A particle moves in a plane along an elliptic | vedclass

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Question

(A) The path of the particle is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Differentiating with respect to time $t$,we get $\frac{2x}{a^2} \frac

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$-b u^2 / a^2$

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(A) The path of the particle is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Differentiating with respect to time $t$,we get $\frac{2x}{a^2} \frac{dx}{dt} + \frac{2y}{b^2} \frac{dy}{dt} = 0$,which simplifies to $\frac{x}{a^2} v_x + \frac{y}{b^2} v_y = 0 \dots (i)$.Differentiating again with respect to time $t$,we get $\frac{1}{a^2} (v_x^2 + x a_x) + \frac{1}{b^2} (v_y^2 + y a_y) = 0 \dots (ii)$.At point $(0, b)$,we have $x = 0$ and $y = b$. Given $v_x = u$ at this point.Substituting $x = 0$ into equation $(i)$,we get $\frac{0}{a^2} u + \frac{b}{b^2} v_y = 0$,which implies $v_y = 0$.Now,substituting $x = 0, y = b, v_x = u, v_y = 0$ into equation $(ii)$,we get $\frac{1}{a^2} (u^2 + 0 \cdot a_x) + \frac{1}{b^2} (0^2 + b \cdot a_y) = 0$.This simplifies to $\frac{u^2}{a^2} + \frac{b a_y}{b^2} = 0$,or $\frac{u^2}{a^2} + \frac{a_y}{b} = 0$.Therefore,$a_y = -\frac{b u^2}{a^2}$.

$A$ particle moves in a plane along an elliptic path given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. At point $(0, b)$,the $x$-component of velocity is $u$. The $y$-component of acceleration at this point is

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