Physics-
General
Easy

Question

If the velocity v of a particle moving along a straight line decreases linearly with its displacement s from 20m s to the power of negative 1 end exponent to a value approaching zero s equals 30 m, then acceleration of the particle at s equals 15m is

  1. fraction numerator 2 over denominator 3 end fraction m s to the power of negative 2 end exponent    
  2. negative fraction numerator 2 over denominator 3 end fraction m s to the power of negative 2 end exponent    
  3. fraction numerator 20 over denominator 3 end fraction m s to the power of negative 2 end exponent    
  4. negative fraction numerator 20 over denominator 3 end fraction m s to the power of negative 2 end exponent    

The correct answer is: negative fraction numerator 20 over denominator 3 end fraction m s to the power of negative 2 end exponent


    Slope of line equals negative fraction numerator 2 over denominator 3 end fraction
    Equation of line is open parentheses v minus 20 close parentheses equals negative fraction numerator 2 over denominator 3 end fraction left parenthesis s minus 0 right parenthesis
    rightwards double arrow blank v equals 20 minus fraction numerator 2 over denominator 3 end fraction s (i)
    Velocity at s equals 15 m blank i e comma
    v equals open fraction numerator d s over denominator d t end fraction close vertical bar subscript s equals 15 m end subscript equals 20 minus fraction numerator 2 over denominator 3 end fraction open parentheses 15 close parentheses equals 10 m s to the power of negative 1 end exponent
    Differentiate Eq. (i) with respect to time,
    acceleration equals fraction numerator d v over denominator d t end fraction equals fraction numerator 2 over denominator 3 end fraction fraction numerator d s over denominator d t end fraction
    therefore blank open fraction numerator d v over denominator d t end fraction close vertical bar subscript s equals 15 m end subscript equals negative fraction numerator 2 over denominator 3 end fraction open fraction numerator d s over denominator d t end fraction close vertical bar subscript s equals 15 m end subscript equals negative fraction numerator 20 over denominator 3 end fraction m s to the power of negative 2 end exponent

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