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WB JEE Medical WB JEE Medical Solved Paper-2009

  • question_answer
    The acceleration a (in\[\text{m}{{\text{s}}^{2}}\]) of a body, starting from rest varies with time t (in s) following the equation \[a=3t+4.\]The velocity of the body at time t = 2 s will be

    A)  \[10\,m{{s}^{-1}}\]                        

    B)  \[18\,m{{s}^{-1}}\]

    C)  \[14\,m{{s}^{-1}}\]                        

    D)  \[26\,m{{s}^{-1}}\]

    Correct Answer: C

    Solution :

                     Given      \[a=3t+4\] \[\Rightarrow \]               \[\frac{dv}{dt}=3t+4\] \[\therefore \] \[\int_{0}^{v}{dv}=\int_{0}^{t}{(3t+4)dt}\] \[\Rightarrow \]               \[v=\frac{3}{2}{{t}^{2}}+4t\] at            \[t=2s\]                 \[v=\frac{3}{2}{{(2)}^{2}}+4(2)=14m{{s}^{-1}}\]


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