Question Number 14816 by tawa tawa last updated on 04/Jun/17

Answered by arnabpapu550@gmail.com last updated on 08/Jun/17

Answer to part 1 Given , x=(5/3)t^3 −(5/2)t^2 −30t+8x differentiating both side with t, ∴(dx/dt)=(5/3)×3t^2 −(5/2)×2t−30+8(dx/dt) or,v=5t^2 −5t−30+8v ∴7v=30+5t−5t^2 when v=0 then, 5t^2 −5t−30=0 or, t^2 −t−6=0 or, t^2 −3t+2t−6=0 or, t(t−3)+2(t−3)=0 or, (t−3)(t+2)=0 ∴ t=3 or, −2 but time can not be negative. Hence, t=3s ∴ x=(5/3)×3^3 −(5/2)×3^2 −30×3+8x or, 7x=67.5 ⇒ x=9.64 ft Again, 7v=30+5t−5t^2 ∴ 7×(dv/dt)=0+5−10t ∴ (a)∣_(t=3) =(1/7)×(5−10×3) =−3.57ft/s^2 ∴ When v=0 , time=3s, position=9.64ft acceleration=−3.57ft.s^(−2) Answer to part 2 Given, x=6t^2 −8+40cosΠt when t=6s, x=216−8+40cos(Π×6) ∴ x=209inch ∴(dx/dt)=12t−40ΠsinΠt=v ∴(v)∣_(t=6) =12×6−40Πsin6Π =72inch.s^(−1) ∴(dv/dt)=12−40Π^2 cosΠt=a ⇒(a)∣_(t=6) =−382.78inch.s^(−2)