2. Solved Papers
  3. EAMCET Medical

EAMCET Medical EAMCET Medical Solved Paper-2000

  • question_answer
    The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06m is \[0.34m{{s}^{-1}}.\] If the change in velocity of the body is \[0.18\text{ }m{{s}^{-}}1,\] during this time its uniform acceleration is:

    A) \[0.01m{{s}^{-2}}\]                         

    B)  \[0.02m{{s}^{-2}}\]

    C) \[0.03m{{s}^{-2}}\]                         

    D)  \[0.04m{{s}^{-2}}\]

    Correct Answer: B

    Solution :

                     Time taken in covering a distance of 3.06 m is \[\text{t =}\frac{\text{Distance}}{\text{Average velocity}}=\frac{3.06}{0.34}=9s\] Also, change in velocity \[v-u=0.18\,m{{s}^{-1}}\] From first law of motion \[v=u+at\] \[\Rightarrow \]               \[a=\frac{v-u}{t}=\frac{0.18}{9}=0.02\,m{{s}^{-2}}\]


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