dv dt = s and t = s a = 28.7 m/s

Transcription

1 COSMOS: Complee Online Soluions Manual Organizaion Sysem Chaper, Soluion. Posiion: Velociy: x = m 3 dx v = = m/s d dv Acceleraion: a = = 0 5 d When v = 0, = 0 Solving he quadraic equaion for, =.686 s and =.86 s =.686 s When =.686, x = (.686) (.686) 0(.686) + 0 x = 4.85 m 3 v = 0 a = 0(.686) 5 a = 8.7 m/s Vecor Mechanics for Engineers: Saics and Dynamics, 7/e, Ferdinand P. Beer (Deceased), E. Russell Johnson, Jr., Ellio R. Eisenberg, William E. Clausen, George H. Saab, 004 The McGraw-Hill Companies.

2 COSMOS: Complee Online Soluions Manual Organizaion Sysem Chaper, Soluion 44. (a) Acceleraion of A. va = A 0 A A A 0 A 0 = v A 0 = 0 and xa 0 = 0 ( v ) + a and x = ( x ) + ( v ) a Using ( ) ( ) gives When cars pass a =, xa = 90 m x = A = va = and xa = ( )( 90) 80 = and va = For 0 5 s, v B = ( v B ) 0 = 96 km/h = m/ s For > 5 s, v = ( v ) + a ( 5) = a ( 5) When vehicles pass, Using Le u =, a A B B 0 B A 6 va = vb = ( 5) = or 7 5 = = gives 5 = 7 80u 5 = 60u or 60u 7 80u + 5 = 0 Solving he quadraic equaion, The corresponding values for are u = 7 80 ± ( 49)( 80) ( 4)( 60)( 5) ( )( 60) = and a A = 85. m/s and u = A ± = m/s = = s, and = = 7.08s Rejec s since i is less han 5 s. Thus, (b) Time of passing. (c) Disance d. 0 5 s, xb = ( xb ) 0 ( vb ) 0 = d A = 5 s, = d (.667)( 5) = d x = d ( v ) ( 5) + a ( ) For > 5 s, B x B B 0 B x B = d ( 5) + ( 5) When = = 7.08 s, x B = x A = = d ( 6.667)(.08) + ( 3.59)(.08) ( )( 6) = 3.59 m/s a A = = 7.08 s d = d = 78 m Vecor Mechanics for Engineers: Saics and Dynamics, 7/e, Ferdinand P. Beer (Deceased), E. Russell Johnson, Jr., Ellio R. Eisenberg, William E. Clausen, George H. Saab, 004 The McGraw-Hill Companies.

3 COSMOS: Complee Online Soluions Manual Organizaion Sysem Chaper, Soluion. k For s 0 s, a = 3 where k is a consan. A = s, v = 5 f/s dv a = d or dv = ad v d k dv = a d k 5 = = 3 ( ) v ( 5) = 0.5 k or v 5 0.5k 4 = 4 A = 0 s, v = 0.36 f/s 5 0.5k ( ) = 0.36 from which 0 4 k = 8 f s Then, 64 v = 5 64 = f/s 4 When v = 0, = 64 or = 8 s For s 8 s, v 0, x is decreasing. 8 s 0 s, v 0, x is increasing. Posiion: x = dx = vd = d = + + C A = s, 64 x = + + C = 34 + C f A = 8 s, 64 x8 = C = 6 + C f 8 A = 0 s, 64 x0 = C = C f 0 Given: x = x0 = ± x0 or 34 + C = ± ( C) Using he plus sign: C =. f, which gives x8 = 7. f, x = 35. f and x 0 = 7.60 f Disance raveled: d = x8 x + x0 x8 = d = 8.40 f Using he minus sign: C =.7 f, which gives x8 = 6.7 f, x =.73 f and x 0 = 5.87 f Disance raveled: d = x8 x + x0 x8 = d = 8.40 f Vecor Mechanics for Engineers: Saics and Dynamics, 7/e, Ferdinand P. Beer (Deceased), E. Russell Johnson, Jr., Ellio R. Eisenberg, William E. Clausen, George H. Saab, 004 The McGraw-Hill Companies.

4 COSMOS: Complee Online Soluions Manual Organizaion Sysem Chaper, Soluion 40. Choose x posiive upward. Consan acceleraion a = g Rocke launch daa: Rocke : Rocke : Velociies: Rocke : Rocke : A 0 x = 0, v = v, = 0 B 0 x = 0, v = v, = B = 4 s A A 0 v = v g B v = v g( ) B 0 B Posiions: Rocke Ax : A = v 0 g Rocke Bx : = v( ) g ( ), For simulaneous explosions a x = x = 80 m when =, B 0 B B B A B E ( ) ( ) v g v g v v g g g 0 E E = 0 E B E B = 0 E 0 B E + E B B Solving for 0, v v0 gb = ge () Then, when = E, gb x A = ge E ge, or xa = 0 g E BE Solving for E, E x ( )( ( ) ( )( )( )( ) )( A + ) B ± B + 4 g 4 ± 9.8 (a) From equaion (), ( 9.8)( 6.507) A ime, E A 0 = = = s vb va gb ( 9.8)( 4) v 0 = v 0 = 44. m/s v v g = v = v g( ) E ( 9.8)( 4) B 0 E B = = v / = 39. m/s BA Vecor Mechanics for Engineers: Saics and Dynamics, 7/e, Ferdinand P. Beer (Deceased), E. Russell Johnson, Jr., Ellio R. Eisenberg, William E. Clausen, George H. Saab, 004 The McGraw-Hill Companies.

