A particle moves along a parabolic path y = 9 | vedclass

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(D) The equation of the path is $y = 9x^2$.Given that the $x$-component of velocity is constant,$v_x = \frac{dx}{dt} = \frac{1}{3} \; m/s$.Since $v_x$

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(D) The equation of the path is $y = 9x^2$.Given that the $x$-component of velocity is constant,$v_x = \frac{dx}{dt} = \frac{1}{3} \; m/s$.Since $v_x$ is constant,the $x$-component of acceleration $a_x = \frac{dv_x}{dt} = 0$.To find the $y$-component of velocity,differentiate $y$ with respect to time $t$:$v_y = \frac{dy}{dt} = \frac{d}{dt}(9x^2) = 18x \frac{dx}{dt}$.Substituting the value of $\frac{dx}{dt} = \frac{1}{3}$:$v_y = 18x 	imes \frac{1}{3} = 6x$.Now,find the $y$-component of acceleration by differentiating $v_y$ with respect to time $t$:$a_y = \frac{dv_y}{dt} = \frac{d}{dt}(6x) = 6 \frac{dx}{dt}$.Substituting $\frac{dx}{dt} = \frac{1}{3}$:$a_y = 6 	imes \frac{1}{3} = 2 \; m/s^2$.Since $a_x = 0$,the total acceleration of the particle is $a = \sqrt{a_x^2 + a_y^2} = \sqrt{0^2 + 2^2} = 2 \; m/s^2$.

$A$ particle moves along a parabolic path $y = 9x^2$ in such a way that the $x$-component of velocity remains constant and has a value of $\frac{1}{3} \; m/s$. The acceleration of the particle is $....... \; m/s^2$.

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