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

Answered by 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)