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(A) Given,velocity of particle is $\vec{v} = x \hat{i} + yt \hat{j}$.The magnitude of velocity is $v = \sqrt{x^2 + y^2 t^2}$.The tangential accelerati

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$\frac{\sqrt{3} y}{2}, \frac{y}{2}$

Vedclass · Answer

(A) Given,velocity of particle is $\vec{v} = x \hat{i} + yt \hat{j}$.The magnitude of velocity is $v = \sqrt{x^2 + y^2 t^2}$.The tangential acceleration $a_t$ is the rate of change of speed:$a_t = \frac{dv}{dt} = \frac{1}{2\sqrt{x^2 + y^2 t^2}} \cdot (2y^2 t) = \frac{y^2 t}{\sqrt{x^2 + y^2 t^2}}$.Substituting $t = \frac{x \sqrt{3}}{y}$:$a_t = \frac{y^2 (x \sqrt{3} / y)}{\sqrt{x^2 + y^2 (3x^2 / y^2)}} = \frac{xy \sqrt{3}}{\sqrt{x^2 + 3x^2}} = \frac{xy \sqrt{3}}{2x} = \frac{\sqrt{3} y}{2}$.The total acceleration vector is $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(x \hat{i} + yt \hat{j}) = y \hat{j}$.The magnitude of total acceleration is $a = |\vec{a}| = y$.The normal acceleration $a_n$ is given by $a_n = \sqrt{a^2 - a_t^2}$.$a_n = \sqrt{y^2 - (\frac{\sqrt{3} y}{2})^2} = \sqrt{y^2 - \frac{3y^2}{4}} = \sqrt{\frac{y^2}{4}} = \frac{y}{2}$.Thus,the tangential and normal accelerations are $\frac{\sqrt{3} y}{2}$ and $\frac{y}{2}$ respectively.

$A$ particle moves in the $xy$-plane with velocity $\vec{v} = x \hat{i} + yt \hat{j}$. At $t = \frac{x \sqrt{3}}{y}$,the tangential and normal accelerations are:

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