Solucionario Beer, Johnton, Mazurek y Eisenberg - Octava Edicion.pdf

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LIBROS UNIVERISTARIOS
Y SOLUCIONARIOS DE
MUCHOS DE ESTOS LIBROS
LOS SOLUCIONARIOS
CONTIENEN TODOS LOS
EJERCICIOS DEL LIBRO
RESUELTOS Y EXPLICADOS
DE FORMA CLARA
VISITANOS PARA
DESARGALOS GRATIS.

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.
( )2
3
2 m
x t t
= − −
( )
2
3 2 2 m/s
dx
v t t
dt
= = − −
2
6 2 m/s
dv
a t
dt
= = −
(a) Time at a = 0.
0
0 6 2 0
t
= − =
0
1
3
t = 0 0.333 s
t = W
(b) Corresponding position and velocity.
3 2
1 1
2 2.741 m
3 3
x
⎛ ⎞ ⎛ ⎞
= − − = −
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2.74 m
x = − W
2
1 1
3 2 2 3.666 m/s
3 3
v
⎛ ⎞ ⎛ ⎞
= − − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
3.67 m/s
v = W

Position: 4 3
5 4 3 2 ft
x t t t
= − + −
Velocity: 3 2
20 12 3 ft/s
dx
v t t
dt
= = − +
Acceleration: 2 2
60 24 ft/s
dv
a t t
dt
= = −
When 2 s,
t =
( )( ) ( )( ) ( )( )
4 3
5 2 4 2 3 2 2
x = − − − 52 ft
x = W
( )( ) ( )( )
3 2
20 2 12 2 3
v = − + 115 ft/s
v = W
( )( ) ( )( )
2
60 2 24 2
a = − 2
192 ft/s
a = W

Position: 4 3 2
6 8 14 10 16 in.
x t t t t
= + − − +
Velocity: 3 2
24 24 28 10 in./s
dx
v t t t
dt
= = + − −
Acceleration: 2 2
72 48 28 in./s
dv
a t t
dt
= = + −
When 3 s,
t =
( )( ) ( )( ) ( )( ) ( )( )
4 3 2
6 3 8 3 14 3 10 3 16
x = + − − + 562 in.
x = !
( )( ) ( )( ) ( )( )
3 2
24 3 24 3 28 3 10
v = + − − 770 in./s
v = !
( )( ) ( )( )
2
72 3 48 3 28
a = + − 2
764 in./s
a = !

Position: 500sin mm
x kt
=
Velocity: 500 cos mm/s
dx
v k kt
dt
= =
Acceleration: 2 2
500 sin mm /s
dv
a k kt
dt
= = −
When 0.05 s, and 10 rad/s
t k
= =
( )( )
10 0.05 0.5 rad
kt = =
( )
500sin 0.5
x = 240 mm
x = !
( )( ) ( )
500 10 cos 0.5
v = 4390 mm/s
v = !
( )( ) ( )
2
500 10 sin 0.5
a = − 3 2
24.0 10 mm/s
a = − × !

Position: ( )
2
1 2
50sin mm
x k t k t
= −
Where 2
1 2
1 rad/s and 0.5 rad/s
k k
= =
Let 2 2
1 2 0.5 rad
k t k t t t
θ = − = −
( )
2
2
2
1 rad/s and 1 rad/s
d d
t
dt dt
θ θ
= − = −
Position: 50sin mm
x θ
=
Velocity: 50cos mm/s
dx d
v
dt dt
θ
θ
= =
Acceleration:
dv
a
dt
=
2
2
2
2
50cos 50sin mm/s
d d
a
dt
dt
θ θ
θ θ
⎛ ⎞
= − ⎜ ⎟
⎝ ⎠
When 0,
v = either cos 0
θ =
or 1 0 1 s
d
t t
dt
θ
= − = =
Over 0 2 s, values of cos are:
t θ
≤ ≤
( )
s
t 0 0.5 1.0 1.5 2.0
( )
rad
θ 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 = ( )( )2
1 0.5 1 0.5 rad
θ = − =
( )
50sin 0.5
x = 24.0 mm
x = W
( )( ) ( )( )
50cos 0.5 1 50sin 0.5 0
a = − − 2
43.9 mm/s
a = − W

Given: 3 2
6 9 5
x t t t
= − + +
Differentiate twice. 2
3 12 9
dx
v t t
dt
= = − +
6 12
dv
a t
dt
= = −
(a) When velocity is zero. 0
v =
( )( )
2
3 12 9 3 1 3 0
t t t t
− + = − − =
1 s and 3 s
t t
= = W
(b) Position at t = 5 s.
( ) ( )( ) ( )( )
3 2
5 5 6 5 + 9 5 + 5
x = − 5 25 ft
x = W
Acceleration at t = 5 s.
( )( )
5 6 5 12
a = − 2
5 18 ft/s
a = W
Position at t = 0.
0 5 ft
x =
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 2
1 1 6 1 9 1 5 9 ft
x = − + + =
Position at t = 3 s.
( ) ( )( ) ( )( )
3 2
3 3 6 3 9 3 5 5 ft
x = − + + =
Distance traveled.
At t = 1 s 1 1 0 9 5 4 ft
d x x
= − = − =
At t = 3 s 3 1 3 1 4 5 9 8 ft
d d x x
= + − = + − =
At t = 5 s 5 3 5 3 8 25 5 28 ft
d d x x
= + − = + − =
5 28 ft
d = W

( )3
2
2 ft
x t t
= − −
( )2
2 3 2 ft/s
dx
v t t
dt
= = − −
(a) Positions at v = 0.
( )2 2
2 3 2 3 14 12 0
t t t t
− − = − + − =
2
14 (14) (4)( 3)( 12)
(2)( 3)
t
− ± − − −
=
−
1 2
1.1315 s and 3.535 s
t t
= =
1
At 1.1315 s,
t = 1 1.935 ft
x = 1 1.935 ft
x = W
2
At 3.535 s,
t = 2 8.879 ft
x = 2 8.879 ft
x = W
(b) Total distance traveled.
0
At 0,
t t
= = 0 8 ft
x =
4
At 4 s,
t t
= = 4 8 ft
x =
Distances traveled.
1
0 to :
t 1 1.935 8 6.065 ft
d = − =
1 2
to :
t t 2 8.879 1.935 6.944 ft
d = − =
2 4
to :
t t 3 8 8.879 0.879 ft
d = − =
Adding, 1 2 3
d d d d
= + + 13.89 ft
d = W

0.2
3 t
a e−
=
0 0
v t
dv a dt
=
∫ ∫
0.2 0.2
0
0
3
0 3
0.2
t
t t t
v e dt e
− −
− = =
−
∫
( ) ( )
0.2 0.2
15 1 15 1
t t
v e e
− −
= − − = −
At t = 0.5 s, ( )
0.1
15 1
v e−
= − 1.427 ft/s
v = W
0 0
x t
dx v dt
=
∫ ∫
( )
0.2 0.2
0
0
1
0 15 1 15
0.2
t
t t t
x e dt t e
− −
⎛ ⎞
− = − = +
⎜ ⎟
⎝ ⎠
∫
( )
0.2
15 5 5
t
x t e−
= + −
At 0.5 s,
t = ( )
0.1
15 0.5 5 5
x e−
= + − 0.363 ft
x = W

Given: 2
0 0
5.4sin ft/s , 1.8 ft/s, 0, 3 rad/s
a kt v x k
= − = = =
0 0 0 0
5.4
5.4 sin cos
t
t t
v v adt kt dt kt
k
− = = − =
∫ ∫
( )
5.4
1.8 cos 1 1.8cos 1.8
3
v kt kt
− = − = −
Velocity: 1.8cos ft/s
v kt
=
0 0 0 0
1.8
1.8 cos sin
t
t t
x x vdt kt dt kt
k
− = = =
∫ ∫
( )
1.8
0 sin 0 0.6sin
3
x kt kt
− = − =
Position: 0.6sin ft
x kt
=
When 0.5 s,
t = ( )( )
3 0.5 1.5 rad
kt = =
1.8cos1.5 0.1273 ft/s
v = = 0.1273 ft/s
v = W
0.6sin1.5 0.5985 ft
x = = 0.598 ft
x = W

Given: 2
3.24sin 4.32cos ft/s , 3 rad/s
a kt kt k
= − − =
0 0
0.48 ft, 1.08 ft/s
x v
= =
( ) ( )
0 0 0 0
0 0
3.24 sin 4.32 cos
3.24 4.32
1.08 cos sin
3.24 4.32
cos 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 kt
k k
kt kt
kt kt
− = = − −
− = −
= − − −
= − −
∫ ∫ ∫
Velocity: 1.08cos 1.44sin ft/s
v kt kt
= −
( ) ( )
0 0 0 0
0 0
1.08 cos 1.44 sin
1.08 1.44
0.48 sin cos
1.08 1.44
sin 0 cos 1
3 3
0.36sin 0.48cos 0.48
t t t
t t
x x vdt kt dt kt dt
x kt kt
k k
kt kt
kt kt
− = = −
− = +
= − + −
= + −
∫ ∫ ∫
Position: 0.36sin 0.48cos ft
x kt kt
= +
When 0.5 s,
t = ( )( )
3 0.5 1.5 rad
kt = =
1.08cos1.5 1.44sin1.5 1.360 ft/s
v = − = − 1.360 ft/s
v = − !
0.36sin1.5 0.48cos1.5 0.393 ft
x = + = 0.393 ft
x = !

Given: 2
mm/s where is a constant.
a kt k
=
At 0,
t = 400 mm/s; at 1 s, 370 mm/s, 500 mm
v t v x
= = = =
2
400 0 0
1
2
v t t
dv a dt kt dt kt
= = =
∫ ∫ ∫
2 2
1 1
400 or 400
2 2
v kt v kt
− = = +
At 1 s,
t = ( )2 3
1
400 1 370, 60 mm/s
2
v k k
= + = = −
Thus 2
400 30 mm/s
v t
= −
At 7 s,
t = ( )( )2
7 400 30 7
v = − 7 1070 mm/s
v = − W
2 2 2
When 0, 400 30 0. Then 13.333 s , 3.651 s
v 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
<
( )
2
500 1 1
400 30
x t t
dx vdt t dt
= = −
∫ ∫ ∫
( )
3 3
1
500 400 10 400 10 390
t
x t t t t
− = − = − −
Position: 3
400 10 110 mm
x t t
= − +
At 0,
t =
0 110 mm
x x
= =
At 3.651 s,
t = ( )( ) ( )( )3
max 400 3.651 10 3.651 110 1083.7 mm
x x
= = − + =
At 7 s,
t = ( )( ) ( )( )3
7 400 7 10 7 110
x x
= = − + 7 520 mm
x = − W
Distances traveled:
Over 0 3.651 s,
t
≤ ≤
1 max 0 973.7 mm
d x x
= − =
Over 3.651 7 s,
t
≤ ≤
2 max 7 1603.7 mm
d x x
= − =
Total distance traveled: 1 2 2577.4 mm
d d d
= + = 2580 mm
d = W

Determine velocity. 0.15 2 2
0.15
v t t
dv adt dt
−
= =
∫ ∫ ∫
( ) ( )( )
0.15 0.15 0.15 2
v t
− − = −
0.15 0.45 m/s
v t
= −
At 5 s,
t = ( )( )
5 0.15 5 0.45
v = − 5 0.300 m/s
v = W
When 0,
v = 0.15 0.45 0 3.00 s
t 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 0.45
x t t
dx v dt t dt
−
= = −
∫ ∫ ∫
( ) ( )
2 2
0
10 0.075 0.45 0.075 0.45
t
x t t t t
− − = − = −
2
0.075 0.45 10 m
x t t
= − −
At 5 s,
t = ( )( ) ( )( )
2
5 0.075 5 0.45 5 10 10.375 m
x = − − = −
5 10.38 m
x = − W
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 mm
x x
= = − − = −
Distances traveled:
Over 0 3.00 s,
t
≤ ≤ 1 0 min 0.675 m
d x x
= − =
Over 3.00 s 5 s,
t
< <
2 5 min 0.300 m
d x x
= − =
Total distance traveled: 1 2 0.975 m
d d d
= + = 0.975 m
d = W

Given: 2
9 3
a t
= −
Separate variables and integrate.
( )
2
0 0
9 3 9
v t
dv a dt t dt
= = − =
∫ ∫ ∫
3
0 9
v t t
− = − ( )
2
9
v t t
= −
(a) When v is zero. 2
(9 ) 0
t t
− =
0 and 3 s (2 roots)
t t
= = 3 s
t = W
(b) Position and velocity at 4 s.
t =
( )
3
5 0 0
9
x t t
dx v dt t t dt
= = −
∫ ∫ ∫
2 4
9 1
5
2 4
x t t
− = −
2 4
9 1
5
2 4
x t t
= + −
At 4 s,
t = ( ) ( )
2 4
4
9 1
5 4 4
2 4
x
⎛ ⎞ ⎛ ⎞
= + −
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
4 13 m
x = W
( )( )
2
4 4 9 4
v = − 4 28 m/s
v = − W
(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 4
3
9 1
5 3 3 25.25 m
2 4
x
⎛ ⎞ ⎛ ⎞
= + − =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
At 3 s
t = 3 3 0 25.25 5 20.25 m
d x x
= − = − =
At 4 s
t = 4 3 4 3 20.25 13 25.25 32.5 m
d d x x
= + − = + − = 4 32.5 m
d = W

