hw19b_sol - ENGINEERING MECHANICS - Dynamics W.F. Riley...

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Unformatted text preview: ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 13-2* The positiOn of a particle moving along the x-axis 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 13-2 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 13-3 fimet(q ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 13-4 The position of a particle moving along the x-axis 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 13-4 Time, t (s) ENGINEERING MECHANICS - Dynamics W.F. Riley & L.D. Sturges 13—9* The velocity of a particle moving along the x-axis 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 13-9 ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 13-25 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 13-51* 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 <— 13-52* 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. Pl3-72 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 13-75* 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 the-length 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.

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hw19b_sol - ENGINEERING MECHANICS - Dynamics W.F. Riley...

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