CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    A body travelling with uniform acceleration crosses two points A and B with velocities \[20\,m{{s}^{-1}}\] and \[30\,m{{s}^{-1}}\] respectively. The speed of the body at the mid-point of A and B is nearest to

    A)  \[25.5\,m{{s}^{-1}}\]       

    B)         \[25\,m{{s}^{-1}}\]

    C)  \[24\text{ }m{{s}^{-1}}\]        

    D)         \[10\sqrt{6}\,m{{s}^{-1}}\]

    E)  \[22\,m{{s}^{-1}}\]

    Correct Answer: A

    Solution :

    A particle moving with uniform acceleration from A to B along straight line has velocities \[{{v}_{1}}\]and\[{{v}_{2}}\]at A and B respectively. If C is the midpoint between A and B then velocity of the particle at C. \[v=\sqrt{\frac{v_{1}^{2}+v_{2}^{2}}{2}}\] \[=\sqrt{\frac{{{(20)}^{2}}+{{(30)}^{2}}}{2}}\] \[=\sqrt{\frac{400+900}{2}}=\sqrt{650}\] \[=25.5\text{ }m{{s}^{-1}}\]


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