Uttarakhand PMT Uttarakhand PMT Solved Paper-2009

  • question_answer
    A car travelling on a straight path moves with uniform velocity \[{{v}_{1}}\] for some time and with velocity \[{{v}_{2}}\] for next equal time, the average velocity is given by

    A)  \[\sqrt{{{v}_{1}}{{v}_{2}}}\]          

    B)  \[\left( \frac{{{v}_{1}}+{{v}_{2}}}{2} \right)\]

    C)  \[\left( \frac{1}{{{v}_{1}}}\frac{1}{{{v}_{2}}} \right)\] 

    D)  \[2\left( \frac{1}{{{v}_{1}}}\frac{1}{{{v}_{2}}} \right)\]

    Correct Answer: B

    Solution :

     When particle moves with different uniform speed\[{{v}_{1}},{{v}_{2}},{{v}_{3}}....\]etc in different time intervals \[{{t}_{1}},{{t}_{2}},{{t}_{3}}....\]etc respectively, its average speed over the total time of journey is given as \[{{v}_{av}}=\frac{total\text{ }distance\text{ }covered}{total\text{ }time\text{ }elapsed}\] Here,   \[{{V}_{av}}=\frac{{{v}_{1}}{{t}_{1}}+{{v}_{2}}{{t}_{2}}}{{{t}_{1}}+{{t}_{2}}}\] \[=\frac{({{v}_{1}}+{{v}_{2}})t}{2t}\] \[[\because {{t}_{1}}={{t}_{2}}]\] Or \[{{v}_{av}}=\frac{{{v}_{1}}+{{v}_{2}}}{2}\]


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