Acceleration-of-a-particle-which-is-at-rest-at-x-0-is-a-4-2x-i-Select-the-correct-alternative-s-a-Maximum-speed-of-the-particle-is-4-units-b-Particle-further-comes-to-rest-at-x-

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Question Number 20842 by Tinkutara last updated on 04/Sep/17

$$\mathrm{Acceleration}\mathrm{of}\mathrm{a}\mathrm{particle}\mathrm{which}\mathrm{is}\mathrm{at}$$$$\mathrm{rest}\mathrm{at}x=0\mathrm{is}\overrightarrow{a}=(4-2x)\stackrel{\wedge }{i}.\mathrm{Select}$$$$\mathrm{the}\mathrm{correct}\mathrm{alternative}\left(\mathrm{s}\right).$$$$\left(a\right)\mathrm{Maximum}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{particle}\mathrm{is}$$$$4\mathrm{units}$$$$\left(b\right)\mathrm{Particle}\mathrm{further}\mathrm{comes}\mathrm{to}\mathrm{rest}\mathrm{at}$$$$x=4$$$$\left(c\right)\mathrm{Particle}\mathrm{oscillates}\mathrm{about}x=2$$$$\left(d\right)\mathrm{Particle}\mathrm{will}\mathrm{continuously}\mathrm{accelerate}$$$$\mathrm{along}\mathrm{the}x-\mathrm{axis}.$$

Answered by ajfour last updated on 09/Sep/17

$$\frac{v_x{dv}_x}{dx}=4-2x$$$$\Rightarrow {\int }_0^{v_x}{v}_x{dv}_x={\int }_0^x(4-2x)dx$$$$\Rightarrow \frac{v_x^2}{2}=4x-x^2$$$$\Rightarrow v_x^2=2[4-{(x-2)}^2]$$$$\Rightarrow particlecomestorestatx=4.$$$$maximumspeedatx=2andis$$$${\left(v_x\right)}_{max}=2\sqrt{2}units.$$$$\Rightarrow {\int }_0^x\frac{dx}{\sqrt{2^2-{(x-2)}^2}}=\pm \sqrt{2}{\int }_0^{t}dt$$$$\Rightarrow {\sin }^{-1}\left(\frac{x-2}{2}\right)-{\sin }^{-1}(-1)=\pm \sqrt{2}t$$$$\Rightarrow {\sin }^{-1}\left(\frac{x-2}{2}\right)=\pm \sqrt{2}t-\pi /2$$$$x=2+2\sin (\pm \sqrt{2}t-\pi /2)$$$$orx=2-\cos \left(\sqrt{2}t\right)$$$$istheequationofmotion.$$$$\Rightarrow particleoscillatesaboutx=2.$$$$\left(\mathit{b}\right)and\left(\mathit{c}\right)arethecorrectoptions.$$

Commented by ajfour last updated on 09/Sep/17

$$thanks,ishallbecareful.$$

Commented by alex041103 last updated on 09/Sep/17

$$Youhaveamistake$$$$\Rightarrow {\int }_0^x\frac{dx}{\sqrt{2^2-{(x-2)}^2}}=\pm \sqrt{2}{\int }_0^{t}dt$$$$$$$$\Rightarrow x\left(t\right)=2-2cos\left(\sqrt{2}t\right)$$

Commented by Tinkutara last updated on 04/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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