NEET AIPMT SOLVED PAPER SCREENING 2007

  • question_answer
                    A car moves from-X to Y with a uniform speed \[{{v}_{u}}\] and returns to Y with a uniform speed \[{{v}_{d}}\]. The average speed for this round trip is:

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

    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.                 \[\text{Average}\,\text{speed=}\frac{\text{Distance}\,\text{travelled}}{\text{Time}\,\text{taken}}\]                 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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