MGIMS WARDHA MGIMS WARDHA Solved Paper-2014

  • question_answer
    If the position of a moving particle, with respect to time be, \[x(t)=1+t-{{t}^{2}}\]. then, the acceleration of particle is given by (assume all measurements in MKS)

    A)  \[-1\text{ }m{{s}^{-2}}\]             

    B)  \[-1.5\text{ }m{{s}^{-2}}\]

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

    D)  \[-2.5\text{ }m{{s}^{-2}}\]

    Correct Answer: C

    Solution :

                    The position of a moving particle \[x({{t}_{1}})=1+t-{{t}^{2}}\] The velocity of the particle                 \[\frac{dx}{dt}=v=1-2t\] The acceleration of the particle                 \[\frac{dv}{dt}=-2m/{{s}^{2}}\]


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