A body starting from rest at $t=0$ moves along a straight line with a constant acceleration. At $t=2 \mathrm{~s}$, the body reverses its direction keeping the acceleration same. The body returns to the initial position at $t=t_0$, then $t_0$ is

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From first equation of the motion, $ v_1=u+a t_1 \Rightarrow v_1=2 a $ Firstly, body decelerate with acceleration to the point $C$ and then reverse it's direction and accelerate with acceleration $a$ to the point $A$. Therefore for distance $B C$, from first equation of the motion, or $ v_2=v_1-a t_2 \Rightarrow 0=2 a-a t_2 $ $ t_2=2 \mathrm{~s} $ Hence, total time taken by body to covered distance $A C, t=2+2=4 \mathrm{~s}$ From second equation one motion, $ \begin{array}{rlrl} & s_1 & =A B=u t_1+\frac{1}{2} a t_1^2=0+\frac{1}{2} a \times 2^2 \\ & & s_1 & =2 a \\ & & s_1 & =s_2=2 a \\ & A C & =s_1+s_2=4 a \end{array} $ Now, body returns from point $C$ to point $A$. So, $\quad u_1=0, A C=u a$ From second equation of the motion, $ \begin{aligned} & \Rightarrow \quad \frac{1}{2} \log _e \phi\left(v^2\right)=\log _e x+\log _e(\sqrt{c})=\log _e(\sqrt{c} x) \\ & \Rightarrow \phi\left(v^2\right)=c x^2 \Rightarrow \phi\left(\frac{y^2}{x^2}\right)=c x^2 \end{aligned} $ Hence, option (d) is correct. $ \begin{aligned} & s=A C=u_1 t+\frac{1}{2} a t^2 \text { or } u a=0+\frac{1}{2} a t^2 \\ & \Rightarrow t^2=8 \Rightarrow t=2 \sqrt{2} \mathrm{~s} \end{aligned} $ Therefore, the total time taken by body, $ t_0=t_2+t=(4+2 \sqrt{2}) \mathrm{s} $

A body starting from rest at $t=0$ moves along a straight line with a constant acceleration. At $t=2…

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