2 0 0.1 0.2 0.3v v A A T T T= + + = + + =
(b) max 4.65 m/sv =
Indicate area 3 4 and A A on the a t− curve.
( )
( )
1 3
4
10.1 0.6 0.052 6
1 0.45 0.03752 6
TA T A T
TA T
= = =
= =
(c) 0 1 3 4 0.1875v v A A A T= + + + = 2.90 m/s v =
By moment-area formula,
( ) ( ) ( )
0 1 3 4
2
2 2 12 2 9 3 6 3 6
50 0.1 0.05 0.0375 0.03541718 9 18
T T T T Tx v A A A
T T TT T T T
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + ⋅ + ⋅⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞= + + + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
( )( )20.035417 15.49= 8.50 m x =
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Chapter 11, Solution 82.
Divide the area of the a t− curve into the four areas 1 2 3 4, , and .A A A A
( )( )
( )( )
( )( )
( )( )
1
2
3
4
2 3 0.2 0.4 m/s35 0.2 1 m/s
1 5 2.5 0.1 0.375 m/s21 2.5 0.1 0.125 m/s2
A
A
A
A
= =
= =
= + =
= =
(a) 0Velocities: 0v =
0.2 0 1 2v v A A= + + 0.2 1.400 m/s v =
0.3 0.2 3v v A= + 0.3 1.775 m/s v =
0.4 0.3 4v v A= + 0.4 1.900 m/s v =
Sketch the v t− curve and divide its area into 5 6 7, , and A A A as shown.
0.3 0.4 0.40.3 or 0.3x t tdx x vdt x vdt= − = = −∫ ∫ ∫
At 0.3 s,t = ( )( )0.3 50.3 1.775 0.1x A= − −
(b) With ( )( )52 0.125 0.1 0.00833 m3
A = = 0.3 0.1142 mx =
At 0.2 s,t = ( )0.2 5 6 70.3x A A A= − + −
With ( )( )5 62 0.5 0.2 0.06667 m3
A A+ = =
and ( )( )7 1.400 0.2 0.28 mA = =
0.2 0.3 0.06667 0.28x = − − 0.2 0.0467 m x = −
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Chapter 11, Solution 83.
Approximate the a t− curve by a series of rectangles of height ,ia each with its centroid at .it t= When equal widths of 0.25 stΔ = are used, the values of and i it a are those shown in the first two columns of the table below.
At 2 s,t = ( )2
0 00 iv v adt v a t= + ≈ + Σ Δ∫
( )( )0 iv a t≈ + Σ Δ
(a) ( )( )00 7.650 0.25v≈ − 0 1.913 ft/s v =
Using moment-area formula,
( ) ( )( )
( )( )( )
20 0 0 00
0 0
2
2
i i i i
i i
x x v t a t t dt x v t a t t
x v t a t t
= + + − ≈ + + Σ − Δ
≈ + + Σ − Δ
∫
(b) ( )( ) ( )( )0 1.913 2 11.955 0.25≈ + − 0.836 ft x =
it ia 2 it− ( )2i ia t−
( )s ( )2ft/s ( )s ( )ft/s
0.125 3.215− 1.875 6.028− 0.375 1.915− 1.625 3.112− 0.625 1.125− 1.375 1.547− 0.875 0.675− 1.125 0.759− 1.125 0.390− 0.875 0.341− 1.375 0.205− 0.625 0.128− 1.625 0.095− 0.375 0.036− 1.875 0.030− 0.125 0.004−
Σ ( )27.650 ft/s− ( )11.955 ft/s−
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Chapter 11, Solution 84.
Approximate the a t− curve by a series of rectangles of height ,ia each with its centroid at .it t= When equal widths of 2 stΔ = are used, the values of and i it a are those shown in the first two columns of table below.
(a) At 8 s,t = ( )88 0 0 0 iv v adt a t= + ≈ + Σ Δ∫
( )( )ia t= Σ Δ
Since 8 s,t = only the first four values in the second column are summed:
217.58 13.41 10.14 7.74 48.87 ft/siaΣ = + + + =
( )( )8 48.87 2v = 8 97.7 ft/s v =
(b) At 20 s,t = ( ) ( )( )2020 0 20 0 20o ix v t a t dt a t t= + − = + Σ − Δ∫
( )( )990.1 2= 20 1980 ft x =
it ia 20 it− ( )20i ia t−
( )s ( )2ft/s ( )s ( )ft/s
1 17.58 19 334.0 3 13.41 17 228.0 5 10.14 15 152.1 7 7.74 13 100.6 9 6.18 11 68.0 11 5.13 9 46.2 13 4.26 7 29.8 15 3.69 5 18.5 17 3.30 3 9.9 19 3.00 1 3.0 Σ ( )990.1 ft/s
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Chapter 11, Solution 85.
The given curve is approximated by a series of uniformly accelerated motions.
For uniformly accelerated motion,
( )2 2
2 2 2 12 1 2 12 or
2v vv v a x x x
a−− = − Δ =
( )2 1 2 1v v a t t− = − or 2 1v vta−Δ =
For the regions shown above,
(a) ( ) 3.19 s t t= Σ Δ =
(b) Assuming 0 0,x = ( )0 62.6 m x x x= + Σ Δ =
Region ( )1 m/sv ( )2 m/sv ( )2m/sa ( )mxΔ ( )stΔ
1 32 30 3− 20.67 0.667 2 30 25 8− 17.19 0.625 3 25 20 11.5− 9.78 0.435 4 20 10 13− 11.54 0.769 5 10 0 14.5− 3.45 0.690 Σ 62.63 3.186
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Chapter 11, Solution 86.
Use dva vdx
= noting that dvdx
= slope of the given curve.
Slope is calculated by drawing a tangent line at the required point, and using two points on this line to
determine and .x vΔ Δ Then, .dv vdx x
Δ=Δ
(a) When 0.25,x =
1.4 m/sv = from the curve
1m/s and 0.25m from the tangent line v xΔ = Δ =
( )( )11 4 s 1.4 42.5
dv adx
−= = = 25.6 m/s a =
(b) When 2.0 m/s,v = 0.5mx = from the curve.
1 m/s and 0.6m from the tangent line.v xΔ = Δ =
( )( )11 1.667s , 2 1.6670.6
dv adx
−= = = 23.33 m/s a =
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Chapter 11, Solution 87.
The a t− curve for uniformly accelerated motion is shown. The area of the rectangle is
.A at=
Its centroid lies at 1 .2
t t=
By moment-area formula,
( ) ( )0 0 0 012
x x v A t t x v t at t⎛ ⎞= + + − = + + ⎜ ⎟⎝ ⎠
20 0
12
x v t at= + +
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Chapter 11, Solution 88.
From the curve,a t− ( )( )1 2 6 12 m/sA = − = −
( )( )2 2 2 4 m/sA = =
Over 6 s 10 s,t< < 4 m/sv = −
0 1 0 0, or 4 12, or 8 m/sv v A v v= + − = − = By moment-area formula,
12 0 0 moment of shaded area about 12sx x v t t= + + =
( )( ) ( )( ) ( )( )12 0 8 12 12 12 3 4 12 11x = + + − − + − 12 8 m x = −
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Chapter 11, Solution 89.
(a) 0.2s.T =
( )( )12 24 0.2 3.2 ft/s3
A = − = −
( )( )2 1
1
24 0.2
24 4.8
A t
t
= − −
= − +
0
10 90 3.2 24 4.8fv v A
t
= + Σ
= − − +
1 3.8167 st =
2 86.80 ft/sA = −
1 3.6167 st T− =
By moment-area formula, 1 0 0 1 moment of areax x v t= + +
( )( ) ( ) ( ) ( )13 3.61670 90 3.8167 3.2 0.2 3.6167 86.808 2
x⎡⎛ ⎞ ⎛ ⎞= + + − + + −⎜ ⎟ ⎜ ⎟⎢⎝ ⎠ ⎝ ⎠⎣
1 174.7 ft x = (b) 0.8 s.T =
( )( )
( )( )
1
2 1 1
0 1 1
2 24 0.8 12.8 ft/s, 3
24 0.8 24 19.2
or 0 90 12.8 24 19.2, 4.0167 sf
A
A t t
v v A t t
= − = −
= − − = − +
= + Σ = − − + =
1 23.2167s 77.2 ft/st T A− = = −
By moment-area formula,
( )( ) ( ) ( ) ( )13 3.21670 90 4.0167 12.8 0.8 3.2167 77.28 2
x ⎡ ⎤ ⎛ ⎞= + + − + + − ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠
1 192.3 ft x =
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Chapter 11, Solution 90.
Data from Prob. 65
0 048 ft, 6 ft/sx v= − =
The a – t curve is just the slope of the v – t curve.
0 10 s,t< < 0a = !
10 s < < 18 s,t 18 6
1.5 ft/s18 10
a−= =−
!
18 s 30 s,t< < 18 18
3 ft/s30 18
a− −= = −
−!
30 s < < 40 st 0a = !
0 0 i ix x v t A t= + + ∑
(a) Position when t = 20 s.
( )( )1 18 10 1.5 12 ft/sA = − =
1 20 14 6st = − =
( )( )2 2 3 6 ft/sA = − = −
2 20 19 1 st = − =
( )( ) ( )( ) ( )( )20 48 6 20 12 6 6 1x = − + + + −
20 138 ftx = !
(b) Maximum value of position coordinate.
x is maximum where 0.v =
From velocity diagram, 24 smt =
( )( )1 18 10 1.5 12 ft/sA = − =
( )1 24 14 10 st = − =
( )( )2 24 18 3 18 ft/sA = − − = −
( )2 24 21 3 st = − =
( )( ) ( )( ) ( )( )48 6 24 12 10 18 3mx = − + + + −
162 ftmx = !
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Chapter 11, Solution 91.
( )21= +x t ( ) 2
4 1−= +y t
( )2 1= = +&xv x t ( ) 38 1
−= = − +&yv y t
2= =&x xa v ( ) 424 1
−= = +&y ya v t
Solve for (t + 1)2 from expression for x. (t + 1)2 = x
Substitute into expression for y. 4
yx
=
Then, 4xy =
This is the equation of a rectangular hyperbola.
(a) t = 0. 2 m/s, 8 m/sx yv v= = −
( ) ( )2 22 8 8.25 m/sv = + − =
1 8tan 76.0
2θ − −= = − °
8.25 m/s=v 76.0 °
( ) ( )
2 2
2 2 2
1
2 m/s , 24 m/s
2 24 24.1 m/s
24tan 85.2
2
x ya a
a
θ −
= =
= + =
= = °
224.1 m/s=a 85.2°
(b) 1
s.2
t = 3 m/s,xv = 2.37 m/s= −yv
( )2 23 (2.37) 3.82 m/sv = + =
1 2.37tan 38.3
3θ − − = = − °
3.82 m/s=v 38.3 °
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2 m/s,xa = 24.74 m/sya =
2 2 22 4.74 5.15 m/sa = + =
1 4.74tan 67.2
2θ − = = °
25.15 m/s=a 67.2°
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Chapter 11, Solution 92.
