A point mass starts moving in a straight line with a constant acceleration a. At a time `t_(1)` a fter the beginning of motion, the acceleration changes sign, remaining the same in magnitude. Determine the time t from the beginning of motion in which the point mass returns to the initial position.

Options

A. `sqrt2t`
B. `(2+sqrt2) t`
C. `(t)/(sqrt2)`
D. Cannot be predicted unless acceleration is given

Answer

B

Explanation

In this problem, point starts moving with uniform acceleration a, after time t (position B) the direction of acceleration get reversed i.e. the retardation of same value works on the point. Due to this velocity of points goes on decreasing and at position C it’s velocity becomes zero. Now the direction of motion of point reverse and it moves from C to A under the effect of acceleration a. We have to calculate the total time in this motion. Starting velocity at position A is equal to zero.
`Velocity at position = `B rArr v = at`

Distance between A and B, `S_(AB) = 1/2 at^2`
As same amount of retardation works on a point and it comes to rest therefore
`S_(BC) = S_(AB) = 1/2 at^2`
` therefore S_(AC) = S_(AB) + S_(BC) = at^2` and time required to cover this distance is also equal to t.
Total time taken for motion between A and C = 2t
Now for the return journey from C to A `(S_(AC) = at^2)`
`S_(AC) = ut + 1/2 at^2 rArr at^2 = 0 + 1/2 at_1^2 rArr t_1 = sqrt2 t`
Hence total time in which point returns to initial point
`T = 2t + sqrt2 t = (2+sqrt2) t`