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Question Number 38365 by Rio Mike last updated on 24/Jun/18

A particle P moves on a straightline from a fixed point O and the distance x from O after t seconds is given as x = (1/(4 )) t^4 − (3/2) t^2 + 2t. Find: a) the velocity of P when t = 2, b) the acceleration of P when t = 2, c) the time at which the speed P is Minimum.

$${A}\:{particle}\:{P}\:{moves}\:{on}\:{a}\:{straightline} \\ $$$${from}\:{a}\:{fixed}\:{point}\:{O}\:{and}\:{the}\:{distance} \\ $$$${x}\:{from}\:{O}\:{after}\:{t}\:{seconds}\:{is}\:{given}\:{as} \\ $$$$\:{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}\:}\:{t}^{\mathrm{4}} \:−\:\frac{\mathrm{3}}{\mathrm{2}}\:{t}^{\mathrm{2}} \:+\:\mathrm{2}{t}. \\ $$$${Find}: \\ $$$$\left.{a}\right)\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}\:=\:\mathrm{2}, \\ $$$$\left.{b}\right)\:{the}\:{acceleration}\:{of}\:{P}\:{when}\:{t}\:=\:\mathrm{2}, \\ $$$$\left.{c}\right)\:{the}\:{time}\:{at}\:{which}\:{the}\:{speed}\:{P}\: \\ $$$${is}\:{Minimum}. \\ $$

Answered by MJS last updated on 25/Jun/18

s(t)=(1/4)t^4 −(3/2)t^2 +2t v(t)=(ds/dt)=t^3 −3t+2 ⇒ v(2)=4 a(t)=(dv/dt)=3t^2 −3 ⇒ a(2)=9 local min(v(t))/max(v(t))=v(x) ⇒ v′(x)=0 ⇒ a(x)=0 v(x)=min(v(t)) ⇒ v′′(x)>0 v(x)=max(v(t)) ⇒ v′′(x)<0 3x^2 −3=0 x^2 =1 x_1 =−1 x_2 =1 v′′(t)=a′(t)=(da/dt)=3t a′(−1)=−3 <0 ⇒ local max(v(t))=v(−1)=4 a′(1)=3 >0 ⇒ local min(v(t))=v(1)=0

$${s}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{4}}{t}^{\mathrm{4}} −\frac{\mathrm{3}}{\mathrm{2}}{t}^{\mathrm{2}} +\mathrm{2}{t} \\ $$$${v}\left({t}\right)=\frac{{ds}}{{dt}}={t}^{\mathrm{3}} −\mathrm{3}{t}+\mathrm{2}\:\Rightarrow\:{v}\left(\mathrm{2}\right)=\mathrm{4} \\ $$$${a}\left({t}\right)=\frac{{dv}}{{dt}}=\mathrm{3}{t}^{\mathrm{2}} −\mathrm{3}\:\Rightarrow\:{a}\left(\mathrm{2}\right)=\mathrm{9} \\ $$$$\mathrm{local}\:\mathrm{min}\left({v}\left({t}\right)\right)/\mathrm{max}\left({v}\left({t}\right)\right)={v}\left({x}\right)\:\Rightarrow\:{v}'\left({x}\right)=\mathrm{0}\:\Rightarrow\:{a}\left({x}\right)=\mathrm{0} \\ $$$${v}\left({x}\right)=\mathrm{min}\left({v}\left({t}\right)\right)\:\Rightarrow\:{v}''\left({x}\right)>\mathrm{0} \\ $$$${v}\left({x}\right)=\mathrm{max}\left({v}\left({t}\right)\right)\:\Rightarrow\:{v}''\left({x}\right)<\mathrm{0} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} =\mathrm{1} \\ $$$${x}_{\mathrm{1}} =−\mathrm{1} \\ $$$${x}_{\mathrm{2}} =\mathrm{1} \\ $$$${v}''\left({t}\right)={a}'\left({t}\right)=\frac{{da}}{{dt}}=\mathrm{3}{t} \\ $$$${a}'\left(−\mathrm{1}\right)=−\mathrm{3}\:<\mathrm{0}\:\Rightarrow\:\mathrm{local}\:\mathrm{max}\left({v}\left({t}\right)\right)={v}\left(−\mathrm{1}\right)=\mathrm{4} \\ $$$${a}'\left(\mathrm{1}\right)=\mathrm{3}\:>\mathrm{0}\:\Rightarrow\:\mathrm{local}\:\mathrm{min}\left({v}\left({t}\right)\right)={v}\left(\mathrm{1}\right)=\mathrm{0} \\ $$

Commented by Rio Mike last updated on 25/Jun/18

perfect!

$${perfect}! \\ $$

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