COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 2.
2
x = t3 − (t − 2) m
(a)
v=
dx
= 3t 2 − 2 ( t − 2 ) m/s
dt
a=
dv
= 6t − 2 m/s 2
dt
Time at a = 0.
0 = 6t0 − 2 = 0
t0 =
(b)
1
3
t0 = 0.333 s W
Corresponding position and velocity.
3
2
⎛1⎞
⎛1
⎞
x = ⎜ ⎟ − ⎜ − 2 ⎟ = − 2.741 m
⎝3⎠
⎝3
⎠
x = − 2.74 m W
2
⎛1⎞
⎛1
⎞
v = 3 ⎜ ⎟ − 2 ⎜ − 2 ⎟ = 3.666 m/s
⎝3⎠
⎝3
⎠
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.
v = 3.67 m/s W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 3.
Position:
x = 5t 4 − 4t 3 + 3t − 2 ft
Velocity:
v=
dx
= 20t 3 − 12t 2 + 3 ft/s
dt
Acceleration:
a=
dv
= 60t 2 − 24t ft/s 2
dt
When t = 2 s,
4
3
x = ( 5 )( 2 ) − ( 4 )( 2 ) − ( 3)( 2 ) − 2
3
2
v = ( 20 )( 2 ) − (12 )( 2 ) + 3
2
a = ( 60 )( 2 ) − ( 24 )( 2 )
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.
x = 52 ft W
v = 115 ft/s W
a = 192 ft/s 2 W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 4.
Position:
x = 6t 4 + 8t 3 − 14t 2 − 10t + 16 in.
Velocity:
v=
dx
= 24t 3 + 24t 2 − 28t − 10 in./s
dt
Acceleration:
a=
dv
= 72t 2 + 48t − 28 in./s 2
dt
When t = 3 s,
4
3
2
x = ( 6 )( 3) + ( 8 )( 3) − (14 )( 3) − (10 )( 3) + 16
3
2
v = ( 24 )( 3) + ( 24 )( 3) − ( 28 )( 3) − 10
2
a = ( 72 )( 3) + ( 48 )( 3) − 28
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.
x = 562 in. !
v = 770 in./s !
a = 764 in./s 2 !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 5.
Position:
x = 500sin kt mm
Velocity:
v=
dx
= 500k cos kt mm/s
dt
Acceleration:
a=
dv
= − 500k 2 sin kt mm /s 2
dt
When t = 0.05 s,
and
k = 10 rad/s
kt = (10 )( 0.05 ) = 0.5 rad
x = 500sin ( 0.5 )
v = ( 500 )(10 ) cos ( 0.5 )
2
a = − ( 500 )(10 ) sin ( 0.5 )
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.
x = 240 mm !
v = 4390 mm/s !
a = − 24.0 × 103 mm/s 2 !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 6.
(
)
x = 50sin k1t − k2t 2 mm
Position:
k2 = 0.5 rad/s 2
Where
k1 = 1 rad/s
Let
θ = k1t − k2t 2 = t − 0.5t 2 rad
dθ
= (1 − t ) rad/s
dt
x = 50sin θ mm
Position:
and
d 2θ
= −1 rad/s 2
dt 2
and
dx
dθ
= 50cosθ
mm/s
dt
dt
dv
a=
dt
v=
Velocity:
Acceleration:
2
a = 50cosθ
When v = 0,
d 2θ
⎛ dθ ⎞
2
− 50sin θ ⎜
⎟ mm/s
dt
dt 2
⎝
⎠
either
cosθ = 0
dθ
=1− t = 0
dt
Over 0 ≤ t ≤ 2 s, values of cosθ are:
t =1s
or
t (s)
0
0.5
1.0
1.5
2.0
θ ( rad )
0
0.375
0.5
0.375
0
cosθ
1.0
0.931
0.878
0.981
1.0
No solutions cosθ = 0 in this range.