Given: 2
dv
a 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 2
10 0
dv kt dt
−
=
∫ ∫
10 3
10
0
1
3
t
v kt
−
=
[ ] ( )3
1
(10) ( 10) 2 0
3
k ⎡ ⎤
− − = −
⎢ ⎥
⎣ ⎦
(a) Solving for k,
( )( )
3 20
8
k = 4
7.5 m/s
k = W
(b) Equations of motion.
Using upper limit of v at t,
( )
3 3
10
0
1 1
10 7.5
3 3
t
v
v kt v t
−
⎛ ⎞
= + = ⎜ ⎟
⎝ ⎠
3
10 2.5 m/s
v t
= − + W
Then,
3
10 2.5
dx
v t
dt
= = − +
Separate variables and integrate using x = 0 when t = 2 s.
( )
3
10 2.5
dx t dt
= − +
( )
3
0 2
10 2.5
x t
dx t dt
= − +
∫ ∫
4
2
0 10 0.625
t
x t t
⎡ ⎤
− = − +
⎣ ⎦
( )( ) ( )( )4
4
10 0.0625 10 2 0.625 2
t t ⎡ ⎤
⎡ ⎤
= − + − − +
⎢ ⎥
⎣ ⎦ ⎣ ⎦
[ ]
4
10 0.625 10
t t
= − + − −
4
10 10 0.625 m
x t t
= − + W

Note that is a given function of .
a x ( )
40 160 160 0.25
a x x
= − = −
( ) Note that is maximum when 0, or 0.25 m
a v a x
= =
( )
Use 160 0.25 with the limits
vdv adx x dx
= = −
max
0.3 m/s when 0.4 m and when 0.25 m
v x v v x
= = = =
( )
max 0.25
0.3 0.4
160 0.25
v
vdv x dx
= −
∫ ∫
( ) ( )
0.25
2 2
2 2
max
0.4
0.25 0.15
0.3
160 160 0 1.8
2 2 2 2
x
v ⎡ ⎤
− −
⎢ ⎥
− = − = − − =
⎢ ⎥
⎣ ⎦
2 2 2
max 3.69 m /s
v = max 1.921 m/s
v = W
( ) Note that is maximum or minimum when 0.
b x v =
( )
Use 160 0.25 with the limits
vdv adx x
= = −
0.3 m/s when 0.4 m, and 0 when m
v x v x x
= = = =
( )
0
0.3 0.4
160 0.25
m
x
vdv x dx
= −
∫ ∫
( ) ( )
( ) ( )( )
2 2
2 2
0.4
0.3 0.25
0 160 80 0.25 80 0.15
2 2
m
x
m
x
x
−
− = − = − − + −
( )2
0.25 0.02306 0.25 0.1519 m
m m
x x
− = − = ±
0.0981 m and 0.402 m
m
x = W

is a function of :
a x ( ) 2
100 0.25 m/s
a x
= −
( )
Use 100 0.25 with limits
vdv a dx x dx
= = − 0 when 0.2 m
v x
= =
( )
0 0.2
100 0.25
v x
vdv x dx
= −
∫ ∫
( )( )2
2
0.2
1 1
0 100 0.25
2 2
x
v x
− = − −
( )2
50 0.25 0.125
x
= − − +
So
( ) ( )
2 2
2
0.25 100 0.25 or 0.5 1 400 0.25
v x v x
= − − = ± − −
Use
( )2
or
0.5 1 400 0.25
dx dx
dx vdt dt
v x
= = =
± − −
Integrate:
( )
0 0.2 2
0.5 1 400 0.25
t x dx
dt
x
= ±
− −
∫ ∫
Let ( )
20 0.25 ; when 0.2 = 1 and 20
u x x u du dx
= − = = −
So 1 1
1 2
1
1 1
sin sin
10 10 2
10 1
u
u du
t u u
u
π
− −
 
= = = −
 
 
−
∫
m m m
Solve for .
u 1
sin 10
2
u t
π
−
= m
( )
sin 10 cos 10 cos10
2
u t t t
π
 
= = ± =
 
 
m
( )
cos 10 20 0.25
u t x
= = −
continued

Solve for and .
x v
1
0.25 cos10
20
x t
= −
1
sin10
2
v t
=
Evaluate at 0.2 s.
t =
( )( )
( )
1
0.25 cos 10 0.2
20
x = − 0.271 m
x = W
( )( )
( )
1
sin 10 0.2
2
v = 0.455 m/s
v = W

Note that is a given function of
a x
Use ( ) ( )
2 3
600 1 600 600
vdv adx 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 3
7.5 0
600 600
vdv x kx dx
= +
∫ ∫
15 0.45
2
2 4
0
7.5
600 600
2 2 4
v
x kx
⎡ ⎤ ⎡ ⎤
= +
⎢ ⎥ ⎢ ⎥
⎣ ⎦
⎣ ⎦
( ) ( )
( )( ) ( ) ( )
2 2
2 4
15 7.5
300 0.45 150 0.45
2 2
k
− = +
84.375 60.75 6.1509k
= +
Solving for ,
k 2
3.84 ft
k −
= W

a x
Use ( )
3
800 3200
vdv adx x x dx
= = +
Using the limit 10 ft/s when 0,
v x
= =
( )
3
10 0
800 3200
v x
vdv x x dx
= +
∫ ∫
( )
2
2
2 4
10
400 800
2 2
v
x x
− = +
2 4 2 2
1600 800 100 Let
v x x u x
= + + =
Then ( )( )
2 2
1 2
1600 800 100 1600 ,
v u u u u u u
= + + = − −
1 2
where and are the roots of
u u 2
1600 800 100 0
u u
+ + =
Solving the quadratic equation,
( ) ( )( )( )
( )( )
2
1,2
800 800 4 1600 100 800 0
0.25 0
2 1600 3200
u
− ± − − ±
= = = − ±
2
1 2 0.25 ft
u u
= = −
So ( ) ( )
2
2
2 2 2 2 2
1600 0.25 1600 0.5 ft /s
v u x
= + = +
Taking square roots, ( )
2 2
40 0.5 ft/s
v x
= ± +

Use
( )
2 2
or
40 0.5
dx dx
dx vdt dt
v x
= = = ±
+
2 2
40 Use limit 0 when 0
0.5
dx
dt x t
x
= ± = =
+
1
2 2
0 0
1
40 tan
0.5 0.5
0.5
t x dx x
dt
x
−
= ± = ±
+
∫ ∫
( ) ( )
1 1
40 2.0tan 2 or tan 2 20
t x x t
− −
= ± = ±
( ) ( )
2 tan 20 or 0.5tan 20
x t x t
= ± = ±
( ) ( ) ( )
2 2
0.5 sec 20 20 10 sec 20
dx
v t t
dt
 
= = ± = ±
 
At 0, 10 ft/s, which agrees with the given data if the minus sign is rejected.
t v
= = ±
Thus, ( ) ( )
2
10 sec 20 ft/s, and 0.5tan 20 ft
v t x t
= =
At 0.05 s,
t = 20 1.0 rad
t =
( )
2
2
10
10sec 1.0
cos 1.0
v = = 34.3 ft/s
v = W
( )
0.5tan 1.0
x = 0.779 ft
x = W

a x 2
7
12 28 12 m/s
3
a x x
 
= − = −
 
 
7
Use 12 with the limits
3
vdv adx x dx
 
= = −
 
 
8 m/s when 0.
v x
= =
2
2
8 0
8 0
7 12 7
12
3 2 2 3
x
v
v x v
vdv x dx x
 
   
= − = −
 
   
 
   
 
∫ ∫
2 2
2 2
8 12 7 7
2 2 2 3 3
v
x
 
   
− = − −
 
   
   
 
 
2 2 2
2 2 7 7 7 4
8 12 12
3 3 3 3
v x x
 
     
= + − − = − −
 
     
     
 
 
2
7 4
12
3 3
v x
 
= ± − −
 
 
Reject minus sign to get 8 m/s at 0.
v x
= =
(a) Maximum value of .
x max
0 when
v x x
= =
2 2
7 4 7 1
12 0 or
3 3 3 9
x x
   
− − = − =
   
   
max max
7 1 8 2
2 m and m 2 m
3 3 3 3
x x x
− = ± = = =
Now observe that the particle starts at 0 with 0 and reaches 2 m. At 2 m, 0 and
x v x x v
= > = = =
2
0, so that becomes negative and decreases. Thus, 2 m is never reached.
3
a v x x
< =
max 2 m
x = !

(b) Velocity when total distance traveled is 3 m.
The particle will have traveled total distance 3 m
d = when max max
d x x x
− = − or 3 2 2 x
− = −
or 1 m.
x =
Using
2
7 4
12
3 3
v x
 
= − − −
 
 
, which applies when x is decreasing, we get
2
7 4
12 1 20
3 3
v
 
= − − − = −
 
 
4.47 m/s
v = − !

Note that is a function of .
a x ( )
1 x
a k e−
= −
( )
Use 1 with the limits 9 m/s when 3 m, and 0 when 0.
x
vdv adx k e dx v x v x
−
= = − = = − = =
( )
0 0
9 3
1 x
vdv k e dx
−
−
= −
∫ ∫
( )
0
2 0
3
9
2
x
v
k x e−
−
⎛ ⎞
= +
⎜ ⎟
⎜ ⎟
⎝ ⎠
( )
2
3
9
0 0 1 3 16.0855
2
k e k
⎡ ⎤
− = + − − − = −
⎣ ⎦
(a) 2.5178
k = 2
2.52 m/s
k = W
( ) ( )
Use 1 2.5178 1 with the limit 0 when 0.
x x
vdv adx k e dx e dx v x
− −
= = − = − = =
( )
0 0
2.5178 1
v x x
vdv e dx
−
= −
∫ ∫
( ) ( )
2
0
2.5178 2.5178 1
2
x
x x
v
x e x e
− −
= + = + −
( ) ( )
1/2
2
5.0356 1 2.2440 1
x x
v x e v x e
− −
= + − = ± + −
(b) Letting 2 m,
x = −
( )
1/ 2
2
2.2440 2 1 4.70 m/s
v e
= ± − + − = ±
Since begins at 2 m and ends at 0, 0.
x x x v
= − = >
Reject the minus sign.
4.70 m/s
v = W

0.00057
6.8 x
dv
a v e
dx
−
= =
0.00057
0 0
6.8
v x x
vdv e dx
−
=
∫ ∫
2
0.00057
0
6.8
0
2 0.00057
x
x
v
e−
− =
−
( )
0.00057
11930 1 x
e−
= −
When 30 m/s.
v =
( )
( )
2
0.00057
30
11930 1
2
x
e−
= −
0.00057
1 0.03772
x
e−
− =
0.00057
0.96228
x
e−
=
0.00057 ln(0.96228) 0.03845
x
− = = −
67.5 m
x = W

Given: 0.4
dv
a v v
dx
= = −
or 0.4
dv
dx
= −
Separate variables and integrate using 75 mm/s when 0.
v x
= =
75 0
0.4 75 0.4
v x
dv v x
= − − = −
∫ ∫
(a) Distance traveled when 0
v =
0 75 0.4x
− = − 187.5 mm
x = W
(b) Time to reduce velocity to 1% of initial value.
(0.01)(75) 0.75
v = =
0.75
2.5ln
75
t = − 11.51 s
t = W

Given:
dv
a v kv
dx
= = − 2
Separate variables and integrate using 9 m/s when 0.
v x
= =
9 0
v x
dv
k dx
v
= −
∫ ∫
ln
9
v
kx
= −
Calculate using 7 m/s when 13 m.
k v x
= =
( )( ) 3 1
7
ln 13 19.332 10 m
9
k k − −
= − = ×
Solve for .
x
1
ln 51.728 ln
9 9
v v
x
k
= − = −
(a) Distance when 3 m/s.
v =
3
51.728 ln
9
x
⎛ ⎞
= − ⎜ ⎟
⎝ ⎠
56.8 m
x = W
(b) Distance when 0.
v =
( )
51.728 ln 0
x = − x = ∞ W

