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Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    A point initially at rest moves along x-axis. Its acceleration varies with time as\[a=(61+5)\]in m/s. If it starts from origin, the distance covered in 2s is

    A)  20m            

    B)  18m

    C)  16 m            

    D)  25 m

    Correct Answer: B

    Solution :

     Given, \[a=\frac{dv}{dt}=6t+5\] \[\int_{0}^{v}{dv}=\int_{0}^{t}{(6t+5)}dt\] \[v=\frac{6{{t}^{2}}}{2}+5t\] \[\frac{ds}{dt}=\left( \frac{6{{t}^{2}}}{2}+5t \right)dt\] \[s=\frac{3{{t}^{2}}}{3}+\frac{5{{t}^{2}}}{2}\] where,      \[t=2\text{ }s,\] \[S=3\times \frac{{{2}^{3}}}{3}+\frac{5\times {{2}^{2}}}{2}=18\,m\]


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