Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2008

  • question_answer
    A car moves from X to Y with a uniform speed V y and returns to Y with a uniform speed \[{{v}_{d.}}\] The average speed for this round trip is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

    A) \[\frac{2v{{  }_{d}}{{v}_{u}}}{{{v}_{d}}+{{v}_{u}}}\]                      

    B) \[\sqrt{v{{  }_{u}}{{v}_{d}}}\]

    C) \[\frac{v{{  }_{d}}{{v}_{u}}}{{{v}_{d}}+{{v}_{u}}}\]                         

    D) \[\frac{v{{  }_{u}}+{{v}_{d}}}{2}\]

    Correct Answer: A

    Solution :

    Key Idea Average speed of a body in a given time interval is defined as the ratio of distance travelled to the time taken. \[Average\text{ }speed=\frac{Distance\text{ }travelled}{Time\text{ }taken}\] Let \[{{t}_{1}}\] and \[{{t}_{2}}\] be times taken by the car to go from X to Y and then from Y to X respectively. Then,   \[{{t}_{1}}+{{t}_{2}}=\frac{XY}{{{v}_{u}}}+\frac{XY}{{{v}_{d}}}=XY\left( \frac{{{v}_{u}}+{{v}_{d}}}{{{v}_{u}}{{v}_{d}}} \right)\] Total distance travelled                 \[=XY+XY=2XY\] Therefore, average speed of the car for this round trip is                 \[{{v}_{av}}=\frac{2XY}{XY\left( \frac{{{v}_{u}}+{{v}_{d}}}{{{v}_{u}}{{v}_{d}}} \right)}\] or            \[{{v}_{av}}=\frac{2{{v}_{u}}{{v}_{d}}}{{{v}_{u}}+{{v}_{d}}}\]


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