Question

# 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

A

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B

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C

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D

Cannot be predicted unless acceleration is provided

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Solution

## The correct option is D 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) SAC=ut−12at2⇒−at2=0−12at12⇒t1=√t Hence total time in which point returns to initial point T = 2t+√2t=(2+√2)t

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