A particle’s velocity v at time t is given by `v=2e^(2t) cos""(pi t)/(3)`. The least value of t at which the acceleration becomes zero, is :

Options

A. 0
B. `(3)/(2)`
C. `(3)/(pi)tan^(-1)((6)/(pi))`
D. `(3)/(pi)cot^(-1)((6)/(pi))`

Answer

C

Explanation

We have,
`v=2e^(2t)cos.(pit)/(3)`
`implies (dv)/(dt)=4e^(2t)cos.(pit)/(3)-(2pi)/(3)e^(2t)sin.(pit)/(3)`
Now,
Acceleration = 0
`implies 4e^(2t)cos.(pit)/(3)-(2pi)/(3)e^(2t)sin.(pit)/(3)=0`
`implies tan.(pit)/(3)=(6)/(pi)impliest=(3)/(pi)tan^(-1)((6)/(pi))`