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JCECE Engineering JCECE Engineering Solved Paper-2004

  • question_answer
    The distance \[s\] (in cm) travelled by a particle in\[t\] seconds is given by\[,\] \[s={{t}^{3}}+2{{t}^{2}}+t\]. The speed of the particle after \[1\,\,s\] will be:

    A) \[2\,\,cm/s\]                     

    B) \[8\,\,cm/s\]

    C) \[6\,\,cm/s\]                     

    D)  None of these

    Correct Answer: B

    Solution :

    Given,\[s={{t}^{3}}+2{{t}^{2}}+t\] \[\therefore \]Speed\[v=\frac{ds}{dt}=3{{t}^{2}}+4t+1\] At\[t=1\],                 \[v=3{{(1)}^{2}}+4(1)+1\]                    \[=8\,\,cm/s\]


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