For t = 1 s,
2
θ = 1 − ( 0.5 )(1) = 0.5 rad
x = 50sin ( 0.5 )
a = 50cos ( 0.5 )( −1) − 50sin ( 0.5 )( 0 )
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.
x = 24.0 mm W
a = − 43.9 mm/s 2 W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 7.
Given:
x = t 3 − 6t 2 + 9t + 5
Differentiate twice.
v=
dx
= 3t 2 − 12t + 9
dt
a=
dv
= 6t − 12
dt
(a)
v=0
When velocity is zero.
3t 2 − 12t + 9 = 3 ( t − 1)( t − 3) = 0
t = 1 s and t = 3 s W
(b)
Position at t = 5 s.
3
2
x5 = ( 5 ) − ( 6 )( 5 ) + ( 9 )( 5 ) + 5
x5 = 25 ft W
Acceleration at t = 5 s.
a5 = ( 6 )( 5 ) − 12
a5 = 18 ft/s 2 W
Position at t = 0.
x0 = 5 ft
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
x1 = (1) − ( 6 )(1) + ( 9 )(1) + 5 = 9 ft
Position at t = 3 s.
3
2
x3 = ( 3) − ( 6 )( 3) + ( 9 )( 3) + 5 = 5 ft
Distance traveled.
At t = 1 s
d1 = x1 − x0 = 9 − 5 = 4 ft
At t = 3 s
d3 = d1 + x3 − x1 = 4 + 5 − 9 = 8 ft
At t = 5 s
d5 = d3 + x5 − x3 = 8 + 25 − 5 = 28 ft
d5 = 28 ft W
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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 8.
3
x = t 2 − ( t − 2 ) ft
v=
(a)
dx
2
= 2t − 3 ( t − 2 ) ft/s
dt
Positions at v = 0.
2
2t − 3 ( t − 2 ) = − 3t 2 + 14t − 12 = 0
t=
−14 ± (14) 2 − (4)(− 3)(−12)
(2)(− 3)
t1 = 1.1315 s and t2 = 3.535 s
(b)
At t1 = 1.1315 s,
x1 = 1.935 ft
x1 = 1.935 ft W
At t2 = 3.535 s,
x2 = 8.879 ft
x2 = 8.879 ft W
Total distance traveled.
At t = t0 = 0,
x0 = 8 ft
At t = t4 = 4 s,
x4 = 8 ft
Distances traveled.
0 to t1:
d1 = 1.935 − 8 = 6.065 ft
t1 to t2:
d 2 = 8.879 − 1.935 = 6.944 ft
t2 to t4:
d3 = 8 − 8.879 = 0.879 ft
Adding,
d = d1 + d 2 + d3
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.
d = 13.89 ft W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 9.
a = 3e− 0.2t
v
t
∫ 0 dv = ∫ 0 a dt
v −0=∫
t
3e− 0.2t dt
0
3
=
e− 0.2t
− 0.2
(
)
(
t
0
v = −15 e− 0.2t − 1 = 15 1 − e− 0.2t
At t = 0.5 s,
(
v = 15 1 − e− 0.1
x
)
)
v = 1.427 ft/s W
t
∫ 0 dx = ∫ 0 v dt
t
0
(
x − 0 = 15∫ 1 − e
(
− 0.2t
)
x = 15 t + 5e− 0.2t − 5
At t = 0.5 s,
(
t
1 − 0.2t ⎞
⎛
dt = 15 ⎜ t +
e
⎟
0.2
⎝
⎠0
)
x = 15 0.5 + 5e− 0.1 − 5
)
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.
x = 0.363 ft W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 10.