0 0
, 0, 25 ft/s
vdv adx k vdx x v
= = − = =
1/2
1
dx v dv
k
= −
0 0
0
3/2
1 2
3
v
x v
x v
v
dx vdv v
k k
= − = −
∫ ∫
( ) ( )3/2
3/2 3/2 3/2 3/2
0 0
2 2 2
or 25 125
3 3 3
x x v v x v v
k k k
⎡ ⎤ ⎡ ⎤
− = − = − = −
⎢ ⎥ ⎣ ⎦
⎣ ⎦
Noting that 6 ft when 12 ft/s,
x v
= =
3/2 3
2 55.62
6 125 12 or 9.27 ft/s
3
k
k k
⎡ ⎤
= − = =
⎣ ⎦
Then,
( )( )
( )
3/2 3/2
2
125 0.071916 125
3 9.27
x v v
⎡ ⎤
= − = −
⎣ ⎦
3/2
125 13.905
v x
= −
( ) When
a 8 ft,
x = ( )( ) ( )3/2
3/2
125 13.905 8 13.759 ft/s
v = − =
5.74 ft/s
v = W
( )
b dv adt k vdt
= = −
1/ 2
1 dv
dt
k v
= −
( )
0
1/2
1/2 1/2
0
1 2
2
v
v
t v v v
k k
⎡ ⎤
= − ⋅ = −
⎣ ⎦
At rest, 0
v =
( )( )1/2
1/2
0 2 25
2
9.27
v
t
k
= = 1.079 s
t = W

x as a function of v.
0.00057
1
154
x
v
e−
= −
2
0.00057
1
154
x v
e− ⎛ ⎞
= − ⎜ ⎟
⎝ ⎠
2
0.00057 ln 1
154
v
x
⎡ ⎤
⎛ ⎞
− = −
⎢ ⎥
⎜ ⎟
⎝ ⎠
⎢ ⎥
⎣ ⎦
2
1754.4 ln 1
154
v
x
⎡ ⎤
⎛ ⎞
= − −
⎢ ⎥
⎜ ⎟
⎝ ⎠
⎢ ⎥
⎣ ⎦
(1)
a as a function of x.
( )
2 0.00057
23716 1
v e−
= −
( )( )
2
0.0005
11858 0.00057
2
x
dv d v
a v e
dx dx
−
⎛ ⎞
= = =
⎜ ⎟
⎜ ⎟
⎝ ⎠
2
0.00057
6.75906 6.75906 1
154
x v
a e−
⎡ ⎤
⎛ ⎞
= = −
⎢ ⎥
⎜ ⎟
⎝ ⎠
⎢ ⎥
⎣ ⎦
(2)
(a) v = 20 m/s.
From (1), x = 29.843 x = 29.8 m e
From (2), a = 6.64506 a = 6.65 m/s2
e
(b) v = 40 m/s.
From (1), x = 122.54 x = 122.5 m e
From (2), a = 6.30306 a = 6.30 m/s2
e

( )0.3
Given: 7.5 1 0.04 with units km and km/h
v x
= −
(a) Distance at 1 hr.
t =
0.3
Using , we get
7.5(1 0.04 )
dx dx
dx vdt dt
v x
= = =
−
Integrating, using 0
t = when 0,
x =
( ) ( ) ( )( )
{ }
0.7
0
0.3
0 0 0
1 1 1
or [ ] 1 0.04
7.5 7.5 0.7 0.04
1 0.04
x
t x t
dx
dt t x
−
= = ⋅ −
−
∫ ∫
( )
{ }
0.7
4.7619 1 1 0.04
t x
= − − (1)
Solving for ,
x ( )
{ }
1/0.7
25 1 1 0.210
x t
= − −
When 1 h,
t = ( )( )
{ }
1/0.7
25 1 1 0.210 1
x ⎡ ⎤
= − −
⎣ ⎦ 7.15 km
x = W
(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 0
t = and 0,
x = 1
7.5 km/h, 0.0900 h
dv
v
dx
−
= −
2
(7.5)( 0.0900) 0.675 km/h
dv
a v
dx
= = − = −
2
2
(0.675)(1000)
m/s
(3600)
= − 6 2
52.1 10 m/s
a −
= − × W
(c) Time to run 6 km.
Using 6 km
x = in equation (1),
( )( )
{ }
0.7
4.7619 1 1 0.04 6 0.8323 h
t ⎡ ⎤
= − − =
⎣ ⎦
49.9 min
t = W

The acceleration is given by
2
2
dv gR
v a
dr r
= = −
Then,
2
2
gR dr
v dv
r
= −
Integrating, using the conditions esc
0 at , and
v r v v
= = ∞ = at r R
=
esc
0 2
2
v R
dr
v dv gR
r
∞
= −
∫ ∫
esc
0
2 2
1 1
2 v R
v gR
r
∞
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
2 2
esc
1 1
0 0
2
v gR
R
⎛ ⎞
− = −
⎜ ⎟
⎝ ⎠
esc 2
v gR
=
6 2
Now, 3960 mi 20.909 10 ft and 32.2 ft/s .
R g
= = × =
Then, ( )( )( )
6
esc 2 32.2 20.909 10
v = × 3
esc 36.7 10 ft/s
v = × W

The acceleration is given by
6
2
20.9 10
32.2
1 y
a
×
−
=
⎡ ⎤
⎛ ⎞
+ ⎜ ⎟
⎢ ⎥
⎝ ⎠
⎣ ⎦
6
2
20.9 10
32.2
1 y
dy
vdv ady
×
−
= =
⎡ ⎤
⎛ ⎞
+ ⎜ ⎟
⎢ ⎥
⎝ ⎠
⎣ ⎦
2
0 max
Integrate, using the conditions at 0 and 0 at . Also, use 32.2 ft/s and
v v y v y y g
= = = = =
6
20.9 10 ft.
R = ×
( ) ( )
0
0 2
2 2
0 0
1
v y
R
dy dy
v dv g gR
R y
∞ ∞
= − = −
+
+
∫ ∫ ∫
max
0
0
2 2
0
1 1
2
y
v
v gR
R y
⎛ ⎞
= ⎜ ⎟
+
⎝ ⎠
( )
2 2 2
max
0 0 max max
max max
1 1 1
0 2
2
gRy
v gR v R y gRy
R y R R y
⎡ ⎤
− = − = − + =
⎢ ⎥
+ +
⎣ ⎦
max
Solving for ,
y
2
0
max 2
0
2
Rv
y
gR v
=
−
Using the given numerical data,
( )( )( )
6 2 6 2
0 0
max 9 2
6 2
0
0
20.9 10 20.9 10
1.34596 10
2 32.2 20.9 10
v v
y
v
v
× ×
= =
× −
× −
0
( ) 2400 ft/s,
a v =
( )( )
( ) ( )
2
6
max 2
9
20.9 10 2400
1.34596 10 2400
y
×
=
× −
3
max 89.8 10 ft
y = × W
0
( ) 4000 ft/s,
b v =
( )( )
( ) ( )
2
6
max 2
9
20.9 10 4000
1.34596 10 4000
y
×
=
× −
3
max 251 10 ft
y = × W
0
( ) 40000 ft/s,
c v =
( )( )
( ) ( )
2
6
max 2
9
20.9 10 40000
negative
1.34596 10 40000
y
×
= =
× −
Negative value indicates that 0
v is greater than the escape velocity.
max
y = ∞ W

( )
( ) Given: sin n
a v v t
ω ϕ
′
= +
At 0,
t = 0
0 sin or sin
v
v v v
v
ϕ ϕ
′
= = =
′ (1)
Let x be maximum at 1
t t
= when 0.
v =
Then, ( ) ( )
1 1
sin 0 and cos 1
n n
t t
ω ϕ ω ϕ
+ = + = ± (2)
Using or
dx
v dx v dt
dt
= =
Integrating, ( )
cos n
n
v
x C t
ω ϕ
ω
′
= − +
At 0,
t = 0 0
cos or cos
n n
v v
x x C C x
ϕ ϕ
ω ω
′ ′
= = − = +
Then, ( )
0 cos cos n
n n
v v
x x t
ϕ ω ϕ
ω ω
′ ′
= + − + (3)
max 0 1
cos using cos 1
n
n
v v
x x t
ϕ ω ϕ
ω ω
′ ′
= + + + = −
Solving for cos ,
ϕ
( )
max 0
cos 1
n
x x
v
ω
ϕ
−
= −
′
max 0
With 2 ,
x x
= 0
cos 1
n
x
v
ω
ϕ = −
′ (4)
Using
2 2
2 2 0 0
sin cos 1, or 1 1
n
v x
v v
ω
ϕ ϕ
   
+ = + − =
   
′ ′
   
Solving for gives
v′
( )
2 2 2
0 0
0
(5)
2
n
n
v x
v
x
ω
ω
+
′ = W

( ) Acceleration:
b ( )
cos
n n
dv
a v t
dt
ω ω ϕ
′
= = +
2
Let be maximum at when 0.
v t t a
= =
Then, ( )
2
cos 0
nt
ω ϕ
+ =
From equation (3), the corresponding value of x is
( )
0
0 0 0
2 2 2 2
0 0 0
0 0 2
0 0
cos 1 2
3 1
2
2 2 2
n
n n n
n
n n n
v v x v
x x x x
v
v x v
x x
x x
ω
ϕ
ω ω ω
ω
ω ω ω
′ ′ ′
 
= + = + − = −
 
′
 
+
= − = −
( )
0
0
2
0
3
2
n
v
x
x
ω
 
−
 
  W

0
( ) 1 sin
dx t
a v v
dt T
π
⎡ ⎤
= = −
⎢ ⎥
⎣ ⎦
0
Integrating, using 0 when 0,
x x t
= = =
0
0 0 0
1 sin
x t t t
dx v dt v dt
T
π
⎡ ⎤
= = −
⎢ ⎥
⎣ ⎦
∫ ∫ ∫
0
0
0
0
cos
t
x v T t
x v t
T
π
π
⎡ ⎤
= +
⎢ ⎥
⎣ ⎦
0 0
0 cos
v T t v T
x v t
T
π
π π
= + − (1)
When 3 ,
t T
= ( )
0 0
0 0
2
3 cos 3 3
v T v T
x v T v T
T
π
π π
⎛ ⎞
= + − = −
⎜ ⎟
⎝ ⎠
0
2.36
x v T
= W
0
cos
dv v t
a
dt T T
π π
= = −
When 3 ,
t T
= 0
cos3
v
a
T
π
π
= − 0
v
a
T
π
= W
( ) Using equation (1) with ,
b t T
=
0 0
1 0 0
2
cos 1
v T v T
x v T v T
π
π π π
⎛ ⎞
= + − = −
⎜ ⎟
⎝ ⎠
Average velocity is
1 0
ave 0
2
1
x x x
v v
t T π
Δ − ⎛ ⎞
= = = −
⎜ ⎟
Δ ⎝ ⎠
ave 0
0.363
v v
= W

10 km/h 2.7778 m/s
= 100 km/h 27.7778 m/s
=
(a) Acceleration during start test.
dv
a
dt
=
8.2 27.7778
0 2.7778
adt vdt
=
∫ ∫
8.2 27.7778 2.7778
a = − 2
3.05 m/s
a = W
(b) Deceleration during braking.
dv
a v
dx
= =
44 0
0 27.7778
a dx v dv
= =
∫ ∫
( ) ( )
0
44 2
0
27.7778
1
2
a x v
=
( )2
1
44 27.7778
2
a = −
2
8.77 m/s
a = − deceleration 2
8.77 m/s
a
= − = W

10 km/h 2.7778 m/s
= 100 km/h 27.7778 m/s
=
(a) Distance traveled during start test.
dv
a
dt
=
0
0
t v
v
a dt dv
=
∫ ∫
0
at v v
= − 0
v v
a
t
−
=
2
27.7778 2.7778
3.04878 m/s
8.2
a
−
= =
0 2.7778 3.04878
v v at t
= + = +
)
8.2
0 0
2.7778 3.04878
t
x v dv t dt
= = +
∫ ∫
( )( ) ( )( )2
2.7778 8.2 1.52439 8.2
= + 125.3 m
x = W
(b) Elapsed time for braking test.
dv
a v
dx
=
0
0
x v
v
adx vdv
=
∫ ∫
2 2
0
2 2
v v
ax = −
( ) ( )( )
( )
2 2 2
0
1 1
0 27.7778
2 2 44
a v v
x
= − = −
2
8.7682 m/s
= −
dv
a
dt
=
0
0
t v
v
a dt dv
=
∫ ∫
0
at v v
= −
0 0 27.7778
8.7682
v v
t
a
− −
= =
−
3.17 s
t = W

Constant acceleration. 0 0
0, 0
A A
v v x x
= = = =
0
v v at at
= + = (1)
2 2
0 0
1 1
2 2
x x v t at at
= + + = (2)
At point ,
B 2700 ft and 30 s
B
x x t
= = =
(a) Solving (2) for a,
( )( )
( )
2 2
2 2700
2
30
x
a
t
= = 2
6 ft/s
a = W
(b) Then, ( )( )
6 30
B
v at
= = 180 ft/s
B
v = W

Constant acceleration. 0 0
x =
0
v v at
= + (1)
2
0 0
1
2
x x v t at
= + + (2)
Solving (1) for a,
0
v v
a
t
−
= (3)
Then, ( ) ( )
2
0
0 0 0 0 0
1 1 1
2 2 2
v v
x x v t t x v v t v v t
t
−
= + + = + + = +
At 6 s,
t = 0 6
1
and 540 ft
2
v v x
= =
( )
0 0 0 0
0
1 1 540
540 6 4.5 or 120 ft/s
2 2 4.5
1
60 ft/s
2
v v v v
v v
⎛ ⎞
= + = = =
⎜ ⎟
⎝ ⎠
= =
Then, from (3), 2 2
60 120 60
ft/s 10 ft/s
6 6
a
−
= = − = −
Substituting into (1) and (2), 120 10
v t
= −
( ) 2
1
0 120 10
2
x t t
= + −
At stopping, 0 or 120 10 0 12 s
s s
v t t
= − = =
( )( ) ( )( )2
1
0 120 12 10 12 720 ft
2
x = + − =
( ) Additional time for stopping 12 s 6 s
a = − 6 s
t
Δ = W
( ) Additional distance for stopping 720 ft 540 ft
b = − 180 ft
d
Δ = W

