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((3V_0)/(2alpha))`
B. `((3V_0)/(2alpha))^(1/3)`
C. `sqrt((3V_0^2)/(3alpha))`
D. `((3V_0^2)/(2alpha))^(1/3)`

Answer

D

Explanation

`((3V_0^2)/(2alpha))^(1/3)`