RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer A body moves for a total of nine second starting from rest with uniform                acceleration and then with uniform retardation, which is twice the value of acceleration and then stops. The duration of uniform acceleration is:

    A)  3 s                                         

    B)  4.5 s

    C)  5s                          

    D)         6 s

    Correct Answer: D

    Solution :

    Here: Total time t = 9 sec acceleration = a retardation = - 2a From the laws of motion                     \[\upsilon =u+a{{t}_{1}}\] or            \[\upsilon =0+a{{t}_{1}}\] or            \[{{t}_{1}}=\frac{\upsilon }{a}\]                                 ?(1) Again,   \[0=\upsilon -2a{{t}_{2}}\] or            \[{{t}_{2}}=\frac{\upsilon }{2a}\]                                               ?(2) From Eqs. (1) and (2), we have            \[{{t}_{1}}+{{t}_{2}}=t\] \[\Rightarrow \]\[\frac{\upsilon }{a}+\frac{\upsilon }{2a}=9\] \[\Rightarrow \]\[\frac{3\upsilon }{2a}=9\] \[\Rightarrow \]\[\frac{\upsilon }{a}=\frac{9\times 2}{3}\]        \[\Rightarrow \]\[\frac{\upsilon }{a}=6\] Hence, duration of acceleration       \[{{t}_{1}}=\frac{\upsilon }{a}=6\,\sec \]


You need to login to perform this action.
You will be redirected in 3 sec spinner