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Question

(B) Let the initial position be $x=0$. The particle starts from rest $(u=0)$ with acceleration $a$ for time $t_0$.At $t=t_0$,the displacement is $s_1

Vedclass · Accepted Answer

$(2+\sqrt{2}) t_0$

Vedclass · Answer

(B) Let the initial position be $x=0$. The particle starts from rest $(u=0)$ with acceleration $a$ for time $t_0$.At $t=t_0$,the displacement is $s_1 = \frac{1}{2} a t_0^2$ and the velocity is $v_1 = a t_0$.After $t_0$,the acceleration becomes $-a$. Let the particle return to the initial position $(x=0)$ at a total time $T = t_0 + t$,where $t$ is the time elapsed after the acceleration change.The displacement equation for the second phase is $s = u t + \frac{1}{2} a t^2$. Here,the initial velocity is $v_1 = a t_0$ and acceleration is $-a$. We want the total displacement from the start to be zero,so the displacement in the second phase must be $-s_1 = -\frac{1}{2} a t_0^2$.Thus,$-\frac{1}{2} a t_0^2 = (a t_0) t - \frac{1}{2} a t^2$.Dividing by $-\frac{1}{2} a$,we get $t_0^2 = -2 t_0 t + t^2$,which rearranges to $t^2 - 2 t_0 t - t_0^2 = 0$.Using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $t = \frac{2 t_0 \pm \sqrt{4 t_0^2 + 4 t_0^2}}{2} = t_0 \pm \sqrt{2} t_0$.Since $t > 0$,we take $t = (1 + \sqrt{2}) t_0$.The total time is $T = t_0 + t = t_0 + (1 + \sqrt{2}) t_0 = (2 + \sqrt{2}) t_0$.

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