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Question Number 16499 by Tinkutara last updated on 23/Jun/17

$$\mathrm{A}\mathrm{particle}\mathrm{is}\mathrm{moving}\mathrm{in}\mathrm{parabolic}\mathrm{path}$$$$x^2=y,\mathrm{with}\mathrm{constant}\mathrm{speed}u.\mathrm{Find}\mathrm{the}$$$$\mathrm{acceleration}\mathrm{of}\mathrm{the}\mathrm{particle}\mathrm{when}\mathrm{it}$$$$\mathrm{crossess}\mathrm{origin}.\mathrm{Also}\mathrm{find}\mathrm{the}\mathrm{radius}\mathrm{of}$$$$\mathrm{curvature}\mathrm{at}\mathrm{origin}.$$

Answered by ajfour last updated on 23/Jun/17

$$x^2=y\Rightarrow \frac{dy}{dx}=2xwhichiszero$$$$attheorigin.And\frac{d^{2}y}{{dx}^2}=2$$$$radiusofcurvature,$$$$r=\mid \frac{{[1+{\left(dy/dx\right)}^2]}^{3/2}}{d^{2}y/{dx}^2}\mid =\frac{1}{2}.$$$${\mathit{x}}^2=\mathit{y}$$$$2x\frac{dx}{dt}=\frac{dy}{dt}=v_y(=0atorigin)..\left(i\right)$$$$u^2={\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2....\left(ii\right)$$$$\Rightarrow {\left(\frac{dx}{dt}\right)}^2=\frac{u^2}{1+4x^2}....\left(iii\right)$$$$so{\left(\frac{dx}{dt}\right)}^2=u^{2}attheorigin$$$$2{\left(\frac{dx}{dt}\right)}^2+2x\frac{d^{2}x}{{dt}^2}=\frac{d^{2}y}{{dt}^2}.....\left(iv\right)$$$$meansatorigin,\frac{d^{2}y}{{dt}^2}=a_y=2u^2$$$$......\left(v\right)$$$$\Rightarrow {\left(\frac{dx}{dt}\right)}^2=\frac{u^2}{1+4x^2}$$$$2\left(\frac{dx}{dt}\right)\left(\frac{d^{2}x}{{dt}^2}\right)=-\frac{8u^{2}x}{{(1+4x^2)}^2}\left(\frac{dx}{dt}\right)$$$$a_x=\frac{d^{2}x}{{dt}^2}=-\frac{4u^{2}x}{{(1+4x^2)}^2},thisiszero$$$$attheorigin.$$$$accelerationa=\frac{d^{2}x}{{dt}^2}+\frac{d^{2}y}{{dt}^2}$$$$atorigina=0+2u^2\left[see\left(v\right)\right].$$

Commented by Tinkutara last updated on 23/Jun/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

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