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Unformatted text preview: ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 132* The positiOn of a particle moving along the xaxis is given as a function of time by x(t) = 15  4t m. Determine the velocity of the particle as a function of time.
Determine the acceleration of the particle as a function of time. Evaluate the position, velocity, and acceleration of the particle at '
= 5 5. Determine the total distance traveled by the particle between t = 0
and t = 5 5. Sketch x(t), V(t), and a(t); 0 < t <.8 a. Solution
Given x(t) = 15  4t m, take two time derivatives to get v(t) = 4 m/s ........................................ Ans. a(t) = 0 m/s2 Evaluating these expressions at
x = —5 m
V = 4 m/s 2
a = 0 m/s Since x(t) is uniformly decreasing s = x(5) — x(0)l = 20 m Problem 132 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 13—3* The position of a particle moving along the x—axis is given as a t/3
function of time by x(t) = 3e ft. Determine the velocity of the particle as a function of time.
Determine the acceleration of the particle as a function of time. Evaluate the position, velocity, and acceleration of the particle at
= 5 s. Determine the total distance traveled by the particle between t = 0
and t = 5 s. Sketch x(t), v(t), and a(t); 0 < t < 8 5. Solution t/3
Given x(t) = 3e ft, take two time derivatives to get —t/3
v(t) = e ft/s .............................. Ans. a(t) = (1/3)e_u3 ft/s2 Evaluating these expressions at
x = 0.567 ft
v = 0.1888 ft/s
a = 0.0630 ft/s2 Since x(t) is uniformly decreasing s = x(5) — x(0) = 2.43 ft Problem 133 ﬁmet(q ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 134 The position of a particle moving along the xaxis is given as a function of time by x(t) = 4 sin wt m (w = 3/2 rad/s). Determine the velocity of the particle as a function of time.
Determine the acceleration of the particle as a function of time. Evaluate the position, velocity, and acceleration of the particle at
t = 5 s. Determine the total distance traveled by the particle between t = O
and t = 5 s. and 0 < t < 8 s. a(t); Sketch x(t), v(t), Solution Given x(t) = 4 sin 3t/2 m, take two time derivatives to get v(t) = 6 cos 3t/2 m/s . . . . . . . . ... . . . . . . . . . . . . . . . . . .. Ans. 2
a(t) = 9 sin 3t/2 m/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Evaluating these expressions at
x = 3.75 m . . . . . ........ . . . . . ...................... Ans. v = 2.08 m/s . . . . ......... . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. 2
a = 8.44 m/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. The velocity is zero (and the particle reverses its direction of motion)
when t = H/3 s, H s, SE/B s .... Therefore, the distance traveled is s = x(n/3) _— x(0) + x(7z) — x(7z/3) + x(5) — x(n) = 15.75 m . . . . . . . . . . . . . . . . . . ....... Ans. Problem 134 Time, t (s) ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 13—9* The velocity of a particle moving along the xaxis is given as a t/3
function of time by V(t) = 30e ft/s and x(3) = 20 ft. Determine the position of the particle as a function of time.
Determine the acceleration of the particle as a function of time. Evaluate the position, velocity, and acceleration of the particle at
= 8 5. Determine the total distance traveled by the particle between t = 5
and t = 8 5. Sketch x(t), V(t), and a(t), O < t < 10 s.
Solution
The position is obtained by integrating the velocity
—t/3
V(t) 30e ft/s —t/3
x(t) 53.109  909 ft where the constant of integration is chosen so that x(3) = 20 acceleration is obtained by differentiating the velocity to get
—t/ 3 2
a(t) = lOe ft/s Evaluating these expressions at t = 8 s
x = 46.9 ft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...... Ans.
v = 2.08 ft/s
a = —O.695 ft/s2 Since x(t) is monotonically increasing, the distance traveled is s = lx(8)  x(5) = 10.75 ft Problem 139 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 1325 The jet shown is
catapulted from the deck of an
aircraft carrier by a hydraulic
ram. Determine the acceleration
of the jet if it accelerates
uniformly from rest to 160 mi/h in 300 ft. Solution Use the chain—rule of differentiation to write the acceleration a_£______v dv
dt dx — dx and then integrate I v dv = f a dx to get 2
V = ax + C = ax
2 where the constant of integration C O is chosen so that v(0) =
Then, when x = 300 ft, V = 160 mi/h 234.67 ft/s, and therefore 2
a 91.8 ft/s E 2.85 g 23 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1351* Train A is traveling eastward at 80 mi/h while train 8 is
traveling westward at 60 mi/h.
Determine The velocity of train A
relative to train 8. The velocity of train
relative to train A. 80 mi/h —% +80 mi/h
60 mi/h @— —60 mi/h = vA  vb = (80)  (69)
= 140 mi/h —9 VB  vA = (60)  (80) —140 mi/h = 140 mi/h <— 1352* Boat A travels down a straight
river at 20 m/s while boat B travels up
the river at 15 m/s. Determine
The velocity of boat A relative to
boat B. The velocity of boat B relative to
boat A. Solution
20 m/s
15 m/s
(—20) — (15) = 35 m/s ¢ (15) ‘ 48 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 13—72* In Fig. Pl372 block B has
a constant downward acceleration of
0.8 m/sz. Determine the speed and
acceleration of block A 5 s after the system starts from rest. Solution
For block B aB = 0.8 m/s2 ¢ SD = constant 0.8t m/s ¢ 5
B But the length of the rope is constant
t = +
25A 5B
and taking two time derivatives gives
o=2é+é
A B
o 23 Then at t 66 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1375* In Fig. P 13—75 block B moves to the right with a speed of 10 ft/s and its speed is decreasing at the rate of 1 ft/sz.
Determine: a. The velocity of block c if
block A is fixed and does not
move. The velocity of block A if
block C is fixed and does not
move. Solution a. Measure the positions of
the two blocks relative to the
stationary block A as shown. Then VB = 10 ft/s % (sB = 10 ft/s) 2 2
aB = 1 ft/s @ = —1 ft/s 2
(SE = 1 ft/s ) Since the length of the cord is constant z = ZSB + sC + constants Taking the derivatives of this relationship gives 0 = 25B + 5C 0 = 25B + 5C v = s = —25 = —2(10) ft/s = 20 ft/s 9— . . . . . . . . .. Ans. 2
2(1) ft/s = 2 ft/s2 —) . . . . . . . . .. b. Measure the positions of
the two blocks relative to the
stationary block C as shown. Again, 53 = 10 ft/s 2
SB — —1 ft/s But now thelength of the cord is AC = 35A + 25B + constants + = o.
0 35A 25B 0 35A + 25B v = s = (2éB/3) = 20/3 ft/s —> . . . . . . . . . . . . . . . . . . .. Ans. 2/3 ft/s2
2/3 ft/s2 <— . . . . . . . . . . . . . . . . . . . . . . . . . . .. '(“253/3) 69 . .x . , A, i , .‘r, .. cs ,. . _,.» ...
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This note was uploaded on 09/22/2009 for the course MGMT 209 taught by Professor Swim during the Spring '08 term at Texas A&M.
 Spring '08
 SWIM

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