A point mass starts moving in a straight line with constant acceleration "a". At a time t after the beginning of motion, the acceleration changes sign, without change in magnitude. Determine the time `t_(0)` from the beginning of the motion in which the point mass returns to the initial position. `((2+sqrt2)t)`

Explanation

Let particle starts at A.
The initial velocity u = 0m/s , Displacement = s m
Acceleration = `am//s^(2)` , Time = t s
Displacement s = `ut+1/2at^(2)rArrs=0*t+1/2*at^(2)`
`rArrs=1/2at^(2)" ".......(1)`
Final velocity `v=u+atrArrv=at" "........(2)`
After time t, acceleration changes its sign but magnitude is same for movement from B to A.
Displacement = -sm, Acceleration = `-am//s^(2)`, Time = `t_(2)s`, Initial velocity at B = at
Displacement - s = ut + `1/2at^(2)`
`rArr-1/2cancel(a)t^(2)=cancel(a)t t_(2)-1/2cancel(a)t_(2)^(2)`
`-t^(2)=2t t_(2)-t_(2)^(2)=0`
`t_(2)^(2)-2t t_(2)-t^(2)=0`
Compare with `ax^(2)+bx+c=0`
`x=(-bpmsqrt(b^(2)-4ac))/(2a)`
`a=1,b=-2t,c=-t^(2)`
`t_(2)=(+2tpmsqrt((-2t)^(2)_(4.1.-t^(2))))/(2.1)`
`t_(2)=(2tpmsqrt(4t^(2)+4t^(2)))/2`
`t_(2)=(2tpmsqrt(8t^(2)))/2`
`t_(2)=(2tpm2sqrt2t)/2`
`t_(2)=(cancel(2)(tpmsqrt2t))/(cancel(2))`
Total time `t_(0)=t+t_(2)`
`=t+t+sqrt2t`
`rArr2t+sqrt2t`
`rArr(2+sqrt2)t`