2
0 0
1
( ) During the acceleration phase
2
a x x v t at
= + +
0 0
Using 0, and 0, and solving for gives
x v a
= =
2
2x
a
t
=
Noting that 130 m when 25 s,
x t
= =
( )( )
( )2
2 130
25
a = 0.416 m/s
a = W
(b) Final velocity is reached at 25 s.
t =
( )( )
0 0 0.416 25
f
v v at
= + = + 10.40 m/s
f
v = W
(c) The remaining distance for the constant speed phase is
400 130 270 m
x
Δ = − =
For constant velocity,
270
25.96 s
10.40
x
t
v
Δ
Δ = = =
Total time for run: 25 25.96
t = + 51.0 s
t = W

Constant acceleration. Choose 0
t = at end of powered flight.
Then, 2
1 27.5 m 9.81 m/s
y a g
= = − = −
(a) When y reaches the ground, 0 and 16 s.
f
y t
= =
2 2
1 1 1 1
1 1
2 2
f
y y v t at y v t gt
= + + = + −
( )( )2
2
1 1
1 2 2
1
0 27.5 9.81 16
76.76 m/s
16
f
y y gt
v
t
− + − +
= = =
1 76.8 m/s
v = W
(b) When the rocket reaches its maximum altitude max,
y
0
v =
( ) ( )
2 2 2
1 1 1 1
2 2
v v a y y v g y y
= + − = − −
2 2
1
1
2
v v
y y
g
−
= −
( )
( )( )
2
max
0 76.76
27.5
2 9.81
y
−
= − max 328 m
y = W

Place origin at 0.
Motion of auto. ( ) ( ) 2
0 0
0, 0, 0.75 m/s
A A A
x v a
= = =
( ) ( ) ( )
2 2
0 0
1 1
0 0 0.75
2 2
A A A A
x x v t a t t
⎛ ⎞
= + + = + + ⎜ ⎟
⎝ ⎠
2
0.375 m
A
x t
=
Motion of bus. ( ) ( )
0 0
?, 6 m/s, 0
B B B
x v a
= = − =
( ) ( ) ( )
0 0 0
6 m
B B B B
x x v t x t
= − = −
At 20 , 0.
B
t s x
= =
( ) ( )( )
0
0 6 20
B
x
= − ( )0
120 m
B
x =
Hence, 120 6
B
x t
= −
When the vehicles pass each other, .
B A
x x
=
2
120 6 0.375
t t
− =
2
0.375 6 120 0
t t
+ − =
( )( )( )
( )( )
2
6 (6) 4 0.375 120
2 0.375
t
− ± − −
=
6 14.697
11.596 s and 27.6 s
0.75
t
− ±
= = −
Reject the negative root. 11.60 s
t = W
Corresponding values of xA and xB.
( )( )2
0.375 11.596 50.4 m
A
x = =
( )( )
120 6 11.596 50.4 m
B
x = − = 50.4 m
x = W

Place the origin at A when t = 0.
Motion of A: ( ) ( ) 2
0 0
0, 15 km/h = 4.1667 m/s, 0.6 m/s
A A A
x v a
= = =
( )0
4.1667 0.6
A A A
v v a t t
= + = +
( ) ( ) 2 2
0 0
1
4.1667 0.3
2
A A A A
x x v t a t t t
= + + = +
Motion of B: ( ) ( ) 2
0 0
25 m, 23 km/h = 6.3889 m/s, 0.4 m/s
B B B
x v a
= = = −
( )0
6.3889 0.4
B B B
v v a t t
= + = −
( ) ( ) 2 2
0 0
1
25 6.3889 0.2
2
B B B B
x x v t a t t t
= + + = + −
(a) When and where A overtakes B. A B
x x
=
2 2
4.1667 0.3 25 6.3889 0.2
t t t t
+ = + −
2
0.5 2.2222 25 0
t t
− − =
( )( )( )
( )( )
2
2.2222 2.2222 4 0.5 25
2 0.5
t
± − −
=
2.2222 7.4120 9.6343 s and 5.19 s
t = ± = −
Reject the negative root. . 9.63 s
t = W
( )( ) ( )( )2
4.1667 9.6343 0.3 9.6343 68.0 m
A
x = + =
( )( ) ( )( )2
25 6.3889 9.6343 0.2 9.6343 68.0 m
B
x = + − =
moves 68.0 m
A W
moves 43.0 m
B W
(b) Corresponding speeds.
( )( )
4.1667 0.6 9.6343 9.947 m/s
A
v = + = 35.8 km/h
A
v = W
( )( )
6.3889 0.4 9.6343 2.535 m/s
B
v = − = 9.13 km/h
B
v = W

Constant acceleration ( )
1 2
and
a a for horses 1 and 2.
Let 0
x = and 0
t = when the horses are at point A.
Then, 2
0
1
2
x v t at
= +
Solving for ,
a
( )
0
2
2 x v t
a
t
−
=
Using 1200 ft
x = and the initial velocities and elapsed times for each horse,
( )( )
( )
2
1 1
1 2 2
1
2 1200 20.4 61.5
0.028872 ft/s
61.5
x v t
a
t
⎡ ⎤
−
− ⎣ ⎦
= = = −
( )( )
( )
2
2 2
2 2 2
2
2 1200 21 62.0
0.053070 ft/s
62.0
x v t
a
t
⎡ ⎤
−
− ⎣ ⎦
= = = −
1 2
Calculating ,
x x
− ( ) ( ) 2
1 2 1 2 1 2
1
2
x x v v t a a t
− = − + −
( ) ( ) ( ) 2
1 2
2
1
20.4 21 0.028872 0.053070
2
0.6 0.012099
x x t t
t t
⎡ ⎤
− = − + − − −
⎣ ⎦
= − +
At point B, 2
1 2 0 0.6 0.012099 0
B B
x x t t
− = − + =
(a)
0.6
49.59 s
0.012099
B
t = =
Calculating B
x using data for either horse,
Horse 1: ( )( ) ( )( )2
1
20.4 49.59 0.028872 49.59
2
B
x = + − 976 ft
B
x = W
Horse 2: ( )( ) ( )( )2
1
21 49.59 0.05307 49.59 976 ft
2
B
x = + − =
When horse 1 crosses the finish line at 61.5 s,
t =
(b) ( )( ) ( )( )2
1 2 0.6 61.5 0.012099 61.5
x x
− = − + 8.86 ft
x
Δ = W

Choose x positive upward. Constant acceleration a g
= −
Rocket launch data: Rocket :
A 0
0, , 0
x v v t
= = =
Rocket :
B 0
0, , 4 s
B
x v v t t
= = = =
Velocities: Rocket :
A 0
A
v v gt
= −
Rocket :
B ( )
0
B B
v v g t t
= − −
Positions: 2
0
1
Rocket :
2
A
A x v t gt
= −
( ) ( )2
0
1
Rocket : ,
2
B B B B
B x v t t g t t t t
= − − − ≥
For simultaneous explosions at 240 ft when ,
A B E
x x t t
= = =
( ) ( )2
2 2 2
0 0 0 0
1 1 1 1
2 2 2 2
E E E B E B E B E E B B
v t gt v t t g t t v t v t gt gt t gt
− = − − − = − − + −
0
Solving for ,
v 0
2
B
E
gt
v gt
= − (1)
Then, when ,
E
t t
= 2
1
,
2 2
B
A E E E
gt
x gt t gt
 
= − −
 
 
or 2 2
0
A
E B E
x
t t t
g
− − =
Solving for ,
E
t
( )( )( ) ( ) ( )( )( )( )
2
2 4 1 2 240
2
32.2
4 1 4 4
6.35 s
2 2
A
x
B B g
E
t t
t
± + ± +
= = =

(a) From equation (1), ( )( )
( )( )
0
32.2 4
32.2 6.348
2
v = − 0 140.0 ft/s
v = W
At time ,
E
t 0
A E
v v gt
= − ( )
0
B E B
v v g t t
= − −
(b) ( )( )
32.2 4
B A B
v v gt
− = = / 128.8 ft/s
B A
v = W

(a) Acceleration of A.
( ) ( )
0 0
, 168 km/h 46.67 m/s
A A A A
v v a t v
= + = =
At 8 s,
t = 228 km/h 63.33 m/s
A
v = =
( )0 63.33 46.67
8
A A
A
v v
a
t
− −
= = 2
2.08 m/s
A
a = W
(b) ( ) ( ) 2
0 0
1
2
A A A A
x x v t a t
= + + ( ) ( ) 2
0 0
1
2
B B B B
x x v t a t
= + +
( ) ( ) ( ) ( ) ( ) 2
0 0 0 0
1
2
A B A B A B A B
x x x x v v t a a t
⎡ ⎤
− = − + − + −
⎣ ⎦
When 0,
t = ( ) ( )
0 0
38 m
A B
x x
− = and ( ) ( )
0 0
0
B A
v v
− =
When 8 s,
t = 0
A B
x x
− =
Hence, ( )( )2
1
0 38 8 , or 1.1875
2
A B A B
a a a a
= + − − = −
1.1875 2.08 1.1875
B A
a a
= + = + 2
3.27 m/s
B
a = W

(a) Acceleration of A.
( ) ( ) ( ) 2
0 0 0
1
and
2
A A A A A A A
v v a t x x v t a t
= + = + =
Using ( ) ( )
0 0
0 and 0 gives
A A
v x
= =
2
1
and
2
A A A A
v a t x a t
= =
When cars pass at 1, 90 m
A
t t x
= =
( )( )
2
1 1
2 90
2 180
and
A
A A
A A A
x
t v a t
a a a
= = = =
For 0 5 s,
t
≤ ≤ ( )0
96 km/h 26.667 m/s
B B
v v
= = − = −
For 5 s,
t > ( ) ( ) ( )
0
1
5 26.667 5
6
B B B A
v v a t a t
= + − = − + −
When vehicles pass, A B
v v
= −
( )
1 1
1
26.667 5
6
A A
a t a t
= − −
1 1
7 5 160
26.667 or 7 5
6 6
A A
A
a t a t
a
− = − =
Using 1
180 7 180 160
gives 5
A A
A
t
a a
a
= − =
Let
1
,
A
u
a
= 2
7 180 5 160
u u
− =
or 2
160 7 180 5 0
u u
− + =
continued

( )( ) ( )( )( )
( )( )
7 180 49 180 4 160 5 93.915 74.967
2 160 320
0.0592125 and 0.52776
u
± − ±
= =
=
2
1
285.2 m/s and 3.590 m/s
A
a
u
= =
The corresponding values for 1
t are
1 1
180 180
0.794 s, and 7.08 s
285.2 3.590
t t
= = = =
Reject 0.794 s since it is less than 5 s.
Thus,
2
3.59 m/s
A
a = W
(b) Time of passing. 1 7.08 s
t t
= = W
(c) Distance d.
( ) ( )
0 0
0 5 s, 26.667
B B B
t x x v t d t
≤ ≤ = − = −
At 5 s,
t = ( )( )
22.667 5 133.33
B
x d d
= − = −
For 5 s,
t > ( ) ( ) ( )2
0
1
133.33 5 5
2
B B B
x d v t a t
= − + − + −
( ) ( )
2
1 3.59
133.33 26.667 5 5
2 6
B
x d t t
 
= − − − + −
 
 
1
When 7.08 s,
t t
= = 90
B A
x x
= =
( )( )
( )( )
( )( )
2
3.59 2.08
90 133.33 26.667 2.08
2 6
d
= − − +
90 133.33 55.47 1.29
d = + + − 278 m
d = W

For 0,
t > ( ) ( ) ( )
2 2 2
0 0
1 1
0 0 6.5 or 3.25
2 2
A A A A A
x x v t a t t x t
= + + = + + =
For 2 s,
t > ( ) ( ) ( ) ( ) ( )( )
2 2
0 0
1 1
2 2 0 0 11.7 2
2 2
B B B B
x x v t a t t
= + − + − = + + −
or ( )2 2
5.85 2 5.85 23.4 23.4
B
x t t t
= − = − +
For ,
A B
x x
= 2 2
3.25 5.85 23.4 23.4,
t t t
= − +
or 2
2.60 23.4 23.4 0
t t
− + =
Solving the quadratic equation, 1.1459 and 7.8541 s
t t
= =
Reject the smaller value since it is less than 5 s.
( )
a 7.85 s
t = W
( )( )2
3.25 7.8541
A B
x x
= = 200 ft
x = W
( )
b ( ) ( )( )
0
0 6.5 7.8541
A A A
v v a t
= + = + 51.1 ft/s
A
v = W
( ) ( ) ( )( )
0
2 0 11.7 7.8541 2
B B B
v v a t
= + − = + − 68.5 ft/s
B
v = W

