# Solution manual vector mechanics engineers dynamics 8th beer chapter 11

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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

v = 3.67 m/s W

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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

x = 52 ft W
v = 115 ft/s W
a = 192 ft/s 2 W

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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

x = 562 in. !
v = 770 in./s !
a = 764 in./s 2 !

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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

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

x = 240 mm !
v = 4390 mm/s !
a = − 24.0 × 103 mm/s 2 !

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Chapter 11, Solution 6.

(

)

x = 50sin k1t − k2t 2 mm

Position:

Where

Let

θ = k1t − k2t 2 = t − 0.5t 2 rad

= (1 − t ) rad/s
dt
x = 50sin θ mm

Position:

and

d 2θ
dt 2

and

dx

= 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

=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

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

x = 24.0 mm W
a = − 43.9 mm/s 2 W

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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

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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

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

d = 13.89 ft W

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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

x = 0.363 ft W

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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

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

v = 0.1273 ft/s W
x = 0.598 ft W

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Chapter 11, Solution 11.
Given:

a = − 3.24sin kt − 4.32 cos kt ft/s 2 ,
x0 = 0.48 ft,

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

v = −1.360 ft/s !
x = 0.393 ft !

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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

d = 2580 mm W

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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

d = 0.975 m W

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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

d 4 = 32.5 m W

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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

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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

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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

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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

x = 0.271 m W
v = 0.455 m/s W

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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

k = 3.84 ft −2 W

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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

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

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

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

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

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

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

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

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