5 COSMOS: Complee Online Soluions Manual Organizaion Sysem Chaper, Soluion 34. Consan acceleraion. x 0 = 0 v = v0 + a () x x v a = () Solving () for a, a v v = 0 (3) Then, v v0 x = x0 + v0 + = x0 + ( v0 + v) = ( v0 + v) A = 6 s, v = v0 and x6 = 80 m Then, from (3), = v0 + v0 ( 6) = 4.5 v0 or v0 = = 40 m/s 4.5 v = v0 = 0 m/s a = = m/s = m/s 6 3 Subsiuing ino () and (), v = ( ) x = A sopping, v = 0 or = 0 = s s s x = 0 + ( 40)( ) ( 3.333)( ) = 40 m ( a ) Addiional ime for sopping = s 6 s = 6 s () b Addiional disance for sopping = 40 m 80 m d = 60 m Vecor Mechanics for Engineers: Saics and Dynamics, 7/e, Ferdinand P. Beer (Deceased), E. Russell Johnson, Jr., Ellio R. Eisenberg, William E. Clausen, George H. Saab, 004 The McGraw-Hill Companies.

6 COSMOS: Complee Online Soluions Manual Organizaion Sysem Chaper, Soluion 6. Noe ha a is a given funcion of x. 3 Use = = ( ) Using he limis v =.5 m/s when x = 0, vdv adx x x dx v x ( ) vdv = x + x dx v.5 = 00x + 800x 4 4 v = 600x + 00x Le u = x Then v = u + u + = ( u u )( u u ) where u and u are he roos of Solving he quadraic equaion, u, , ( ) ( )( )( ) ( )( ) 600u + 00u = 0 00 ± ± 0 = = = ± u = u = f So v = ( u + ) = ( x + ) Taking square roos, v =± 40( x ) f/s Use f /s dx dx dx = v d or d = = ± v ( x + ) dx 40 d =± Use limi x = 0 when = 0 x x dx x 40 d =± an 0 =± 0 x ( ) ( ) ( ) ( ) 40 =± 4.0 an 4 x or an 4x =± 0 4x =± an 0 or x =± 0.5an 0 dx v = = ± 0.5 sec ( 0) ( 0) = ±.5 sec ( 0) d A = 0, v = ±.5 m/s, which agrees wih he given daa if he minus sign is rejeced. Thus, v = ( ) x = ( ).5 sec 0 m/s, and 0.5an 0 m A = 0.05 s, 0 = 0.5 rad.5 v =.5sec ( 0.5) = v = 3.5 m/s cos 0.5 x = x = m 0.5an ( 0.5) Vecor Mechanics for Engineers: Saics and Dynamics, 7/e, Ferdinand P. Beer (Deceased), E. Russell Johnson, Jr., Ellio R. Eisenberg, William E. Clausen, George H. Saab, 004 The McGraw-Hill Companies.

7 Problem.C3 The moioned of a paricle is defined by he equaions x = 30 0 and y = , where x and y are expressed in millimeers and in seconds. Derive expressions for he velociy and acceleraion of he paricle as a funcion of. Consider he ime inerval 0 s and plo (a) he pah of he paricle in he xy plane, (b) he componens of he velociy v x and v y and he magniude of he velociy v, (c) he componens of he acceleraion a x and a y and he magniude of he acceleraion. From he moion of he paricle we find Soluion (a) 3 x= 30 0 y = 0 40 v = x = 60 0 v = y = 40 0 x a = x= 60 a = y = x y y (b) Plo he rajecory of he paricle Program % Problem.C3 % = [0:0.5:0]; x = 30*.^-0*; y = 0*.^-40*.^3; v_x = 60*-0; v_y = 40*-0*.^; v = sqr(v_x.^+v_y.^); a_x = 60; a_y = 40-40*; a = sqr(a_x.^+a_y.^); figure() plo(x,y); xlabel('x (f)') ylabel('y (f)') legend('trajecory') grid on axis([ e5 0]) figure()

8 plo(,v_x,,v_y,,v) xlabel(' (sec)') ylabel('v_x, v_y, v (fps)') legend('v_x','v_y','v',) grid on figure(3) plo(,a_x,,a_y,,a) xlabel(' (sec)') ylabel('a_x, a_y, a (fps^)') legend('a_x','a_y','a',) grid on Program Oupu

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