Let x be the position relative to point P.
Then, ( ) ( )
0 0
0 and 0.62 mi 3273.6 ft
A B
x x
= = =
Also, ( ) ( )
0 0
68 mi/h 99.73 ft/s and 39 mi/h 57.2 ft/s
A B
v v
= = = − = −
(a) Uniform accelerations.
( ) ( )
( ) ( )
0 0
2
2
0 0
2
1
or
2
A A A
A A A A A
x x v t
x x v t a t a
t
⎡ ⎤
− −
⎣ ⎦
= + + =
( )( )
( )
2
2
2 3273.6 0 99.73 40
0.895 ft/s
40
A
a
⎡ ⎤
− −
⎣ ⎦
= = − 2
0.895 ft/s
A
a = W
( ) ( )
( ) ( )
0 0
2
2
0 0
2
1
or
2
B B B
B B B B B
x x v t
x x v a t a
t
⎡ ⎤
− −
⎣ ⎦
= + + =
( )( )
( )
2
2
2 0 3273.6 57.2 42
0.988 ft/s
42
B
a
⎡ ⎤
− − −
⎣ ⎦
= = − 2
0.988 ft/s
B
a = W
(b) When vehicles pass each other .
A B
x x
=
( ) ( ) ( ) ( )
2 2
0 0 0 0
1 1
2 2
A A A B B B
x v t a t x v t a t
+ + = + +
( ) ( )
2 2
1 1
0 99.73 0.895 3273.6 57.2 0.988
2 2
t t t t
+ + − = − + −
2
0.0465 156.93 3273.6 0
t t
− − + =
Solving the quadratic equation, 20.7 s
t = and 3390 s
−
Reject the negative value. Then, 20.7 s
t = W
(c) Speed of B.
( ) ( )( )
0
57.2 0.988 20.7 77.7 ft/s
B B B
v v a t
= + = − + − = −
77.7 ft/s
B
v = W

Let x be positive downward for all blocks and for point D.
1 m/s
A
v =
Constraint of cable supporting A: ( ) constant
A A B
x x x
+ − =
( )( )
2 0 or 2 2 1 2 m/s
A B B A
v v v v
− = = = =
Constraint of cable supporting B: 2 constant
B C
x x
+ =
( )( )
2 0 or 2 2 2 4 m/s
C B C B
v v v v
+ = = − = − = −
(a) 4 m/s
C =
v W
(b) / 2 1
B A B A
v v v
= − = − / 1 m/s
B A =
v W
(c) constant, 0
D C D C
x x v v
+ = + =
4 m/s
D C
v v
= − =
/ 4 1
D A D A
v v v
= − = − / 3 m/s
D A =
v W

Let x be positive downward for all blocks.
Constraint of cable supporting A: ( ) constant
A A B
x x x
+ − =
2 0 or 2 and 2
A B B A B A
v v v v a a
− = = =
Constraint of cable supporting B: 2 constant
B C
x x
+ =
2 0, or 2 , and 2 4
B C C B C B A
v v v v a a a
+ = = − = − = −
Since C
v and C
a are down, A
v and A
a are up, i.e. negative.
( ) ( )
2
2
0 0
2
A A A A A
v v a x x
⎡ ⎤
− = −
⎣ ⎦
( )
( )
( )
( )( )
2 2
2
2
0
0
0.2 0
( ) 0.04 m/s
2 0.5
2
A A
A
A A
v v
a a
x x
− −
= = = −
⎡ ⎤ −
−
⎣ ⎦
2
0.04 m/s
A
a = W
4
C A
a a
= − 2
0.16 m/s
C
a = W
( )( ) 2
( ) 2 2 0.04 0.08 m/s
B A
b a a
= = − = −
( )( )
0.08 2 0.16 m/s
B B
v a t
Δ = = − = − 0.16 m/s
B
v
Δ = W
( )( )2
2
1 1
0.08 2 0.16 m
2 2
B B
x a t
Δ = = − = − 0.16 m
B
x
Δ = W

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: constant
A B
x x
+ =
0
A B
v v
+ = B A
v v
= −
Constraint of cable BED: 2 constant
B D
x x
+ =
1 1
2 0 or
2 2
B D D B A
v v v v v
+ = = − =
Constraint of cable BCD: ( ) ( ) constant
C B C D
x x x x
− + − =
1
2 0 or 2 0
2
C B D C A A
v v v v v v
− − = + − =
(a) Velocity of block A.
1
2 (2)(4)
2
A C
v v
= − = − 8 ft/s
A
v = − 8 ft/s
A
v = W
(b) Velocity of block D.
1
4 ft/s
2
D A
v v
= = − 4 ft/s
D
v = W

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: constant
A B
x x
+ =
0
A B
v v
+ = B A
v v
= −
0
A B
a a
+ = B A
a a
= −
Constraint of cable BED: 2 constant
B D
x x
+ =
1 1
2 0
2 2
B D D B A
v v v v v
+ = = − =
1 1
2 0
2 2
B D D A A
a a a a a
+ = = − =
Constraint of cable BCD: ( ) ( ) constant
C B C D
x x x x
− + − =
2 0 2 0
C B D C A
v v v v v
− − = + =
1
2 0 2 0
2
C B D C A
a a a a a
− − = + =
1
4
C A
a 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
2
A D A D A
v v v v
= − =
/
1
2
A D A D A
a a a a
= − =

(a) Acceleration of block C.
/ 2
/
2 (2)(8)
2 3.2 ft/s
5
A D
A A D
v
a a
t
= = = =
2
3.2 ft/s
A
a = −
2
1
0.8 ft/s
4
C A
a a
= − = 2
0.8 ft/s
C
a = W
Constraint of cable portion BE: constant
B E
x x
+ =
0
B E
v v
+ = 0
B E
a a
+ =
(b) Acceleration of point E.
2
3.2 ft/s
E B A
a a a
= − = = − 2
3.2 ft/s
E
a = W

Let x be position relative to the right supports, increasing to the left.
Constraint of entire cable: ( )
2 constant
A B B A
x x x x
+ + − =
2 0 2
B A A B
v v v v
+ = = −
Constraint of point C of cable: 2 constant
A C
x x
+ =
2 0 2
A C C A
v v v v
+ = = −
(a) Velocity of collar A.
( )( )
2 2 300 600 mm/s
A B
v v
= − = − = − 600 mm/s
A
v = W
(b) Velocity of point C of cable.
( )( )
2 2 600 1200 mm/s
C A
v v
= − = − − = 1200 mm/s
C
v = W
(c) Velocity of point C relative to collar B.
/ 1200 300 900 mm/s
C B C B
v v v
= − = − = / 900 mm/s
C B
v = W

Let x be position relative to the right supports, increasing to the left.
Constraint of entire cable: ( )
2 constant,
A B B A
x x x x
+ + − =
1 1
2 0, or , and
2 2
B A B A B A
v v v v a a
+ = = − = −
(a) Accelerations of A and B.
/ /
1 2
2 3
B A B A A A A B A
v v v v v v v
= − = − − = −
( )
2
610 406.67 mm/s
3
A
v = − = −
( )
( ) 2
0
A
0
406.67 0
, or 50.8 mm/s
8
A A
A A A
v v
v v a t a
t
− − −
− = = = = −
2
50.8 mm/s
A
a = W
( )
1 1
50.8
2 2
B A
a a
= − = − − 2
25.4 mm/s
B
a = W
(b) Velocity and change in position of B after 6 s.
( ) ( )( )
0
0 25.4 6
B B B
v v a t
= + = + 152.5 mm/s
B
v = W
( ) ( ) ( )( )2
2
0 0
1 1
25.4 6
2 2
B B B B
x x v t a t
− = + = 458 mm
B
x
Δ = W

Let x be position relative to left anchor. At the right anchor, .
x d
=
Constraint of cable: ( ) ( )
2 constant
B B A A
x x x d x
+ − + − =
2 2
2 3 0 or and
3 3
B A A B A B
v v v v a a
− = = =
Constraint of point D of cable: ( ) constant
A D
d x d x
− + − =
0 or and
A D D A D A
v v v v a a
+ = = − = −
(a) Accelerations of A and B.
( ) ( ) ( )
0 0
2
6 in./s 6 4 in./s
3
B A
v v
= = =
( ) ( )
2
2
0 0
2
A A A A A
v v a x x
⎡ ⎤
− = −
⎣ ⎦
( )
( )
( ) ( )
( )( )
2 2 2
2
2
0
0
2.4 4
0.512 in./s
2 10
2
A A
A
A A
v v
a
x x
− −
= = = −
⎡ ⎤
−
⎣ ⎦
2
0.512 in./s
A
a = W
( ) 2
3 3
0.512 0.768 in./s
2 2
B A
a a
= = = − 2
0.768 in./s
B
a = W
(b) Acceleration of point D. ( )
0.512
D A
a a
= − = − − 2
0.512 in./s
D
a = W
(c) Velocity of block B after 4 s.
( ) ( )( )
0
6 0.768 4
B B B
v v a t
= + = + − 2.93 in./s
B
v = W
Change in position of block B.
( ) ( ) ( )( ) ( )( )2
2
0 0
1 1
6 4 0.768 4
2 2
B B B B
x x v t a t
− = + = + − 17.86 in.
B
x
Δ = W

Let x be position relative to left anchor. At right anchor .
x d
=
Constraint of entire cable: ( ) ( )
2 constant
B B A A
x x x d x
+ − + − = 2 3 0
B A
v v
− =
(a) Velocity of A: ( )
2 2
12
3 3
A B
v v
= = 8.00 in./s
A
v = W
Constraint of point C of cable: constant
B B C
x x x
+ − = 2 0
B C
v v
− =
(b) Velocity of C: ( )
2 2 12
C B
v v
= = 24 in./s
C
v = W
Constraint of point D of cable: constant
A C
d x d x
− + − = 0,
A D
v v
+ =
(c) Velocity of D: 8.00 in./s
D A
v v
= − = − 8.00 in./s
D
v = W
(d) Relative velocity. / 24 8
C A C A
v v v
= − = − / 16.00 in./s
C A
v = W

Let x be position relative to the anchor, positive to the right.
3 constant
B C B C A
x x x x x
− + − + − =
4 2 3 0 4 2 3 0
C B A C B A
v v v a a a
− − = − − = (1, 2)
When 0,
t = ( )0
50 mm/s and 100 mm/s
B a
v v
= − =
(a) ( ) ( ) ( )( ) ( )( )
0
0
1 1
2 3 2 50 3 100
4 4
C B A
v v v
⎡ ⎤ ⎡ ⎤
= + = − +
⎣ ⎦
⎣ ⎦ ( )0
50 mm/s
C
v = W
Constraint of point D: ( ) ( ) ( ) constant
D A C A C B B
x x x x x x x
− + − + − − =
2 2 2 0
D C A B
v v v v
+ − − =
(b) ( ) ( ) ( ) ( )( ) ( )( ) ( )( )
0 0 0
2 2 2 2 100 + 2 50 2 50
D A B C
v v v v
= + − = − − ( )0
0
D
v = W
( ) ( ) 2
0 0
1
2
C C C C
x x v t a t
− = +
(c)
( ) ( ) ( )( )
( )
0 0 2
2 2
2 2 40 50 2
30 mm/s
2
C C C
C
x x v t
a
t
⎡ ⎤
− − ⎡ ⎤
−
⎣ ⎦ ⎣ ⎦
= = = −
2
30 mm/s
C
a = W
Solving (2) for aA
( ) ( )( ) ( )( ) 2
1 1
4 2 4 30 2 0 40 mm/s
3 3
A C B
a a a ⎡ ⎤
= − = − − = −
⎣ ⎦
2
40 mm/s
A
a = W

Let x be position relative to the anchor, positive to the right.
3 constant
B C B C A
x x x x x
− + − + − =
4 2 3 0 and 4 2 3 0
C B A C B A
v v v a a a
− − = − − =
(a) Accelerations of B and C.
At 2 s,
t = 420 mm/s and 30 mm/s
A B
v v
= = −
( ) ( )( ) ( )( )
1 1
2 3 2 30 3 420 300 mm/s
4 4
C B A
v v v ⎡ ⎤
= + = − + =
⎣ ⎦
( )0
0
C
v =
( )0
C C C
v v a t
= +
( )0 300 0
2
C C
C
v v
a
t
− −
= = 2
150 mm/s
C
a = W
( ) ( )( ) ( )( ) 2
1 1
4 3 4 150 3 270 105 mm/s
2 2
B C A
a a a ⎡ ⎤
= − = − = −
⎣ ⎦
2
105 mm/s
B
a = W
(b) Initial velocities of A and B.
( )0
A A A
v v a t
= − ( ) ( )( )
0
420 270 2 120 mm/s
A A A
v v a t
= − = − = −
( )0
120 mm/s
A
v = W
( )0
B B B
v v a t
= − ( ) ( )( )
0
30 105 2
B B B
v v a t
= − = − − − ( )0
180 mm/s
B
v = W
Constraint of point E: ( ) ( )
2 constant
C A E A
x x x x
− + − =
3 2 0
E A C
v v v
− + =
(c) ( ) ( ) ( ) ( )( ) ( )( )
0 0 0
3 2 3 120 2 0 360 mm/s
E A C
v v v
= − = − − = −
( )0
360 mm/s
E
v = W

