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Home / Doubts and Solutions/JEE/Physics

Pls solve this?

Asked by dasratna_72 11th June 2017, 5:00 PM
Answered by Expert
Answer:
begin mathsize 12px style Space minus average space velocity space is space given space as space fraction numerator integral straight v space ds over denominator integral ds end fraction space and space time minus average space velocity space is space fraction numerator integral straight v space dt over denominator integral dt end fraction Given space that space the space initial space velocity space is space zero. space therefore space straight v space equals space at therefore space Time minus average space open angle brackets straight v close angle brackets subscript time equals space fraction numerator integral straight v space dt over denominator integral dt end fraction space Putting space the space limits space from space straight T space to space 0 comma space we space get therefore space open angle brackets straight v close angle brackets subscript time space equals space space fraction numerator integral subscript 0 superscript straight T at space dt over denominator integral subscript 0 superscript straight T dt end fraction space equals space aT over 2 therefore space Space minus average space open angle brackets straight v close angle brackets subscript space equals space fraction numerator integral straight v space ds over denominator integral ds end fraction space equals fraction numerator integral straight v space begin display style ds over dt end style dt over denominator integral ds over dt dt end fraction space rightwards arrow open angle brackets straight v close angle brackets subscript space equals space fraction numerator integral subscript 0 superscript straight T straight v squared dt over denominator integral subscript 0 superscript straight T vdt end fraction space space equals space fraction numerator integral subscript 0 superscript straight T straight a squared straight t squared space dt over denominator integral subscript 0 superscript straight T at space dt end fraction space space therefore space open angle brackets straight v close angle brackets subscript space equals space 2 over 3 aT Now space taking space the space ratio space of space open angle brackets straight v close angle brackets subscript space space and space open angle brackets straight v close angle brackets subscript time space comma space we space get fraction numerator open angle brackets straight v close angle brackets subscript space over denominator space open angle brackets straight v close angle brackets subscript time end fraction space equals space fraction numerator begin display style bevelled aT over 2 end style over denominator begin display style bevelled fraction numerator 2 aT over denominator 3 end fraction end style end fraction space equals space 4 over 3 end style
Answered by Expert 12th June 2017, 11:25 AM
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