A point mass starts moving in a st. line with constant acceleration. After time t, the acceleration changes its sign but magnitude remains constant. Determine time from beginning of motion in which the point mass returns to the initial position?

after time t velocity of particle is v

v=u+at           (u=0)

v=at......1

now particle is having some velocity and  the direction of accleration is changed so after time t1 it will come to rest....

v=u-at1      (v=0)

t1=u/a           (after putting value of v from eq 1)

t1=t.................2

total displacement during this period is S

S=St+St1=at^2/2 + ut1-at1^2/2=ut1=ut=at^2 ..........3 (from eq 1)

now again particle will move but opposite direction with a accleration a ...

let after time t3 particle reaches the original position..

S=at3^2/2............4

from 3 and 4

t3=t(sqrt2)

total time of this journey is T=t1+t2+t3

                    T=t+t+t(sqrt2)

                    T=t(2+sqrt2)sec