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AMU Medical AMU Solved Paper-2009

  • question_answer
    A body starting from rest, accelerates at a constant rate a\[\text{m}{{\text{s}}^{-2}}\]for some time after which it decelerates at a constant rate b \[\text{m}{{\text{s}}^{\text{-2}}}\]to come to rest finally. If the total time elapsed is t sec, the maximum velocity attained by the body is given by

    A)  \[\frac{ab}{a+b}t\,m{{s}^{-1}}\]                              

    B)  \[\frac{ab}{a-b}t\,m{{s}^{-1}}\]

    C)   \[\frac{2ab}{a+b}t\,m{{s}^{-1}}\]                           

    D)  \[\frac{2ab}{a-b}t\,m{{s}^{-1}}\]

    Correct Answer: A

    Solution :

                     When the body accelerates initial velocity u = 0 acceleration = a and time \[={{t}_{1}}\] then final velocity \[v=0+a\,{{t}_{1}}\]        ... (i) Now, as the body decelerates initial velocity, = v deceleration = - b and time \[=(t-{{t}_{1}})\] [t = total time elapsed] final velocity here \[{{v}_{2}}=0\] Again     \[0=v-b(t-{{t}_{1}})\]                      ... (ii) From Eqs. (i) and (ii), we get                 \[a{{t}_{1}}=b(t-{{t}_{1}})\] or            \[(a+b){{t}_{1}}=bt\] \[\Rightarrow \]               \[{{t}_{1}}=\frac{bt}{(a+b)}\] Substituting the value of \[{{t}_{1}}\] in Eq. (i), we get                 \[v=\frac{abt}{(a+b)}m{{s}^{-1}}\]


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