Given:
a = − 5.4sin kt ft/s 2 ,
t
v0 = 1.8 ft/s, x0 = 0,
t
v − v0 = ∫ 0 a dt = − 5.4 ∫ 0 sin kt dt =
v − 1.8 =
Velocity:
0
v = 1.8cos kt ft/s
t
x−0=
When t = 0.5 s,
t
5.4
( cos kt − 1) = 1.8cos kt − 1.8
3
t
x − x0 = ∫ 0 v dt = 1.8 ∫ 0 cos kt dt =
Position:
5.4
cos kt
k
k = 3 rad/s
1.8
sin kt
k
t
0
1.8
( sin kt − 0 ) = 0.6sin kt
3
x = 0.6sin kt ft
kt = ( 3)( 0.5 ) = 1.5 rad
v = 1.8cos1.5 = 0.1273 ft/s
x = 0.6sin1.5 = 0.5985 ft
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.
v = 0.1273 ft/s W
x = 0.598 ft W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 11.
Given:
a = − 3.24sin kt − 4.32 cos kt ft/s 2 ,
x0 = 0.48 ft,
k = 3 rad/s
v0 = 1.08 ft/s
t
t
t
v − v0 = ∫ 0 a dt = − 3.24 ∫ 0 sin kt dt − 4.32 ∫ 0 cos kt dt
v − 1.08 =
=
t
3.24
cos kt
k
0
−
t
4.32
sin kt
k
0
3.24
4.32
( cos kt − 1) −
( sin kt − 0 )
3
3
= 1.08cos kt − 1.08 − 1.44sin kt
Velocity:
v = 1.08cos kt − 1.44sin kt ft/s
t
t
t
x − x0 = ∫ 0 v dt = 1.08 ∫ 0 cos kt dt − 1.44 ∫ 0 sin kt dt
x − 0.48 =
1.08
sin kt
k
t
0
+
1.44
cos kt
k
t
0
1.08
1.44
( sin kt − 0 ) +
( cos kt − 1)
3
3
= 0.36sin kt + 0.48cos kt − 0.48
=
Position:
When t = 0.5 s,
x = 0.36sin kt + 0.48cos kt ft
kt = ( 3)( 0.5 ) = 1.5 rad
v = 1.08cos1.5 − 1.44sin1.5 = −1.360 ft/s
x = 0.36sin1.5 + 0.48cos1.5 = 0.393 ft
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.
v = −1.360 ft/s !
x = 0.393 ft !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 12.
a = kt mm/s 2
Given:
At t = 0,
v = 400 mm/s;
at t = 1 s,
where k is a constant.
v = 370 mm/s,
x = 500 mm
1
v
t
t
2
∫ 400 dv = ∫ 0 a dt = ∫ 0 kt dt = 2 kt
v − 400 =
1 2
kt
2
v = 400 +
or
1
2
k (1) = 370,
2
At t = 1 s,
v = 400 +
Thus
v = 400 − 30t 2 mm/s
v7 = 400 − ( 30 )( 7 )
At t = 7 s,
When v = 0,
400 − 30t 2 = 0.
Then t 2 = 13.333 s2 ,
1 2
kt
2
k = − 60 mm/s3
2
v7 = −1070 mm/s W
t = 3.651 s
For 0 ≤ t ≤ 3.651 s,
v>0
and
x is increasing.
For t > 3.651 s,
v<0
and
x is decreasing.
x
t
t
2
∫ 500 dx = ∫ 1 v dt = ∫ 1 ( 400 − 30t ) dt
(
x − 500 = 400t − 10t 3
)
t
1
= 400t − 10t 3 − 390
Position:
x = 400t − 10t 3 + 110 mm
At t = 0,
x = x0 = 110 mm
At t = 3.651 s,
x = xmax = ( 400 )( 3.651) − (10 )( 3.651) + 110 = 1083.7 mm
At t = 7 s,
x = x7 = ( 400 )( 7 ) − (10 )( 7 ) + 110
3
3
x7 = − 520 mm W
Distances traveled:
Over 0 ≤ t ≤ 3.651 s,
d1 = xmax − x0 = 973.7 mm
Over 3.651 ≤ t ≤ 7 s,
d 2 = xmax − x7 = 1603.7 mm
Total distance traveled:
d = d1 + d 2 = 2577.4 mm
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.
d = 2580 mm W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 13.
