• # question_answer A particle moves in space along the path $z=a{{x}^{3}}+6{{y}^{2}}$ in such a way that $\frac{dx}{dt}=c=\frac{dy}{dt}$ where a, b and c are constants. The acceleration of the particle is - A) $\pi$                B) $(2a{{x}^{2}}x+6b{{y}^{2}})\,\,\hat{k}$ C) $(4b{{c}^{2}}x+3a{{c}^{2}})\,\,\hat{k}$      D) $(4{{c}^{2}}x+2b{{y}^{2}})\,\,\hat{k}$

Solution :

[A]  Given that $\frac{dx}{dt}=\frac{dy}{dt}=c$ $\therefore \frac{{{d}^{2}}x}{d{{t}^{2}}}$ $\therefore \frac{{{d}^{2}}x}{d{{t}^{2}}}=\frac{{{d}^{2}}y}{d{{t}^{2}}}=0$ Further $z=a{{x}^{3}}+b{{y}^{2}}$ $\therefore \frac{dz}{dt}=3\,a{{x}^{2}}\frac{dx}{dt}+2\,by\frac{dy}{dt}$$=3\,ac{{x}^{2}}+2\,bcy\left( \frac{dx}{dt}=c=\frac{dt}{dt} \right)$ $\therefore \frac{{{d}^{2}}x}{d{{t}^{2}}}=6\,acx\left( \frac{dx}{dt} \right)+2\,bc\left( \frac{dy}{dt} \right)$$=6a{{c}^{2}}x+2b{{c}^{2}}$ Now acceleration of particle is $\vec{a}=\frac{{{d}^{2}}x}{d{{t}^{2}}}\,\,\hat{i}+\frac{{{d}^{2}}x}{d{{t}^{2}}}\,\,\hat{j}+\frac{{{d}^{2}}x}{d{{t}^{2}}}\,\,\hat{k}$$=(6a{{c}^{2}}x+2b{{c}^{2}})\,\,\hat{k}$

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