Define positions as positive downward from a fixed level.
Constraint of cable. ( ) ( ) ( )
2 constant
B A C A C B
x x x x x x
− + − + − =
3 2 constant
C B A
x x x
− − =
3 2 0
C B A
v v v
− − =
3 2 0
C B A
a a a
− − =
Motion of block C.
( ) ( )
2
0 0
0, 3.6 in./s , 18 in./s, 0
A A B B B
v a v v a
= = − = = =
( ) ( ) ( )
0 0
0
1
2 6 in./s
3
C B A
v v v
 
= + =
 
( ) ( )( ) 2
1 1
2 0 2 3.6 2.4 in./s
3 3
C B A
a a a  
= + = + − = −
 
( )0
6 1.2
C C C
v v a t t
= + = −
( ) ( ) 2 2
0 0
1
6 0.6
2
C C C C
x x v t a t t t
− = + = −
(a) Time at vC = 0.
0 6 2.4t
= − 2.5 s
t = W
(b) Corresponding position of block C.
( ) ( )( ) ( )( )
2
0
1
6 2.5 2.4 2.5
2
C C
x x
 
− = + −
 
 
( )0
7.5 in.
C C
x x
− = W

Define positions as positive downward from a fixed level.
Constraint of cable: ( ) ( ) ( )
2 constant
B A C A C B
x x x x x x
− + − + − =
3 2 constant
C B A
x x x
− − =
3 2 0
C B A
v v v
− − =
3 2 0
C B A
a a a
− − =
Motion of block C.
( )0
0,
A
v = 2
2.5 in./s ,
A
a t
= − ( )0
0,
B
v = 2
15 in./s
B
a =
( ) ( ) ( )
0 0
0
1
2 0
3
C B A
v v v
⎡ ⎤
= + =
⎣ ⎦
( ) 2
1 1
2 (15 5 ) in./s
3 3
C B A
a a a t
= + = −
( ) 0
0
t
C C C
v v a dt
= + ∫
( )
2
1
0 15 2.5 in./s
3
t t
= + −
( ) ( )
2 3
0
1
7.5 0.83333 in.
3
C C
x x t t
− = −
( ) Time at 0
C
a v =
( )
2
1
0 15 2.5 0
3
t t
+ − = 0 and 6 s
t t
= = 6 s
t = W
(b) Corresponding position of block C.
( ) ( )( ) ( )( )
2 3
0
1
0 7.5 6 0.83333 6
3
C C
x x ⎡ ⎤
− = + −
⎢ ⎥
⎣ ⎦
( )0
30 in.
C C
x x
− = W

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 0
A B C A B C
x x x v v v
+ + = + + =
2 2 0
A B C
a a a
+ + = (1)
Constraint of cable supporting block D:
( ) ( ) constant, 2 0
D A D B D A B
x x x x v v v
− + − = − − =
2 0
D B A
a a a
− − = (2)
Given: / 120 or 120
C B C B C B
a a a a a
= − = − = − (3)
Given: / 220 or 220
D A D A D A
a a a a a
= − = = + (4)
Substituting (3) and (4) into (1) and (2),
( )
2 2 120 0 or 2 3 120
A B B A B
a a a a a
+ + − = + = (5)
( )
2 220 0 or 440
A A B A B
a a a a a
+ − − = − = − (6)
Solving (5) and (6) simultaneously,
2 2
240 mm/s and 200 mm/s
A B
a a
= − =
From (3) and (4), 2 2
80 mm/s and 20 mm/s
C D
a a
= = −
(a) Velocity of C after 6 s.
( ) ( )( )
0
0 80 6
C C C
v v a t
= + = + 480 mm/s
C
v = W
(b) Change in position of D after 10 s.
( ) ( ) ( )( )2
2
0 0
1 1
0 20 10 1000 mm
2 2
D D D D
x x v t a t
− = + = + − = −
1.000 m
D
x
Δ = W

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 C
x x x
+ + = 2 2 0,
A B C
v v v
+ + = 2 2 0
A B C
a a a
+ + =
( ) ( ) ( ) ( ) ( ) ( )
0 0 0 0
0 0
0, ,
A B C A B C
v v v x x x
= = = = = ( ) ( )
/ /
0 0
0, 0
B A B A
x v
= =
( ) ( ) ( )
2 2
/ / / / /
0 0
2
B A B A P A B A B A
v v a x x
⎡ ⎤
− = −
⎣ ⎦
( )
2
/ /
0 2 0
B A B A B A
v a x x
− = − −
( ) ( )
2 2
/ 2
/
40
10 mm/s
2 2 160 80
B A
B A
B A
v
a
x x
= = =
− −
( ) ( ) 2 2
/ / / / /
0 0
1 1
0 0
2 2
B A B A B A B A B A
x x v t a t a t
= + + = + +
( ) ( )
/
2
/ /
2 2 2 160 80
, or 4 s
10
B A B A
B A B A
x x x
t t
a a
− −
= = = =
( ) ( ) 2
0 0
1
2
A A A A
x x v t a t
− = +
(a)
( ) ( ) ( )
( )
0 0
2 2
2 2 80 0
4
A A A
A
x x v t
a
t
⎡ ⎤
− − −
⎣ ⎦
= = 2
10 mm/s
A
a = W
/ 10 10
B A B A
a a a
= + = + 2
20 mm/s
B
a = W
( ) ( )( ) ( )( )
2 2 2 20 2 10 60 mm/s
C B A
a a a ⎡ ⎤
= − + = − + = −
⎣ ⎦
( )
( )0
0
300 0
5 s
60
C C
C C C
C
v v
v v a t t
a
− − −
= + = = =
−
Constraint of cable supporting block D:
( ) ( ) constant, 2 0
D A D B D A B
x x x x v v v
− + − = − − =
( ) ( )
1 1
2 0, 10 20 15 mm/s
2 2
D A B D A B
a a a a a a
− − = = + = + =
(b) ( ) ( ) ( )( )2
2
0 0
1 1
0 15 5
2 2
D D D D
x x v t a t
− = + = + 187.5 mm
D
x
Δ = W

curve
a t
−
1 2
12 m/s, 8 m/s
A A
= − =
(a) curve
v t
−
6 4 m/s
v = −
( )
0 6 1 4 12
v v A
= − = − − − 8 m/s
=
10 4 m/s
v = −
(b) 14 10 2 4 8
v v A
= + = − + 14 4 m/s
v = W
3 4
16 m, 4 m
A A
= = −
5 6
16 m, 4 m
A A
= − = −
7 4 m
A =
(a) curve
x t
−
0 0
x =
4 0 3 16 m
x x A
= + =
6 4 4 12 m
x x A
= + =
10 6 5 4 m
x x A
= + = −
12 10 6 8 m
x x A
= + = −
(b) 14 12 7
x x A
= + 14 4 m
x = − W
Distance traveled:
0 4 s,
t
≤ ≤ 1 16 0 16 m
d = − =
4 s 12 s,
t
≤ ≤ 2 8 16 24 m
d = − − =
12 s 14 s,
t
≤ ≤ ( )
3 4 8 4 m
d = − − − =
Total distance traveled: 16 24 4
d = + + 44 m
d = W

(a) Construction of the curves.
curve
a t
−
1 2
12 m/s, 8 m/s
A A
= − =
curve
v t
−
0 8 m/s
v =
( )
6 0 1 8 12 4 m/s
v v A
= + = + − = −
10 6 4 m/s
v v
= = −
14 10 2 4 8 4 m/s
v v A
= + = − + =
3 4
16 m, 4 m
A A
= = −
5 6
16 m, 4 m
A A
= − = −
7 4 m
A =
curve
x t
−
0 0
x =
4 0 3 16 m
x x A
= + =
6 4 4 12 m
x x A
= + =
10 6 5 4 m
x x A
= + = −
12 10 6 8 m
x x A
= + = −
14 12 7 4 m
x x A
= + = −
(b) Time for 8 m.
x >
From the x t
− diagram, this is time interval 1 2
to .
t t
Over 0 6 s,
t
< < 8 2
dx
v t
dt
= = −
continued

Integrating, using limits 0
x = when 0
t = and 8 m
x = when
1
t t
=
8 2 2
1 1
0 0
8 or 8 8
t
x t t t t
 
= − = −
 
or 2
1 1
8 8 0
t t
− + =
( ) ( )( )( )
( )( )
2
1
8 8 4 1 8
4 2.828 1.172 s and 6.828 s
2 1
t
± −
= = ± =
The larger root is out of range, thus 1 1.172 s
t =
Over 6 10,
t
< < ( )
12 4 6 36 4
x t t
= − − = −
Setting 8,
x = 2 2
8 36 4 or 7 s
t t
= − =
Required time interval: ( )
2 1 5.83 s
t t
− = W

The a–t curve is just the slope of the v–t curve.
0 10 s,
t
< < 0
a = W
10 s < 18 s,
t < 2
18 6
1.5 ft/s
18 10
a
−
= =
−
W
18 s < 30 s,
t < 2
18 18
3 ft/s
30 18
a
− −
= = −
−
W
30 s < 40 s
t < 0
a = W
Points on the x–t curve may be calculated using areas of the v–t
curve.
1 (10)(6) 60 ft
A = =
2
1
(6 18)(18 10) 96 ft
2
A = + − =
3
1
(18)(24 18) 54 ft
2
A = − =
4
1
( 18)(30 24) 54 ft
2
A = − − = −
5 ( 18)(40 30) 180 ft
A = − − = −
0 48 ft
x = − W
0
1 0 1 12 ft
x x A
= + = W
8
1 10 2 108 ft
x x A
= + = W
24 18 3 162 ft
x x A
= + = W
30 24 4 108 ft
x x A
= + = W
40 30 5 72 ft
x x A
= + = − W
continued

(a) Maximum value of x.
Maximum value of x occurs
When 0,
v = i.e. 24 s.
t =
max 162 ft
x = W
(b) Time s when 108 ft.
x =
From the x–t curve,
18 s and 30 s
t t
= = W

Data from problem 11.65: 0 48 ft
x = −
The a–t curve is just the slope of the v–t curve.
0 10 s,
t
< < 0
a = !
10 s < 18 s,
t < 2
18 6
1.5 ft/s
18 10
a
−
= =
−
!
18 s < 30 s,
t < 2
18 18
3 ft/s
30 18
a
− −
= = −
−
!
30 s < 40 s,
t < 0
a = !
Points on the x–t curve may be calculated using areas of the v–t
curve. !
1 (10)(6) 60 ft
A = =
2
1
(6 18)(18 10) 96 ft
2
A = + − =
3
1
(18)(24 18) 54 ft
2
A = − =
4
1
( 18)(30 24) 54 ft
2
A = − − = −
5 ( 18)(40 30) 180 ft
A = − − = −
0 48 ft
x = − !
0 0
1 1 12 ft
x x A
= + = !
10
18 2 108 ft
x x A
= + = !
24
x = 18 3
x A
+ = 162 ft !
30
x = 24 4
x A
+ = 108 ft !
40 30 5 72 ft
x x A
= + = − !
continued

(a) Total distance traveled during 0 30 s
t
≤ ≤ .
For 0 24 s
t
≤ ≤ 1 24 0 210 ft
d x x
= − =
For 24 s 30 s
t
≤ ≤ 2 30 24 54 ft
d x x
= − =
Total distance. 1 2
d d d
= + 264 ft
d = !
(b) Values of t for which 0.
x =
In the range 0 10 s
t
≤ ≤
0 0 48 6
x x v t t
= + = − +
Set 0.
x = 1
48 6 0
t
− + = 1 8 s
t = !
In the range 30 s 40 s,
t
< <
30 30 ( 30)
x x v t
= + −
108 ( 18)( 30)
t
= + − −
648 18t
= −
Set 0.
x = 2
648 18 0
t
− = 2 36 s
t = !

