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JEE Main & Advanced Sample Paper JEE Main Sample Paper-28

  • question_answer
    A particle is moving on straight path. If initial velocity is \[7\,m{{s}^{-1}}\] and acceleration is \[-2\,m{{s}^{-2}}\] then distance travelled by the particle in fourth second will be:

    A)  \[0.25\,m\]                                       

    B)  \[0.8\,m\]

    C)  \[0.5\,m\]                                         

    D)  \[1\,\,m\]

    Correct Answer: C

    Solution :

    \[v=u+at\] \[0=7-2t\] \[t=3.5\] Hence velocity of the particle will be zero at \[3.5\,\,\sec \] Hence distance travelled between 3 to 3.5 sec will be same    as    distance    travelled    between3.5 to 4 sec due to symmetry of motion. \[S=\frac{1}{2}\times 2\times {{(0.5)}^{2}}=0.25\,\,m\] Hence distance travelled in fourth second will be\[2\times 0.25=0.5\,\,m\]


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