Determine velocity.
v
t
t
∫ − 0.15 dv = ∫ 2 a dt = ∫ 2 0.15 dt
v − ( −0.15 ) = 0.15t − ( 0.15 )( 2 )
v = 0.15t − 0.45 m/s
At t = 5 s,
When v = 0,
v5 = ( 0.15 )( 5 ) − 0.45
0.15t − 0.45 = 0
t = 3.00 s
For 0 ≤ t ≤ 3.00 s,
v ≤ 0,
x is decreasing.
For 3.00 ≤ t ≤ 5 s,
v ≥ 0,
x is increasing.
Determine position.
v5 = 0.300 m/s W
x
t
t
∫ −10 dx = ∫ 0 v dt = ∫ 0 ( 0.15t − 0.45) dt
(
x − ( −10 ) = 0.075t 2 − 0.45t
)
t
0
= 0.075t 2 − 0.45t
x = 0.075 t 2 − 0.45t − 10 m
2
x5 = ( 0.075 )( 5 ) − ( 0.45 )( 5 ) − 10 = −10.375 m
At t = 5 s,
x5 = −10.38 m W
At t = 0,
x0 = −10 m (given)
At t = 3.00 s,
x3 = xmin = ( 0.075 )( 3.00 ) − ( 0.45 )( 3.00 ) − 10 = −10.675 mm
2
Distances traveled:
Over 0 ≤ t ≤ 3.00 s,
d1 = x0 − xmin = 0.675 m
Over 3.00 s < t < 5 s,
d 2 = x5 − xmin = 0.300 m
Total distance traveled:
d = d1 + d 2 = 0.975 m
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.
d = 0.975 m W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 14.
a = 9 − 3t 2
Given:
Separate variables and integrate.
v
t
2
∫ 0 dv = ∫ a dt = ∫ 0 ( 9 − 3t ) dt = 9
(
v − 0 = 9 t − t3
(a)
When v is zero.
v = t 9 − t2
)
t (9 − t 2 ) = 0
t = 0 and t = 3 s (2 roots)
(b)
t =3sW
Position and velocity at t = 4 s.
x
t
t
3
∫ 5 dx = ∫ 0 v dt = ∫ 0 ( 9t − t ) dt
x−5=
9 2 1 4
t − t
2
4
x=5+
At t = 4 s,
⎛9⎞ 2 ⎛1⎞ 4
x4 = 5 + ⎜ ⎟ ( 4 ) − ⎜ ⎟ ( 4 )
⎝2⎠
⎝4⎠
(
v4 = ( 4 ) 9 − 42
(c)
9 2 1 4
t − t
2
4
)
x4 = 13 m W
v4 = − 28 m/s W
Distance traveled.
Over 0 < t < 3 s,
v is positive, so x is increasing.
Over 3 s < t ≤ 4 s,
v is negative, so x is decreasing.
At t = 3 s,
⎛9⎞ 2 ⎛1⎞ 4
x3 = 5 + ⎜ ⎟ ( 3) − ⎜ ⎟ ( 3) = 25.25 m
⎝2⎠
⎝4⎠
At t = 3 s
d3 = x3 − x0 = 25.25 − 5 = 20.25 m
At t = 4 s
d 4 = d3 + x4 − x3 = 20.25 + 13 − 25.25 = 32.5 m
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.
d 4 = 32.5 m W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 15.
Separate variables
Integrate using
dv
= kt 2
dt
a=
Given:
dv = kt2 dt
v = –10 m/s when t = 0
and v = 10 m/s when t = 2 s.
