A particle is kept fixed on as turntable rotating uniformly. As seen from the ground the particle goes in a circle,its speed is `30 cm//s` and acceleration is `30 cm//s^2 `The particle is now shifted to a new positon to make the radius half of the original value. The new values of the speed and acceleration will be

Answer

A

Explanation

Let x is the distance of point P from O, the from figure
`tan phi=x/h` or `x=htan phi`
`rArr (dx)/(dt)=h sec^(2) phi (d theta)/(dt)`
`[(dphi)/(dt)=omega]rArr v=h sec^(2) phi omega`
So putting values
`h=3, phi=180-(90+45)=45^(@)`
we get `v=(3sqrt(2))^(2)xx0.1=0.6 m//s`