Let ( )2 3 29 18 9 18u t t t t t t= − + = − +
Then, 2
223 18 18, and 6 18du d ut t t
dt dt= − + = −
6 0.8 mx u= − 4 0.6 my u= − +
0.8dx dudt dt
= − 0.6dy dudx dt
= +
0.6 0.75 constant0.8
= = − = − =dydtdxdt
dydx
Since dydx
does not change, the path is straight.
(a) At 2 s,t = 2
26, and 6.= − = −du d udt dt
( )( ) ( )( )0.8 6 4.8 m/s, 0.6 6 3.6 m/s= = − − = = = − = −x ydx dyv vdt dt
( )( ) ( )( )2
2 22 0.8 6 4.8 m/s , 0.6 6 3.6 m/s= = − − = = − = −x y
d xa adt
6.0 m/s=v 36.9 ,° 26.0 m/s=a 36.9 °
(b) At 3 s,t = 2
29, and 0du d udt dt
= − =
( )( ) ( )( )0.8 9 7.2 m/s, 0.6 9 5.4 m/sx yv v= − − = = − = −
0, 0x ya a= =
9.0 m/s=v 36.9 ,° 0 =a
(c) At 4 s,t = 2
26, and 6du d udt dt
= − =
( )( ) ( )( )0.8 6 4.8 m/s, 0.6 6 3.6 m/s= − − = = − = −x yv v
( )( ) ( )( )2 20.8 6 4.8 m/s , 0.6 6 3.6 m/sx ya a= − = − = =
6.0 m/s=v 36.9 ,° 26.0 m/s=a 36.9°
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Chapter 11, Solution 93.
Substitute the given expressions for x and y into the given equation of the ellipse, and note that the equation is satisfied.
( )( ) ( )
( ) ( )
22 2 2
2 2
2 2 2
2 2
16cos 16cos 4 9sin4 3 4 2 cos 3 2 cos
4cos 4cos 1 3sin 4 4cos cos 12 cos 2 cos
t tx y tt t
t t t t tt t
π π ππ π
π π π π ππ π
− ++ = +
− −
− + + − += = =− −
Calculate x& and y& by differentiation.
( )( )( )
( ) ( )
( )( )
( )( )
( )
2 2
2 2
4cos 2 sin4 sin 6 sin2 cos 2 cos 2 cos
3sin sin 3 2cos 13 cos2 cos 2 cos 2 cos
t tt txt t t
t t ttyt t t
π π ππ π π ππ π π
π π π π ππ ππ π π
−− −= − =− − −
−= − =
− − −
&
&
(a) When 0 s,t = 0 and 3 ,x y π= =& & 9.42 m/s=v
(b) When 1 s,3
t = ( )
( )
32
212
6 4 3, 032
x yπ
π−
= = − − =−
& & 7.26 m/s=v
(c) When 1 s,t = ( )( )2
3 30 and ,
3x y
ππ
−= = = −& & 3.14 m/s=v
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Chapter 11, Solution 94.
Sketch the path of the particle, i.e. plot of y versus x.
Using 6 sin , and 6 3cosx t t y t= − = − obtain the values in the table below. Plot as shown.
(a) Differentiate with respect to t to obtain velocity components.
( ) ( )
( )
2 22 2 2 2
2
6 3cos and 3sin
6 3cos 9sin 45 36cos m/s
36sin 0 0, , and 2 in the range 0 2 .
x y
x y
dx dyv t v t
dt dx
v v v t t t
d vt t t
dtπ π π
= = − = =
= + = − + = −
= = = ≤ ≤
When 0 or 2 ,t π= 2cos 1, and is minimum.t v=
When ,t π= 2cos 1, and is maximum.t v= −
( ) ( )22
min45 36 9 m/s ,v = − = min 3 m/s v =
( )t s ( )x m ( )y m
0 0 3
2
π 6.42 6
π 18.85 9
32
π 31.27 6
2π 37.70 3
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( ) ( )2
max45 36 81 m/s ,v = + = max 9 m/s v =
(b) 0, 0, 3 m, 3 m/s, 0x yt x y v v= = = = =
0t =
( )3 m=r j
tan 0y
x
v
vθ = = 0θ =
2 s, 12 m, = 3 m, 3 m/s, 0x yt x y v vπ π= = = =
2 st π=
( ) ( )12 m + 3 mπ=r i j
tan y
x
v
vθ = 0θ =
s, 6 m, = 9 m, 9 m/s, 0x yt x y v vπ π= = = =
st π=
( ) ( )6 m + 9 mπ=r i j
tan y
x
v
vθ = 0 θ = °
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Chapter 11, Solution 95.
Given: ( ) ( )cos sin sin cosA t t t A t t t= + + −r i j
( ) ( )
( ) ( )
( ) ( )
sin sin cos cos cos sin
cos sin
cos sin sin cos
d A t t t t A t t t tdt
A t t A t t
d A t t t A t t tdt
= = − + + + − +
= +
= = − + +
rv i j
i j
va i j
(a) When r and a are perpendicular, 0⋅ =r a
( ) ( ) ( ) ( )cos sin sin cos cos sin sin cos 0A t t t t t t A t t t t t t⎡ ⎤ ⎡ ⎤+ + − ⋅ − + + =⎣ ⎦ ⎣ ⎦i j i j
( )( ) ( )( )2 cos sin cos sin sin cos sin cos 0A t t t t t t t t t t t t⎡ ⎤+ − + − + =⎣ ⎦
( ) ( )2 2 2 2 2 2cos sin sin cos 0t t t t t t− + − =
21 0t− = 1 st =
(b) When r and a are parallel, 0× =r a
( ) ( ) ( ) ( )cos sin sin cos cos sin sin cos 0A t t t t t t A t t t t t t⎡ ⎤ ⎡ ⎤+ + − × − + + =⎣ ⎦ ⎣ ⎦i j i j
( )( ) ( )( )2 cos sin sin cos sin cos cos sin 0A t t t t t t t t t t t t⎡ ⎤+ + − − − =⎣ ⎦k
( ) ( )2 2 2 2 2 2sin cos sin cos sin cos sin cos cos sin sin cos 0t t t t t t t t t t t t t t t t t t+ + + − − − + =
2 0t = 0t =
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Chapter 11, Solution 96.
Given: ( )/2130 1 20 cos 21
te tt
π π−⎡ ⎤= − +⎢ ⎥+⎣ ⎦r i j
Differentiating to obtain v and a,
( )/2 /2
2130 20 cos 2 2 sin 2
21t td e t e t
dt tπ ππ π π π− −⎛ ⎞= = + − −⎜ ⎟
⎝ ⎠+rv j
( )
/22
30 120 cos 2 2sin 221
te t tt
ππ π π−⎡ ⎤⎛ ⎞= − +⎜ ⎟⎢ ⎥⎝ ⎠+ ⎣ ⎦
i j
( )( )/2 /2
32 130 20 cos 2 2sin 2 sin 2 4 cos 2
2 21t td e t t e t t
dt tπ πππ π π π π π π− −⎡ ⎤⎛ ⎞= = − − − + + − +⎜ ⎟⎢ ⎥
⎝ ⎠+ ⎣ ⎦
va i j
( )
( )2 /23
60 10 4sin 2 7.5cos 21
te t tt
ππ π π−−= − −+
i j
(a) At 0,t = ( )130 1 20 11
⎛ ⎞= − +⎜ ⎟⎝ ⎠
r i j 20 in.=r
( )1 130 20 1 01 2
π ⎡ ⎤⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦
v i j 43.4 in./s=v 46.3 °
( )( )260 10 1 0 7.51
π= − − −a i j 2743 in./s=a 85.4 °
(b) At 1.5 s,t = 0.25130 1 20 cos32.5
e π π−⎛ ⎞= − +⎜ ⎟⎝ ⎠
r i j
( ) ( )18 in. 1.8956 in.= + −i j 18.10 in.=r 6.0 °
( )
0.752
30 120 cos3 022.5
e ππ π− ⎛ ⎞= − +⎜ ⎟⎝ ⎠
v i j
( ) ( )4.80 in./s 2.9778 in./s= +i j 5.65 in./s=v 31.8 °
( )
( )2 0.753
60 10 0 7.5cos32.5
e ππ π−= − + −a i j
( ) ( )2 23.84 in./s 70.1582 in./s= − +i j 270.3 in./s=a 86.9 °
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Chapter 11, Solution 97.
Given: ( ) ( )cos sinn nRt t ct Rt tω ω= + +r i j k
Differentiating to obtain v and a.
( ) ( )cos sin sin cosn n n n n nd R t t t c R t t tdt
ω ω ω ω ω ω= = − + + +rv i j k
( ) ( )( ) ( )
2 2
2 2
sin sin cos cos cos sin
2 sin cos 2 cos sin
n n n n n n n n n n n n
n n n n n n n n
d R t t t t R t t t tdt
R t t t t t t
ω ω ω ω ω ω ω ω ω ω ω ω
ω ω ω ω ω ω ω ω
= = − − − + + −
⎡ ⎤= − − + −⎣ ⎦
va i k
i k
Magnitudes of v and a.
( ) ( ) ( )
2 2 2 2
2 22
2 2 2 2 2 2
2 2 2 2 2
cos sin sin cos
cos 2 sin cos sin
sin 2 sin cos cos
x y z
n n n n n n
n n n n n n
n n n n n n
v v v v
R t t t c R t t t
R t t t t t t c
R t t t t t t
ω ω ω ω ω ω
ω ω ω ω ω ω
ω ω ω ω ω ω
= + +
⎡ ⎤ ⎡ ⎤= − + + +⎣ ⎦ ⎣ ⎦
⎡ ⎤= − + +⎣ ⎦
⎡ ⎤+ + +⎣ ⎦
( )2 2 2 21 nR t cω= + + ( )2 2 2 21 nv R t cω= + +
( ) ( )
2 2 2 2
2 22 2 2
2 2 2 3 4 2 2 2 2
3 4 2 2
2 sin cos 2 cos sin
4 sin 4 sin cos cos 4 cos
4 sin cos sin
x y z
n n n n n n n n
n n n n n n n n n
n n n n n
a a a a
R t t t t t t
R t t t t t t t
t t t t t
ω ω ω ω ω ω ω ω
ω ω ω ω ω ω ω ω ω
ω ω ω ω ω
= + +
⎡ ⎤= − − + −⎢ ⎥⎣ ⎦
⎡= + + +⎣
⎤− + ⎦
( )2 2 4 24 n nR tω ω= + 2 24 n na R tω ω= +
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Chapter 11, Solution 98.
Given: ( ) ( ) ( )2cos 1 sin from whichAt t A t Bt t= + + +r i j k
2cos , 1, sinx At t y A t z Bt t= = + = 2
2cos sin 1x z y
t t tAt Bt A
= = = −
2 2 2 22 2 2cos sin 1 1 or
x z x yt t t
At At A B + = ⇒ + = = +
Then, 2 2 2
1y x z
A A B − = +
2 2 2
1 y x z
A A B − − =
!