10
2 2
∫ −10 dv = ∫ 0 kt dt
v
10
− 10
1 3
kt
3
=
t
0
1
[(10) − (−10)] = 3 k ⎡⎣⎢( 2 )3 − 0⎤⎦⎥
(a)
Solving for k,
(b)
Equations of motion.
k=
( 3)( 20 )
k = 7.5 m/s 4 W
8
Using upper limit of v at t,
v
v
−10
1
= kt 3
3
t
⎛1⎞
v + 10 = ⎜ ⎟ ( 7.5 ) t 3
⎝3⎠
0
v = −10 + 2.5 t 3 m/s W
dx
= v = −10 + 2.5 t 3
dt
Then,
Separate variables and integrate using x = 0 when t = 2 s.
(
)
dx = −10 + 2.5 t 3 dt
∫
x
dx
0
t
2
(
)
= ∫ −10 + 2.5 t 3 dt
x − 0 = ⎡⎣ −10 t + 0.625 t 4 ⎤⎦
t
2
4
= ⎡⎣ −10 t + 0.0625 t ⎦ − ⎡⎢( −10 )( 2 ) + ( 0.625 )( 2 ) ⎤⎥
⎣
⎦
4⎤
= −10 t + 0.625 t 4 − [ −10]
x = 10 − 10t + 0.625t 4 m W
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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 16.
a = 40 − 160 x = 160 ( 0.25 − x )
Note that a is a given function of x.
(a) Note that v is maximum when a = 0, or x = 0.25 m
Use v dv = a dx = 160 ( 0.25 − x ) dx with the limits
v = 0.3 m/s when x = 0.4 m and v = vmax when x = 0.25 m
vmax
0.25
∫ 0.3 v dv = ∫ 0.4 160 ( 0.25 − x ) dx
2
( 0.25 − x )
vmax
0.32
−
= −160
2
2
2
2
0.25
0.4
⎡
( − 0.15)2 ⎤⎥ = 1.8
= −160 ⎢0 −
2
⎢⎣
⎥⎦
2
vmax
= 3.69 m 2 /s 2
vmax = 1.921 m/s W
(b) Note that x is maximum or minimum when v = 0.
Use v dv = a dx = 160 ( 0.25 − x ) with the limits
v = 0.3 m/s
when x = 0.4 m,
and
v = 0 when x = xm
xm
0
∫ 0.3 v dv = ∫ 0.4 160 ( 0.25 − x ) dx
2
0.3)
(
0−
2
2
0.25 − x )
(
= −160
2
xm
2
= − 80 ( 0.25 − xm ) + ( 80 )( − 0.15 )
2
0.4
( 0.25 − xm )2 = 0.02306
0.25 − xm = ± 0.1519 m
xm = 0.0981 m and 0.402 m W
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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 17.
a = 100 ( 0.25 − x ) m/s 2
a is a function of x:
Use v dv = a dx = 100 ( 0.25 − x ) dx with limits v = 0 when x = 0.2 m
∫ 0 v dv = ∫ 0.2100 ( 0.25 − x ) dx
v
x
x
1 2
1
2
v − 0 = − (100 )( 0.25 − x )
2
2
0.2
= − 50 ( 0.25 − x ) + 0.125
2
So
v 2 = 0.25 − 100 ( 0.25 − x )
Use
Integrate:
dx = v dt
t
x
∫ 0 dt = ± ∫ 0.2
Let u = 20 ( 0.25 − x ) ;
So
Solve for u.
or
dt =
2
v = ± 0.5 1 − 400 ( 0.25 − x )
or
2
dx
dx
=
2
v
± 0.5 1 − 400 ( 0.25 − x )
dx
0.5 1 − 400 ( 0.25 − x )
when x = 0.2 u = 1
2
and du = − 20dx
1
1 −1
π
= m sin −1 u = m
t = m∫
sin u −
2
10
10
2
10 1 − u
1
u
1
u
du
sin −1 u =
π
2
m 10t
π
u = sin m 10t = cos ( ± 10t ) = cos10t
2
u = cos 10t = 20 ( 0.25 − x )
continued
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.
