Physics-
General
Easy

Question

A particle travels 10m in first 5 blank s e c and 10 m in next 3 blank s e c. Assuming constant acceleration what is the distance travelled in next 2 blank s e c

  1. 8.3 blank m  
  2. 9.3blank m  
  3. 10.3 m  
  4. None of above  

The correct answer is: 8.3 blank m


    Let initial left parenthesis t equals 0 right parenthesis velocity of particleequals u
    For first 5 blank s e c of motion s subscript 5 end subscript equals 10 blank m e t r e
    s equals u t plus fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent rightwards double arrow 10 equals 5 u plus fraction numerator 1 over denominator 2 end fraction a open parentheses 5 close parentheses to the power of 2 end exponent
    2 u plus 5 a equals 4
    For first 8 s e c of motion s subscript 8 end subscript equals 20 blank m e t r e
    20 equals 8 u plus fraction numerator 1 over denominator 2 end fraction a open parentheses 8 close parentheses to the power of 2 end exponent rightwards double arrow 2 u plus 8 a equals 5
    By solving u equals fraction numerator 7 over denominator 6 end fraction m divided by sand a equals fraction numerator 1 over denominator 3 end fraction m divided by s to the power of 2 end exponent
    Now distance travelled by particle in Total 10 blank s e c
    s subscript 10 end subscript equals u cross times 10 plus fraction numerator 1 over denominator 2 end fraction a open parentheses 10 close parentheses to the power of 2 end exponent
    By substituting the value of u and a blankwe will get s subscript 10 end subscript equals 28.3 blank m
    so the distance in last 2 blank s e c equals s subscript 10 end subscript minus s subscript 8 end subscript
    equals 28.3 minus 20 equals 8.3 m

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