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(D) The particle moves under a constant force $F$ directed toward $E$. Let $E$ be at $x = 0$. The force is $F = -F$ for $x > 0$ and $F = +F$ for $x <

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(D) The particle moves under a constant force $F$ directed toward $E$. Let $E$ be at $x = 0$. The force is $F = -F$ for $x > 0$ and $F = +F$ for $x Starting from $x = -A$ with initial velocity $v = 0$,the acceleration is $a = F/m$. The equation of motion is $x(t) = -A + \frac{1}{2}at^2 = -A + \frac{F}{2m}t^2$. At $x = -A/2$,we have $-A/2 = -A + \frac{F}{2m}t_1^2$,which gives $\frac{A}{2} = \frac{F}{2m}t_1^2$,so $t_1 = \sqrt{\frac{mA}{F}}$. The velocity at $x = -A/2$ is $v_1 = at_1 = \frac{F}{m} \sqrt{\frac{mA}{F}} = \sqrt{\frac{FA}{m}}$. From $x = -A/2$ to $x = 0$,the particle moves with initial velocity $v_1$ and acceleration $a = F/m$. Using $x = v_1 t_2 + \frac{1}{2}at_2^2$,we have $A/2 = \sqrt{\frac{FA}{m}} t_2 + \frac{F}{2m}t_2^2$. Let $k = \sqrt{\frac{mA}{F}}$. Then $A/2 = \frac{A}{k} t_2 + \frac{A}{2k^2}t_2^2$. Dividing by $A/2$,we get $1 = \frac{2}{k} t_2 + \frac{1}{k^2}t_2^2$,or $t_2^2 + 2kt_2 - k^2 = 0$. Solving for $t_2$,$t_2 = \frac{-2k \pm \sqrt{4k^2 + 4k^2}}{2} = k(\sqrt{2}-1)$. The total time is $t = t_1 + t_2 = k + k(\sqrt{2}-1) = k\sqrt{2} = \sqrt{\frac{2mA}{F}}$. Since this result is not in the options,the correct answer is $D$.

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