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NEET Physics Motion in a Straight Line / सरल रेखा में गति NEET PYQ-One Dimensional Motion

  • question_answer
    A particle moving along x-axis has acceleration \[f,\] at time t, given by \[f={{f}_{0}}\left( 1-\frac{t}{T} \right),\] where \[{{f}_{0}}\] and T are constants. The particle at \[t=0\] has zero velocity. In the time interval between \[t=0\] and the instant when \[f=0,\] the particle's velocity \[({{v}_{x}})\] is:                                     [AIPMT (S) 2007]

    A) \[{{f}_{0}}T\]  

    B) \[\frac{1}{2}{{f}_{0}}{{T}^{2}}\]      

    C) \[{{f}_{0}}{{T}^{2}}\]           

    D) \[\frac{1}{2}{{f}_{0}}T\]

    Correct Answer: D

    Solution :

    Acceleration
                            \[f={{f}_{0}}\left( 1-\frac{t}{T} \right)\]
                or         \[f=\frac{dv}{dt}={{f}_{0}}\left( 1-\frac{t}{T} \right)\]  \[\left[ \because \,f=\frac{dv}{dt} \right]\]             
                or         \[dv={{f}_{0}}\left( 1-\frac{t}{T} \right)dt\]                     …(i)
                Integrating Eq. (i) on both sides,
                            \[\int{dv=\int{{{f}_{0}}\left( 1-\frac{t}{T} \right)\,dT}}\]
                \[\therefore \]      \[v={{f}_{0}}t-\frac{{{f}_{0}}}{T}.\frac{{{t}^{2}}}{2}+C\]              …(ii)
                Where C is constant of integration.
                Now, when \[t=0,\text{ }v=0\]
                So, from Eq. (ii), we get \[C=0\]
                \[\therefore \]      \[v={{f}_{0}}t-\frac{{{f}_{0}}}{T}.\frac{{{t}^{2}}}{2}\]                               …(iii)
    As,        \[f={{f}_{0}}\left( 1-\frac{t}{T} \right)\]
    When,    \[f=0,\,\,0={{f}_{0}}\left( 1-\frac{t}{T} \right)\]
    As, \[{{f}_{0}}\ne 0\] so, \[1-\frac{t}{T}=0\,\therefore \,\,t=T\]
                Substituting, \[t=T\] in Eq. (iii), then velocity
                            \[{{v}_{x}}={{f}_{0}}T-\frac{{{f}_{0}}}{T}.\frac{{{T}^{2}}}{2}\]
                            \[={{f}_{0}}T-\frac{{{f}_{0}}T}{2}=\frac{1}{2}\,{{f}_{0}}T\]


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