Sketch v t
− curve as shown. Label areas 1 2
, ,
A A and 3
A
( )( )
1 3 20 60 in.
A = =
1 1
2 in./s
v at t
Δ = =
( ) 2
2 1 1
1
in.
2
A v t t
= Δ =
( )( ) ( )
3 1 1 1
20 2 20 in.
A v t t t
= Δ − = −
Distance traveled: 12 ft 144 in.
x
Δ = =
( )
2
1 1 1
total area, 144 60 2 20
x t t t
Δ = = + + −
or 2
1 1
40 84 0
t t
− + =
( )( )( )
( )( )
2
1
40 40 4 1 84
2.224 s and 37.8 s
2 1
t
± −
= =
Reject the larger root. 1 2.224 s
t =
1
2 4.45 in./s
v t
Δ = =
max 3 3 4.45
v v
= + Δ = + max 7.45 in./s
v = W

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 3
t as
shown.
Use constant slopes in the v t
− curve for the constant
acceleration stages.
Areas of v t
− curve:
( )
1 1 1
1
180 44 112 ft
2
A t t
= − + = −
2 2
44
A t
= −
( )
3 3 3
1
44 22
2
A t t
= − = −
Changes in position: 1 1800 1900 100 ft
x
Δ = − = −
2 100 1800 1700 ft
x
Δ = − = −
3 0 100 100 ft
x
Δ = − = −
Using i i
x A
Δ = gives 1
100
0.893 s
112
t
−
= =
−
2
1700
38.64 s
44
t
−
= =
−
3
100
4.55 s
22
t
−
= =
−
(a) Total time: 1 2 3 44.1 s
t t t
+ + = W
(b) Initial acceleration.
( ) ( )
44 180
0.893
v
a
t
− − −
Δ
= =
Δ
2
152.3 ft/s
a = W

Sketch the v t
− curve
Data: 0 64 km/h 17.778 m/s
v = =
3
2 4.8 km 4.8 10 m
x = = ×
1 32 km/hr 8.889 m/s
v = =
3 3
1 4.8 10 800 4.0 10 m
x = × − = ×
2 450 s
t =
(a) Time 1
t to travel first 4 km.
( ) ( )
3
1 1 0 1 1 1
1 1
4.0 10 17.778 8.889
2 2
x A v v t t
= × = = + = + 1 300 s
t = W
(b) Velocity 2.
v
( )( ) ( )( )
2 1 2 1 2 2 1 1 2
1 1
800 450 300
2 2
x x A v v t t v v
− = = = + − = + −
2 1 10.667 m
v v
+ =
2 10.667 8.889
v = − 2 1.778 m/s
v = W
(c) Final deceleration.
2
2 1
12
2 1
1.778 8.889
0.0474 m/s
450 300
v v
a
t t
− −
= = = −
− −
2
12 0.0474 m/s
a = W

10 20
10 min 20 s 0.1722 h
60 3600
= + =
Sketch the v t
− curve
60
25
35
a
b
c
t
a
t
a
t
a
=
=
=
( )( ) ( )
1 1 1
1 1 1 1
60 60 25 60 1800 312.5
2 2
a b
A t t t t
a a
= − − = − −
But 1 5 mi
A =
1
1
60 2112.5 5
t
a
− = (1)
( )
2 1 1
1
35 0.1722 35 6.0278 35 612.5
c
A t t t
a
= − − = − −
But 2 8 5 3 mi
A = − =
1
1
35 612.5 3.0278
t
a
+ = (2)
1
1
Solving equations (1) and (2) for and ,
t
a
3
1 85.45 10 h 5.13 min
t −
= × =
6 2
1
60.23 10 h /mi
a
−
= ×
( )( )
( )
3
3 2
2
16.616 10 5280
16.616 10 mi/h
3600
a
×
= × = 2
6.77 ft/s
a = W

Sketch the curve as shown
a t
−
0 1
20 ft/s, 6 ft/s
v v
= = −
( )
1 1
2 1
6
1
40 6 17
2
A t
1
A t t
= −
= − − = −
1 0 1
v v A A2
= + +
1 1
6 20 6 17
t t
= − −
(a) 1 0.6087 s
t = 1 0.609 s
t = W
2 1.4 s
t =
2 1 0.7913 s
t t
− =
( )( )
1 3 6 1.4 8.4 ft/s
A A
+ = − = −
( )( )
2 17 0.6087 10.348 ft/s
A = − = −
2 0 1 3 2 20 8.4 10.348
v v A A A
= + + + = − − 2 1.252 ft/s
v = W
(b) ( )
2 0 0 2 1 3 13 2 2 by moment-area method
x x v t A A x A x
= + + + +
( )
0 2 1 3 2 2 2 1
1 1
0
2 3
v t A A t A t t
⎛ ⎞ ⎛
= + + + + −
⎜ ⎟ ⎜
⎝ ⎠ ⎝
⎞
⎟
⎠
( )( ) ( ) ( ) ( )
1 0.6087
0 20 1.4 8.4 1.4 10.348 1.4
2 3
⎛ ⎞ ⎛ ⎞
= + − − −
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
2 9.73 ft
x = W
)

Note that
1 5280
mile 660 ft
8 8
= =
Sketch v t
− curve for first 660 ft.
Runner A: 1 2
4 s, 25 4 21 s
t t
= = − =
( )( ) ( )
1 max max
1
4 2
2
A A
A v v
= =
( )
2 max
21 A
A v
=
1 2
5280 ft
= 660 ft
8
A A x
+ = ∆ =
( ) ( )
max max
23 660 or 28.696 ft/s
A A
v v
= =
Runner B: 1 2
5 s, 25.2 5 20.2 s
t t
= = − =
( )( ) ( )
1 max max
1
5 2.5
2
B B
A v v
= =
( )
2 max
20.2 B
A v
=
1 2 660 ft
A A x
+ = ∆ =
( ) ( )
max max
22.7 660 or 29.075 ft/s
B B
v v
= =
Sketch v t
− curve for second 660 ft. 3 3
0.3
v a t t
∆ = =
2
3 max 3 3 3 max 3
1
660 or 0.15 660 0
2
A v t vt t v t
= − ∆ = − + =
( ) ( )( )( )
( )( )
( )
2
2
max max
3 max max
4 0.15 660
3.3333 396
2 0.15
v v
t v v
± −  
= = ± −
 
 
Runner A:( )
max 28.696,
A
v = ( )
3 164.57 s and 26.736 s
A
t =
Reject the larger root. Then total time (a) 25 26.736 51.736 s
A
t = + =
51.7 s
A
t = W

Runner B: ( )
max 29.075,
B
v = ( )
3 167.58 s and 26.257 s
B
t =
Reject the larger root. Then total time 25.2 26.257 51.457 s
B
t = + =
51.5 s
B
t = W
Velocity of A at 51.457 s:
t =
( )( )
1 28.696 0.3 51.457 25 20.759 ft/s
v = − − =
Velocity of A at 51.736 s:
t =
( )( )
2 28.696 0.3 51.736 25 20.675 ft/s
v = − − =
Over 51.457 s 51.736 s, runner covers a distance
t A x
≤ ≤ ∆
(b) ( ) ( )( )
ave
1
20.759 20.675 51.736 51.457
2
x v t
∆ = ∆ = + − 5.78 ft
x
∆ = W

Sketch the v t
− curves.
At 12 min 720 s,
t = =
( )( )
truck
bus
bus
19.44 720 14000 m
14000 1200 15200 m
area under curve
x
x
x v t
= =
= + =
= −
( )( ) ( )( )
1 1
1
120 27.78 720 27.78 15200
2
t t
− + − =
1 225.8 s
t =
(a) When bus truck,
x x
= areas under the v t
− curves are equal.
( )( ) ( )
1 2 1 2
1
27.78 120 27.78 19.44
2
t t t t
− + − =
With 1 225.8 s,
t = 2 576 s
t = W
( )( )
truck 19.44 576 11200 m
x = = truck 11.20 km
x = W
(b) 0
bus
1
27.78 0
120 225.8 120
v v
a
t
− −
= =
− −
2
bus 0.262 m/s
a = W

( )0
32 km/h 8.889 m/s
24 km/h 6.667 m/s
A
B
v
v
= =
= =
Sketch the v t
− curves.
( )( )
( )( ) ( )
( )
( )
1
2 /
/
1 2
0
1
0
6.667 45 300 m
1 1
2.222 45 45
2 2
50 22.5
A B
A B
A A
B B
A
A v
v
x x A A
x x A
= =
= +
= +
= + +
= +
( )
/ / 2
0
B A B A
x x A
= −
( )
b /
0 60 50 22.5 A B
v
= − − / 0.444 m/s
A B
v = W
/ 6.667 0.444 7.111 m/s
A B A B
v v v
= + = + =
(a)
( )0 7.111 8.889
45
A A
A
v v
a
t
− −
= = 2
0.0395 m/s
A
a = − W

( )
( )
0
0
22 mi/h 32.267 ft/s
13 mi/h 19.067 ft/s
A
B
v
v
= =
= =
Sketch the v t
− curves.
Slope of v t
− curve for car A.
( )( )
2
1
1
2
13.2
0.14 ft/s
13.2
94.29 s
0.14
1
13.2 94.29 622.3 m
2
a
t
t
A
= − = −
= =
= =
( )
( )
1
0
1 2
0
B B
A A
x x A
x x A A
= +
= + +
( ) ( )
/ 2 2
0 0
, or 0
B A B A B A
x x x x x A d A
= − = − − = −
2
d A
= 622 m
d = W

Construct the a t
− curves for the elevator and the ball.
Limit on 1
A is 24 ft/s. Using 1 4
A t
=
2 2
4 24 6 s
t t
= =
Motion of elevator.
For 1
0 6 s,
t
≤ ≤ ( ) ( )
0 0
0 0
E E
x v
= =
Moment of 1
A about 1:
t t
= 2
1
1 1
4 2
2
t
t t
=
( ) ( ) 2 2
1 1 1
0 0
2 2
E E E
x x v t t t
= + + =
Motion of ball. At 2,
t = ( ) ( )
0 0
40 ft 64 ft/s
B B
x v
= =
For 1 2 s,
t > ( )
2 1
32.2 2 ft/s
A t
= − −
Moment of 2
A about 2 :
t t
= ( ) ( )2
1
1 1
2
32.2 2 16.1 2
2
t
t t
−
⎛ ⎞
− − = − −
⎜ ⎟
⎝ ⎠
( ) ( ) ( ) ( )
( ) ( )
2
1 1
0 0
2
1 1
2 16.1 2
40 64 2 16.1 2
B B B
x x v t t
t t
= + − − −
= + − − −
When ball hits elevator, B E
x x
=
( ) ( )2 2
1 1 1
2
1 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 s
t =
The smaller root is out of range, hence 1 5.59 s
t = W
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
v
dt
= and
dv
a
dt
= .
The motion of the car relative to the truck occurs in 3 phases, lasting t1, t2, and t3 seconds, respectively.
Phase 1, acceleration. 2
1 2 m/s
a =
Phase 2, constant speed. 2 90 km/h 54 km/h
v = −
36 km/h = 10 m/s
=
Phase 3, deceleration. 2
3 8 m/s
a = −
Time of phase 1. 2
1
1
0 10 0
5 s
2
v
t
a
− −
= = =
Time of phase 3. 2
3
2
0 0 10
1.25 s
8
v
t
a
− −
= = =
Sketch the a t
− curve.
Areas: 1 1 2 10 m/s
A t v
= =
3 3 10 m/s
A t v
= = −
Initial and final positions.
0 30 16 46 m
x = − − = −
30 5 35 m
f
x = + =
Initial velocity. 0 0
v =

Final time. 1 2 3
f
t t t t
= + +
0 0
f f i i
x x v t A t
= + + ∑
1 1
1
2
f
t t t
= −
2
5 1.25 2.5
t
= + + −
2
3.75 t
= +
2 3
1
0.625 s
2
t t
= =
( )( ) ( )( )
2
35 46 0 10 3.75 10 0.625
t
= − + + + + −
2
49.75
4.975 s
10
t = =
1 2 3 11.225 s
f
t t t t
= + + =
Total time. 11.23 s
f
t = W
1 2 9.975 s
t t
+ =

Let x be the position of the front end of the car relative to the front end of the truck.
Let
dx
v
dt
= and
dv
a
dt
= .
The motion of the car relative to the truck occurs in two phases, lasting t1 and t2 seconds,
respectively.
Phase 1, acceleration. 2
1 2 m/s
a =
Phase 2, deceleration. 2
2 8 m/s
a = −
Sketch the a–t curve.
Areas: 1 1
2
A t
=
2 2
8
A t
= −
Initial and final positions
0 30 16 46 m
x = − − = −
30 5 35 m
f
x = + =
Initial and final velocities.
0 0
f
v v
= =
0 1 2
f
v v A A
= + +
1 2
0 0 2 8
t t
= + −
1 2
4
t t
=
0 0
f f i i
x x v t A t
= + + ∑
1 2 1 2
1
3
2
t t t t
= + =
2 2
1
2
t t
=

( )( ) ( )
2 2 2 2
1
35 46 0 2 4 3 8
2
t t t t
 
= − + + + −  
 
2
2
81 20 t
=
2 2.0125 s
t =
1 8.05 s
t =
1 2 10.0625 s.
f
t t t
= + =
Maximum relative velocity.
( )( )
1 1 2 8.05 16.10 m/s
m
v a t
= = =
60.0 km/h
m
v =
Maximum velocity relative to ground.
max 54 60.0
T
v v v
= + = +
max 112.0 km/h
v = !

