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NEET NEET SOLVED PAPER 2016 Phase-I

  • question_answer
    If the velocity of a particle is \[v=At=B{{t}^{2}},\] where A and B are constants, then the distance travelled by it between 1s and 2s is :-                                                                                                                                 

    A)  \[\frac{3}{2}A+4B\]                       

    B)  \[3A+7B\]

    C)   \[\frac{3}{2}A+\frac{7}{3}B\]                   

    D)   \[\frac{A}{2}+\frac{B}{3}\]

    Correct Answer: C

    Solution :

                     \[V=At+B{{t}^{2}}\Rightarrow \frac{dx}{dt}=At+B{{t}^{2}}\] \[\Rightarrow \]\[\int\limits_{0}^{x}{dx}=\int\limits_{2}^{2}{(At+B{{t}^{2}})dt}\] \[\Rightarrow \] \[x=\frac{A}{2}({{2}^{2}}-{{1}^{2}})+\frac{B}{3}({{2}^{3}}-{{1}^{3}})=\frac{3A}{2}+\frac{7B}{3}\]


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