Problem 11.56 B a Given Xa YB E.EE x d 74 Is c n Collars A and B start from rest so at ᵗx atePIE mm pfrt.to time d Xa Xa Xa d 24A Xp 4 2X 0 PserposfYfErtdownanwdarads the length of the cable which is i 4 Xc be the position constant is given by B c Differentiate for velocity and again for acceleration 25A TB 4T 26A Of 48C a We can now solve for acceleration of block C 22A ñB 4a 4 E 28A IB At Tec 32 4 1.5 J If Bdx FEB E ñ a TB TB 16 TB XB XB constant acceleration f Idt 52 L Ñ TBo a IB Is 16mm s Using the acceleration found above for block b fift's feat fax E at Vc 4t f du t 42 for its distance 8 F fia.at fidu 4 we can solve 4t at v 2 3 2 213 dx̅ dt t ft x ft Xc 3 É Xc Xc 0 xc 7.875m Problem 11.125 Given EEEi's 8m f Horizontal motion of the ball Fit Tx Tx x̅B x̅o Txt exot a EE At ymax 8m 2 5 219.81 0 8 E 556.96 ix 12.53m s 0 i from Ty 0 12.534 0 12.53 we 8 0 At the time of the catch y 0 Vy Zay T 4.9055 9 81 2 o 0 i from J T.tvy.tt fayt 12.53 4.9052 474 2.55s 53 get we get Motion of the deck ix txottb xp xottxottf p notionofbanreiativet.th JB Jp TBD x̅ x Box apt t.ie IB ID ix t ñpt IBD Tx tapt Box TBDx E f dean ix At the time of the catch d x̅Bp a d age t thrown qq.EE IIg E Ñpt 1 21 0.3 2.55 0.9756 1 b The relative velocity of the ball with respect to the deck is given Box TBDx by apt 0.3 2.55 TBDx 0.765mn Problem 11.142 If 400m Ta JB A 450km h aa 8m52 300m at ten a A af 160 v rm 3ms tal and normal components 450km h 197 0s 540kmh 197 0s 8m s feet 3m15 125m15 150m s I a Velocity of B relative to A TBA TB ÑB TBet a break the velocities down into VBA VBCOS60 VA Vola 15000560 VBA 129.9m s 129.90 m s TBA 1200 TBA 1505in60 0 VBA UBlax 200m15 and y componets Vesin Got Vay VBA 125 x Tt29.9 ÑBA 238.48m b Acceleration of B relative to A Tpa Ip Ga et IB en are of TB is given need to find are air 75m15 Tangental component In Bla guff abt app abtcosbo Galax 300560 aBax 58.45m52 GBA EBIA en normal component x and y components abncoszotteaxtblay abtsin60 abns.in30 day 750530 58.451 34.90 m s 452 34.92 ÑBA 68.08m aBn da Break acceleration down into ABlax et 8 GBA 35in60 755in30 0 GBa 34.9mm Problem 11.158 Normal Acceleration of the Moon Earth r Efsatellite r 384103km moon 63okm E ñn g in a.at EY In 2.7 103m sa j We know that an is also given by at where p r in this case can i we find the speed of the Moons orbit using this equation 8 a 32 air I 7 105384 106 w̅ 1018.23m s 8 3665.630116 The speed of the Moon relative to the Earth is 3665.63km h Problem 11.163 fr Transverse and Radial components a zoom n IT min e 9 13 so 0 8.33m's zoom so 0.0349rad s 0 25 2 velocity of parasailer relative to boat Jam rertrifeo TaB Oer 200 0.0349 VAB 6.9816ms 1850m 3 05 Acceleration of parasailer relative to boat 2 r zero eo TAB CO 200 0.0349 ert 200 i TAB r er 20 0.0349 o 0.2436hm52 GAB Convert to rectangular components Ja 6.9815in30 i Tay 3.4911 6.0460 m s Since i Jay TA JB JA 8.3331 3.4911 JA and b c ñA Gay 0.2436L cos30 0.24365in300 QaB10.2110 0.12180 m s2 6.98120530 1.8241 6.0460 TB 0 at we can write Ja JAB JB 6.0460 m s GAB 0.21101 0.12180 m s2 TA T 4 Ñ 13.28n 6.0462 velocity of parasailer Tea IA 10512182 0.2436m1 accelerationofparasailer