JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    A particle moves towards East from a point A to a point B at the rate of 4 km/h and then towards North from B to C at rate of 5 km/h. If AB = 12 km and BC = 5 km, then its average speed for its journey from A to C and resultant average velocity direct from A to C are respectively

    A) \[\frac{17}{14}km/h\text{ }and\frac{13}{4}km/h\]

    B) \[\frac{13}{4}km/h\text{ }and\frac{17}{4}km/h\]

    C) \[\frac{17}{9}km/h\text{ }and\frac{13}{9}km/h\]

    D) \[\frac{13}{9}km/h\text{ }and\frac{17}{9}km/h\]

    Correct Answer: A

    Solution :

    Given\[AB=12\text{ }km\text{ }({{s}_{1}})\] and \[BC=5\,km\,({{s}_{2}})\] Speed from A to\[B=4\text{ }km/h\] Time taken, \[{{t}_{1}}=\frac{12}{4}=3h\] Speed from B to\[C=5\text{ }km/h\] Time taken to complete distance from B to C \[{{t}_{2}}=\frac{5}{5}=1h\] \[Average\,speed=\frac{Total\,dis\tan ce}{Total\,time}=\frac{{{s}_{1}}+{{s}_{2}}}{{{t}_{1}}+{{t}_{2}}}\] \[=\frac{12+5}{3+1}=\frac{17}{4}km/h\] In \[\Delta BAC,\] \[AC=\sqrt{{{(AB)}^{2}}+{{(BC)}^{2}}}\] \[=\sqrt{144+25}=\sqrt{169},AC=13\,km\] Average velocity \[=\frac{\text{Distance}\,\text{AC}}{\text{Total}\,\text{time}}\,=\frac{13}{4}km/h\]   

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