JEE Main & Advanced Sample Paper JEE Main Sample Paper-25

  • question_answer
    A particle crossing the origin of coordinates at time \[t=0\]. moves in the xy-plane with a constant acceleration a in y-direction. If, its equation of motion is \[y=b{{x}^{2}}\] (where, is b is a constant), its velocity component in the x-direction is :

    A)  \[\sqrt{\frac{2b}{a}}\]                          

    B)  \[\sqrt{\frac{a}{2b}}\]

    C)  \[\sqrt{\frac{a}{b}}\]                            

    D)  \[\sqrt{\frac{b}{a}}\]

    Correct Answer: B

    Solution :

    At origin acceleration a is centripetal acceleration v is along the tangent along x axis. Hence dy/dx=2bx


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