For 3 and 1,A B= = ( ) ( ) ( )23 cos 3 1 sint t t t t= + + +r i j k
Differentiating to obtain v and a.
( ) ( )
( )( )
( )32
2
2
3 cos sin 3 sin cos1
13 2sin cos 3 2cos sin
1
d tt t t t t t
dt t
dt t t t t t
dtt
= = − + + ++
= = − − + + −+
rv i j k
va i j k
(a) At 0,t = ( ) ( ) ( )3 1 0 0 0= − + +v i j k 3 ft/s v = !
And ( ) ( ) ( )3 0 3 1 2 0= − + + −a i j h
Then, ( ) ( )2 22 3 2 13a = + = 23.61 ft/s a = !
(b) If and are perpendicular, 0⋅ =r v r v
( ) ( ) ( ) ( )( )2
2
33 cos 3 cos sin 3 1 sin sin cos 0
1
tt t t t t t t t t t t
t
− + + + + = +
or ( ) ( ) ( )2 2 2 29 cos 9 sin cos 9 sin sin cos 0t t t t t t t t t t t− + + + =
continued
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With 0,t ≠ 2 29cos 8 sin cos 9 sin 0t t t t t− + + =
210 8 sin cos 8cos 0t t t t− + =
or 7 2cos2 2 sin 2 0t t t+ − =
The smallest root is 2 7.631 st = 3.82 s t = !
The next root is 4.38 st =
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Chapter 11, Solution 99.
(a) At the landing point, tan 30y x= − °
Horizontal motion: ( )0 00xx x v t v t= + =
Vertical motion: ( ) 2 20 0
1 12 2yy y v t gt gt= + − = −
from which 2 02 2 tan 30 2 tan 30y x v ttg g g
° °= − = =
Rejecting the 0t = solution gives ( )( )0 2 25 tan 302 tan 309.81
vtg
°°= = 2.94 s t =
(b) Landing distance: ( )( )0 25 2.94cos30 cos30 cos30
x v td = = =° ° °
84.9 m d =
(c) Vertical distance: tan 30h x y= ° +
or 20
1tan 302
h v t gt= ° −
Differentiating and setting equal to zero,
0tan 30tan 30 0 or odh vv gt t
dt g°= ° − = =
Then, ( ) ( ) 20 0 0
maxtan 30 tan 30 1 tan 30
2v v vh g
g g° ° ⎛ ⎞°= − ⎜ ⎟
⎝ ⎠
( ) ( )( )( )
2 22 20 25 tan 30tan 30
2 2 9.81v
g°°= = max 10.62 m h =
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Chapter 11, Solution 100.
Horizontal motion: ( )0 000
, or xxx x v t v t tv
= + = =
Vertical motion: ( )2
2 20 0 0 20
0
1 1 or 2 2 2y
gxy y v t gt y gt y yv
= + − = − = −
At ground level, 0,y = so that 2
0 202
gxyv
=
At 50 m,x = ( )( )( )( )
2
0 29.81 50
13.625 m2 30
y = =
0 13 0.625 mh y= − =
At 53 m,x = ( )( )( )( )
2
0 29.81 53
15.31 m2 30
y = =
0 13 2.31 mh y= − =
Range to avoid: 0.625 m 2.31 mh< <
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Chapter 11, Solution 101.
Horizontal motion. 0 0xv v x v t= =
Vertical motion. 212
y h gt= −
Eliminate t. 2
20 02x gxt y hv v
= = −
Solve for v0. ( )2
0 2=
−gxvh y
Data: h = 3 ft, g = 32.2 ft/s2
(a) To strike corner C. 15 ft, 0x d y= = =
( )( )( )( )
2
032.2 152 3 0
=−
v 0 34.7 ft/s v =
To strike point B. 15 ft, 1 ftx y= =
( )( )( )( )
2
032.2 152 3 1
=−
v 0 42.6 ft/sv =
To strike point D. 15 1 14 ft, 0x y= − = =
( )( )( )( )
2
032.2 142 3 0
=−
v 0 32.4 ft/sv =
(b) Range to strike corner BCD. 032.4 ft/s < 42.6 ft/s v <
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Chapter 11, Solution 102.
Place origin of coordinates at point A.
Horizontal motion: ( )0 90 mi/h 132 ft/s= =xv
( )0 0 0 132 ft= + = +xx x v t t
At point B where 6.5 s,Bt =
( )( )132 6.5 858 ft= =Bx
(a) Distance AB.
From geometry 858cos 10
d =°
871 ftd =
Vertical motion: ( ) 20 0
12
= + −yy y v t gt
At point B
( )( )21tan 10 0 32.2 6.52
− ° = + −Bx h
(b) Initial height. 529 fth =
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Chapter 11, Solution 103.
Data: 20 25 ft/s, 90 55 35 , 32.2 ft/sv gα= = ° − ° = ° =
Horizontal motion. ( )0 cos=x v tα
Vertical motion. ( ) 20
1sin2
= + −y h v t gtα
Eliminate t. 0 cos
xtv α
=
2 20
tan2 cos
gxy h xv
αα
2= + −
Solve for h. 2
2 20
tan2 cos
gxh y xv
αα
= − +
To hit point B. 20 ft, 0x y= =
( )( )( )( )
2
232.2 20
0 20 tan 35 1.352 ft2 25cos35
= − ° + =°
h
To hit point C. 24 ft, 0x y= =
( )( )( )( )
2
232.2 24
0 24 tan 35 5.31 ft2 25cos35
= − ° + =°
h
Range of values of h. 1.352 ft < 5.31 fth <
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Chapter 11, Solution 104.
Place the origin at A. Let β be the direction of the discharge velocity measured counterclockwise from the x-axis
Horizontal motion. ( ) ( )0 00 cos cos= =xv v x v tβ β
Solve for t. 0 cos
xtv β
=
Vertical motion. ( ) 00sin=yv v β
( ) 20
1sin2
= −y v t gtβ
2
2 20
tan2 cos
gxxv
ββ
= −
Geometry. At points B and C tany x α=
Hence, 2
2 20
tan tan2 cos
gxx xv
α ββ
= −
Solve for x. ( )2 202 cos tan tan= −vx
gβ β α
To water point B. 090 90 40 50β φ= °− = °− ° = °
( )( ) ( )2 22 24 cos 50
tan 50 tan10 15.01 ft32.2
°= ° − ° =Bx
15.01 ftBd =
To water point C. 090 90 40 130= ° + = ° + ° = °β φ
( )( ) ( )2 22 24 cos 130
tan130 tan10 20.2 ft32.2
°= ° − ° = −Cx
20.2 ftC Cd x= − =
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Chapter 11, Solution 105.
0 0 013 m/s, 33 , 0, 0.6 mv x yα= = ° = =
Vertical motion: 0 sinyv v gtα= −
( ) 20 0
1sin2
y y v t gtα= + −
At maximum height, 0 sin0 or yvv t
gα= =
(a) 13sin 33 0.7217 s9.81
t °= =
( )( ) ( )( )2max
10.6 13sin 33 0.7217 9.81 0.72172
y = + ° − max 3.16 m y =
1.8 m 3.16 m 3.7 m< < yes
Horizontal motion: ( ) 00 0
0cos or
cosx xx x v t t
vα
α−= + =
At 15.2 m,x = 15.2 0 1.3941 s13cos33
t −= =°
(b) Corresponding value of :y ( )( ) ( )( )210.6 13sin 33 1.3941 9.81 1.39412
y = + ° −
0.937 m y =
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Chapter 11, Solution 136.
Velocities:
/ 1 m/sA B A B= − =v v v
Accelerations:
2/ 0.25 m/sA B A B= − =a a a
(a)
( )
( )
2 2
22
22
/
100
1
96
10.25
100 96
A AA
A
ABB
B
AAA B
v va
vva
vva
ρ
ρ
= =
−= =
−= − =
2 50 625 0A Av v− + =
25Av = ± 25 m/sAv = !
(b) 25 1 24Bv = − = 24 m/sBv = !
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Chapter 11, Solution 137.
2
2max, 0, n t n
va a v aρρ
= = =
( )( ) ( )( )( )2 2 2max 25 3 25 3 9.81 735.35 m /sv g= = =
max 27.125 m/sv = max 97.6 km/hv = !
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Chapter 11, Solution 138.
( ) ( )
( ) ( )
( )( )
( )( )
2 2
2
,
0.66 0.097066.8
0.09706 0.09706 60 5.8235 mm
c cc cn nA AA A
c A c B cn nA B
c nB A
A c n B
B A
v va a
v a a
a
a
ρ ρ
ρ ρ
ρρ
ρ ρ
= =
= =
= = =
= = =
2 11.65 mmB Bd ρ= = !
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Chapter 11, Solution 139.
Initial speed. 0 72 km/h 20 m/sv = =
Tangential acceleration. 21.25 m/sta = −
(a) Total acceleration at 0.t =
( )2220 20
1.14286 m/s350n
vaρ
= = =
( ) ( )2 22 2 1.25 1.14286t na a a= + = − + 21.694 m/s a = !
(b) Total acceleration at 4 s.t =
( )( )0 20 1.25 4 15 m/stv v a t= + = + − =
( )22215
0.6426 m/s350n
vaρ
= = =
( ) ( )2 22 2 1.25 0.6426t na a a= + = − + 21.406 m/sa = !
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Chapter 11, Solution 140.
Length of run. 130 metersL Dπ π= = (1)
Radius of circle. 1 65m2
Dρ = =
Tangential acceleration of starting portion of run.
( )( )1 4 4 m/sm t t tv a t a a= = = (2)
( )( )221 1
1 1 4 8 m2 2t t ts a t a a= = = (3)
Constant speed portion of run. mv v=
( )1 1ms s v t t= + − (4)
Substituting (1), (2) and (3) into (4)
( )130 8 4 54 4t ta aπ = + −
Solving for .ta 2130 1.9635 m/s8 200ta π= =
+
From (2) ( )( )4 1.9635 7.854 m/smv = =
Normal acceleration during constant speed portion of run.
( )2227.854
0.9490 m/s65
mn
vaρ
= = =
Maximum total acceleration.
( ) ( )2 22 2 1.9635 0.9490t na a a= + = + 22.18 m/sa = !
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Chapter 11, Solution 141.
For uniformly decelerated motion: 0 tv v a t= +
At 9 s,t = ( ) 20 150 9 , or 16.667 ft/st ta a= − = −
Total acceleration: 2 2 2t na a a= +
( ) ( )1/21/2 2 22 2 2130 16.667 128.93 ft/sn ta a a = − = − − =
Normal acceleration: 2 1 5, where diameter ft
2 12nva ρρ
= = =
( )2 2 25 128.93 53.72 ft /s , 7.329 ft/s12nv a vρ = = = =
Time: 0 7.329 15016.667t
v vta− −= =
− 8.56 st = !
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Chapter 11, Solution 142.