COSMOS: Complete Online Solutions Manual Organization System
Solve for x and v.
x = 0.25 −
v=
Evaluate at t = 0.2 s.
1
sin10t
2
x = 0.25 −
v=
1
cos10t
20
1
cos ( (10 )( 0.2 ) )
20
1
sin ( (10 )( 0.2 ) )
2
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.
x = 0.271 m W
v = 0.455 m/s W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 18.
Note that a is a given function of x
Use
(
)
(
)
v dv = a dx = 600 x 1 + kx 2 dx = 600 x + 600kx3 dx
Using the limits
v = 7.5 ft/s
and
v = 15 ft/s
15
0.45
∫ 7.5 v dv = ∫ 0
⎡ v2 ⎤
⎢ ⎥
⎣2⎦
(15)2
2
−
15
7.5
when x = 0.45 ft,
( 600x + 600kx ) dx
3
⎡ 600 2 600 4 ⎤
=⎢
x +
kx ⎥
4
⎣ 2
⎦
( 7.5)2
2
when x = 0,
2
0.45
0
= ( 300 )( 0.45 ) + (150 ) k ( 0.45 )
4
84.375 = 60.75 + 6.1509k
Solving for k ,
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.
k = 3.84 ft −2 W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 19.
Note that a is a given function of x.
(
)
v dv = a dx = 800 x + 3200 x3 dx
Use
Using the limit v = 10 ft/s when x = 0,
v
x
3
∫10 v dv = ∫ 0 ( 800 x + 3200 x ) dx
v 2 (10 )
−
= 400 x 2 + 800 x 4
2
2
2
v 2 = 1600 x 4 + 800 x 2 + 100
Let u = x 2
v 2 = 1600u 2 + 800u + 100 = 1600 ( u − u1 )( u − u2 ) ,
Then
1600u 2 + 800u + 100 = 0
where u1 and u2 are the roots of
Solving the quadratic equation,
u1,2 =
− 800 ±
(800 )2 − ( 4 )(1600 )(100 )
( 2 )(1600 )
=
− 800 ± 0
= − 0.25 ± 0
3200
u1 = u2 = − 0.25 ft 2
So
Taking square roots,
(
)
(
)
v 2 = 1600 ( u + 0.25 ) = 1600 x 2 + 0.52
2
2
ft 2 /s 2
v = ± 40 x 2 + 0.52 ft/s
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.
COSMOS: Complete Online Solutions Manual Organization System
Use
dx = v dt
40dt = ±
t
dt =
or
dx
x + 0.52
x
x=0
)
when
t=0
dx
1
x
=±
tan −1
2
0.5
0.5
x + 0.5
2
40t = ± 2.0 tan −1 ( 2 x )
2 x = ± tan ( 20t )
v=
(
Use limit
2
40∫ 0 dt = ± ∫ 0
dx
dx
=±
2
v
40 x + 0.52
tan −1 ( 2 x ) = ± 20t
or
x = ± 0.5 tan ( 20t )
or
dx
= ± 0.5 sec2 ( 20t ) ( 20 ) = ± 10 sec2 ( 20t )
dt
At t = 0, v = ± 10 ft/s, which agrees with the given data if the minus sign is rejected.
Thus,
At t = 0.05 s,
v = 10 sec 2 ( 20t ) ft/s,
and
x = 0.5 tan ( 20t ) ft
20t = 1.0 rad
v = 10sec2 (1.0 ) =
10
cos 2 1.0
x = 0.5 tan (1.0 )
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.
v = 34.3 ft/s W
x = 0.779 ft W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 20.
7
a = 12 x − 28 = 12 x − m/s 2
3
Note that a is a given function of x.
7
Use v dv = a dx = 12 x − dx with the limits v = 8 m/s when x = 0.
