BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer A person is at a distance x from a bus when the bus begins 10 move with a constant acceleration a. What is the minimum velocity with which [he person should run towards the bus so as to catch it?

    A)  \[2ax\]                                

    B)  \[\sqrt{2ax}\]

    C)  \[ax\]                  

    D)         \[\sqrt{ax}\]

    Correct Answer: B

    Solution :

    Let the person catches bus in t second. Distance covered by bus, \[s=\frac{1}{2}a{{t}^{2}}\] Distance covered by person \[=vt\] According to condition, \[s+x=vt\] \[\frac{1}{2}a{{t}^{2}}+x=vt\] \[a{{t}^{2}}+2x=2vt\] \[a{{t}^{2}}-2vt+2x=0\]                 For t to be real,                                 \[{{b}^{2}}-4ac\ge 0\] \[{{(-2v)}^{2}}-4\times a\times 2x\ge 0\] \[4{{v}^{2}}-8ax\ge 0\] \[{{v}^{2}}-2ax\ge 0\] \[{{v}_{\min }}=\sqrt{2ax}\]

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