BCECE Medical BCECE Medical Solved Papers-2004

  • question_answer
    An object accelerates from rest to a velocity 27.5 m/s in 10 s, then the distance covered after next 10 s is :

    A)  550 m         

    B)  137.5 m

    C)  412.5 m

    D)  175 m

    Correct Answer: C

    Solution :

    Since, the object accelerates from rest, its initial velocity (u) is zero. That is, \[u=0\] From first equation of motion                 \[v=u+at\] \[\therefore \] \[27.5=0+a\times 10\] or            \[a=2.75\,\,m/{{s}^{2}}\] Hence, distance covered in first 10s                 \[{{s}_{1}}=ut+\frac{1}{2}a{{t}^{2}}\]                 \[=0+\frac{1}{2}\times 2.75\times {{(10)}^{2}}\]                 \[=137.5\,\,m\] Distance covered in next 10 s with uniform velocity of 27.5 m/s                 \[{{s}_{2}}=27.5\times 10=275\,m\] Total distance covered \[s=137.5+275=412.5\,\,m\]


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