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A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of ´ ´ P ´ ´ is such that it sweeps out a length s equals t to the power of 3 end exponent plus 5, where s is in metres and t is in seconds. The radius of the path is 20 blank m. The acceleration of ´ ´ P ´ ´ when t equals 2 s is nearly

  1. 14 blank m divided by s to the power of 2 end exponent    
  2. 13 blank m divided by s to the power of 2 end exponent    
  3. 12 blank m divided by s to the power of 2 end exponent    
  4. 7.2 blank m divided by s to the power of 2 end exponent    

The correct answer is: 14 blank m divided by s to the power of 2 end exponent


    As S equals t to the power of 3 end exponent plus 5
    fraction numerator d s over denominator d t end fraction equals 3 t to the power of 2 end exponent equals v
    therefore a subscript t end subscript equals fraction numerator d v over denominator d t end fraction equals 6 t
    at t equals 2 blank s e c
    open vertical bar stack a with rightwards arrow on top close vertical bar equals square root of a subscript c end subscript superscript 2 end superscript plus a subscript t end subscript superscript 2 end superscript end root
    equals square root of open parentheses fraction numerator v to the power of 2 end exponent over denominator R end fraction close parentheses to the power of 2 end exponent plus a subscript t end subscript superscript 2 end superscript end root equals square root of open parentheses fraction numerator 4 t to the power of 4 end exponent over denominator R end fraction close parentheses to the power of 2 end exponent plus open parentheses fraction numerator d v over denominator d t end fraction close parentheses to the power of 2 end exponent end root
    equals square root of open parentheses 7.2 close parentheses to the power of 2 end exponent plus 144 end root
    open vertical bar stack a with rightwards arrow on top close vertical bar equals 14 m divided by s to the power of 2 end exponent

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