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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    A particle moving with a uniform acceleration travels 24 m and 64 m in the first two consecutive intervals of 4s each. Its initial velocity is

    A)  1m/s                                    

    B)  10m/s

    C)  5 m/s                                   

    D)  2 m/s

    Correct Answer: A

    Solution :

    \[s=ut+\frac{1}{2}a{{t}^{2}}\] for \[\varepsilon t\,t=4s,\] \[s=24m\] \[\therefore \]  \[24=4u+\frac{1}{2}a{{(4)}^{2}}\] or            \[8a+4u=24\] or            \[2a+u=6\]          ??(i) For         \[t=8s,\]                 \[s=24+64=88m\] \[\therefore \]  \[88=8u+\frac{1}{2}a{{(8)}^{2}}\] or            \[32a+8u=88\] or            \[4a+u=11\]       ?..(ii) Solving Eqs. (i) and (ii), we get initial velocity \[u=1\text{ }m/s\].


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