Question

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

• A. 4 s
• B. $(4+2\sqrt{2})s$
• C. $(2+2\sqrt{2})s$
• D. $(4+4\sqrt{2})s$

Answer 1

Answer: (B)

According to the question,

From first equation of the motion,
$v_1=u+a{t}_1$
$\Rightarrow v_1=2a$
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 $BC$, from first equation of the motion,
$v_2=v_1-a{t}_2$
$\Rightarrow 0=2a-a{t}_2$
or $t_2=2s$
Hence, total time taken by body to covered distance $AC,t=2+2=4s$
From second equation one motion,
$s_1=AB=u{t}_1+\frac{1}{2}a{t}_1^2=0+\frac{1}{2}a\times 2^2$
$s_1=2a$
$\because s_1=s_2=2a$
$\therefore AC=s_1+s_2=4a$
Now, body returns from point $C$ to point $A$. So,
$u_1=0,AC=ua$
From second equation of the motion,
$s=AC=u_{1}t+\frac{1}{2}a{t}^2$ or $ua=0+\frac{1}{2}a{t}^2$
$\Rightarrow t^2=8$
$\Rightarrow t=2\sqrt{2}s$
Therefore, the total time taken by body,
$t_0=t_2+t=(4+2\sqrt{2})s$