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Question Number 123767 by oustmuchiya@gmail.com last updated on 28/Nov/20

Answered by physicstutes last updated on 28/Nov/20

  s = 2t + 3 sin 2t  (i) Initiat position occurs at t = 0 s  ⇒ s_0  = 2(0) + 3 sin 2(0)        s_0  = 0 m  (ii) v = (ds/dt) = 2 + 6 cos 2t         a = (dv/dt) = 0 − 12 sin 2t  ⇒ v =( 2 + 6 cos 2t) m s^(−1)  and a = (−12 sin 2t) m s^(−2)   (iii) At rest v = 0  ⇒ 2 + 6 cos 2t = 0     cos 2t = −(1/3)       2t = 109.5 ⇒ t = 54.75 s  b. F(x) = 250 e^(0.09x) , x ≥ 0      ((dF(x))/dx) = 22.5 e^(0.09x)    ((dF(x))/dx)∣_(8 months)  = 22.5 e^(0.09×8)  = 46.2

s=2t+3sin2t(i)Initiatpositionoccursatt=0ss0=2(0)+3sin2(0)s0=0m(ii)v=dsdt=2+6cos2ta=dvdt=012sin2tv=(2+6cos2t)ms1anda=(12sin2t)ms2(iii)Atrestv=02+6cos2t=0cos2t=132t=109.5t=54.75sb.F(x)=250e0.09x,x0dF(x)dx=22.5e0.09xdF(x)dx8months=22.5e0.09×8=46.2

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