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

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Cannot be predicted unless acceleration is provided

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Solution

The correct option is B

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 $\Rightarrow v = a t$ [As u = 0]

Distance between A and B, $S A B = \frac{1}{2} a t^{2}$

As same amount of retardation works on a point and it comes to rest therefore $S_{B C} = S_{A B} = \frac{1}{2} a t^{2}$

$∴$ Total time taken for motion between A and C = 2t

Now for the return journey from C to A $(S_{A C} = a t^{2})$

$S_{A C} = u t - \frac{1}{2} a t^{2} \Rightarrow - a t^{2} = 0 - \frac{1}{2} a t_{1}^{2} \Rightarrow t_{1} = \sqrt{t}$

Hence total time in which point returns to initial point T = $2 t + \sqrt{2} t = (2 + \sqrt{2}) t$

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