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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

  • question_answer
    A bus starts from rest with an acceleration\[1\,m{{s}^{-2}}\]. A man who is 48 m behind the bus starts with a uniform velocity of\[10\text{ }m{{s}^{-1}}\]. Then the minimum rime after which the man will reach the bus

    A)  4 s            

    B)  8 s

    C)  10 s           

    D)  15 s

    Correct Answer: B

    Solution :

     Let the minimum time after which the man catches the bus be t second. Then, For bus, \[s=ut+\frac{1}{2}a{{t}^{2}}\] i.e.,   \[{{s}_{b}}=(0)(t)+\frac{1}{2}(1){{(t)}^{2}}=\frac{{{t}^{2}}}{2}\] For man \[{{s}_{m}}=(10)(t)+\frac{1}{2}(0){{t}^{2}}=10t\] From the question\[{{s}_{m}}={{s}_{b}}+48\] \[\Rightarrow \] \[10t=\frac{{{t}^{2}}}{2}+48\] \[\Rightarrow \] \[{{t}^{2}}-20t+96=0\] \[\Rightarrow \] \[(t-8)(t-12)=0\] \[\therefore \]        \[t=8\text{ }s\]


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