Speeds: 0 10 65 mi/h 95.33 ft/sv v= = =
Distance: ( )450 300 1006.86 ft2
s π= + =
Tangential component of acceleration: 2 21 0 2 tv v a s= +
( )( )( )
22 221 0 95.33 0
4.5133 ft/s2 2 1006.86t
v vas
+−= = =
At point B, 2 20 2B t Bv v a s= + where ( )450 706.86 ft
2Bs π= =
( )( )( )2 2 20 2 4.5133 706.86 6380.5 ft /sBv = + =
(a) 79.88 ft/sBv = 54.5 mi/h Bv = !
At 15 s,t = ( )( )0 0 4.5133 15 67.70 ft/stv v a t= + = + =
Since ,Bv v< the car is still on the curve. 450 ftρ =
Normal component of acceleration: ( )22267.70
10.185 ft/s450n
vaρ
= = =
(b) Magnitude of total acceleration: ( ) ( )2 22 2 4.5133 10.185t na a a= + = + 211.14 ft/s a = !
COSMOS: Complete Online Solutions Manual Organization System
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Chapter 11, Solution 143.
(a) 420 km/hA =v , 520 km/hB =v 60°
/B A B A= +v v v or ( )/B A B A B A= − = + −v v v v v
Sketch the vector addition as shown.
( ) ( ) ( )( )( )
2 2 2/
2 2
2 cos60
420 520 2 420 520 cos60
B A A B A Bv v v v v= + − °
= + − °
or / 477.9 km/hB Av =
sin sin 60520 477.9
α °= or 70.4α = °
/ 478 km/hB A =v 70.4 °!
(b) 26 m/sAa = ( ) 22 m/sB ta = 60°
520 km/h 144.44 m/sBv = =
( ) ( )222144.44
104.32 m/s200
BB n
vρ
= = =a 30°
( ) ( )/B A B A B B At n= − = + −a a a a a a
[2= ] [60 104.32° + ] [30 6° − ]
( ) ( )2 cos60 sin 60 104.32 cos30 sin 30 6= − ° + ° + − ° − ° −i j i j i
( ) ( )2 297.34 m/s 50.43 m/s= − −i j
2/ 109.6 m/sB Aa = 27.4 °!
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Chapter 11, Solution 144.
(a) 180 km/h 50 m/sA = =v 30 , 162 km/h 45 m/sB° = =v 45°
( ) ( )/ 45 cos 45 sin 45 50 cos120 sin120B A B A= − = ° − ° − ° + °v v v i j i j
56.82 75.12 94.2 m/s= − =i j 52.9°
/ 339 km/hB Av = 52.9 °!
(b) ( ) 28 m/sA t =a ( ) 260 , 3 m/sB t° =a 45°
( ) ( )22250
6.25 m/s400
AA n
A
vρ
= = =a 30°
( ) ( )22245
6.75 m/s300
BB n
B
vρ
= = =a 45°
( ) ( ) ( ) ( )/B A B A B B A At n t n= − = + − −a a a a a a a
( ) ( )3 cos 45 sin 45 6.75 cos 45 sin 45= ° − ° + ° + °i j i j
( ) ( )8 cos60 sin 60 6.25 cos30 sin 30− ° − ° − − ° − °i j i j
( ) ( )2 28.31 m/s 12.07 m/s= +i j
or 2/ 15.18 m/sB A =a 56.8° !
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Chapter 11, Solution 145.
(a) As water leaves nozzle.
8 m/sv =
2sin 55 9.81 sin 55 8.04 m/sna g= ° = ° =
2
nvaρ
=
( )22 88.04n
va
ρ = = 7.96 m ρ = !
(b) At maximum height of stream.
( )0 8 sin 55 6.55 m/sxv v= = ° =
29.81 m/sna g= =
2
nvaρ
=
( )22 6.559.81n
va
ρ = = 4.38 m ρ = !
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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 146.
Horizontal motion. 0 0cos cosxv v x v tα α= =
Vertical motion. 0 sinyv v gtα= −
20 0
1sin2
y y v t gtα= + −
Eliminate t. 2
0 2 20
tan2 cos
gxy y xv
αα
= + − (1)
Solving (1) for 0v and applying result at point B
( )( )( )
( )( )( )22
0 2 20
9.81 62 tan cos 2 1.5 6 tan 3 0.97 cos 3
gxvy x yα α
= =+ − + ° − °
(a) Magnitude of initial velocity. 0 14.48 m/s=v !
(b) Minimum radius of curvature of trajectory.
2 2 2
cosnn
v v va ga g
ρρ θ
= = = = (2)
where θ is the slope angle of the trajectory.
The minimum value of ρ occurs at the highest point of the trajectory where cos 1=θ
and 0 cos= =xv v v α Then
( )2 22 20
min14.48 cos 3cos
9.81°
= =vg
αρ
min 21.3 m=ρ !
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Chapter 11, Solution 147.
(a) At point A, 0 120 ft/s 120 ft/sv v= = =v 60°
A g=a 232.2 ft/s=
( )2
sin 30 AA n
A
va gρ
= ° =
( )22 120sin 30 32.2sin 30
AA
vg
ρ = =° °
894 ftAρ = !
(b) At the point where velocity is parallel to incline,
0 sin 30 120 sin 30 60 ft/sxv v= ° = ° =
tan 30 60 tan 30 34.64 ft/sy xv v= ° = ° =
( ) ( )2 260 34.64 69.282 ft/sv = + =
2
sin 60 Bn
B
va gρ
= ° =
( )22 69.282sin 60 32.2sin 60
BB
vg
ρ = =° °
172.1 ft Bρ = !
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Chapter 11, Solution 148.
Compute x- and y-components of velocity and acceleration.
( )2
2cos 1 3 sin, ,2 cos 2 cos
t tx xt t
π π ππ π− −= =
− −&
( )
( )( )
2
2 36 sin sin3 cos
2 cos 2 cos
t ttxt t
π π π ππ ππ π
−= +− −
&&
( )( )2
1.5 2cos 11.5sin , ,2 cos 2 cos
tty yt t
π πππ π
−= =
− −&
( )( )( )
( )2
2 33 2cos 1 sin3 sin
2 cos 2 cos
t ttyt t
π π π ππ ππ π
−−= −− −
&&
(a) 0,t = 21, 0, 0, 1.5 , 3 ,x y x y xπ π= = = = = −& & &&
1.5 ,v y π= =&& 23 ,na x π= − =&& ( )22
21.53n
va
πρ
π= = 0.75 ftρ = !
(b) 1 ,3
t = 23 2 20, , , 0, ,
2 3 3x y x y yπ π= = = − = = −& & &&
2 ,3
v x π= − = −& 22 ,
3na y π= − =&& 2 2
24 33.2n
va
πρπ
= = 1.155 ftρ = !
(c) 1,t = 2
1, 0, 0, , ,2 3
x y x y xπ π= − = = = − =& & &&
,2
v y π= − =& 2
,3na x π= =&&
2 2
23
4n
va
πρπ
= = ⋅ 0.75 ftρ = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 149.
Given: ( )324
m6
tx t
−= + ( )23 1
m6 4
tty−
= −
Differentiating twice
( ) ( )( )22
2 2 6 m/s2xv x
−= = + =&
( )242 m/s
2t
x t−
= +& ( )2 1m/s
2 2tty
−= −&
24 2 2 m/sx t t= − + = −&& 21 m/s2
y t= −&&
At 2 s.t =
( ) ( )22 11.5 m/s
2 2yv y= = − =&
2 2 0xa x= = − =&&
212 1.5 m/s2ya y= = − =&&
(a) Acceleration. ( )21.5 m/s=a j!
(b) Radius of curvature of path.
1.5tan6
y
x
vv
θ = =
14.036θ = °
2 2 2 2 26 1.5x yv v v= + = +
2 238.25 m /s= cos 1.5 cos14.036na a θ= = °
21.45522 m/s=
2
nvaρ
=
2 38.25
1.45522n
va
ρ = = 26.3 mρ = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 150.
x Av v=
At point B ( )B Axv v=
( )cos cos
B AxB
v vvθ θ
= =
cos A
B
vv
θ =
cos cosn Ba a gθ θ= =
A
B
vgv
=
2 2B B B
Bn A
v v va gv
ρ = = 3B
BA
vgv
ρ = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 151.
Let θ be the slope angle of the trajectory at an arbitrary point C.
Then, ( )2 2
cos , or cos
C CC Cn
C
v va gg
θ ρρ θ
= = =
But, the horizontal component of velocity is constant, ( ) ( )C A xxv v=
where ( ) ( )0 cos cosA C Cx xv v v vα θ= =
Then, 0 cos cosCv vα θ=
or 0coscosCv vα
θ=
so that 2 2 2
00 3
1 cos coscos cos cosC
vvg g
α αρθ θ θ
= =
(a) Since 0, ,v α and g are constants, Cρ is a minimum at point B where cosθ is a maximum or 0.θ =
Then, 2 20
mincos Q.E.D.B
vg
αρ ρ= = !
(b) 2 20
31 cos
cosCv
gαρ
θ
=
or min3 Q.E.D.
cosCρρ
θ= !
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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 152. Let θ be the slope angle of the trajectory at an arbitrary point C.
Then, ( )2 2
cos or cos
C CC Cn
C
v va gg
θ ρρ θ
= = =
But the horizontal component of velocity is constant, ( ) ( )C A xxv v=
( ) ( ) ( ) ( )0 0 cosA Cx x xv v x v t v tα= = = 0
or (1)cosxt
v α=
where ( ) ( )0 0cos and cosA Cx xv v v vα θ= =
Then, 0 cos cosCv vα θ=
so that 3
0 cosC
Cv
gvρ
α= (2)
The vertical motion is uniformly accelerated
( ) ( )0 00
sincosC y ygxv v gt v
vα
α= − = − (3)
( ) ( ) ( )2
2 2 220 0 0 0
0
2 220 2 4 2
0 0
But cos sincos
2 tan1cos
C x yxv v v v v g
v
gx g xvv v
α αα
αα
= + = + −
= − +
or 3/22 2
3 30 2 4 2
0 0
2 tan1cosC
gx g xv vv v
αα
= − +
(4)
Finally, substituting (4) into (2) gives
3/22 2 2
02 4 20 0
2 tan1cos cosv gx g x
g v vαρ
α α
= − +
!
COSMOS: Complete Online Solutions Manual Organization System
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Chapter 11, Solution 153.
Given: ( ) ( )cos sinn nRt t ct Rt tω ω= + +r j ki
Differentiating to obtain v and a,
( ) ( )
( ) ( )( ) ( )
2 2
2 2
cos sin sin cos
sin sin cos cos cos sin
2 sin cos 2 cos sin
n n n n n n
n n n n n n n n n n n n
n n n n n n n n
d R t t t c R t t tdtd R t t t t R t t t tdt
R t t t t t t
ω ω ω ω ω ω
ω ω ω ω ω ω ω ω ω ω ω ω
ω ω ω ω ω ω ω ω
= = − + + +
= = − − − + + −
= − − + −
rv i j k
va i k
i k
Magnitudes of v and a.
