solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 11

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 =




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 =




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 = !




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 = − × !




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 = −




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 =




( )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 =




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 =




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 =




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 = !




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 =




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 =




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 =




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= − +




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 =




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



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 =




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 −=




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= ± +



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 =




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 = !



(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 = − !




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 =




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 =




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 =




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 = ∞




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 =




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




( )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 =




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 = ×




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 = ∞




( )( ) 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

+′ =



( ) 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

−




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=




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= − =




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 =



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 =



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 =




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 =



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 =



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 =




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 =



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 =



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

± + ± += = =



(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 =



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 =



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




( )( ) ( )( )( )( )( )

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 =




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 =



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 =



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



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 =




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 =




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= − =



(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 =




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 =




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 =




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 =




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 =




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 =




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 =




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− =




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− =




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 =




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 =




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 =




(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



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− + =


( ) ( )( )( )( )( )

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− =




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



(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= =




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



(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 = !




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 =




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 =




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 =




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 =



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.




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 =



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∆ =




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 =




( )0 32 km/h 8.889 m/s

24 km/h 6.667 m/sA

B

v

v

= =

= =


( )( )

( )( ) ( )

( )( )

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 = −




( )( )

0

0

22 mi/h 32.267 ft/s

13 mi/h 19.067 ft/sA

B

v

v

= =

= =


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 =




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.




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 =



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+ =




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=



( )( ) ( )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 = !




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




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 =




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 =


( ) ( ) ( )

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 =




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 = −




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−




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




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




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 =




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=


( ) ( )0 0 0 012

x x v A t t x v t at t⎛ ⎞= + + − = + + ⎜ ⎟⎝ ⎠

20 0

12

x v t at= + +




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 = −




(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− = = −


( )( ) ( ) ( ) ( )13 3.21670 90 4.0167 12.8 0.8 3.2167 77.28 2

x ⎡ ⎤ ⎛ ⎞= + + − + + − ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

1 192.3 ft x =




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 = !




( )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 °



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°




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°




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




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



( ) ( )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 = °




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 =




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 °




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


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

= + +

⎡ ⎤= − − + −⎢ ⎥⎣ ⎦

⎡= + + +⎣

⎤− + ⎦

( )2 2 4 24 n nR tω ω= + 2 24 n na R tω ω= +




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


( ) ( )

( )( )

( )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



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 =




(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 =




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< <




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 <




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 =




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 <




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= − =




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 =




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 = !




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 = !




( ) ( )

( ) ( )

( )( )

( )( )

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= = !




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 = !




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 = !




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 = !




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 = !




(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 °!




(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° !




(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 = !




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 !




(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 = !




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 = !




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 = !




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 = !




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= !



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

= − +

!




Given: ( ) ( )cos sinn nRt t ct Rt tω ω= + +r j ki


( ) ( )

( ) ( )( ) ( )

2 2

2 2

cos sin sin cos


2 sin cos 2 cos sin

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



= = − + + +

= = − − − + + −

= − − + −

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


= + +

= + +

= − − + −

= + +

+ − +

= ( )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+= !




With 3 and 1,A B= = the position vector is

( ) ( ) ( )23 cos 3 1 sinr t t t t t= + + +i j k


( ) ( )

( )

( )

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 = !




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 = !




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 = !




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 = !





2ngRar

=


nvar

=


=

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 = !





2ngRar

=


nvar

=


=

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 = !




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 = !




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 = !




From Problems 155 through 156, 2

2ngR

ar

=


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



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 = !




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&& !




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 !




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.





( )

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





( ) 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




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 = ∞




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=

&




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= +&& &




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=

−&




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 =&&




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

−× ×= =

°




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 = °




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= + &




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= + &




( )

( ) ( )

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= +




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= + +




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=& &




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 =




From problem 11.97, the position vector is ( ) ( )cos sin .n nRt t ct Rt tω ω= + +r i j k


( ) ( )

( ) ( )2 2

cos sin sin cos


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


= = − + + +

= = − − − + + −

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

= − + + − −

− − −

+ − − −

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−

+=

+ !




For 3 and 1,A B= = ( ) ( ) ( )23 cos 3 1 sint t t t t= + + +r i j k


( ) ( )

( )( )

( )

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



( ) ( ) ( )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= ° = ° = ° !




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= +




(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 = − + −



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 =




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 =




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 =




(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 = − =



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 =




( ) ( )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 =




(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 = °




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 =




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 °




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 =




(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 =





( ) 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.

solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 11

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Transcript of solucionario mecanica vectorial para ingenieros - beer & johnston (dinamica) 7ma edicion Cap 11