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Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
    The velocity of a particle which starts from rest is given by the following table.
    t (in second) 0 2 4 6 8 10
    v ( in m/s) 0 12 16 20 35 60
    The total distance travelled (in metre) by the particles in 10 s, using Trapezoidal rule is given by

    A)  113                                       

    B)  226

    C)  143                       

    D)         246

    Correct Answer: B

    Solution :

    Given table is
    t 0 2 4 6 8 10
    v 0 12 16 20 35 60
    Here,     \[h=\frac{10-0}{5}=2\] \[\therefore \]Total distance                 \[=\frac{h}{2}[f({{x}_{0}})+2\{f({{x}_{1}})+f({{x}_{2}})\]                                 \[+f({{x}_{3}})+f({{x}_{4}})\}+f({{x}_{5}})]\]                 \[=\frac{2}{2}[0+2(12+16+20+35)+60]\]                 \[=166+60=226\]


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