( ) ( ) ( )
( )
2 2 2 2
2 22
2 2 2 2 2 2
2 2 2 2 2
2 2 2 2
cos sin sin cos
cos 2 sin cos sin
sin 2 sin cos cos
1
x y z
n n n n n n
n n n n n n
n n n n n n
n
v v v v
R t t t c R t t t
R t t t t t t c
R t t t t t t
R t c
ω ω ω ω ω ω
ω ω ω ω ω ω
ω ω ω ω ω ω
ω
= + +
= − + + +
= − + +
+ + +
= + + ( )
( ) ( )
2 2 2 2
2 2 2 2
2 22 2 2
2 2 2 3 4 2 2
2 2 3 4 2 2
2
or 1
2 sin cos 2 cos sin
4 sin 4 sin cos cos
4 cos 4 sin cos sin
4
n
x y z
n n n n n n n n
n n n n n n n
n n n n n n n
v R t c
a a a a
R t t t t t t
R t t t t t t
t t t t t t
R
ω
ω ω ω ω ω ω ω ω
ω ω ω ω ω ω ω
ω ω ω ω ω ω ω
= + +
= + +
= − − + −
= + +
+ − +
= ( )2 4 2 2 2 or 4n n n nt a R tω ω ω ω+ = +
Tangential component of acceleration: ( )
2 2
1/22 2 2 21t
n
dv R n tadt R t c
ω
ω= =
+ +
At 2 2 20, , 2 , 0n tt v R c a R aω= = + = =
Normal component of acceleration: 2 2 2n t na a a Rω= − = 2
But nvaρ
=
or 2
n
va
ρ = 2 2
2 n
R cR
ρω+= !
COSMOS: Complete Online Solutions Manual Organization System
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Chapter 11, Solution 154.
With 3 and 1,A B= = the position vector is
( ) ( ) ( )23 cos 3 1 sinr t t t t t= + + +i j k
Differentiating to obtain v and a,
( ) ( )
( )
( )
2
22
2
3 3 cos sin sin cos1
11
3 sin sin cos 31
cos cos sin
3 2sin
d tt t t t t tdt t
tt ttd t t t t
dt t
t t t t
t t
= = − + + + +
+ −
+ = = − − − ++
+ + −
= − +
rv i j k
va i j
k
( )( )
( )3/22
3cos 2cos sin1
t t t tt
+ + −+
i j k
Magnitude of 2.v
( ) ( )2
2 22 2 2 2299 cos sin sin cos
1x y ztv v v v t t t t t t
t= + + = − + + +
+
Differentiating,
( )( )( )
( )( )
22
182 18 cos sin 2sin cos1
2 sin cos 2cos sin
dv tv t t t t t tdt t
t t t t t t
= − − − ++
+ + −
2When 0, 3 2 , 9, 2 0dvt v vdt
= = + = =a j k
2 2 23 2 13a = + =
Tangential acceleration: 0tdvadt
= =
Normal acceleration: 2 2 2 13 or 13n t na a a a= − = =
But 2 2 9or
13nn
v vaa
ρρ
= = = 2.50 ftρ = !
COSMOS: Complete Online Solutions Manual Organization System
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Chapter 11, Solution 155.
For the sun, 2274 m/s ,g =
and ( )9 91 1 1.39 10 0.695 10 m2 2
R D = = × = ×
Given that2
2ngRar
= and that for a circular orbit 2
nvar
=
Eliminating na and solving for r, 2
2gRrv
=
For the planet Earth, 6 3107 10 m/h 29.72 10 m/sv = × = ×
( )( )( )
299
2
274 0.695 10Then, 149.8 10 m
29.72r
×= = × 149.8 Gmr = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 156.
For the sun, 2274 m/sg =
and ( )9 91 1 1.39 10 0.695 10 m2 2
R D = = × = ×
Given that2
2ngRar
= and that for a circular orbit: 2
nvar
=
Eliminating na and solving for r, 2
2gRrv
=
For the planet Saturn, 6 334.7 10 m/h 9.639 10 m/sv = × = ×
Then, ( )( )
( )
2912
2
274 0.695 101.425 10 m
9.639r
×= = × 1425 Gmr = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 157.
From Problems 11.155 and 11.156, 2
2ngRar
=
For a circular orbit, 2
nvar
=
Eliminating na and solving for v, gv Rr
=
For Venus, 229.20 ft/sg =
63761 mi 19.858 10 ft.R = = ×
63761 100 3861 mi 20.386 10 ftr = + = = ×
Then, 6 36
29.2019.858 10 23.766 10 ft/s20.386 10
v = × = ××
16200 mi/hv = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 158.
From Problems 11.155 and 11.156, 2
2ngRar
=
For a circular orbit, 2
nvar
=
Eliminating na and solving for v, gv Rr
=
For Mars, 212.24 ft/sg =
62070 mi 10.930 10 ftR = = ×
32070 100 2170 mi 11.458 10 ftr = + = = ×
Then, 6 36
12.2410.930 10 11.297 10 ft/s11.458 10
v = × = ××
7700 mi/hv = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 159.
From Problems 11.155 and 11.156, 2
2ngRar
=
For a circular orbit, 2
nvar
=
Eliminating na and solving for v, gv Rr
=
For Jupiter, 275.35 ft/sg =
644432 mi 234.60 10 ftR = = ×
644432 100 44532 mi 235.13 10 ftr = + = = ×
( )6 36
75.35Then, 234.60 10 132.8 10 ft/s235.13 10
v = × = ××
90600 mi/hv = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 160.
Radius of Earth ( )( ) 63960 mi 5280 ft/mi 20.908 10 ftR = = ×
Radius of orbit ( )( ) 63960 10900 5280 78.4608 10 ftr = + = ×
Normal acceleration 2
2ngRar
= and 2
nvar
=
Thus, 2 2
2v gRr r
= or 2
2 gRvr
=
( )( )26
2 6 2 26
32.2 20.908 10179.40 10 ft /s
78.4608 10v
×= = ×
×
313.3941 10 ft/sv = ×
Time T for one orbit. 2vT rπ=
( )6
33
2 78.4608 102= 36.806 10 s13.3941 10
rTv
ππ ×= = ×
×
10.22 hT = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 161.
Normal acceleration. 2
2ngRar
= and 2 2
nv va
rρ= =
Solve for v2. 2
2n
gRv rar
= =
Data: 29.81 m/s ,g = 66370 km = 6.370 10 mR = ×
3 6384 10 km = 384 10 mr = × ×
( )( )26
2 6 2 26
9.81 6.370 101.0366 10 m /s
384 10v
×= = ×
×
= 1.018 m/sv 3670 km/hv = !
COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 162.
From Problems 155 through 156, 2
2ngR
ar
=
For a circular orbit, 2
nv
ar
=
Eliminating na and solving for v, g
v Rr
=
For one orbit the distance traveled is 2 ;rπ hence, the time is 2 r
tv
π=
or 3 2
1 2
2 rt
Rg
π=
For satellites A and B, 3 2 3 2
1 2 1 2
2 2 and A B
A Br r
t tRg Rg
π π= =
Let number of orbits of .n B= For the next alignment,
( )3 2
3 2
11 or
11
B BA B
A A
B
A
n t rn t nt
n t r
r
n r
++ = = =
= −
Data: 36370 km 6.370 10 mR = = ×
36370 190 6560 km 6.560 10 mAr = + = = ×
36370 320 6690 km 6.690 10 mBr = + = = ×
Then,
3/23
3
1 6.690 101 0.02987 or 33.475
6.560 10n
n
×= − = = ×
continued
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Time for orbit of satellite B is
( )( )( )
3 263
1 26
2 6.690 105.449 10 s 1.5137 h
6.370 10 9.81Bt
π ×= = × =
×
Time for next alignment is
( )( )33.475 1.5137Bnt = 50.7 hBnt = !
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Chapter 11, Solution 163.
Differentiate the expressions for r and θ with respect to time. 2 3
2
0.8
0.8 0.8
0.8 0.8
1 2 6 8
2 12 2412 48
0.5 sin 3
0.4 sin 3 1.5 cos3
0.32 sin 3 1.2 cos3
t
t t
t t
r t t t
r t tr t
e t
e t e t
e t e t
θ π
θ π π π
θ π π π
−
− −
− −
= + − +
= − += − +
=
= − +
= −
&
&&
&
&&
0.8 2 0.81.2 cos3 4.5 sin 3t te t e tπ π π π− −− −
At 0.5 s,t = 21.5 ft, 2.00 ft/s, 12 ft/s ,r r r= = =& &&
0.8
2
0.67032, sin 3 1, cos 3 0
0.33516 rad, 0.26812 rad/s, 29.56 rad/s
te t tπ π
θ θ θ
− = = − =
= − = =& &&
(a) Velocity of the collar.
rr r θθ= +v e e&& ( ) ( )2.00 ft/s 0.402 ft/sr θ= +v e e !
2 ft/s, 0.402 ft/srv vθ= = !
(b) Acceleration of the collar.
( ) ( )2 2r r rr r r r a aθ θ θθ θ θ= − + + = +a e e e e& && &&& &
( )( )212 1.5 0.26812ra = − 211.89 ft/sra = !
( )( ) ( )( )( )1.5 29.56 2 2 0.26812aθ = + 245.41 ft/saθ = !
( ) ( )2 211.89 ft/s 45.41 ft/sr θ= +a e e !
(c) Acceleration of the collar relative to the rod.
( )212 ft/sr rr =e e&& !
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Chapter 11, Solution 164.
Differentiate the expressions for r and θ with respect to time.
( ) ( )2
2 310 10 20mm, mm/s, mm/s
6 6 6r r r
t t t= = − =
+ + +& &&
24 sin rad, 4cos rad/s 4 sin rad/st t tθ π θ π θ π ππ
= = =& &&
At 1s,t = 210 10 20mm; mm/s, mm/s7 49 343
r r r= = − =& &&
0, 4 rad/s, 0θ θ θ= = − =& &&
(a) Velocity of the collar.
0.204 mm/s, 5.71 mm/srv r v rθ θ= = = = −&&
( ) ( )0.204 mm/s 5.71 mm/sB r θ= −v e e !
(b) Acceleration of the collar.
( )
( ) ( ) ( )
22 2
2
20 10 4 22.8 mm/s343 7
10 102 0 2 4 1.633 mm/s7 49
ra r r
a r rθ
θ
θ θ
= − = − − = −
= + = + − − =
&&&
&& &&
( ) ( )2 222.8 mm/s 1.633 mm/sB r θ= − +a e e !
(c) Acceleration of the collar relative to the rod.
/20
343B OA r rr= =a e e&& ( )2/ 0.0583 mm/sB OA r=a e !
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Chapter 11, Solution 165.
Given ( )2 cos /2r B At B= /2At Bθ =
Differentiating twice
( )sin /2r A At B= −& /2A Bθ =&
( ) ( )2/2 cos /2r A B At B= −&& 0θ =&&
Components and magnitude of velocity.
