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General
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Question

A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the 5ths e c to that covered in 5 blank s e c is

  1. 9 divided by 15  
  2. 3 divided by 5  
  3. 25 divided by 9  
  4. 1 divided by 25  

The correct answer is: 9 divided by 15


    Distance covered in 5 to the power of t h end exponent second
    S subscript 5 to the power of t h end exponent end subscript equals u plus fraction numerator a over denominator 2 end fraction open parentheses 2 n minus 1 close parentheses equals 0 plus fraction numerator a over denominator 2 end fraction open parentheses 2 cross times 5 minus 1 close parentheses equals fraction numerator 9 a over denominator 2 end fraction
    and distance covered in 5 blank s e c o n d,
    S subscript 5 end subscript equals u t plus fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent equals 0 plus fraction numerator 1 over denominator 2 end fraction cross times a cross times 25 equals fraction numerator 25 a over denominator 2 end fraction
    therefore fraction numerator S subscript 5 to the power of t h end exponent end subscript over denominator S subscript 5 end subscript end fraction equals fraction numerator 9 over denominator 25 end fraction

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