RAJASTHAN PMT Rajasthan - PMT Solved Paper-2011

  • question_answer
    The distance \[x\] in metre, travelled by a panicle in time t second is given by \[x=0.5{{t}^{3}}+2{{t}^{2}}\]. The acceleration of the particle at t = 2 s is

    A)  \[2\,m/{{s}^{2}}\]                          

    B)  \[4\,m/{{s}^{2}}\]

    C)  \[8\,m/{{s}^{2}}\]                          

    D)  \[10\,m/{{s}^{2}}\]

    Correct Answer: D

    Solution :

    Given,\[x=0.5{{t}^{3}}+2{{t}^{2}}\] and        \[t=2s\]                 \[v=\frac{dx}{dt}=0.5\times 3{{t}^{2}}+4t\]                 \[a=\frac{{{d}^{2}}x}{d{{t}^{2}}}=1.5\times 2t+4\]                 \[a=3\times 2+4\]                            \[(\because \,\,t=2s)\]                 \[a=10\,\,m/{{s}^{2}}\]


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