3
∫
v
v dv
8
v2
2
x
x
0
7
= 12∫
− dx
3
v
8
12
7
= x −
2
3
2
x
0
2
2
v 2 82 12
7
7
−
=
x − −
2
2
2
3
3
2
2
2
7
7
4
7
v = 8 + 12 x − − = 12 x − −
3
3
3
3
2
2
2
7
4
v = ± 12 x − −
3
3
Reject minus sign to get v = 8 m/s at x = 0.
(a) Maximum value of x.
v = 0 when x = xmax
2
7
4
12 x − − = 0
3
3
x−
7
1
=±
3
3
2
or
7
1
x − 3 = 9
xmax = 2 m
and
xmax =
8
2
m=2 m
3
3
Now observe that the particle starts at x = 0 with v > 0 and reaches x = 2 m. At x = 2 m, v = 0 and
2
a < 0, so that v becomes negative and x decreases. Thus, x = 2 m is never reached.
3
xmax = 2 m !
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.
COSMOS: Complete Online Solutions Manual Organization System
(b) Velocity when total distance traveled is 3 m.
The particle will have traveled total distance d = 3 m when d − xmax = xmax − x or 3 − 2 = 2 − x
or x = 1 m.
2
7
4
Using v = − 12 x − − , which applies when x is decreasing, we get
3
3
2
7
4
v = − 12 1 − − = − 20
3
3
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.
v = − 4.47 m/s !
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 21.
(
a = k 1 − e− x
Note that a is a function of x.
(
)
)
Use v dv = a dx = k 1 − e− x dx with the limits v = 9 m/s when x = −3 m, and v = 0 when x = 0.
0
0
−x
∫ 9 v dv = ∫ − 3 k (1 − e ) dx
⎛ v2 ⎞
⎜⎜ ⎟⎟
⎝ 2⎠
0−
0
(
= k x + e− x
9
)
0
−3
92
= k ⎡⎣0 + 1 − ( − 3) − e3 ⎤⎦ = −16.0855k
2
k = 2.52 m/s 2 W
k = 2.5178
(a)
(
)
(
)
Use v dv = a dx = k 1 − e− x dx = 2.5178 1 − e− x dx with the limit v = 0 when x = 0.
∫ 0 v dv = ∫ 0 2.5178 (1 − e
v
x
−x
v2
= 2.5178 x + e− x
2
(
)
) dx
(
x
(
)
= 2.5178 x + e− x − 1
0
)
(
v 2 = 5.0356 x + e− x − 1
)
v = ± 2.2440 x + e− x − 1
1/2
(b) Letting x = −2 m,
(
)
v = ± 2.2440 − 2 + e2 − 1
1/ 2
= ± 4.70 m/s
Since x begins at x = − 2 m and ends at x = 0, v > 0.
Reject the minus sign.
v = 4.70 m/s W
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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 22.
a=v
v
dv
= 6.8 e−0.00057 x
dx
x
∫ 0 v dv = ∫ 0 6.8 e
−0.00057 x
dx
v2
6.8
e−0.00057 x
−0=
2
− 0.00057
(
= 11930 1 − e−0.00057 x
x
0
)
When v = 30 m/s.
( 30 )2
2
(
= 11930 1 − e−0.00057 x
)
1 − e−0.00057 x = 0.03772
e−0.00057 x = 0.96228
− 0.00057 x = ln (0.96228) = − 0.03845
x = 67.5 m W
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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 23.
a=v
Given:
dv
= − 0.4v
dx
dv
= − 0.4
dx
or
Separate variables and integrate using v = 75 mm/s when x = 0.
v
x
∫ 75 dv = − 0.4∫ 0
v − 75 = − 0.4 x
(a) Distance traveled when v = 0
0 − 75 = − 0.4x
x = 187.5 mm W
(b) Time to reduce velocity to 1% of initial value.
v = (0.01)(75) = 0.75
t = − 2.5ln
0.75
75
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.
t = 11.51 s W