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NEET Sample Paper NEET Sample Test Paper-67

  • question_answer
    A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle is given by \[\operatorname{x}= 40 + 12t -{{t}^{3}}\] How long would the particle travel before coming to rest?

    A) 24 m                

    B) 40 m

    C) 56 m                

    D) 16 m

    Correct Answer: C

    Solution :

    Distance travelled by the particle is \[x=40+12t-{{t}^{3}}\] We know that velocity is rate of change of distance \[\therefore \,\,\,\,\,\,\,\,\,\,\nu =\frac{d}{dt}(40+12t-{{t}^{3}})=0+12-3{{t}^{2}}\] but final velocity \[\nu = 0\] \[12=3{{t}^{2}}=0\,\,\,or\,\,\,{{t}^{2}}=\frac{12}{3}=4\] Or     \[t=2\,s\] Hence, distance travelled by the particle before coming to rest is given by \[\operatorname{x}=40+12\left( b \right)-{{\left( b \right)}^{3}}=56\,\,m\]


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