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(B) Given the trajectory $y = \frac{x^2}{2}$.At $t=0$,the particle is at $(0, 0)$ with speed $v = 1 	ext{ m/s}$.Differentiating $y = \frac{x^2}{2}$ w

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$A, B, C, D$

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(B) Given the trajectory $y = \frac{x^2}{2}$.At $t=0$,the particle is at $(0, 0)$ with speed $v = 1 	ext{ m/s}$.Differentiating $y = \frac{x^2}{2}$ with respect to $t$,we get $\frac{dy}{dt} = x \frac{dx}{dt}$,which means $v_y = x v_x$.Differentiating again with respect to $t$,we get $a_y = v_x^2 + x a_x$.$(A)$ If $a_x = 1 	ext{ m/s}^2$ and the particle is at the origin $(x=0)$,then $a_y = v_x^2 + (0)(1) = v_x^2$. Since the speed is $1 	ext{ m/s}$ and at the origin $v_y = x v_x = 0$,we have $v_x = 1 	ext{ m/s}$. Thus,$a_y = 1^2 = 1 	ext{ m/s}^2$. This is correct.$(B)$ If $a_x = 0$,then $v_x$ is constant. Since $v_x(0) = 1 	ext{ m/s}$,$v_x = 1 	ext{ m/s}$ for all $t$. Then $a_y = v_x^2 + x a_x = 1^2 + x(0) = 1 	ext{ m/s}^2$ for all $t$. This is correct.$(C)$ At $t=0$,$x=0$. Since $v_y = x v_x$,we have $v_y = 0 \cdot v_x = 0$. Thus,the velocity vector is $\vec{v} = v_x \hat{i} + 0 \hat{j}$,which points in the $x$-direction. This is correct.$(D)$ If $a_x = 0$,then $v_x = 1 	ext{ m/s}$ and $a_x = 0$. From $a_y = v_x^2 + x a_x$,we get $a_y = 1^2 + 0 = 1 	ext{ m/s}^2$. Since $a_y$ is constant,$v_y = a_y t = 1 \cdot t = t$. At $t=1 	ext{ s}$,$v_y = 1 	ext{ m/s}$. The angle $	heta$ with the $x$-axis is given by $	an 	heta = \frac{v_y}{v_x} = \frac{1}{1} = 1$,so $	heta = 45^{\circ}$. This is correct.Therefore,all statements $(A), (B), (C), (D)$ are correct.

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