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Question Number 14816 by tawa tawa last updated on 04/Jun/17
Answered by arnabpapu550@gmail.com last updated on 08/Jun/17
Answertopart1Given,x=53t3−52t2−30t+8xdifferentiatingbothsidewitht,∴dxdt=53×3t2−52×2t−30+8dxdtor,v=5t2−5t−30+8v∴7v=30+5t−5t2whenv=0then,5t2−5t−30=0or,t2−t−6=0or,t2−3t+2t−6=0or,t(t−3)+2(t−3)=0or,(t−3)(t+2)=0∴t=3or,−2buttimecannotbenegative.Hence,t=3s∴x=53×33−52×32−30×3+8xor,7x=67.5⇒x=9.64ftAgain,7v=30+5t−5t2∴7×dvdt=0+5−10t∴(a)∣t=3=17×(5−10×3)=−3.57ft/s2∴Whenv=0,time=3s,position=9.64ftacceleration=−3.57ft.s−2Answertopart2Given,x=6t2−8+40cosΠtwhent=6s,x=216−8+40cos(Π×6)∴x=209inch∴dxdt=12t−40ΠsinΠt=v∴(v)∣t=6=12×6−40Πsin6Π=72inch.s−1∴dvdt=12−40∏2cosΠt=a⇒(a)∣t=6=−382.78inch.s−2
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