( )sin /2 sinrv r A At B A θ= = − = −&
( ) ( )2 cos /2 /2 cosv r B At B A B Aθ θ θ = = = &
(a) 2 2 2 2 2 2sin cosrv v v A A Aθ θ θ= + = + = v A= !
Components and magnitude of acceleration.
( ) ( ) ( ) [ ]22 2/2 cos /2 2 cos /2 /2ra r r A B At B At B A Bθ = − = − + &&&
( )2/ cosA B θ= −
( ) ( )2 0 (2) sin /2 /2a r r A At B A Bθ θ θ = + = + − && &&
2/ sinA B θ= −
( ) ( )2 2 4 2 2 4 2 2/ cos / sinra a a A B A Bθ θ θ= + = +
= 2/A B 2/a A B= !
From the figure a is perpendicular to v
Thus, 2/na a A B= =
2
nvaρ
= 2
n
va
ρ =
(b) ( )2
2/A B
A Bρ = = Bρ = !
Since ρ is constant, the path is a circle of radius B.
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Chapter 11, Solution 166.
Differentiate the expressions for r and θ with respect to time.
( )
2
2 cos ,
sin ,
cos,
,
0
r b t
r b t
r b tt
ππ π
π πθ π
θ π
θ
= +
= −
= −=
=
=
&
&&
&
&&
(a) At 2 s,t = sin 0, cos 1t tπ π= =
23 , 0, , 2 rad, rad/sr b r r bπ θ π θ π= = = − = =&& &&
0 , 3 ,rv r v r bθ θ π= = = =&& 3 b θπ=v e
( )2 2 2 23 4ra r r b b bθ π π π= − = − − = −&&&
2 0,a r rθ θ θ= + =&& && 24 rbπ= −a e
(b) Values of θ for which v is maximum.
( )( )
( )
22 2 2 2 2 2
2 2 2 2
2 2
sin
2 cos
sin 2 cos
sin 4 4cos cos
5 4cos
r
r
v r b t
v r b t
v v v b t t
b t t t
b t
θ
θ
π π
θ π π
π π π
π π π π
π π
= = −
= = − +
⎡ ⎤= + = + +⎢ ⎥⎣ ⎦
⎡ ⎤= + + +⎣ ⎦
= +
&
&
2v is maximum when cos 1 or 0, 2 , 4 , 6 , etct tπ π π π π= =
But , hencetθ π= 2 , 0,1, 2,N Nθ π= = K
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Chapter 11, Solution 167.
Differentiate the expressions for r and θ with respect to time.
( ) 1 22 2 2 26 1 4 , 6 1 4 24 1 4r t t r t t t−
= + = + + +&
( ) ( )1 2 3 22 3 272 1 4 96 1 4 ,r t t t t− −
= + − +&&
( ) 12arctan 2 2 1 4 ,t tθ θ−
= = +&
( ) 2216 1 4t tθ−
= − +&&
(a) At 0,t = 0, 6 ft/s, 0r r r= = =& &&
0, 2 rad/s, 0θ θ θ= = =& &&
6 ft/s, 0,rv r v rθ θ= = = =&& ( )6 ft/s r=v e
2 0,ra r rθ= − =&&& 22 24 ft/s ,a r rθ θ θ= + =&& && ( )2 24 ft/s θ=a e
(b) At 0.5 s,t = 23 2 ft, 9 2 ft/s, 15 2 ft/sr r r= = =& &&
2rad, 1 rad/s, 2 rad/s4πθ θ θ= = = −& &&
12.73 ft/s, 4.243 ft/srv r v rθ θ= = = =&&
( ) ( )12.73 ft/s 4.24 ft/sr θ= +v e e
( )22 215 2 3 2 1 16.97 ft/sra r rθ= − = − =&&&
( ) ( )( )( ) 22 3 2 2 2 9 2 1 16.97 ft/sa r rθ θ θ= + = − + =&& &&
( ) ( )2 216.97ft/s 16.97ft/sr θ= +a e e
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Chapter 11, Solution 168.
Change to rectangular coordinates. cos and sinx yr r
θ θ= =
Equation of the path: 3 3 3sin cos
rr y x y xr r
θ θ= = =
− −−
from which 3 or 3.y x y x− = = +
Also, 23 3 1tan 1 1y x
x x x tθ += = = + = +
from which 2 23 and 3 1x t y t= = +
Differentiating, 6 , 6x yv x t v y t= = = =& &
6, 6x ya x a y= = = =&& &&
(a) Magnitudes: 2 2x yv v v= + 6 2 ft/sv t=
2 2x ya a a= + 26 2 ft/sa =
(b) 3y x= + is the equation of a straight line.
Hence, ρ = ∞
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Chapter 11, Solution 169.
Sketch the directions of the vectors v and .θe
cosv vθ θ θ= ⋅ = −v e
But v rθ θ= &
Hence, cosr vθ θ= −&
But from geometry, cos
brθ
=
2cos or cos cosb bv vθ θθ
θ θ= − = −
& &
Speed is the absolute value of v.
2cosbv θ
θ=
&
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Chapter 11, Solution 170.
From geometry, cos
brθ
=
Differentiating with respect to time, 2sin
cosbr θθ
θ=
&&
Transverse component of acceleration
2
22 sin2
cos cosb ba r rθθ θθθ θθ θ
= + = +&& &
&& && (1)
Sketch the directions of the vectors a and .θe
cosa aθ θ θ= ⋅ = −a e (2)
Matching from (1) and (2) and solving for a,
( )
2
2 3
22
2 sincos cos
2 tancos
b ba
b
θ θθθ θ
θ θθθ
= − −
= − +
&& &
&& &
Since magnitude of a is sought, 22| | 2 tan
cosba θ θθ
θ= +&& &
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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 171.
Sketch the geometry.
( )180 180θ β α+ ° − + = °
α β θ= −
( )sin 180 sinr d
β α=
° −
sinsin
dr βα
=
Sketch the velocity vectors.
( )cos 90v vθ θ α= ⋅ = ° −v e
sinv α=
But sin or sin ,sin
dv r vθβθ α θ
α= =& &
or 2sin
sindv β θ
α= &
( )2sin
sindv β θ
β θ=
−&
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Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.
Chapter 11, Solution 172.
Looking at d and β as polar coordinates with 0,d =&
2 2
, 0
2 0,
d
d
v d d v d
a d d a d d d
β
β
β ω
β β β ω
= = = =
= + = = − = −
&&
& &&&& & &
Geometry analysis: 3r d= for angles shown.
(a) Velocity analysis:
Sketch the directions of v, and .r θe e
cos120r rv r dω= = ⋅ = °v e&
12
r dω= −&
cos30v r dθ θθ ω= = ⋅ = °v e&
3
2cos303
ddr d
ωωθ °= =&
12
θ ω=&
(b) Acceleration analysis:
Sketch the directions of a, and .r θe e
23cos150
2r ra a a dω= ⋅ = ° = −e
2 23
2r r dθ ω− = −&&&
2
2 2 23 3 132 2 2
r d r d dω θ ω ω⎛ ⎞= − + = − + ⎜ ⎟⎝ ⎠
&&&
23
4r dω= −&&
2 21cos1202
2
a d d
a r r
θ θ
θ
ω ω
θ θ
= ⋅ = ° = −
= +
a e
&& &&
( ) ( )21 1 1 1 12 22 2 23
a r d dr dθθ θ ω ω ω⎡ ⎤⎛ ⎞⎛ ⎞= − = − − −⎜ ⎟⎜ ⎟⎢ ⎥
⎝ ⎠⎝ ⎠⎣ ⎦&& && 0θ =&&
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Chapter 11, Solution 173.
Rate of change of .θ 348.0 47.0 1.0 17.453 10 radθ −Δ = ° − ° = ° = ×
0.5 stΔ =
3
317.453 10 34.907 10 rad/s0.5t
θθ−
−Δ ×≈ = = ×Δ
&
Let r be a polar coordinate with origin at A.
34 km 4 10 mb = = ×
3
34 10 5.921 10 mcos cos 47.5
brθ
×= = = ×°
( )( )3 35.921 10 34.907 10 206.68 m/sv rθ θ −= = × × =&
From geometry, 206.68cos cos 47.5
vv θθ
= =°
306 m/sv =
Alternate solution. tanx b θ=
22sec
cosbv x b θθθ
θ= = =
&&&
( )( )3 3
2
4 10 34.907 10306 m/s
cos 47.5v
−× ×= =
°
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Chapter 11, Solution 174.
Changes in values over the interval 13600 12600 1000 ftrΔ = − =
328.3 31.2 2.9 5.0615 10 radθ −Δ = ° − ° = − ° = − ×
2t sΔ =
Rates of change. 1000 500 ft/s2
rrt
Δ= = =Δ
&
3
35.0615 10 2.5307 10 rad/s2t
θθ−
−Δ − ×= = = − ×Δ
&
Mean values. 12600 13600 13100 ft2
r += =
31.2 28.3 29.752
θ ° + °= = °
Velocity components.
500 ft/srv r= =&
( )( )313100 2.5307 10 331.53 ft/sv rθ θ −= = − × = −&
( ) ( )2 22 2 500 331.53 600 ft/srv v vθ= + = + − =
409 mi/hv =
cos sinx rv v vθθ θ= −
( )500cos 29.75 331.53 sin 29.75 598.61 ft/s= ° − − ° =
sin cosy rv v vθθ θ= +
( )( )500sin 29.75 331.53 cos 29.75 39.73 ft/s= ° + − ° = −
39.73tan 0.06636598.61
y
x
vv
α−
= = = 3.80α = °
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Chapter 11, Solution 175.
2 21/2 1/2, r be r beθ θ θθ= = &&
( ) ( )
2 2
2
1/2 1/2
22 2 2 1/2 2 2
,
1
r
r
v r be v r be
v v v be
θ θθ
θθ
θθ θ θ
θ θ
= = = =
= + = +
& & &&
&
( )2 1/21 2 2 1v be θ θ θ= + &
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Chapter 11, Solution 176.
2 32, b br r θ
θ θ= = − &&
3 22 , rb bv r v rθθ θ θ
θ θ= = − = =& & &&
( )2 3 2
2 2 2 2 2 2 26 4 6
4 4rb b bv v vθ θ θ θ θ
θ θ θ= + = + = +& & &
( )1 223 4bv θ θ
θ= + &
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Chapter 11, Solution 177.
( )
( ) ( )
2 2 2
2 2
2 2 2
21/2 1/2 1/2 2
2 22 1/2 2 2 1/2
1/2 1/2 2 1/2 2
, ,
2 2 2
r
r be r be r be
a r r be be
a r r be be be
θ θ θ
θ θ
θ θ θθ
θθ θθ θ θθ
θ θθ θ θθ θ θθ θθ
θ θ θ θθ θ θθ
⎡ ⎤= = = + +⎢ ⎥⎣ ⎦
⎡ ⎤ ⎡ ⎤= − = + + − = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
⎡ ⎤= + = + = +⎣ ⎦
& & & &&& &&
& & & && & & &&&&
&& & && & && &&
But and 0θ ω θ= =& &&
( ) ( )( ) ( )
2 2
2
21/2 1/2 2
22 2 2 1/2 4 2 4
and 2
4
r
r
a be a be
a a a be
θ θθ
θθ
θω θω
θ θ ω
= =
= + = +
( )2 1/21/2 2 24a be θ θ θ ω= +
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Chapter 11, Solution 178.
