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CBSE Class 11-science Answered

A particle moves in a straight line so that s=root t, then its acceleration iis proportional to
Asked by Anil | 06 Jun, 2017, 01:44: PM
answered-by-expert Expert Answer
begin mathsize 12px style Given space that space straight s space equals space square root of straight t We space know space that space velocity space straight v space equals space ds over dt therefore space straight v space equals space fraction numerator straight d open parentheses square root of straight t close parentheses over denominator dt end fraction therefore space straight v space space equals space fraction numerator 1 over denominator 2 square root of straight t end fraction rightwards arrow square root of straight t equals fraction numerator 1 over denominator 2 straight v end fraction or space straight t to the power of bevelled fraction numerator negative 1 over denominator 2 end fraction end exponent space equals space 2 straight v space......... space Equation space 1 We space know space that space acceleration space straight a space equals space dv over dt therefore space straight a space equals space fraction numerator straight d open parentheses straight v close parentheses over denominator dt end fraction space equals space fraction numerator straight d open parentheses fraction numerator 1 over denominator 2 square root of straight t end fraction close parentheses over denominator dt end fraction or space space straight a space equals space 1 half fraction numerator straight d open parentheses straight t to the power of bevelled fraction numerator negative 1 over denominator 2 end fraction end exponent close parentheses over denominator dt end fraction or space space straight a space equals space fraction numerator negative 1 over denominator 4 straight t begin display style bevelled 3 over 2 end style end fraction or space space straight a space equals space fraction numerator negative 1 over denominator 4 end fraction space open square brackets fraction numerator 1 over denominator square root of straight t end fraction close square brackets cubed From space equation 1. space we space get or space space straight a space equals space fraction numerator negative 1 over denominator 4 end fraction space open square brackets 2 straight v close square brackets cubed rightwards arrow space straight a space proportional to space straight v cubed end style
Answered by Yashvanti Jain | 06 Jun, 2017, 03:33: PM

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CBSE 11-science - Physics
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