J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    At time t, the distance x cm of a moving particle in a horizontal line is given by\[x=12\,{{t}^{3}}-7\,{{t}^{2}}+14t+2.\]. The acceleration when \[t=1\text{ }s\]is

    A)  \[48\,\,cm/{{s}^{2}}\]      

    B)  \[68\text{ }cm/{{s}^{2}}\]

    C)  \[58\text{ }cm/{{s}^{2}}\]       

    D)  \[56\text{ }cm/{{s}^{2}}\]

    Correct Answer: C

    Solution :

    We have,  \[x=12{{t}^{3}}-7{{t}^{2}}+14t\,+2\] \[\frac{dx}{dt}=36\,{{t}^{2}}-14\,t+14\] Acceleration \[=\frac{{{d}^{2}}x}{d{{t}^{2}}}=72\,t-14\] \[{{\left( \frac{{{d}^{2}}x}{d{{t}^{2}}} \right)}_{t=1}}\,=72-14=58\,cm/{{s}^{2}}\]


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