Sketch acceleration curve.
Let jerk
da
j
dt
= =
Then, ( )
max
a j t
= Δ
( ) ( )
( )
1 max max
2
1
2
2
A a t a t
j t
= Δ = Δ
= Δ
0 1 2
1 2
2 1
0 0
f
v v A A
A A
A A
= + −
= + −
=
( ) ( )( ) ( )
( ) ( ) ( )
( )( )
0 1 2
3 3 3
3 3
4 3
0 3 2
0.36
0.4932
2 2 1.5
x v t A t A t
j t j t j t
x
t
j
Δ = Δ + Δ − Δ
= + Δ − Δ = Δ
Δ
Δ = = =
(a) Shortest time: ( )( )
4 4 0.4932 1.973 s
t
Δ = = W
(b) Maximum velocity: ( )2
max 0 1 0
v v A j t
= + = + Δ
( )( )2
1.5 0.4932 0.365 m/s
= = W
Average velocity: ave
0.36
0.1825 m/s
4 1.973
x
v
t
Δ
= = =
Δ
W

Sketch the a t
− curve.
From the jerk limit, ( )
1 max
j t a
Δ = or ( ) max
1
1.25
5 s.
0.25
a
t
j
Δ = = =
( )( )
1
1
5 1.25 3.125 m/s
2
A = =
( )( )
max 1 2
2 max 1
2
2
max
32 km/hr 8.889 m/s 2
2 8.889 2 3.125 2.639 m/s
2.639
2.111 s
1.25
v A A
A v A
A
t
a
= = = +
= − = − =
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 1
2 2
2 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 2
max
5000
2 2 5 2.111 10 2.111 562.5 575 s
8.889
f
f
x
t t t
v
= Δ + Δ + = + + = + + =
9.58 min
f
t = W
(b) ave
5000
8.70 m/s
575
f
f
x
v
t
= = = ave 31.3 km/h
v = W

Indicate areas 1 2
and
A A on the a t
− curve.
( )
1
1
0.6 0.1 m/s
2 3
T
A T
= =
( )
2
1 2
0.6 0.2 m/s
2 3
T
A T
= =
By moment-area formula,
( )
( )( )
0 1 2
2 2 2 2
2 2
7 4
9 9
7 8 15 1
40 0
90 90 90 6
40 6 240 s
x v t A T A T
T T T T
T
⎛ ⎞ ⎛ ⎞
= + +
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= + + = =
= =
(a) 15.49 s
T = W
max 0 1 2 0 0.1 0.2 0.3
v v A A T T T
= + + = + + =
(b) max 4.65 m/s
v = W
Indicate area 3 4
and
A A on the a t
− curve.
( )
( )
1 3
4
1
0.1 0.6 0.05
2 6
1
0.45 0.0375
2 6
T
A T A T
T
A T
= = =
= =
(c) 0 1 3 4 0.1875
v v A A A T
= + + + = 2.90 m/s
v = W
( ) ( ) ( )
0 1 3 4
2
2 2 1
2 2 9 3 6 3 6
5
0 0.1 0.05 0.0375 0.035417
18 9 18
T T T T T
x v A A A
T T T
T T T T
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
= + − + ⋅ + ⋅
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞ ⎛ ⎞
= + + + =
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
( )( )2
0.035417 15.49
= 8.50 m
x = W

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/s
3
5 0.2 1 m/s
1
5 2.5 0.1 0.375 m/s
2
1
2.5 0.1 0.125 m/s
2
A
A
A
A
= =
= =
= + =
= =
(a) 0
Velocities: 0
v =
0.2 0 1 2
v v A A
= + + 0.2 1.400 m/s
v = W
0.3 0.2 3
v v A
= + 0.3 1.775 m/s
v = W
0.4 0.3 4
v v A
= + 0.4 1.900 m/s
v = W
Sketch the v t
− curve and divide its area into 5 6 7
, , and
A A A as
shown.
0.3 0.4 0.4
0.3 or 0.3
x t t
dx x vdt x vdt
= − = = −
∫ ∫ ∫
At 0.3 s,
t = ( )( )
0.3 5
0.3 1.775 0.1
x A
= − −
(b) With ( )( )
5
2
0.125 0.1 0.00833 m
3
A = = 0.3 0.1142 m
x = W
At 0.2 s,
t = ( )
0.2 5 6 7
0.3
x A A A
= − + −
With ( )( )
5 6
2
0.5 0.2 0.06667 m
3
A A
+ = =
and ( )( )
7 1.400 0.2 0.28 m
A = =
0.2 0.3 0.06667 0.28
x = − − 0.2 0.0467m
x = − W

Approximate the a t
− curve by a series of rectangles of height ,
i
a each with its centroid at .
i
t t
= When equal
widths of 0.25 s
t
Δ = are used, the values of and
i i
t a are those shown in the first two columns of the table
below.
At 2 s,
t = ( )
2
0 0
0 i
v v adt v a t
= + ≈ + Σ Δ
∫
( )( )
0 i
v a t
≈ + Σ Δ
(a) ( )( )
0
0 7.650 0.25
v
≈ − 0 1.913 ft/s
v = W
Using moment-area formula,
( ) ( )( )
( )
( )( )
2
0 0 0 0
0
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 = W
i
t i
a 2 i
t
− ( )
2
i i
a t
−
( )
s ( )
2
ft/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 ( )
2
7.650 ft/s
− ( )
11.955 ft/s
−

Approximate the a t
− curve by a series of rectangles of height ,
i
a each with its centroid at .
i
t t
= When equal
widths of 2 s
t
Δ = are used, the values of and
i i
t a are those shown in the first two columns of table below.
(a) At 8 s,
t = ( )
8
8 0 0
0 i
v v adt a t
= + ≈ + Σ Δ
∫
( )( )
i
a t
= Σ Δ
Since 8 s,
t = only the first four values in the second column are summed:
2
17.58 13.41 10.14 7.74 48.87 ft/s
i
a
Σ = + + + =
( )( )
8 48.87 2
v = 8 97.7 ft/s
v = W
(b) At 20 s,
t = ( ) ( )( )
20
20 0
20 0 20
o i
x v t a t dt a t t
= + − = + Σ − Δ
∫
( )( )
990.1 2
= 20 1980 ft
x = W
i
t i
a 20 i
t
− ( )
20
i i
a t
−
( )
s ( )
2
ft/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 1
2 1 2 1
2 or
2
v v
v v a x x x
a
−
− = − Δ =
( )
2 1 2 1
v v a t t
− = − or 2 1
v v
t
a
−
Δ =
For the regions shown above,
(a) ( ) 3.19 s
t t
= Σ Δ = W
(b) Assuming 0 0,
x = ( )
0 62.6 m
x x x
= + Σ Δ = W
Region ( )
1 m/s
v ( )
2 m/s
v ( )
2
m/s
a ( )
m
x
Δ ( )
s
t
Δ
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
dv
a v
dx
= noting that
dv
dx
= 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 v
Δ Δ Then, .
dv v
dx x
Δ
=
Δ
(a) When 0.25,
x =
1.4 m/s
v = from the curve
1m/s and 0.25m from the tangent line
v x
Δ = Δ =
( )( )
1
1
4 s 1.4 4
2.5
dv
a
dx
−
= = = 2
5.6 m/s
a = W
(b) When 2.0 m/s,
v = 0.5m
x = from the curve.
1 m/s and 0.6m from the tangent line.
v x
Δ = Δ =
( )( )
1
1
1.667s , 2 1.667
0.6
dv
a
dx
−
= = = 2
3.33 m/s
a = W

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 0
1
2
x x v A t t x v t at t
⎛ ⎞
= + + − = + + ⎜ ⎟
⎝ ⎠
2
0 0
1
2
x v t at
= + +

From the curve,
a t
− ( )( )
1 2 6 12 m/s
A = − = −
( )( )
2 2 2 4 m/s
A = =
Over 6 s 10 s,
t
< < 4 m/s
v = −
0 1 0 0
, or 4 12, or 8 m/s
v v A v v
= + − = − =
12 0 0 moment of shaded area about 12s
x x v t t
= + + =
( )( ) ( )( ) ( )( )
12 0 8 12 12 12 3 4 12 11
x = + + − − + − 12 8 m
x = − W

(a) 0.2s.
T =
( )( )
1
2
24 0.2 3.2 ft/s
3
A = − = −
( )( )
2 1
1
24 0.2
24 4.8
A t
t
= − −
= − +
0
1
0 90 3.2 24 4.8
f
v v A
t
= + Σ
= − − +
1 3.8167 s
t =
2 86.80 ft/s
A = −
1 3.6167 s
t T
− =
By moment-area formula, 1 0 0 1 moment of area
x x v t
= + +
( )( ) ( ) ( ) ( )
1
3 3.6167
0 90 3.8167 3.2 0.2 3.6167 86.80
8 2
x
⎡⎛ ⎞ ⎛ ⎞
= + + − + + −
⎜ ⎟ ⎜ ⎟
⎢
⎝ ⎠ ⎝ ⎠
⎣
1 174.7 ft
x = W
(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 s
f
A
A t t
v v A t t
= − = −
= − − = − +
= + Σ = − − + =
1 2
3.2167s 77.2 ft/s
t T A
− = = −
( )( ) ( ) ( ) ( )
1
3 3.2167
0 90 4.0167 12.8 0.8 3.2167 77.2
8 2
x
⎡ ⎤ ⎛ ⎞
= + + − + + − ⎜ ⎟
⎢ ⎥
⎣ ⎦ ⎝ ⎠
1 192.3 ft
x = W

Data from Prob. 65
0 0
48 ft, 6 ft/s
x v
= − =
The a – t curve is just the slope of the v – t curve.
0 10 s,
t
< < 0
a = !
10 s < < 18 s,
t
18 6
1.5 ft/s
18 10
a
−
= =
−
!
18 s 30 s,
t
< <
18 18
3 ft/s
30 18
a
− −
= = −
−
!
30 s < < 40 s
t 0
a = !
0 0 i i
x x v t A t
= + + ∑
(a) Position when t = 20 s.
( )( )
1 18 10 1.5 12 ft/s
A = − =
1 20 14 6s
t = − =
( )( )
2 2 3 6 ft/s
A = − = −
2 20 19 1 s
t = − =
( )( ) ( )( ) ( )( )
20 48 6 20 12 6 6 1
x = − + + + −
20 138 ft
x = !
(b) Maximum value of position coordinate.
x is maximum where 0.
v =
From velocity diagram, 24 s
m
t =
( )( )
1 18 10 1.5 12 ft/s
A = − =
( )
1 24 14 10 s
t = − =
( )( )
2 24 18 3 18 ft/s
A = − − = −
( )
2 24 21 3 s
t = − =
( )( ) ( )( ) ( )( )
48 6 24 12 10 18 3
m
x = − + + + −
162 ft
m
x = !

( )2
1
= +
x t ( ) 2
4 1
−
= +
y t
( )
2 1
= = +
&
x
v x t ( ) 3
8 1
−
= = − +
&
y
v y t
2
= =
&
x x
a v ( ) 4
24 1
−
= = +
&
y y
a v t
Solve for (t + 1)2
from expression for x. (t + 1)2
= x
Substitute into expression for y.
4
y
x
=
Then, 4
xy =
This is the equation of a rectangular hyperbola.
(a) t = 0. 2 m/s, 8 m/s
x y
v v
= = −
( ) ( )
2 2
2 8 8.25 m/s
v = + − =
1 8
tan 76.0
2
θ −  
−
= = − °
 
 
8.25 m/s
=
v 76.0° W
( ) ( )
2 2
2 2 2
1
2 m/s , 24 m/s
2 24 24.1 m/s
24
tan 85.2
2
x y
a a
a
θ −
= =
= + =
 
= = °
 
 
2
24.1 m/s
=
a 85.2°W
(b)
1
s.
2
t = 3 m/s,
x
v = 2.37 m/s
= −
y
v
( )2 2
3 (2.37) 3.82 m/s
v = + =
1 2.37
tan 38.3
3
θ − −
 
= = − °
 
 
3.82 m/s
=
v 38.3° W

2 m/s,
x
a = 2
4.74 m/s
y
a =
2 2 2
2 4.74 5.15 m/s
a = + =
1 4.74
tan 67.2
2
θ −  
= = °
 
 
2
5.15 m/s
=
a 67.2°W

Let ( )
2 3 2
9 18 9 18
u t t t t t t
= − + = − +
Then,
2
2
2
3 18 18, and 6 18
du d u
t t t
dt dt
= − + = −
6 0.8 m
x u
= − 4 0.6 m
y u
= − +
0.8
dx du
dt dt
= − 0.6
dy du
dx dt
= +
0.6
0.75 constant
0.8
= = − = − =
dy
dt
dx
dt
dy
dx
Since
dy
dx
does not change, the path is straight.
(a) At 2 s,
t =
2
2
6, and 6.
= − = −
du d u
dt dt
( )( ) ( )( )
0.8 6 4.8 m/s, 0.6 6 3.6 m/s
= = − − = = = − = −
x y
dx dy
v v
dt dt
( )( ) ( )( )
2
2 2
2
0.8 6 4.8 m/s , 0.6 6 3.6 m/s
= = − − = = − = −
x y
d x
a a
dt
6.0 m/s
=
v 36.9 ,
° 2
6.0 m/s
=
a 36.9° W
(b) At 3 s,
t =
2
2
9, and 0
du d u
dt dt
= − =
( )( ) ( )( )
0.8 9 7.2 m/s, 0.6 9 5.4 m/s
x y
v v
= − − = = − = −
0, 0
x y
a a
= =
9.0 m/s
=
v 36.9 ,
° 0
=
a W
(c) At 4 s,
t =
2
2
6, and 6
du d u
dt dt
= − =
( )( ) ( )( )
0.8 6 4.8 m/s, 0.6 6 3.6 m/s
= − − = = − = −
x y
v v
( )( ) ( )( )
2 2
0.8 6 4.8 m/s , 0.6 6 3.6 m/s
x y
a a
= − = − = =
6.0 m/s
=
v 36.9 ,
° 2
6.0 m/s
=
a 36.9° W

Solucionario Beer, Johnton, Mazurek y Eisenberg - Octava Edicion.pdf

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