22 3 3 4
2 2 6, , b b b br r rθ θ θθ θ θ θ
= = − = − +& && && &&
( )2 2 2 2 2 23 4 2 4
2 6 2 6rb b b ba r rθ θ θ θ θθ θ θ θ
θ θ θ θ= − = − + − = − + −& & & & && & &&&
( ) ( )2 22 3 3
22 2 4b b ba r rθ θ θ θ θ θθ θθ θ θ
⎛ ⎞= + = = − = −⎜ ⎟⎝ ⎠
&& & && & && &&
But and 0θ ω θ= =& &&
( )2 2 24 3
46 and rb ba aθθ ω ω
θ θ= − = −
( )2 2
2 2 2 2 4 2 28 6
1636 12rb ba a aθ θ θ ω ωθ θ
= + = − + +
( )2
2 4 28 36 4b θ θ ω
θ= + +
( )1 22 4 24 36 4ba θ θ ω
θ= + +
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Chapter 11, Solution 179.
Sketch the geometry.
Law of cosines: 2 2 2 2 cosr d h dh ϕ= + −
Differentiating with respect to time and noting that d and h are constant,
2 2 sinrr dh ϕϕ=& &
sindhrr
ϕ ϕ=& &
Law of sines: sin sinr dϕ θ=
so that sin Q.E.Dr h θϕ=& &
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Chapter 11, Solution 180.
Given: , , 1 1
A CtR Bt zt t
θ= = =+ +
Differentiating with respect to time,
( )( )( ) ( )
( ) ( )
2 2 2
3 3
1, ,
1 1 1
2 2, 0,1 1
C t CtA CR B zt t t
A CR zt t
θ
θ
+ −= − = = =
+ + +
= = = −+ +
&& &
&&&& &
(a) 0.t = , 0, 0R A zθ= = =
, ,
2 , 0, 2
R A B z C
R A z C
θ
θ
= − = =
= = = −
&&& &
&&&& &&
, , R zv R A v R AB v z Cθ θ= = − = = = =&& &
2 2 2 2 2 2 2 2R zv v v v A A B Cθ= + + = + + 2 2 2 2v A A B C= + +
2 2 2 2 2 2 2 42 4 4R Ra R R A AB a A A B A Bθ= − = − = − +&&&
2 0 2a R R ABθ θ θ= + = −&& && 2 24a A Bθ =
2za z c= = −&& 2 24za C=
2 2 2 2 2 2 4 24 4R za a a a A A B Cθ= + + = + + 2 2 4 24 4a A A B C= + +
(b) .t = ∞ 0, , , 0, , 0,R z C R B zθ θ= = ∞ = = = =&& &
0, 0, 0R zθ= = =&&&& &&
0, 0, 0,r zv R v R v zθ θ= = = = = =&& & 0v =
2 20, 0, 0,r za R R a R R a zθθ θ θ= − = = − = = =& && &&& &&
0a =
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Chapter 11, Solution 182.
From problem 11.97, the position vector is ( ) ( )cos sin .n nRt t ct Rt tω ω= + +r i j k
Differentiating to obtain v and a,
( ) ( )
( ) ( )2 2
cos sin sin cos
sin sin cos cos cos sin
n n n n n n
n n n n n n n n n n n n
dR t t t c R t t t
dtd
R t t t t R t t t tdt
ω ω ω ω ω ω
ω ω ω ω ω ω ω ω ω ω ω ω
= = − + + +
= = − − − + + −
rv i j k
va i k
( ) ( )2 22 sin cos 2 cos sinn n n n n n n nR t t t t t tω ω ω ω ω ω ω ω = − − + − i k
( ) ( ) ( )x y z y z z y z x x z x y y x
x y z
v v v v a v a v a v a v a v a
a a a
× = = − + − + −i j k
v a i j k
( ) ( )( )( )( )
( )
2 2 2
2 2
2
2 cos sin sin cos 2 sin cos
cos sin 2 cos sin
2 sin cos
n n n n n n n n n n n
n n n n n n n
n n n n
cR t t t R t t t t t t
R t t t t t t
cR t t t
ω ω ω ω ω ω ω ω ω ω ω
ω ω ω ω ω ω ω
ω ω ω ω
= − + + − −
− − −
+ − − −
i
j
k
( ) ( ) ( )2 2 22cos sin 2 2sin cosn n n n n n n n n ncR t t t R t cR t t tω ω ω ω ω ω ω ω ω ω= − − + + +i j k
( ) ( )1/222 2 2 2 2 4 2 2 2| | 4 2n n n nc R R tω ω τ ω ω × = + + +
v a
The binormal unit vector b e is given by | |b
×=×
v a e
v a
Let α be the angle between the y-axis and the binormal.
( ) ( )( ) ( )
2
1 22 2 2 2 2 4 2 2 2
2cos
| | 4 2
n nb
n n n n
R t
v a c R t R t
ω ωα
ω ω ω ω
+× ⋅= ⋅ = =
× + + +
v a je j
( ) ( )1 22 2 2 2 2Let 2 , 4 ,n n n nA R t B cR tω ω ω ω= + = + 2 2 C A B= + so that cos as
A
Cα =
shown in the sketch. The angle that the osculating plane makes with the y-axis is the angle .β
( )
( )2 2
1 22 2
2tan
4
n
n
R tA
B c t
ωβ
ω
+= =
+
( )( )
2 21
1 22 2
2tan
4
n
n
R t
c t
ωβ
ω−
+=
+ !
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Chapter 11, Solution 183.
For 3 and 1,A B= = ( ) ( ) ( )23 cos 3 1 sint t t t t= + + +r i j k
Differentiating to obtain v and a.
( ) ( )
( )( )
( )
2
3/22
3 cos sin 3 sin cos1
13 2sin cos 3 2cot sin
1
d tt t t t t t
dt t
dt t t t t t
dt t
= = − + + ++
= = − − + + −+
rv i j k
va i j k
(a) At 0,t = ( ) ( ) ( )3 1 0 0 0 3= − + + =v i j k i
( )3(0) 3(1) 2 0 3 2= − + + − = +a i j k j k
3 0 0 6 9
0 3 2
× = = − +i j k
v a j k
2 2| | 6 9 10.817× = + =v a
0.55470 0.83205| |b
×= = − +×
v ae j k
v a
2cos 0, cos 0.55470, cos 0.83205x yθ θ θ= = − =
z90 , 123.7 , 33.7x yθ θ θ= ° = ° = ° !
(b) At s,2
tπ= 4.71239 2.53069v = − + +i j k
6 0.46464 1.5708= − + −a i j k
4.71239 2.53069 1
6 0.46464 1.5708
4.43985 13.4022 12.9946
× = −−
= − − +
i j k
v a
i j k
continued
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( ) ( ) ( )1/22 2 2
| | 4.43985 13.4022 12.9946 19.1883 × = + + = v a
0.23138 0.69846 0.67721| |b
×= = − − +×
v ae i j k
v a
cos 0.23138, cos 0.69846, cos 0.67721x y zθ θ θ= − = − =
103.4 , 134.3 , 47.4x y zθ θ θ= ° = ° = ° !
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Chapter 11, Solution 184.
Given: 2 2
0 9 9ft/s , 36 ft, 144 ft, 27 ft/sa kt x x v= = = =
2 30 0 0
13
t tv v a dt kt dt kt− = = =∫ ∫
Velocity: 3
013
v v kt= +
40 00
112
tx x v dt v t kt− = = +∫
Position: 4 4
0 0 01 136
12 12x x v t kt v t kt= + + = + +
When 9 s,t = 144 ft and 27 ft/sx v= =
( ) ( )40
136 9 9 14412
v k+ + =
or 09 546.75 108v k+ = (1)
( )30
1 9 273
v k+ =
0 243 27v k+ = (2)
Solving equations (1) and (2) simultaneously yields:
40 7 ft/s and 0.082305 f t/sv k= =
Then, 4 36 7 0.00686 ftx t t= + +
37 0.0274 ft/sv t= +
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Chapter 11, Solution 185.
(a) Determination of k.
From ( ), 0.6 1
dv dvdv a dt dt
a kv= = =
−
Integrating, using the condition 0v = when 0,t =
( ) ( ) ( )0 0 0 0
1 1 or ln 1 ln 1
0.6 1 0.6 0.6
vtt v dvdt t kv t kv
kv k k = = − − = − − −∫ ∫ (1)
Using 20 s when 6 mm/s,t v= = ( )120 ln 1 6
0.6k
k= − −
Solving by trial, 0.1328 s/mk =
(b) Position when 7.5 m/s.v =
From ,v dv a dx= ( )0.6 1
v dv v dvdx
a kv= =
−
Integrating, using the condition 6 mx = when 0,v = ( )6 0 0.6 1x v v dvdx
kv=
−∫ ∫
( )00
1 1 1 16 1 ln 1
0.6 1 0.6
vv
x dv v kvk kv k k
− = − + = − − − − ∫
( )1 16 ln 1
0.6x v kv
k k = − + −
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Using 7.5 m/sv = and the determined value of k:
( )( ) ( )( )( )1 16 7.5 ln 1 0.1328 7.5
0.6 0.1328 0.1328x
= − + − 434 mx =
(c) Maximum velocity occurs when a = 0. max1 1
0.1328v
k= = max 7.53 m/s v =
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Chapter 11, Solution 186.
Constant acceleration. 0 25 mi/h 36.667 ft/sv = =
65 mi/h 95.333 ft/sfv = =
0 0 and 0.1 mi 528 ftfx x= = =
( )2 20 02f fv v a x x= + −
(a) Acceleration. ( ) ( )2 2 2 2
0 2
0
95.333 36.667 7.3333 ft/s2 528 02
f
f
v va
x x− −= = =
−−
27.33 ft/sa =
(b) Time to reach 65 mph. 0f fv v at= +
0 95.333 36.667
7.3333f
fv v
ta− −= = 8.00 sft =
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Chapter 11, Solution 187.
Let x be position relative to the fixed supports, taken positive if downward.
Constraint of cable on left: 2 3 constantA Bx x+ =
2 22 3 0, or , and 3 3A B B A B Av v v v a a+ = = − = −
Constraint of cable on right: 2 constantB Cx x+ =
1 1 12 0, or , and 2 3 3B C C B A C Av v v v v a a+ = = − = =
Block C moves downward; hence, block A also moves downward.
(a) Accelerations.
( ) ( ) 200
456 0 or 38.0 mm/s12
A AA A A A
v vv v a T a
t− −= + = = =
238.0 mm/sA =a
( ) 22 2 38.0 25.3 mm/s3 3B Aa a ⎛ ⎞= − = − = −⎜ ⎟
⎝ ⎠ 225.3 mm/sB =a
( ) 21 1 38.0 12.67 mm/s3 3C Aa a ⎛ ⎞= = =⎜ ⎟
⎝ ⎠ 212.67 mm/sC =a
(b) Velocity and change in position of B after 8 s.
