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(D) Given,the velocity of the particle is $\vec{v} = v_x \hat{i} + v_y \hat{j} = (2 \hat{i} + 2 \hat{j}) 	ext{ m s}^{-1}$.Thus,$v_x = \frac{dx}{dt} =

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$32$

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(D) Given,the velocity of the particle is $\vec{v} = v_x \hat{i} + v_y \hat{j} = (2 \hat{i} + 2 \hat{j}) 	ext{ m s}^{-1}$.Thus,$v_x = \frac{dx}{dt} = 2 	ext{ m s}^{-1}$ and $v_y = \frac{dy}{dt} = 2 	ext{ m s}^{-1}$.The acceleration in the $x$-direction is $a_x = \frac{dv_x}{dt} = 0$ (since $v_x$ is constant).The acceleration in the $y$-direction is $a_y = \frac{dv_y}{dt} = a$.The trajectory is given by $y = 4x^2$.Differentiating with respect to time $t$,we get $\frac{dy}{dt} = 8x \frac{dx}{dt}$,which means $v_y = 8x v_x$.At the origin $(0,0)$,$x=0$. Substituting $v_y = 2$ and $v_x = 2$ into the equation $v_y = 8x v_x$ is not directly possible to find $x$ at the given velocity,but we differentiate $v_y = 8x v_x$ again with respect to $t$:$\frac{dv_y}{dt} = 8 \left( x \frac{dv_x}{dt} + v_x \frac{dx}{dt} ight)$.Substituting $a_y = a$,$a_x = 0$,$v_x = 2$,and $\frac{dx}{dt} = v_x = 2$:$a = 8(x \cdot 0 + 2 \cdot 2) = 8(4) = 32 	ext{ m s}^{-2}$.

$A$ particle is moving in the $xy$-plane and crosses the origin at time $t=0$. The equation of motion of the particle is $y=4x^2$. If the velocity of the particle is $\vec{v}=(2\hat{i}+2\hat{j}) \text{ m s}^{-1}$ and acceleration is $\vec{a}=(a\hat{j}) \text{ m s}^{-2}$,then the magnitude of $a$ is

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