Question

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

• 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} $
$\Rightarrow v_{1}=2\, a$
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,
$v_{2}=v_{1}-a t_{2} $
$\Rightarrow 0=2 a-a t_{2}$
or $t_{2}=2 \,s$
Hence, total time taken by body to covered distance $A C, \,t=2+2=4 \,s$
From second equation one motion,
$s_{1}=A B=u t_{1}+\frac{1}{2} a t_{1}^{2}=0+\frac{1}{2} a \times 2^{2} $
$s_{1}=2\,a$
$\because s_{1}=s_{2}=2\, a$
$\therefore A C=s_{1}+s_{2}=4\, a$
Now, body returns from point $C$ to point $A$. So,
$u_{1}=0,\, A C=u a$
From second equation of the motion,
$s=A C=u_{1}\, t+\frac{1}{2} a t^{2}$ or $u a=0+\frac{1}{2} a t^{2}$
$\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$