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Question Number 37972 by Rio Mike last updated on 20/Jun/18

 The distance S metres is   given as a funtion   f(t) where is time taken...  if S = t^3  + t^2  + 4  find the velocity and acceleration

$$\:{The}\:{distance}\:{S}\:{metres}\:{is}\: \\ $$$${given}\:{as}\:{a}\:{funtion}\: \\ $$$${f}\left({t}\right)\:{where}\:{is}\:{time}\:{taken}... \\ $$$${if}\:{S}\:=\:{t}^{\mathrm{3}} \:+\:{t}^{\mathrm{2}} \:+\:\mathrm{4} \\ $$$${find}\:{the}\:{velocity}\:{and}\:{acceleration} \\ $$

Answered by Joel579 last updated on 20/Jun/18

s(t) = t^3  + t^2  + 4  v(t) = (ds/dt) = 3t^2  + 2t  a(t) = (dv/dt) = 6t

$${s}\left({t}\right)\:=\:{t}^{\mathrm{3}} \:+\:{t}^{\mathrm{2}} \:+\:\mathrm{4} \\ $$$${v}\left({t}\right)\:=\:\frac{{ds}}{{dt}}\:=\:\mathrm{3}{t}^{\mathrm{2}} \:+\:\mathrm{2}{t} \\ $$$${a}\left({t}\right)\:=\:\frac{{dv}}{{dt}}\:=\:\mathrm{6}{t} \\ $$

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