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 time in which the point returns to the initial point is

(1) 2t

(2) (2+2)t

(3) t2

(4) Cannot be predicted unless acceleration is given

Concept Videos :-

#16-Acceleration
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#18 Solved Examples 10
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Concept Questions :-

Acceleration

(2) In this problem point starts moving with uniform acceleration a and 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 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

SAC=SAB+SBC=at2 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 (SAC=at2)

SAC=ut+12at2at2=0+12at12t1=2t

Hence total time in which point returns to initial point

T=2t+2t=(2+2)t

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