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A particle is projected with velocity v(...

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

A

`sqrt((3V_0)/(2alpha))`

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
D
`((3V_0^2)/(2alpha))^(1/3)`
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