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General
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Question

A particle moves along a straight line such that its displacement at any time t is given by s equals t to the power of 3 end exponent minus 6 t to the power of 2 end exponent plus 3 t plus 4. The velocity when its acceleration is zero is

  1. 2 blank m s to the power of negative 1 end exponent  
  2. 12 blank m s to the power of negative 1 end exponent  
  3. negative 9 blank m s to the power of negative 1 end exponent  
  4. 2 blank m s to the power of negative 1 end exponent  

The correct answer is: negative 9 blank m s to the power of negative 1 end exponent


    Given, s equals t to the power of 3 end exponent minus 6 t to the power of 2 end exponent plus 3 t plus 4
    Velocity v equals fraction numerator d s over denominator d t end fraction equals 3 t to the power of 2 end exponent minus 12 t plus 3 blank open parentheses i close parentheses
    Acceleration a equals fraction numerator d v over denominator d t end fraction equals 6 t minus 12 blank open parentheses i i close parentheses
    Since, acceleration is zero, so, 6 t minus 12 equals 0 comma blank o r blank t equals 2 blank s
    So, velocity v a t
    t equals 2 blank s comma blank i s equals 3 cross times 2 to the power of 2 end exponent minus 12 cross times 2 plus 3 equals negative 9 blank m s to the power of negative 1 end exponent

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