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-acce

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

$$AparticlePmovesonastraightline$$$$fromafixedpointOandthedistance$$$$xfromOaftertsecondsisgivenas$$$$x=\frac{1}{4}{t}^4-\frac{3}{2}{t}^2+2t.$$$$Find:$$$$a)thevelocityofPwhent=2,$$$$b)theaccelerationofPwhent=2,$$$$c)thetimeatwhichthespeedP$$$$isMinimum.$$

Answered by MJS last updated on 25/Jun/18

$$s\left(t\right)=\frac{1}{4}{t}^4-\frac{3}{2}{t}^2+2t$$$$v\left(t\right)=\frac{ds}{dt}=t^3-3t+2\Rightarrow v\left(2\right)=4$$$$a\left(t\right)=\frac{dv}{dt}=3t^2-3\Rightarrow a\left(2\right)=9$$$$\mathrm{local}\min \left(v\left(t\right)\right)/\max \left(v\left(t\right)\right)=v\left(x\right)\Rightarrow v^{\prime }\left(x\right)=0\Rightarrow a\left(x\right)=0$$$$v\left(x\right)=\min \left(v\left(t\right)\right)\Rightarrow v^{″}\left(x\right)>0$$$$v\left(x\right)=\max \left(v\left(t\right)\right)\Rightarrow v^{″}\left(x\right)<0$$$$3x^2-3=0$$$$x^2=1$$$$x_1=-1$$$$x_2=1$$$$v^{″}\left(t\right)=a^{\prime }\left(t\right)=\frac{da}{dt}=3t$$$$a^{\prime }(-1)=-3<0\Rightarrow \mathrm{local}\max \left(v\left(t\right)\right)=v(-1)=4$$$$a^{\prime }\left(1\right)=3>0\Rightarrow \mathrm{local}\min \left(v\left(t\right)\right)=v\left(1\right)=0$$

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

$$perfect!$$

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