A particle starts travelling on a circle with constant tangential acceleration. The angle between velocity vector and acceleration vector, at the moment when particle completes half the circular track. Is

Options

A. `tan^(-1)(2pi)`
B. `tan^(-1)(pi)`
C. `tan^(-1)(3pi)`
D. zero

Answer

A

Explanation

`v=sqrt(2a_(t)s)=sqrt(2a_(t)(piR))`

`therefore" "a_(n)=(v^(2))/(R)=2pia_(t)`
or `(a_(n))/(a_(t))=2pirArrtantheta=(a_(n))/(a_(t))=2pi`
`therefore" "theta=tan^(-1)(2pi)`