Q: A particle is moving on a circular path. The angle turned by radius vector is θ = (1 – *e*^{-t/3}), where θ is in radian and t is in second. The angular acceleration at t = 0 is

(a) 1/3 rad/s²

(b) -1/9 rad/s²

(c) 10 rad/s²

(d) -1/6 rad/s²

Ans: (b)

θ = (1 – *e*^{-t/3})

differentiating w.r.t. time

$ \displaystyle \omega = \frac{d\theta}{dt} = \frac{e^{-t/3}}{3} $

$ \displaystyle \alpha = \frac{d\omega}{dt} = -\frac{e^{-t/3}}{9} $

At , t = 0

α = -1/9 rad/s^{2}