-->

Motion In Straight Line

Question
CBSEENPH11020462

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to

  • x2

  • ex

  • x

  • logex

Solution

A.

x2

In this problem acceleration (a) is given in terms of displacement (x)  to determine the velocity with respect to position or displacement we have to apply integration method.
From given information a =-kx, where a is acceleration, x is displacement and k is proportionality constant. 
vdx over dx space equals space minus space kx space open square brackets therefore space dv over dt space equals space fraction numerator begin display style dv end style over denominator dx end fraction open parentheses dx over dt close parentheses space equals space straight v dv over dx close square brackets
straight v space dv space equals space minus kx space dx
Let for any displacement from 0 to x , the velocity changes from vo to v
integral subscript straight v subscript straight o end subscript superscript straight v space vdv space equals space minus integral subscript 0 superscript straight x space kx space dx
rightwards double arrow space integral subscript 0 superscript straight x space kx space dx
rightwards double arrow space fraction numerator straight v squared minus straight v subscript 0 superscript 2 over denominator 2 end fraction space equals space minus space kx squared over 2
rightwards double arrow space straight m space open parentheses fraction numerator straight v squared minus space straight v subscript 0 superscript 2 over denominator 2 end fraction close parentheses space equals space fraction numerator negative mkx squared over denominator 2 end fraction
increment straight K space proportional to space straight x squared