A point object moves along an arc of a circle of radius 'R'. Its velocity depends upon the distance covered 'S' as V = `Ksqrt(S)` where 'K' is a constant. If ' theta' is the angle between the total acceleration and tangential acceleration, then

Options

A. `tan theta =sqrt((S)/( R)`
B. `tan theta = sqrt((S)/(2R))`
C. `tan theta= (S)/(2R)`
D. `tan theta =(2S)/( R)` .

Answer

D

Explanation

Given `v=k sqrt(S)`
`tan theta=(a_("radial"))/(a_("tangential"))=((v^2)/( R))/(a_("tangential"))`
` rArr tan theta =(1)/(a_("tangential"))[(1)/( R) xx k^2 S]`
`rArr tan theta =(1)/(a_("tangent"))[(k^2 S)/( R)]`
`a_("tangential")=(dv)/(dt)`
`=(d)/(dt)=[ksqrt(S)]`
`a_("tangential")=(k)/(2sqrt(S))xx v=(k)/(2sqrt(S))xx k sqrt(S) xx k sqrt(S)=(k^2)/(2)`.
From equation (1) and (2) , we get ,
`tan theta =(1)/((k^2)/(2)) [(k^2S)/( R)]`.
`=(2k^2S)/(k^2 S)`
`tan theta =(2S)/( R)`.