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NEET AIPMT SOLVED PAPER SCREENING 2008

  • question_answer
    A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 \[\text{m}{{\text{s}}^{\text{-1}}}\] to 20 \[\text{m}{{\text{s}}^{\text{-1}}}\] while passing through a distance 135 m in t- second. The value of t is

    A) 10                 

    B)        1.8

    C) 12                                          

    D) 9

    Correct Answer: D

    Solution :

    Key Idea: The problem requires kinematics    equations of motion. Let u and v be the First and final velocities of particle and a & s be the constant acceleration and distance covered, by it. From third equation of motion \[{{v}^{2}}={{u}^{2}}+2as\] \[\Rightarrow \]\[{{(20)}^{2}}={{(10)}^{2}}+2a\times 135\] or\[a=\frac{300}{2\times 135}=\frac{10}{9}m{{s}^{-2}}\] Now using first equation of motion, \[t=\frac{v-u}{a}=\frac{20-10}{(10/9)}=\frac{10\times 9}{10}=9s\]


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