Physics-
General
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Question

A particle moves along a straight line such that its position x at any time t is x equals 6 t to the power of 2 end exponent minus t to the power of 3 end exponent. Where x in metre ant t is in second, then

  1. At t equals 0 acceleration is 12 blank m s to the power of negative 2 end exponent  
  2. x minus t curve has maximum at 4 s  
  3. Both [a] and [b] are wrong  
  4. Both [a] and [b] are correct  

The correct answer is: Both [a] and [b] are correct


    Given, x equals 6 t to the power of 2 end exponent minus t to the power of 3 end exponent
    fraction numerator d x over denominator d t end fraction equals 12 t minus 3 t to the power of 2 end exponent blank open parentheses i close parentheses
    fraction numerator d x over denominator d t end fraction equals 0 rightwards double arrow t equals 4 blank s
    Now, again differentiating Eq. (i), we get
    fraction numerator d to the power of 2 end exponent x over denominator d t to the power of 2 end exponent end fraction equals 12 minus 6 t equals 12 minus 6 open parentheses 4 close parentheses equals negative 12
    Since, fraction numerator d to the power of 2 end exponent x over denominator d t to the power of 2 end exponent end fraction is negative, hence t equals 4 blank s gives the maximum value for x minus t curve.
    Moreover, acceleration a equals fraction numerator d to the power of 2 end exponent x over denominator x t to the power of 2 end exponent end fraction comma blank a t blank t equals 0 comma fraction numerator d to the power of 2 end exponent x over denominator d t to the power of 2 end exponent end fraction equals 12 blank m s to the power of negative 2 end exponent

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