Physics-
General
Easy

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 time in which the point returns to the initial point is

  1. square root of 2 t end root  
  2. left parenthesis 2 plus square root of 2 right parenthesis t  
  3. fraction numerator t over denominator square root of 2 end fraction  
  4. Cannot be predicted unless acceleration is given  

The correct answer is: left parenthesis 2 plus square root of 2 right parenthesis t


    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 blankto 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 rightwards double arrow v equals a t [A s blank u equals 0]

    Distance between A and B comma blank S subscript A B end subscript equals fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent
    As same amount of retardation works on a point and it comes to rest therefore
    S subscript B C end subscript equals S subscript A B end subscript equals fraction numerator 1 over denominator 2 end fraction a blank t to the power of 2 end exponent
    therefore S subscript A C end subscript equals S subscript A B end subscript plus S subscript B C end subscript equals a blank t to the power of 2 end exponentand time required to cover this distance is also equal to t.
    therefore T o t a l blank t i m e blank t a k e n blank f o r blank m o t i o n blank b e t w e e n blank A blank a n d blank C equals 2 t
    Now for the return journey from C blank t o blank A blank open parentheses S subscript A C end subscript equals a t to the power of 2 end exponent close parentheses
    S subscript A C end subscript equals u blank t plus fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent rightwards double arrow a t to the power of 2 end exponent equals 0 plus fraction numerator 1 over denominator 2 end fraction a t subscript 1 end subscript superscript 2 end superscript rightwards double arrow t subscript 1 end subscript equals square root of 2 t
    Hence total time in which point returns to initial point
    T equals 2 t plus square root of 2 blank end root t equals left parenthesis 2 plus square root of 2 right parenthesis t

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