Introduction:
In this problem, we are given that a point mass starts moving in a straight line with constant acceleration 'a' from rest at t=0. After 2 seconds, the acceleration changes sign but remains the same in magnitude. We need to find the time 't1' at which the mass returns to its initial position.

Understanding the problem:
To solve this problem, we need to understand the concept of motion with constant acceleration. When a body moves with constant acceleration 'a', its displacement as a function of time can be given by the equation:
s = ut + (1/2)at^2
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

Solution:
To find the time 't1' at which the mass returns to its initial position, we need to determine the displacement of the mass at time t=2s and set it equal to zero.

Step 1: Finding displacement at t=2s:
Using the equation of motion, we substitute t=2s and u=0 (since the mass starts from rest) to find the displacement at t=2s:
s = ut + (1/2)at^2
s = 0 + (1/2)a(2)^2
s = 2a

Step 2: Finding displacement at t=t1:
Next, we need to find the displacement at time t=t1, when the mass returns to its initial position. Since the acceleration changes sign but remains the same in magnitude, the displacement equation remains the same for both positive and negative time intervals. Therefore, the displacement at t=t1 will be equal to the displacement at t=2s multiplied by a factor of -1:
s = -2a

Step 3: Setting the displacements equal and solving for t1:
Setting the displacements at t=2s and t=t1 equal to each other, we have:
2a = -2a
Simplifying the equation, we get:
4a = 0

Step 4: Solving for t1:
Since the acceleration 'a' cannot be zero (otherwise, the mass would not change its direction), we can conclude that the time 't1' required for the mass to return to its initial position is infinite or undefined. This means that the mass will never return to its initial position.

Conclusion:
After analyzing the given problem, we find that the mass will never return to its initial position. The time 't1' required for the mass to return to its initial position is infinite or undefined.

Initially u=0 then it is accelerated till 2s so v=2a then distance covered will be v^2/2a so distance =2a now for the next part of motion accln=-a now speed will go on decreasing till it gets 0 and particle stops momentarily but accln is still acting in -x direction so now it will turn in the direction of accln so calculate the total distance till it goes in +x dirn and then that same distance it has to travel in the opp dirn to reach initial point if you have confusion in solving then tell me