Tecnlogia de materiales

Chapter 11, Solution 2.
x = t3 - (t - 2)2 m
( ) v dx 3t2 2 t 2 m/s
dt
= = - -
a dv 6t 2 m/s2
dt
= = -
(a) Time at a = 0.
0 = 6t0 - 2 = 0
0
1
3
t = t0 = 0.333 s
(b) Corresponding position and velocity.
3 2 1 1 2 2.741m
3 3
x = ⎛⎜ ⎞⎟ - ⎛⎜ - ⎞⎟ = -
⎝ ⎠ ⎝ ⎠ 2.74 m x = -
2 3 1 2 1 23.666 m/s
3 3
v = ⎛⎜ ⎞⎟ - ⎛⎜ - ⎞⎟ =
⎝ ⎠ ⎝ ⎠ 3.67 m/s v =

Chapter 11, Solution 3.
Position:
x = 5t4 - 4t3 + 3t - 2 ft
Velocity:
v dx 20t3 12t2 3 ft/s
dt
= = - +
Acceleration:
a dv 60t2 24t ft/s2
dt
= = -
When t = 2 s,
( )( ) ( )( ) ( )( ) 4 3 x = 5 2 - 4 2 - 3 2 - 2 x = 52 ft
( )( ) ( )( ) 3 2 v = 20 2 - 12 2 + 3 v = 115 ft/s
( )( ) ( )( ) 2 a = 60 2 - 24 2 a = 192 ft/s2Chapter 11, Solution 4.
Position:
x = 6t4 + 8t3 - 14t2 - 10t + 16 in.
Velocity:
v dx 24t3 24t2 28t 10 in./s
dt
= = + - -
Acceleration:
a dv 72t2 48t 28 in./s2
dt
= = + -
When t = 3 s,
( )( ) ( )( ) ( )( ) ( )( ) 4 3 2 x = 6 3 + 8 3 - 14 3 - 10 3 + 16 x = 562 in.!
( )( ) ( )( ) ( )( ) 3 2 v = 24 3 + 24 3 - 28 3 - 10 v = 770 in./s !
( )( ) ( )( ) 2 a = 72 3 + 48 3 - 28 a = 764 in./s2!Chapter 11, Solution 5.
Position: x = 500sin kt mm
Velocity: v dx 500k cos kt mm/s
dt
= =
Acceleration:
a dv 500k2 sin kt mm/s2
dt
= =-
When t = 0.05 s, and k = 10 rad/s
kt = (10)(0.05) = 0.5 rad
x = 500sin (0.5) x = 240 mm !
v = (500)(10)cos(0.5) v = 4390 mm/s !
( )( ) ( ) 2 a = - 500 10 sin 0.5 a = -24.0 × 103mm/s2 !
Chapter 11, Solution 6.
Position: ( 2 )
x = 50sin k1t - k2t mmWhere
2
k1 = 1 rad/s and k2 = 0.5 rad/s
Let

2 2
θ = k1t - k2t = t - 0.5t rad
( )
2
2
2 d 1 t rad/s and d 1 rad/s
dt dt
θ = - θ = -
Position: x = 50sinθ mm
Velocity: v dx 50cos d mm/s
dt dt
= = θ θ
Acceleration:
a dv
dt
=
2 2
2
2 a 50cos d 50sin d mm/s
dt dt
= θ θ - θ ⎛ θ ⎞ ⎜ ⎟
⎝ ⎠
When v = 0, either cosθ = 0
or d 1 t 0 t 1 s
dt
θ = - = =
Over 0 = t = 2 s, values ofcosθ are:
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) x = 24.0 mm
a = 50cos(0.5)(-1) - 50sin (0.5)(0) a = -43.9 mm/s2
COSMOS: Complete Online Solutions Manual Chapter 11, Solution 7.
Given:
x = t3 - 6t2 + 9t + 5
Differentiate twice.
v dx 3t2 12t 9dt
= = - +
6 12 a dv t
dt
= = -
(a) When velocity is zero. v = 0
3t2 - 12t + 9 = 3(t - 1)(t - 3) = 0
t = 1 s and t = 3 s
(b) Position at t = 5 s.
( ) ( )( ) ( )( ) 3 2
x5 = 5 - 6 5 + 9 5 + 5 x5 = 25 ft
Acceleration at t = 5 s.
a5 = (6)(5) - 12
2
a5 = 18 ft/s
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
Chapter 11, Solution 8.
x = t2 - (t - 2)3 ft( )2 v dx 2t 3 t 2 ft/s
dt
= = - -
(a) Positions at v = 0.
2t - 3(t - 2)2 = -3t2 + 14t - 12 = 0
14 (14)2 (4)( 3)( 12)
(2)( 3)
t
- ± - - -
=
-
t1 = 1.1315 s and t2 = 3.535 s
At t1 = 1.1315 s, x1 = 1.935 ft x1 = 1.935 ft
At t2 = 3.535 s, x2 = 8.879 ft x2 = 8.879 ft
(b) 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: d2 = 8.879 - 1.935 = 6.944 ft
t2 to t4: d3 = 8 - 8.879 = 0.879 ft
Adding, d = d1 + d2 + d3 d = 13.89 ft
Chapter 11, Solution 9.
a = 3e- 0.2t
0 0
v t ∫ dv = ∫ a dt
0.2 0.2
0
0
0 3 3
0.2
t
t t t v e dt e - - - = =-
∫
( ) ( ) v = -15 e-0.2t - 1 = 15 1 - e- 0.2t
At t = 0.5 s, ( ) v = 15 1 - e-0.1 v = 1.427 ft/s
0 0
x t ∫ dx = ∫ v dt
( ) 0.2 0.2...