Velocity and Complete Online Solutions
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 2. x = t3 − (t − 2) m
2
v= a= (a) Time at a = 0.
dx = 3t 2 − 2 ( t − 2 ) m/s dt dv = 6t − 2 m/s 2 dt
0 = 6t0 − 2 = 0 t0 = 1 3
t0 = 0.333 s
(b)
Corresponding position and velocity.
⎛1⎞ ⎛1 ⎞ x = ⎜ ⎟ − ⎜ − 2 ⎟ = − 2.741 m ⎝3⎠ ⎝3 ⎠
3 2
x = − 2.74 m
⎛1⎞ ⎛1 ⎞ v = 3 ⎜ ⎟ − 2 ⎜ − 2 ⎟ = 3.666 m/s ⎝3⎠ ⎝3 ⎠
2
v = 3.67 m/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
Chapter 11, Solution 3.
Position:
Velocity: x = 5t 4 − 4t 3 + 3t − 2 ft v= a= dx = 20t 3 − 12t 2 + 3 ft/s dt dv = 60t 2 − 24t ft/s 2 dt
Acceleration:
When t = 2 s,
x = ( 5 )( 2 ) − ( 4 )( 2 ) − ( 3)( 2 ) − 2 v = ( 20 )( 2 ) − (12 )( 2 ) + 3 a = ( 60 )( 2 ) − ( 24 )( 2 )
2 3 2
4
3
x = 52 ft v = 115 ft/s a = 192 ft/s 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 4.
Position:
Velocity: x = 6t 4 + 8t 3 − 14t 2 − 10t + 16 in.
v= a= dx = 24t 3 + 24t 2 − 28t − 10 in./s dt dv = 72t 2 + 48t − 28 in./s 2 dt
Acceleration:
When t = 3 s,
x = ( 6 )( 3) + ( 8 )( 3) − (14 )( 3) − (10 )( 3) + 16 v = ( 24 )( 3) + ( 24 )( 3) − ( 28 )( 3) − 10 a = ( 72 )( 3) + ( 48 )( 3) − 28
2 3 2
4
3
2
x = 562 in. ! v = 770 in./s ! a = 764 in./s 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.
COSMOS: Complete Online
Chapter 11, Solution 2. x = t3 − (t − 2) m
2
v= a= (a) Time at a = 0.
dx = 3t 2 − 2 ( t − 2 ) m/s dt dv = 6t − 2 m/s 2 dt
0 = 6t0 − 2 = 0 t0 = 1 3
t0 = 0.333 s
(b)
Corresponding position and velocity.
⎛1⎞ ⎛1 ⎞ x = ⎜ ⎟ − ⎜ − 2 ⎟ = − 2.741 m ⎝3⎠ ⎝3 ⎠
3 2
x = − 2.74 m
⎛1⎞ ⎛1 ⎞ v = 3 ⎜ ⎟ − 2 ⎜ − 2 ⎟ = 3.666 m/s ⎝3⎠ ⎝3 ⎠
2
v = 3.67 m/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
Chapter 11, Solution 3.
Position:
Velocity: x = 5t 4 − 4t 3 + 3t − 2 ft v= a= dx = 20t 3 − 12t 2 + 3 ft/s dt dv = 60t 2 − 24t ft/s 2 dt
Acceleration:
When t = 2 s,
x = ( 5 )( 2 ) − ( 4 )( 2 ) − ( 3)( 2 ) − 2 v = ( 20 )( 2 ) − (12 )( 2 ) + 3 a = ( 60 )( 2 ) − ( 24 )( 2 )
2 3 2
4
3
x = 52 ft v = 115 ft/s a = 192 ft/s 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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 11, Solution 4.
Position:
Velocity: x = 6t 4 + 8t 3 − 14t 2 − 10t + 16 in.
v= a= dx = 24t 3 + 24t 2 − 28t − 10 in./s dt dv = 72t 2 + 48t − 28 in./s 2 dt
Acceleration:
When t = 3 s,
x = ( 6 )( 3) + ( 8 )( 3) − (14 )( 3) − (10 )( 3) + 16 v = ( 24 )( 3) + ( 24 )( 3) − ( 28 )( 3) − 10 a = ( 72 )( 3) + ( 48 )( 3) − 28
2 3 2
4
3
2
x = 562 in. ! v = 770 in./s ! a = 764 in./s 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.
COSMOS: Complete Online