A particle is projected with velocity `V_(0)`along axis x . The deceleration on the particle is proportional to the square of the distance from the origin i.e., `a=omegax^(2).`distance at which the particle stops is

Options

A. `sqrt((3nu_(0))/(2alpha))`
B. `((3nu_(0))/(2alpha))^((1)/(3))`
C. `sqrt((2nu_(0)^(2))/(3alpha))`
D. `((3nu_(0)^(2))/(2alpha))^((1)/(3))`

Answer

D

Explanation

`a=-alphax^(2)`
`V*(dv)/(dx)=-alphaX^(2)`
`int_(V_(0))^(0)v.dv=-int_(0)^(X)alphaX^(2)dx`
`(v_(0)^(2))/(2)=(alphaX^(3))/(3) implies X=[(3)/(2)(V_(0)^(2))/(alpha)]^(1//3)`