Manipal Medical Manipal Medical Solved Paper-2003

  • 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)  5 s              

    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 \[{{t}_{1}}=\frac{\upsilon }{a}\]                 ...(i) Again,   \[0=\upsilon -2a{{t}_{2}}\] ?.(ii) From Eqs. (i) and (ii), 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 \[{{t}_{1}}=\frac{\upsilon }{a}=6\text{ }sec\]


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