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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2005

  • question_answer
    The distance covered by a particle in t second is\[s=a{{e}^{t}}+b{{e}^{-t}}\]. At time t the acceleration of the particle is

    A)  \[s\,m/{{s}^{2}}\]           

    B)  \[as\,m/{{s}^{2}}\]

    C)  \[bs\text{ }m/{{s}^{2}}\]          

    D)  \[-bs\text{ }m/{{s}^{2}}\]

    Correct Answer: A

    Solution :

     Given, \[s=a{{e}^{t}}-b{{e}^{-t}}\] \[\frac{ds}{dt}=a{{e}^{t}}-b{{e}^{-t}}\] Acceleration, \[\frac{{{d}^{2}}s}{d{{t}^{2}}}=a{{e}^{t}}+b{{e}^{t}}=s\,m/{{s}^{2}}\]


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