A point starts moving in a straight line with a certain acceleration. At a time t after beginning of motion the acceleration suddenly becomes retardation of the same value. The total time in which the point returns to the initial position is
In this problem point starts moving with a uniform acceleration a and after time t (Position B) the direction of acceleration get reversed i.e, the retradation of same value works on the point.Due to this velocity of the point goes on decreasing and at position C its velocity becomes zero.Now the direction of motion of point reversed 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⇒v=at [As u = 0]
Distance between A and B, SAB=12at2
As same amount of retardation works on a point and it comes to rest therefore SBC=SAB=12at2
∴ Total time taken for motion between A and C = 2t
Now for the return journey from C to A(SAC=at2)
Hence total time in which point returns to initial point T = 2t+√2t=(2+√2)t