A point mass starts moving in straight line with constant acceleration `a` from rest at `t=0`. At time `t=2s`, the acceleration changes the sign remaining the same in magnitude. The mass returns to the initial position at time `t=t_(0)` after start of motion. Here `t_(0)` is

Options

A. ` 4 s `
B. ` ( 4+2sqrt2) s`
C. ` (2+2sqrt(2))s`
D. ` (4+4sqrt2) s `

Answer

B

Explanation

Let initial direction of motion is positive, after 2 sec, ` v= alpha t =2alpha , " " s=1 //2 alpha t^(2) =2alpha `
From t `= 2 sec " to " t ="to " ("using "s= u t +1//2 at ^(2)) `
` rArr -2alpha =2alpha (t_0 -2 ) -(alpha )/(2) ( t_0-2)^(2) rArr t_0 = 4+-2 sqrt 2 as 4-2sqrt 2 lt 2 t_0 =4+2sqrt 2 sec. `