Physics-
General
Easy

Question

Two boys are standing at the ends A and B of a ground where A B equals a. The boy at B starts running in a direction perpendicular to A B with velocity v subscript 1 end subscript. The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is

  1. a divided by square root of v to the power of 2 end exponent plus v subscript 1 end subscript superscript 2 end superscript end root  
  2. square root of a to the power of 2 end exponent divided by left parenthesis v to the power of 2 end exponent minus v subscript 1 end subscript superscript 2 end superscript right parenthesis end root  
  3. a divided by left parenthesis v minus v subscript 1 end subscript right parenthesis  
  4. a divided by left parenthesis v plus v subscript 1 end subscript right parenthesis  

The correct answer is: square root of a to the power of 2 end exponent divided by left parenthesis v to the power of 2 end exponent minus v subscript 1 end subscript superscript 2 end superscript right parenthesis end root


    Let two boys meet at point C after time ‘t’ from the starting. Then A C equals v t comma blank B C equals v subscript 1 end subscript t

    open parentheses A C close parentheses to the power of 2 end exponent equals open parentheses A B close parentheses to the power of 2 end exponent plus open parentheses B C close parentheses to the power of 2 end exponent rightwards double arrow v to the power of 2 end exponent t to the power of 2 end exponent equals a to the power of 2 end exponent plus v subscript 1 end subscript superscript 2 end superscript t to the power of 2 end exponent
    By solving we get t equals square root of fraction numerator a to the power of 2 end exponent over denominator v to the power of 2 end exponent minus v subscript 1 end subscript superscript 2 end superscript end fraction end root

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