A particle move in a straight line with retardation proportional to its displacement its loss of kinectic energy for any displacement `x` is proportional to

Options

A. `x^2`
B. `e^(x)`
C. `x`
D. `log_(e)x`

Answer

A

Explanation

`aporp-x`
`a=-lamdax`,`lamda:` constant
`v(dv)/(dx)=-lamdax`
`int_(v1)^(v2)vdv=-lamdaint_(0)^(x)dx`
`(|v^2|_(v1)^(v2))/(2)=-(lamdax^2)/(2)`
`(1)/(2)mv_2^2-(1)/(2)mv_1^2=-(lamdamx^2)/(2)`
(loss in K.E.)`propx^2`