BCECE Engineering BCECE Engineering Solved Paper-2008

  • question_answer
    A point initially at rest moves along \[x-\]axis. Its acceleration varies with time as \[a=(6t+5)m/{{s}^{2}}.\]If it starts from origin, the distance covered in 2 s is

    A)  20 m                    

    B)         18 m

    C)  16 m                    

    D)         25 m

    Correct Answer: B

    Solution :

    Given, \[a=\frac{dv}{dt}=6t+5\] or            \[dv=(6t+5)dt\] Integrating, we get \[\int_{0}^{v}{dv}=\int_{0}^{t}{(6t+5)}dt\]                          or            \[v=\left( \frac{6{{t}^{2}}}{2}+5t \right)\]                 Again     \[v=\frac{ds}{dt}\]                 \[\therefore \]  \[ds=\left( \frac{6{{t}^{2}}}{2}+5t \right)dt\] Integrating again, we get                 \[\int_{0}^{s}{ds}=\int_{0}^{t}{\left( \frac{6{{t}^{2}}}{2}+5t \right)}dt\] \[\therefore \]  \[s=\frac{3{{t}^{3}}}{3}+\frac{5{{t}^{2}}}{2}\] when,   \[t=2s,s=3\times \frac{{{2}^{3}}}{3}+\frac{5\times {{2}^{2}}}{2}\]                 \[=3\times \frac{8}{3}+\frac{5\times 4}{2}\]                 \[=8+10=18m\]                


You need to login to perform this action.
You will be redirected in 3 sec spinner