BCECE Medical BCECE Medical Solved Papers-2011

  • question_answer
    A constant power P is applied to a particle of mass m. The distance travelled by the particle when its velocity increases from \[{{v}_{1}}\] to \[{{v}_{2}}\]  is (neglect friction)

    A)  \[\frac{m}{3P}(v_{2}^{3}-v_{1}^{3})\]     

    B)  \[\frac{m}{3P}({{v}_{2}}-{{v}_{1}})\]

    C)  \[4.32\times {{10}^{6}}\]     

    D)  \[\frac{m}{3P}(v_{2}^{2}-v_{1}^{2})\]

    Correct Answer: A

    Solution :

    \[p=Fv=mav\]                 \[a=\frac{p}{mv}\] or            \[v\frac{dv}{ds}=\frac{p}{mv}\] or            \[{{v}^{2}}=dv=\frac{p}{m}ds\] or            \[\frac{p}{m}\int_{0}^{s}{ds}=\int_{{{v}_{1}}}^{{{v}_{2}}}{{{v}^{2}}\,\,dv}\] or            \[\frac{p}{m}s=\frac{1}{3}\,(v_{2}^{3}-v_{1}^{3})\] \[s=\frac{m}{3p}\,\,(v_{2}^{3}-v_{1}^{3})\]


You need to login to perform this action.
You will be redirected in 3 sec spinner