( ) ( )( )0 0 25.3 8 203 mm/sB B Bv v a t= + = + − = −
203 mm/sB =v
( ) ( ) ( )( )220 0
1 10 25.3 8 811 mm2 2B B B Bx x v t a t− = + = + − = −
811 mmBxΔ =
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Chapter 11, Solution 188.
(a) Construction of the curves.
Construct the a t− curve. slope of curvea v t= −
0 10 s:t< < 10 s,t∆ = 0v∆ = 0v
at
∆= =∆
10 s 26 s:t< < 16 s,t∆ = 80 m/sv∆ = − 25 m/sv
at
∆= = −∆
26 s 41 s:t< < 15 s,t∆ = 0v∆ = 0v
at
∆= =∆
41 s 46 s:t< < 5 s,t∆ = 15 m/sv∆ = 23 m/sv
at
∆= =∆
46 s 50 s:t< < 4 s,t∆ = 0v∆ = 0v
at
∆= =∆
Construct the curve.x t− area of curve.x v t∆ = −
x is maximum or minimum where 0.v =
For 10 s 26 s,t≤ ≤ ( )60 5 10v t= − −
0v = when 60 5 50 0 or 22 st t− + = =
Also 0 540 mx = −
0 to 10 s ( )( )10 60 600 mx∆ = = 10 540 600 60 mx = − + =
10 s to 22 s ( )( )112 60 360 m
2x∆ = = 22 60 360 420 mx = + =
22 s to 26 s ( )( )14 20 40 m
2x∆ = − = − 26 420 40 380 mx = − =
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26 s to 41 s ( )( )15 20 300 mx∆ = − = − 41 380 300 80 mx = − =
41 s to 46 s ( ) 20 55 62.5 m
2x
− − ∆ = = −
46 80 62.5 17.5 mx = − =
46 s to 50 s ( )( )4 5 20 mx∆ = − = − 50 17.5 20 2.5 mx = − = −
(b) Total distance traveled.
( )1 22 00 22 s, 420 540 960 mt d x x≤ ≤ = − = − − =
2 50 2222 s 50 s, 2.5 420 422.5t d x x≤ ≤ = − = − − =
Total: 1 2 1382.5 md d d= + = 1383 md =
(c) Times when 0.x =
For 0 10 s,t≤ ≤ 540 60 mx t= − +
At 0,x = 540 60 0t− + = 9 s t =
For 46 s 50,t≤ ≤ ( )17.5 5 46 mx t= − −
At 0,x = ( )17.5 5 46 0 46 3.5t t− − = − = 49.5 st =
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Chapter 11, Solution 189.
( ) ( )0 0100 km/h 27.778 m/s 25 km/h 6.944 m/sA Bv v= = = =
Sketch acceleration curve for car B over 0 5 s.t< <
Using moment-area formula at 5 s.t =
( ) ( ) ( )( )( )( )( )
0
2
5 2.5
70 6.944 5 12.5
2.822 m/s
B B o B
B
B
x x v t a
a
a
− = +
= +
=
Determine when reaches 100 km/h.B
( ) ( ) 20
27.778 6.944 2.8227.38 s
B Bf
B
B
v v A
tt
= +
= +=
( )( )2 2.822 7.38 20.83 m/sA = =
Then, ( ) ( ) 20 0 2B
B B B Btx x v t A= + + by moment-area formula
and ( ) ( )0 0A A A Bx x v t= +
Subtracting, ( ) ( ) ( ) ( ) 20 0 0 0 2B
B A B A B A Btx x x x v v t A⎡ ⎤− = − + − +⎣ ⎦
Then, ( )( ) ( ) 7.38120 6.944 27.778 7.38 20.832B Ax x ⎛ ⎞− = + − + ⎜ ⎟
⎝ ⎠
/Car is ahead of car . 43.1 mB AB A x =
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Chapter 11, Solution 190.
(a) Vertical motion: 0 1.5 m,y = ( )00yv =
( ) ( )020 0
21 or 2y
y yy y v t gt t
g−
= + − =
At point B, ( )02 or B
y hy h t
g−
= =
When 788 mm 0.788 m,h = = ( )( )2 1.5 0.7880.3810 s
9.81Bt−
= =
When 1068 mm 1.068 m,h = = ( )( )2 1.5 1.0680.2968 s
9.81Bt−
= =
Horizontal motion: ( )0 000, ,xx v v= =
0 0 or B
B
x xx v t vt t
= = =
With 12.2 m,Bx = 012.2we get 32.02 m/s
0.3810v = =
012.2and 41.11 m/s
0.2968v = =
032.02 m/s 41.11 m/sv≤ ≤ or 0115.3 km/h 148.0 km/hv ≤≤
(b) Vertical motion: ( )0y yv v gt gt= − = −
Horizontal motion: 0xv v=
( )( ) 0
tany BB
x B
vdy gtdx v v
α = − = − =
For 0.788 m,h = ( )( )9.81 0.3810tan 0.11673,
32.02α = = 6.66α = °
For 1.068 m,h = ( )( )9.81 0.2968tan 0.07082,
41.11α = = 4.05α = °
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Chapter 11, Solution 191.
The horizontal and vertical components of velocity are
0
0
sin15cos15
x
y
v vv v gt
= °= ° −
At point B,
0
0
sin15 tan12cos15
x
y
v vv v gt
°= = − °° −
or 0 0sin15 cos15 tan12 tan12v v gt° + ° ° = °
00.46413 tan12v gt= °
02.1836 vtg
=
Vertical motion:
( )
20 0
2220 0
20
1cos152
12.1836cos15 2.18362
0.27486
y y v t gt
v vgg g
vg
− = ° −
⎛ ⎞= ° − ⎜ ⎟
⎝ ⎠
= −
( ) ( )( )20 0
2 2
83.638 3.638 32.2 012
78.10 ft / s
v g y y ⎛ ⎞= − − = − − −⎜ ⎟⎝ ⎠
=
0 8.84 ft /s v =
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Chapter 11, Solution 192.
First determine the velocity Cv of the coal at the point where the coal impacts on the belt.
Horizontal motion: ( ) ( )0
1.8cos50C Cx xv v⎡ ⎤= = − °⎣ ⎦
1.1570 m/s= −
Vertical motion: ( ) ( ) ( )22
00
2C Cy yv v g y y⎡ ⎤= − −⎣ ⎦
( ) ( )( )( )
( )
2
2 2
1.8sin 50 2 9.81 1.5
31.331 m / s
5.5974 m/sC yv
= ° − −
=
= −
( ) ( )2 22 2 2
5.5974tan 4.8379, 78.321.1570
32.669 m /sC C Cx yv v v
β β−= = = °−
= + =
5.7156 m/s, 5.7156 m/sC Cv = =v 78.32°
or ( ) ( )1.1570 m/s 5.5974 m/sC = − + −v i j
Velocity of the belt: ( )cos10 sin10B Bv= − ° + °v i j
Relative velocity: ( )/C B C B C B= − = + −v v v v v
(a) /C Bv is vertical. ( )/ 0C B xv =
( ) ( )/ 1.1570 cos10 0, 1.175 m/sC B B Bxv v v= − − − ° = =
1.175 m/sB =v 10 °
(b) /C Bv is minimum. Sketch the vector addition as shown.
2 2 2/ 2 cos88.32B C B C B Cv v v v v= + − °
Set the derivative with respect to Bv equal to zero.
2 2 cos88.32 0B Cv v− ° =
cos88.32 0.1676 m/sB Cv v= ° = 0.1676 m/sB =v 10 °
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Chapter 11, Solution 193.
Given: ( ) ( ) 20 0, 0.8 in./sA
A A tdvv adt
= = =
Then, ( ) ( )0 0.8 A A A tv v a t t= + =
(a) 0,t = ( )2
0, 0AA A n
vv aρ
= = =
( )A A ta a= 20.800 in./sAa =
(b) 2 s,t = ( )( )0 0.8 2 1.6 in./sAv = + =
( ) ( )2221.6
0.731 in./s3.5
AA n
vaρ
= = =
( ) ( ) ( ) ( )1/2 1/22 2 2 20.8 0.731A A At na a a⎡ ⎤ ⎡ ⎤= + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦
21.084 in./sAa =
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Chapter 11, Solution 194.
(a) At point A. Aa g= 29.81 m/s=
Sketch tangential and normal components of acceleration at A.
( ) cos50A na g= °
( )( )22 2
9.81cos50A
AA n
va
ρ = =°
0.634 mAρ =
(b) At point B, 1 meter below point A.
Horizontal motion: ( ) ( ) 2cos50 1.286 m/sB Ax xv v= = ° =
Vertical motion: ( ) ( ) ( )2 2 2B A y B Ay yv v a y y= + −
( ) ( )( )( )2
2 2
2cos 40 2 9.81 1
21.97 m /s
= ° + − −
=
( ) 4.687 m/sB yv =
( )( )
4.687tan , or 74.61.286
B y
B x
v
vθ θ= = = °
cos74.6Ba g= °
( )( ) ( )2 2
2
cos74.6B Bx yB
BB n
v vva g
ρ+
= =°
( )21.286 21.979.81cos74.6
+=
° 9.07 mBρ =
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Chapter 11, Solution 195.
Differentiate the expressions for r and θ with respect to time.
( ) 26 4 2 ft, 12 ft/s, 12 ft/st t tr e r e r e− − −= − = = −& &&
( ) ( )2 2 2 22 2 4 rad, 2 2 8 rad/s 32 rad/st t tt e e eθ θ θ− − −= + = − =& &&
(a) At 0 s,t = 212 ft, 12 ft/s, 12 ft/sr r r= = = −& &&
28 rad, 12 rad/s, 32 rad/sθ θ θ= = − =& &&
12 ft/s, 144 ft/srv r v rθ θ= = = = −&&
( ) ( )12 ft/s 144 ft/sr θ= −v e e
( )( )( )( ) ( )( )( )
22 2
2
12 12 12 1740 ft/s
2 12 32 2 12 12 96 ft/s
ra r r
a r rθ
θ
θ θ
= − = − − = −
= + = + − =
&&&
&& &&
( ) ( )2 21740 ft/s 96 ft/sr θ= − +a e e
(b) At t ,∞ te− 20 and te− 0
24 ft, 0, 0 r r r≈ ≈ ≈& &&
4 rad, 4 rad/s, 0tθ θ θ≈ ≈ ≈& &&
0, 96 rad/srv r v rθ θ= ≈ = ≈&&
( )96 ft/s θ=v e
( )( )22 224 4 384 ft/s , 0ra r r aθθ≈ − = − = − ≈&&&
( )2384 ft/s r= −a e
The particle is moving on a circular path of radius of 24 ft and with a speed of 96 ft/s. The acceleration is the
normal acceleration ( )22 296/ 384 ft/s
24v r = = directed toward the center of the circle.