Physics-
General
Easy

Question

A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The distance covered by the boggy and distance covered by the train in the same time has relation

  1. Both will be equal  
  2. First will be half of second  
  3. First will be 1/4 of second  
  4. No definite ratio  

The correct answer is: First will be half of second


    Let ‘a’ be the retardation of boggy then distance covered by it be S. If u is the initial velocity of boggy after detaching from train (i. e. uniform speed of train)
    v to the power of 2 end exponent equals u to the power of 2 end exponent plus 2 a s rightwards double arrow 0 equals u to the power of 2 end exponent minus 2 a s blank rightwards double arrow s subscript b end subscript equals fraction numerator u to the power of 2 end exponent over denominator 2 a end fraction
    Time taken by boggy to stop
    v equals u plus a t rightwards double arrow 0 equals u minus a t rightwards double arrow t equals fraction numerator u over denominator a end fraction
    In this time t distance travelled by train = s subscript t end subscript equals u t equals fraction numerator u to the power of 2 end exponent over denominator a end fraction
    Hence ratio fraction numerator s subscript b end subscript over denominator s subscript 1 end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction

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